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HIGHER NATIONAL DIPLOMA
IN
INFORMATION TECHNOLOGY
STUDY PACK
OPERATIONS RESEARCH
(Version: January 2004)
COPYRIGHTS RESERVED
1
HIGHER NATIONAL DIPLOMA IN INFORMATION TECHNOLOGY
SUBJECT:
OPERATIONS RESEARCH
CODE:
710 / 04 /S02
AIM OF THE SUBJECT:
1. To provide various mathematical tools for analysis of systems in the Business
environment.
2. To provide techniques which would be needed in analysis of Business Systems
to managerial decision-making.
3. To provide tools to solve complex business problems.
4. To represent the concepts of “efficiency” and “scarcity in well defined
mathematical model of a given situation.
5. To provide Science Techniques of the derivation of computational methods for
solving such models.
DESIGN LENGTH:
THEORY
LABORATORY
TOTAL
190: HOURS
10: HOURS
200: HOURS
2
HIGHER NATIONAL DIPLOMA IN INFORMATION TECHNOLOGY
SUBJECT:
OPERATIONS RESEARCH
CODE:
710 / 04 /S02
UNIT 1
APPROXIMATIONS
HOURS:
20
OBJECTIVE
At the end of the unit the student should be able to:
Apply given Technologies in estimating solutions in Business environment.
1.1 Newton-Raphson iteration method for solving polynomial equations.
1.2 Trapezium Rule for approximating a definite integral.
1.3 Simpson Rule for approximating a definite integral
1.4 Maclaurin Series expansion.
3
HIGHER NATIONAL DIPLOMA IN INFORMATION TECHNOLOGY
SUBJECT:
OPERATIONS RESEARCH
CODE:
710 / 04 /S02
UNIT 2
LINEAR PROGRAMMING
HOURS:
20
OBJECTIVE
To formulate and solve linear programming models using the Graphical and other
techniques.
2.1 Model formulation.
2.2 Solution of L.P model by graphical and simplex method.
2.3 Duality and the Dual Simplex method.
2.4 Sensitivity Analysis using the Simplex method.
2.5 Assignment problem.
2.6 Transportation problem.
UNIT 3:
NON-LINEAR FUNCTIONS:
HOURS:
20.
OBJECTIVE
To develop an optimization theory using differential calculus to determine maximum and
minimum points for Non Linear Functions.
3.1 Overview of differentiation and integration.
3.2 Continuous functions and partial derivatives.
3.3 Necessary and sufficient conditions for extreme points.
3.3.1
The gradient vector
3.3.2
The Hessian matrix
4
HIGHER NATIONAL DIPLOMA IN INFORMATION TECHNOLOGY
SUBJECT:
OPERATIONS RESEARCH
CODE:
710 / 04 /S02
UNIT 4
PROJECT MANAGEMENT WITH PERT/CPM
HOURS:
20
OBJECTIVE
To select a method which is appropriate in a given situation to find the shortest route between
two given nodes in a network and determine the route and its length.
4.1
Introduction.
4.2
The project network diagram.
4.3
Critical Path Analysis.
4.4
Determination of the critical path.
4.5
Project activity crushing.
4.6
Probabilistic time duration of activities.
UNIT 5:
RANDOM VARIABLES AND THEIR PROBABILITY:
HOURS:
20.
OBJECTIVE
To appreciate the properties of a probability function and be able to apply it in Business
environment.
5.1 Introduction
5.2 Probability density functions (pdf)
5.3 Discrete random variables.
5.4 Continuous random variables.
5.5 Cumulative density functions (CDF)
5.6
Relations among probability distributions
5.7
Joint probability distributions with two variables
5
HIGHER NATIONAL DIPLOMA IN INFORMATION TECHNOLOGY
SUBJECT:
OPERATIONS RESEARCH
CODE:
710 / 04 /S02
UNIT 6
DECISION THEORY
HOURS:
20
OBJECTIVE
Apply the concept of optimistic approach, conservative approach, and minimax regret
approach to decision-making problems.
6.1 Decisions under risk
6.1.1 Expected value criterion
6.1.2 Expected profit with perfect information
6.1.3 Minimizing expected loss.
6.1.4
Expected value of perfect information
6.1.5
Decision trees in multistage decision-making
6.2 Decisions under uncertainty
6.2.1 Laplace criterion
6.2.2 Minimax/Maximin criterion
6.2.3 Savage Minimax Regret criterion
6.2.4
Hurwicz criterion
6
UNIT 7:
THEORY OF GAMES:
HOURS:
20
OBJECTIVES:
Use analytical criteria for decisions under uncertainty with the assumptions that “nature” is
the opponent.
7.1
Pure strategy games
7.1.1 Two-Person Zero Sum games
7.1.2 Saddle point solutions
7.2
Mixed strategy games
7.2.1 Solution of games by Linear Programming.
7.2.2 Graphical solution of (2*n) and (m*2) games.
7.2.3 Solution of (m*n) games by the Simplex methods.
UNIT 8:
INVENTORY MODELLING:
HOURS:
20
OBJECTIVES:
To determine the reorder point and cost of inventory for a given model.
8.1
Generalized inventory model
8.2
Deterministic Models.
8.3
The Economic Order Quantity (EOQ) model
8.4
Inventory with backordering
8.5
Economic Production Quantity model
8.6
Probabilistic models
8.7
Continuous Review model.
7
UNIT 8:
REGRESSION ANALYSIS AND CORRELATION:
HOURS:
20
OBJECTIVES:
To identify the relationships between variables and forecast future trends.
9.1
9.2
Simple Linear Regression Analysis.
9.1.1
Scatter plots and Line of best fit.
9.1.2
Least Squares equation regression model.
9.1.3
Forecasting using Regression techniques.
9.1.4
Making inferences about parameters a and b.
Correlation Coefficient.
9.2.1
Coefficient of determination.
9.2.2
Interpret the value of the correlation coefficient.
UNIT 10
INTRODUCTION TO QUEUEING THEORY
HOURS:
20
OBJECTIVE:
To identity a Queuing problem and calculate parameters related to the queue.
10.1
Queuing processes of Single Server models.
10.2
Definitions and Notations.
10.3
Relationships between
10.3.1 Expected Waiting Time per Customer in the system.
10.3.2 Expected Waiting Time per Customer in the queue.
10.3.3 Expected Number of Customer in the system.
10.3.4 Expected Number of Customer in the Queue.
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UNIT 2
LINEAR PROGRAMMING
HOURS:
20
LINEAR PROGRAMMING
Is a technique used to determine how best to allocate personnel, equipment, materials,
finance, land, transport e.t.c. , So that profit are maximized or cost are minimized or other
optimization criterion is achieved.
Linear programming is so called because all equations involved are linear. The variables in
the problem are Constraints. It is these constraints, which gives rise to linear equations or
Inequalities.
The expression to the optimized is called the Objective function usually represented by an
equation.
Question 1:
A furniture factory makes two products: Chairs and tables. The products pass through 3
manufacturing stages; Woodworking, Assembly and Finishing.
The Woodworking shop can make 12 chairs an hour or 6 tables an hour.
The Assembly shops can assembly 8 chairs an hour or 10 tables an hour.
The Finishing shop can finish 9 chairs or 7 tables an hour.
The workshop operates for 8 hours per day. If the contribution to profit from each Chair is $4
and from each table is $5, determine by Graphical method the number of tables and chairs
that should be produced per day to maximize profits.
Solution:
Let number of chairs be X.
Let number of tables be Y.
Objective Function is:
P = 4X + 5Y.
Constraints:
WW:
AW:
FNW:
X/12 + Y/6 <= 8
X + 2Y <= 96
when X=0 Y= 48
when Y=0 X= 96
X/8 + Y/10 <= 8
5X + 4Y <= 320
when X=0 Y=80
when Y=0 X=64
X/9 + Y/7 <= 8
7X + 9Y <= 504
when X=0 Y=56
when Y=0 X=72
X>=0 Y>=0
9
100
90
80
70
60
50
40
30
20
P------ this gives the maximum point
10
0
10
20
30
40
50
60
70
80
90
100
Question 2:
Mr. Chabata is a manager of an office in Guruwe; he decides to buy some new desk and
chairs for his staff.
He decides that he need at least 5 desk and at least 10 chairs and does not wish to have more
than 25 items of furniture altogether. Each desk will cost him $120 and each chair will cost
him $80. He has a maximum of $2400 to spend altogether.
Using the graphical method, obtain the maximum number of chairs and desk Mr. Chabata
can buy.
Solution:
Let X represents number of Desk.
Let Y represents number of Chairs.
Objective Function is:
P = 120X + 80Y.
Constraints:
X >= 5
Y>= 10
X + Y <= 25
10
30
25
20
PPoint P (10,15) gives the maximum point
15
10
5
0
5
10
15
20
25
30
35
40
The Optimum Solution = 120 * 10 + 80 * 15
= 1200 + 1200
= 2400
Question 3:
A manufacturer produces two products Salt and Sugar. Salt has a contribution of $30 per
unit and Sugar has $40 per unit. The manufacturer wishes to establish the weekly production,
which maximize the contribution. The production data are shown below:
Salt
Sugar
Total available per unit
Machine Hours
4
2
100
Production Unit
Labour Hours
4
6
180
Materials in Kg
1
1
40
Because of the trade agreement sales of Salt are limited to a weekly maximum of 20 units and
to honor an agreement with an old established customer, at least 10 units of Sugar must be
sold per week.
Solution:
Let X represents Salt.
Let Y represents Sugar.
Objective Function is:
P = 30X + 40Y.
Constraints:
4X + 2Y <= 100 {machine hours}
4X + 6Y <= 180 {Labour hours}
X + Y <= 40 {material}
Y>= 10
X<= 20
X>=0; Y>=0;
11
 SOLVING LINEAR PROGRAMMING USING SIMPLEX METHOD:
The graphical outlined above can only be applied to problems containing 2 variables. When 3
or more variables are involved we use the Simplex method. Simplex comprises of series of
algebraic procedures performed to determine the optimum solution.
In Simplex method we first convert inequalities to equations by introducing a Slack variable.
A Slack variable represents a spare capacity in the limitation.
 STEPS TO FOLLOW IN SIMPLEX METHOD:
i.
ii.
iii.
iv.
v.
vi.
Obtain the pivot column as the column with the most positive indicator row.
Obtain pivot row by dividing elements in the solution column by their corresponding
pivot column entries to get the smallest ratio. Element at the intersection of the pivot
column and pivot row is known as pivot elements.
Calculate the new pivot row entries by dividing pivot row by pivot element. This new
row is entered in new tableau and labeled with variables of new pivot column.
Transfer other row into the new tableau by adding suitable multiplies of the pivot row
(as it appears in the new tableau) to the rows so that the remaining entries in the pivot
column becomes zeroes.
Determine whether or not this solution is optimum by checking the indicator row
entries of the newly completed tableau to see whether or not they are any positive
entries. If they are positive numbers in the indicator row, repeat the procedure as from
step 1.
If they are no positive numbers in the indicator row, this tableau represents an
optimum solution asked for, the values of the variables together with the objective
function could then be stated.
Question 1:
Maximize Z = 40X + 32Y
Subject to:
40X +20Y <= 600
4X + 10Y <= 100
2X + 3Y <= 38 Using the Simplex method
Solution:
To obtain the initial tableau, we rewrite the Objective Function as:
Z = 40X + 32Y
Introducing Slack variables S1, S2, S3 in the 3 inequalities above we get:
40X + 20Y + S1 = 600
4X + 10Y + S2 = 100
2X + 3Y + S3 = 38
X >= 0
Y >=0.
12
TABLEAU 1:
Pivot Element
X
Y
S1
S2
S3
Solution
S1
40
20
1
0
0
600
S2
4
10
0
1
0
100
S3
2
3
0
0
1
38
Z
40
32
0
0
0
0
S2
S3
Indicator Row
Pivot Column
TABLEAU 2:
X
Y
S1
Solution
X
1
0.5
0.025 0
0
15
S2
0
8
-0.1
1
0
40
S3
0
2
-0.05 0
1
8
Z
0
12
-1
0
0
-600
S2
S3
Solution
TABLEAU 3:
X
Y
S1
X
1
0
0.0375 0
-0.25 13
S2
0
0
0.1
-4
8
Y
0
0.025 0
0.5
4
Z
0
-0.7
-6
-648
1
0
1
0
Indicator Row
Pivot Column
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum
Z, X = 13 and Y = 4 giving Z = 648.
13
NB:
a) If when selecting a pivot column we have ties in the indicator row, we then
select the pivot column arbitrary.
b) If all the entries in the selected pivot column are negative then the objective
function is unbound and the maximum problem has no solution.
c) A minimization problem can be worked as a maximization problem after
multiply the objective function and the inequalities by –1.
d) Inequalities change their signs when multiplied by negative number.
Question 2:
Maximize Z = 5X1 + 4X2
Subject to:
2X1 +3X2 <= 17
X1 + X2 <= 7
3X1 + 2X2 <= 18 Using the Simplex method
Solution:
Max Z = 5X1 + 4X2
Subject to:
2X1 + 3X2 + S1 = 17
X1 + X2 + S2 = 7
3X1 + 2X2 + S3 = 18
X1 >= 0
X2>=0.
TABLEAU 1:
X1
X2
S1
S2
S3
Solution
S1
2
3
1
0
0
17
S2
1
1
0
1
0
7
S3
3
2
0
0
1
18
Z
5
4
0
0
0
0
Indicator Row
Pivot Column
14
TABLEAU 2:
X1
X2
S1
S2
S3
Solution
S1
0
5/3
1
0
-2/3
5
S2
0
1/3
0
1
-1/3
1
X1
1
2/3
0
0
1/3
6
Z
0
2/3
0
0
-5/3
-30
X2
S1
S2
S3
3/5
0
-2/5
3
TABLEAU 3:
X1
Solution
X2
0
S2
0
0
-1/5
1
-1/5
0
X1
1
0
-2/5
0
9/15
4
Z
0
0
-2/5
0
-7/5
-32
1
Pivot Column
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum
Z, X1= 4 and X2 = 3 giving Z = 32.
Question 3:
A company can produce 3 products A, B, C. The products yield a contribution of $8, $5
and $10 respectively. The products use a machine, which has 400 hours capacity in the
next period. Each unit of the products uses 2, 3 and 1 hour respectively of the machine’s
capacity.
There are only 150 units available in the period of a special component, which is used
singly in products A and C.
200 kgs only of a special Alloy is available in the period. Product A uses 2 kgs per unit and
Product C uses 4kgs per units. There is an agreement with a trade association to produce
no more than 50 units of product in the period.
The Company wishes to find out the production plan which maximized contribution.
15
Solution:
Maximize Z = 8X1 + 5X2+ 10X3
Subject to:
2X1 + 3X2 + X3 <= 400
X1 + X3 <= 150
2X1 + 4X3 <= 200
X2 <=50
X1 >= 0
X2 >= 0
X3 >= 0.
{machine hour}
{component}
{Alloy}
{Sales}
Introducing slack variables:
Maximize Z = 8X1 + 5X2+ 10X3
Subject to:
2X1 + 3X2 + X3 + S1= 400
X1 + X3 + S2 = 150
2X1 + 4X3 + S3 = 200
X2 + S4 =50
X1 >= 0
X2 >= 0
X3 >= 0.
TABLEAU 1:
X1
X2
X3
S1
S2
S3
S4
Solution
S1
2
3
1
1
0
0
0
400
S2
1
0
1
0
1
0
0
150
S3
2
0
4
0
0
1
0
200
S4
0
1
0
0
0
0
1
50
Z
8
5
10
0
0
0
0
0
Indicator Row
Pivot Column
NB: Ignore S4 in finding pivot row.
16
TABLEAU 2:
X1
X2
X3
S1
S2
S3
S4
Solution
S1
3/2
3
0
1
0
-1/4
0
350
S2
1/2
0
0
0
1
-1/4
0
100
X3
1/2
0
1
0
0
1/4
0
50
S4
0
1
0
0
0
0
1
50
Z
3
5
0
0
0
-5/2
0
-500
Pivot Column
TABLEAU 3:
X1
X2
X3
S1
S2
S3
S4
Solution
S1
3/2
0
0
1
0
-1/4
-3
200
S2
1/2
0
0
0
1
-1/4
0
100
X3
1/2
0
4
0
0
1/4
0
50
X2
0
1
0
0
0
0
1
50
Z
3
0
0
0
0
-5/2
-5
-750
Pivot Column
TABLEAU 4:
X1
X2
X3
S1
S2
S3
S4
Solution
S1
0
0
-3
1
0
-1
-3
50
S2
0
0
-1
0
1
-1/2
0
50
X1
1
0
2
0
0
1/2
0
100
X2
0
1
0
0
0
0
1
50
Z
0
0
-6
0
0
-4
-5
-1050
Pivot Column
17
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum
Z, X1= 100 and X2 = 50 giving Z = 1050.
Two slack variable S1 =0 and S2 =0. This means that there is no value to be gained by altering
the machine hours and component constraints.

GENERAL RULE:
Constraints only have a valuation when they are fully utilized.
These valuations are known as the SHADOW Prices or Shadow Costs or Dual Prices or Simplex
Multipliers.
A constraint only has a Shadow price when it is binding i.e. fully utilized and the Objective
function would be increased if the constraint were increased by 1 unit.
When solving Linear Programming problems by Graphical means the Shadow price have to be
calculated separately. When using Simplex method they are an automatic by product.
 MIXED CONSTRAINTS:
This involves constraints containing a mixture of <= and >= varieties. Using Maximization
problem we use “Less than or equal to” type. (<=).
Faced with a problem which involves a mixture of <= and >= variety. The alternative solution
to deal with “Greater than or equal to” (>=) type is to multiply both sides by –1 and change
the inequality sign.
Question 4:
Maximize Z = 5X1 + 3X2+ 4X3
Subject to:
3X1 + 12X2 + 6X3 <= 660
6X1 + 6X2 + 3X3 <= 1230
6X1 + 9X2 + 9X3 <= 900
X3 >=10
Solution:
The only constraint that need to be changed is X3 >=10 by multiply by –1 both sides and we
get:
-X3 <= -10
Maximize Z = 5X1 + 3X2+ 4X3
Subject to:
3X1 + 12X2 + 6X3 + S1 = 660
6X1 + 6X2 + 3X3 + S2 = 1230
6X1 + 9X2 + 9X3 <+ S3 = 900
-X3+ S4 = -10
18
TABLEAU 1:
X1
X2
X3
S1
S2
S3 Solution
S1
3
12
6
1
0
0
600
S2
6
6
3
0
1
0
1200
S3
6
9
9
0
0
1
900
Z
5
3
4
0
0
0
0
Indicator Row
Pivot Column
FINAL TABLEAU:
X1
X2
X3
S1
S2
S3 Solution
S1
0
15/2
3/2
1
0
-1/2
150
S2
0
-3
-6
0
1
-1
300
X1
1
3/2
3/2
0
0
1/6
150
Z
0
-9/2
-7/2
0
0
-5/6
-750
Pivot Column
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum
Z, X1= 150 producing Z = $750. Plus production to satisfy constrain (d) 20 units of X3
producing $ 40 contribution.
Therefore Total solution is 150 units of X1 and 10 units of X3 giving $790.
NB: Maximize Z = 5(150) + 3(0)+ 4(10)
= $790.
Question 5:
Maximize Z = 3X1 + 4X2
Subject to:
4X1 + 2X2 <= 100
4X1 + 6X2 <= 180
X1 + X2 <= 40
X1 <= 20
X2 >=10
19
Solution:
The only constraint that need to be changed is X2 >=10 by multiply by –1 both sides and we
get: -X2 <= -10
Maximize Z = 3X1 + 4X2
Subject to:
4X1 + 2X2 + S1 = 100 {1}
4X1 + 6X2 + S2 = 180 {2}
X1 + X2 + S3 = 40
{3}
X1 + S4 = 20 {4}
-X2 + S5 =-10 {5}
TABLEAU 1:
X1
X2
S1
S2
S3
S4
S5
Solution
S1
4
1
0
0
0
0
100
S2
4
6
0
1
0
0
0
180
S3
1
1
0
0
1
0
0
40
S4
1
0
0
0
0
1
0
20
S5
0
-1
0
0
0
0
1
-10
Z
3
4
0
0
0
0
0
0
2
Indicator Row
Pivot Column
The problem is then solved by the usual Simplex iterations. Each iteration improves on the
one before and the process continues until optimum is reached.
TABLEAU 2:
X1
X2
S1
S2
S3
S4
S5
Solution
X2
0
0
0
0
0
-1
10
S2
4
0
1
0
0
0
2
80
S3
4
0
0
1
0
0
6
120
S4
1
0
0
0
1
0
1
30
S5
1
0
0
0
0
1
0
20
Z
4
0
0
0
0
0
4
1
-40
20
This shows 10X2 being produced and $40 contribution. The first four constraints have
surpluses of 80, 120, 30 and 20 respectively. Not optimums as there are still positive values
in Z row.
TABLEAU 3:
X1
X2
S1
S2
X2
0.667 1
0
S2
2.667 0
S3
S3
S4
S5
Solution
0.167 0
0
0
30
1
-0.333 0
0
0
40
-0.333 0
0
-0.167 0
0
0
10
S4
1
0
0
0
1
1
0
20
S5
0.333 0
0
0.167 0
0
1
20
Z
0.333 0
0
-0.667 0
0
0
-120
This shows 30X2 being produced and $120 contribution. All constraints have surpluses
except Labour hours. Not optimum as there is a positive value in Z row.
TABLEAU 4:
X1
X2
S1
X1
1
0
X2
0
S3
S2
S3
S4
S5
Solution
0.375 0.125 0
0
0
15
1
-0.25 0.25
0
0
20
0
0
-0.125 -0.125 1
0
0
5
S4
0
0
-0.375 0.125 0
1
0
5
S5
0
0
-0.25 0.25
0
0
1
10
Z
0
0
-0.125 -0.625 0
0
0
-125
0
Conclusion:
Since the indicator row is negative the solution is optimum with 15X1 and 20X2 giving $125
contribution.
Shadow prices are X1 = $0.125 and X2 = $0.625.
Non-binding constraints are {3}, {4}, {5} with 5, 5 and 10 spare respectively.
21
 DUALITY:
There is a dual or inverse for every Linear Programming problem. Because solving Simplex
problem in Maximization is quite simple and straightforward, it is usually to convert a
Minimization problem into Maximization problem using dual.
The dual or inverse of Linear Programming problem is obtained by making the constraints in
the inequalities coefficient of the new objective function.
The cofficiences of the original inequalities are combined with the cofficiences of the original
objective function as the constraints.
Question 6:
Minimize Z = 40X1 + 50X2
Subject to:
3X1 + 5X2 >= 150
5X1 + 5X2 >= 200
3X1 + X2 >= 60
X1, X2 >=0
Solution:
The Dual Linear Programming problem is as follows:
Maximize P = 150Y1 + 200Y2 +60 Y3
Subject to:
3Y1 + 5Y2 + 3Y3 <= 40
5Y1 + 5Y2 + Y3 <= 50
Y1>=0, Y2>=0, Y3>=0.
Y1
Y2
5
Y3
S1
S2 Solution
3
1
0
40
S1
3
S2
5
5
1
0
1
50
P
150
200
60
0
0
0
Indicator Row
Pivot Column
Y1
Y2
Y3
S1
S2 Solution
Y2
3/5
1
3/5
1/5
0
8
S2
2
0
-2
-1
1
10
P
30
0
-60
-40
0
-1600
22
Y1
Y2
Y3
S1
S2 Solution
Y2
0
1
1.2
0.5
-0.3
5
Y1
1
0
-1
-0.5
0.5
5
P
0
0
-30
-25
-15
-1750
Conclusion:
Since the indicator row is negative the solution is optimum with 5Y1 and 5Y2 giving
$1750 contribution.
Question 7:
Minimize Z = 16X1 + 11X2
Subject to:
2X1 + 3X2 >= 3
5X1 + X2 >= 8
X1, X2 >=0 Using the Dual problem.
Solution:
The Dual Linear Programming problem is as follows:
Maximize P = 3Y1 + 8Y2
Subject to:
2Y1 + 5Y2 <= 16
3Y1 + Y2 <= 11
Y1>=0, Y2>=0.
Y1
Y2
S1
S2
Solution
1
0
16
S1
2
S2
3
1
0
1
11
Z
3
8
0
0
0
5
Indicator Row
Pivot Column
Y1
Y2
S1
S2
Solution
0.2
0
3.2
Y1
0.4
S2
2.6
0
-0.2
1
7.8
Z
-0.2
0
-1.6
0
-25.6
1
23
Conclusion:
Since the indicator row is negative the solution is optimum. Hence P = 3Y1 + 8Y2 is
maximum when Y1 = 3.2 and Y2= 0 and P = 25.6
In the primary problem, the solution correspond the slack variable values in the final
tableau.
i.e.
X1= S1 = 1.6
X2= S2 = 0.
Hence Z = 16*1.6 + 11*0 => 25.6
 TERMS USED WITH LINEAR PROGRAMMING:
 FEASIBLE REGION:
Represents all combinations of values of the decision variables that satisfy every
restriction simultaneous.
The corner point of the feasible region gives what is known as BASIC
FEASIBLE SOLUTION i.e. the solution that is given by the coordinates at the
intersection of any two binding constraints.
 BINDING CONSTRAINTS:
Is an inequality whose graph forms the bounder of the feasible region.
 NON BINDING CONSTRAINTS:
Is an inequality, which does not conform to the feasible region.
 DUAL PRICE / SHADOW PRICES:
It is important that management information to value the scarce resources.
These are known as Dual price / Shadow price. Derived from the amount of
increase (or decrease) in contribution that would arise if one more (or one less)
unit of scare resource was available.
24
 ASSIGNMENT PROBLEM:
This is the problem of assigning any worker to any job in such a way that only one worker is
assigned to each job, every job has one worker assigned to it and the cost of completing all
jobs is minimized.
 STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:
a. Layout a two way table containing the cost for assigning a worker to a job.
b. In each row subtract the smallest cost in the row from every cost in the row. Make
a new table.
c. In each column of the new table, subtract the smallest cost from every cost in the
column. Make a new table.
d. Draw horizontal and vertical lines only through zeroes in the table in such a way
that the minimum number of lines is used.
e. If the minimum number of lines that covers zeroes is equal to the number of rows
in the table the problem is finished.
f. If the minimum number of lines that covers zeroes is less than the number of rows
in the table the problem is not finished go to step g.
g. Find the smallest number in the table not covered by a line.
i. Subtract that number from every number that is not covered by a line.
ii. Add that number to every number that is covered by two lines.
iii. Bring other numbers unchanged. Make a new table.
h. Repeat step d through step g until the problem is finished.
Question 1:
Use the assignment method to find the minimum distance assignment of Sales
representative to Customer given the table below: What is the round trip distance of the
assignment.
Sales Representative
Customer
Distance (km)
A
1
200
A
2
400
A
3
100
A
4
500
B
1
1000
B
2
800
B
3
300
B
4
400
C
1
100
C
2
50
C
3
600
C
4
200
D
1
700
D
2
300
D
3
100
D
4
250
25
A
B
C
D
1
200
1000
100
700
2
400
800
50
300
3
100
300
600
100
4
500
400
200
250
2
300
500
0
200
3
0
0
550
0
4
400
100
150
150
2
300
500
0
200
3
0
0
550
0
4
300
0
50
50
2
250
450
0
150
3
0
0
600
0
4
300
0
100
50
TABLEAU 2:
A
B
C
D
1
100
700
50
600
TABLEAU 3:
A
B
C
D
1
50
650
0
550
TABLEAU 4:
A
B
C
D
1
0
600
0
500
Conclusion:
Since the number of lines is now equal to number of rows, the problem is finished with
the following assignment:
SALES REP
CUSTOMER
DISTANCE
A
1
200
B
4
400
C
2
50
D
3
100
750 km
Therefore total round Trip distance = 750 km * 2 => 1500 km
26
Question 2:
A foreman has 4 fitters and has been asked to deal with 5 jobs. The times for each job are
estimated as follows.
A
B
C
D
1
6
12
20
12
2
22
18
15
20
3
12
16
18
15
4
16
8
12
20
5
18
14
10
17
Allocate the men to the jobs so as to minimize the total time taken.
Solution:
Insert a Dummy fitter so that number of rows will be equal to number of column.
TABLEAU 1:
1
2
3
4
5
A
6
22
12
16
18
B
12
18
16
8
14
C
20
15
18
12
10
D
12
20
15
20
17
DUMMY
0
0
0
0
0
B
4
10
8
0
6
C
10
5
8
2
0
D
0
8
3
8
5
DUMMY
0
0
0
0
0
B
4
7
5
0
6
C
10
2
5
2
0
D
0
5
0
8
5
DUMMY
3
0
0
3
3
TABLEAU 2:
1
2
3
4
5
A
0
16
6
10
12
TABLEAU 3:
1
2
3
4
5
A
0
13
3
10
12
27
Conclusion:
Since the number of lines is now equal to number of rows, the problem is finished with
the following assignment:
FITTERS
JOBS
TOTALS
A
1
6
B
4
8
C
5
10
D
2
15
Dummy
2
0
39

THE ASSIGNMENT TECHNIQUE FOR MAXIMIZING PROBLEMS:
Maximizing assignment problem typically involves making assignments so as to
maximize contributions.
 STEPS INVOLVED:
a) Reduce each row by largest figure in that row and ignore the resulting
minus signs.
b) The other procedures are the same as applied to minimization problems.
Question 3:
A foreman has 4 fitters and has been asked to deal with 4 jobs. The times for each job are
estimated as follows.
A
B
C
D
W
X
Y
Z
25
38
15
26
18
15
17
28
23
53
41
36
14
23
30
29
Allocate the men to the jobs so as to maximize the total time taken.
Solution:
A
B
C
D
TABLEAU 1:
W
X
0
7
15
38
26
24
10
8
Y
2
0
0
0
Z
7
30
11
7
28
TABLEAU 2:
A
B
C
D
W
0
15
26
10
X
0
31
17
1
Y
2
0
0
0
Z
0
23
4
0
X
0
30
16
0
Y
3
0
0
0
Z
1
23
4
0
X
0
26
12
0
Y
7
0
0
4
Z
1
19
0
0
TABLEAU 3:
A
B
C
D
W
0
14
25
9
TABLEAU 4:
A
B
C
D
W
0
10
21
9
Conclusion:
Since the number of lines is now equal to number of rows, the problem is finished with
the following assignment:
A
B
C
D
W
Y
Z
X
25
53
30
28
$136
39
29
Question 4:
A Company has four salesmen who have to visit four clients. The profit records from
previous visits are shown in the table and it is required to Maximize profits by the best
assignment.
1
2
3
4
A
B
C
D
6
22
12
16
12
18
16
8
20
15
18
12
12
20
15
20
Solution:
TABLEAU 1:
W
X
1
6
12
2
22
18
3
12
16
4
16
8
Y
20
15
18
12
Z
12
20
15
20
1
2
3
4
TABLEAU 2:
W
X
14
8
0
4
6
2
4
10
Y
0
7
0
8
Z
8
2
3
0
1
2
3
4
TABLEAU 3:
W
X
14
6
0
2
6
0
4
10
Y
0
7
0
8
Z
8
2
3
0
Conclusion:
Since the number of lines is now equal to number of rows, the problem is finished with
the following assignment:
4
2
1
3
D
A
C
B
20
22
20
16
$78
39
30

TRANSPORTATION PROBLEM:
This is the problem of determining routes to minimize the cost of shipping commodities from
one point to another.
The unit cost of transporting the products from any origin to any destination is given. Further
more, the quantity available at each origin and quantity required at each destination is
known.
 STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:
 Arrange the problem in a table with row requirements on the right and
column requirements at the bottom. Each cell should contain the unit cost
approximates to the shipment.
 Obtain an initial solution by using the North West Corner rule. By this
method one begins at the up left corner cell and works up to the lower right
corner. Place the quantity of goods in the first cell equal to the smallest of the
rows or column totals in the table. Balance the row and column respectively
until you reach the lower right hand cell.
 Find cell values for every empty cell by adding and subtract around the
closing loop.
 If all empty cell have + values the problem is finished. If not pick the cell with
most – (negative) value. Allocate a quantity of goods to that cell by adding
and subtract the small value of the column or row entries in the closed loop.
The closed loop techniques involves the following steps:
 Pick an empty cell, which has no quantity of goods in it.
 Place a + sign in the empty cell.
 Use only occupied cells for the rest of the closed loop.
 Find an occupied cell that has occupied values in the same row or
same column and place a – (negative) sign in this cell.
 Go to the next occupied cell and place + sign in it.
 Continue in this manner until you return to the unoccupied cell in
which you started.
 A closed loop exists for every empty cell as long as they are occupied
cell equal to number of rows + number of column – 1.
Question 1:
A firm has 3 factory (A, B, C) and 4 warehouses (1, 2, 3, 4). The capacities of the factories
and the requirements of the warehouse are in the table below.
FACTORY
A
B
C
CAPACITY
220
300
380
WAREHOUSE
1
2
3
4
REQUIREMENTS
160
260
300
180
31
The cost of shipping one unit from each factory to each warehouse is given below.
FACTORY
WAREHOUSE
COST $
A
1
3
A
2
5
A
3
6
A
4
5
B
1
7
B
2
4
B
3
9
B
4
6
C
1
5
C
2
12
C
3
10
C
4
8
Using the Transportation method, find the least cost shipping schedule and state what it is
?.
Solution:
TABLEAU 1:
1
2
3
4
Capacity
3
5
6
5
-3
-4
A
160
60
220
B
5
7
200
C
2
5
7
Req
160
4
100 9
12
260
-1
6
300
200 10
180
8
380
300
180
900
This is the initial solution, which costs
(160 * 3) + (60 * 5) + (200 * 4) + (100 * 9) + (200 * 10) + (180 * 8)
= $5920.00
TABLEAU 2:
1
3
A
160
2
5
5
7
260
C
-2
5
7
160
4
6
4
12
260
Capacity
5
1
60
4
B
Req
3
-1
6
300
200 10
180
8
380
300
180
40
9
220
900
The costs = (160 * 3) + (60 * 6) + (260 * 4) + (40 * 9) + (200 * 10) + (180 * 8)
= $5680.00
32
TABLEAU 3:
1
3
A
2
2
5
4
6
3
7
260
C
160
5
7
4
160
12
5
9
40
10
40
260
Capacity
1
220
4
B
Req
3
300
220
-1
6
300
180
8
380
180
900
The costs = (160 * 5) + (40 * 10) + (260 * 4) + (40 * 9) + (220 * 6) + (180 * 8)
= $5360.00
TABLEAU 4:
1
3
A
2
2
5
4
7
260
C
160
5
6
160
4
6
4
12
1
9
5
260
40
10
80
Capacity
1
220
3
B
Req
3
140
300
180
220
6
300
8
380
900
The costs = (160 * 5) + (80 * 10) + (260 * 4) + (40 * 6) + (220 * 6) + (140 * 8)
= $5320.00
Conclusion:
Since all the cell values are positive the solution is optimum with the following
allocations:
A
B
C
C
C
Supplies
Supplies
Supplies
Supplies
Supplies
220
260
160
80
140
to
to
to
to
to
3
2
1
3
4
With a minimum cost of $ 5320.00
33
HEXCO NOV’93:
Below is a transportation problem where costs are in thousand of dollars.
SOURCES
DESTINATIONS
X
Y
Z
REQUIREMENTS
i.
ii.
iii.
iv.
A
B
C
14
16
20
13
15
15
15
12
16
700
300
500
CAPACITIES
500
400
600
Solve this problem fully indicating the optimum delivery allocations and the
corresponding total delivery cost. [6 marks]
There are two optimum solutions. Find the second one [4 marks].
Solve the same problem considering XA is an infeasible (prohibited / impossible)
route and find the new total transportation cost [7 marks].
If under consideration is a road network in a war zone, what is the simple economic
effect of bombing a bridge between X and A? [3 marks].
Solution:
PART (i)
TABLEAU 1:
A
B
3
5
0
X
500
Y
200
7
Z
4
5
Req
700
200
100
300
4
12
C
6
1
Capacity
500
9
400
500 10
600
500
1500
-4
The initial solution = (500 * 14) + (200 * 16) + (15 * 200) + (100 * 15) + (500 * 16)
= $22700 0000
34
TABLEAU 2:
A
3
X
500
B
C
6
5
4
Y
200
7
4
Z
0
5
300
Req
700
Capacity
5
4
12
300
200
500
9400
400
300 10
600
500
1500
Hence delivery allocations are:
X
Y
Y
Z
Z
Supplies
Supplies
Supplies
Supplies
Supplies
500
200
200
300
300
to
to
to
to
to
A
A
C
B
C
With a minimum cost of (500 * 14) + (200 * 16) + (15 * 300) + (200 * 12) + (300 * 16)
= $21900 0000
PART (ii)
The existence of an alternative least cost solution is indicated by a value of zero in an
unoccupied cell in the final table. We add and subtract the smallest quantity in the column
or row of the zero to get the alternative.
TABLEAU 1:
A
3
X
500
7
B
C
6
5
4
Y
0
Z
200 5
300
Req
700
300
4
5
4
12
400
Capacity
500
9400
400
100 10
600
500
1500
35
Hence delivery allocations are:
X
Y
Z
Z
Z
Supplies
Supplies
Supplies
Supplies
Supplies
500
400
200
300
100
to
to
to
to
to
A
C
A
B
C
With a minimum cost of (500 * 14) + (200 * 20) + (15 * 300) + (400 * 12) + (100 * 16)
= $21900 0000
PART (iii)
TABLEAU 1:
A
3
X
---
B
5
300
400
7
5
4
Z
300
5
1
12
700
Capacity
200
Y
Req
C
6
300
0
500
9
400
400
300 10
600
500
1500
Hence delivery allocations are:
X
X
Y
Z
Z
Supplies
Supplies
Supplies
Supplies
Supplies
300
200
400
300
300
to
to
to
to
to
B
C
A
A
C
With a minimum cost of (300 * 13) + (200 * 15) + (16 * 400) + (300 * 20) + (300 * 16)
= $24100 0000
PART (iv)
The simple economic effect of bombing the bridge between X and A
= 24100 0000 – 21900 0000
= 2200 000
36
 DUMMIES:
This is an extra row or column in a transportation table with zero cost in each cell and
with a total equal to the difference between total capacity and total demand.
In an unbalance transportation problem a dummy source or destination is introduced.
HEXCO NOV’98:
The transport manager of a company has 3 factories A, B and C and four warehouse I, II,
III and IV is faced with a problem of determining the way in which factories should supply
warehouses so as to minimize the total transportation costs.
In a given month the supply requirements of each warehouse, the production capacities of
the factories and the cost of shipping one unit of product from each factory to each
warehouse in $ are shown below.
FACTORY
WAREHOUSES
A
B
C
REQUIREMENTS
I
II
III
IV
PRO AVAIL
12
63
33
23
23
1
43
33
63
3
53
13
6
53
17
4
7
6
14
31
You are required to determine the minimum cost transportation plan [20 marks].
Solution:
TABLEAU 1:
I
12
A
4
2
II
23
B
51
63
5
C
21
33
-22
Req
4
III
43
1
7
33
6
63
30
6
3
- 51
10
23
IV
14
-40
14
Dummy Capacity
0
0
6
53
28
13
17
0
0
45
53
17
76
This is the initial solution, which costs
(4 * 12) + (2 * 23) + (5 * 23) + (6 * 33) + (14 * 53) + (28 * 0) + (17 * 0)
= $1149.00
37
TABLEAU 2:
I
12
A
4
II
23
60
50
B
1
63
7
C
-29
33
-22
Req
III
43
4
23
1
63
30
7
Dummy Capacity
0
50
6
3
2
33
6
IV
6
12
-40
53
28
13
17
14
0
0
45
53
17
76
The costs=
(4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (12 * 53) + (28 * 0) + (17 * 0)
= $1049.00
TABLEAU 3:
I
12
A
4
II
23
20
16
B
41
63
7
C
11
33
-22
Req
III
43
4
23
1
6
30
7
IV
Dummy Capacity
0
10
6
3
2
33
40
53
40
63
12
13
5
6
0
0
14
45
53
17
76
The costs=
(4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (40 * 0) + (12 * 13) + (5 * 0)
= $569.00
TABLEAU4:
I
12
A
4
II
23
19
63
2
C
11
33
5
4
42
32
B
Req
III
43
23
1
7
6
52
6
IV
3
2
33
18
53
63
12
13
14
Dummy Capacity
0
32
6
0
45
22
0
45
53
17
76
The costs=
(4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1)
= $459.00
38
Hence delivery allocations are:
Factory
Factory
Factory
Factory
Factory
Factory
Factory
A
A
B
B
B
C
C
Supplies
Supplies
Supplies
Supplies
Supplies
Supplies
Supplies
Warehouse
Warehouse
Warehouse
Warehouse
Warehouse
Warehouse
Warehouse
I
IV
II
III
Dummy
II
IV
With a minimum cost of (4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1)
= $459.00
HEXCO MARCH’2000:
A well-known organization has 3 warehouse and 4 Shops. It requires transporting its goods
from the warehouse to the shops. The cost of transporting a unit item from a warehouse to
a shop and the quantity to be supplied are shown below.
TO
DESTINATION
SOURCE
SOURCE
SOURCE
A
B
C
TOTAL DEMAND
I
II
III
IV TOTAL SUPPLY
10
12
0
0
7
14
20
9
16
11
20
18
5
15
15
10
15
25
5
Use any method to find the optimum transportation schedule and indicate the cost
[14marks].
 DEGENERATE SOLUTION:
It involves working a transportation problem if the number of used routes is equal to:
Number of rows + Number of column – 1.
However if the number of used routes can be less than the required figure we pretend that
an empty route is really used by allocating a zero quantity to that route.
 MAXIMIZATION PROBLEMS:
Transportation algorithm assumes that the objective is to minimize cost. However it is
possible to use the method to solve maximization problem by either:
 Multiply all the units’ contribution by – 1.
 Or by subtracting each unit contribution from the maximum contribution in
the table.
39
UNIT 3:
NON-LINEAR FUNCTIONS:
HOURS:
20.
 NON-LINEAR FUNCTIONS:
 MARGINAL DISTRIBUTION:
 PARTIAL INTEGRATION:
Partial integration is a function with more than one variable or finding the probability of a
function with more than one variables i.e. f(X1, X2, X3, ….Xn) and is just the rate at which
the values of a function change as one of the independent variables change and all others
are held constant.
Question 1:
If f(x, y) = 2(x + y –2xy) given the intervals 0<= x<=1, 0<=y<=1.
Find the marginal distribution of x = f(x).
Find the marginal distribution of y = f(x).
Solution:
Pr {0<=x<=1} = 01 2(x + y –2xy)x
= 2 01 (x + y –2xy)x
= 2 [x2/2 + xy + x2y]01
= 2 [½ + y – y]
= 2[½]
=1
Pr {0<=y<=1} = 01 2(x + y –2xy)y
= 2 01 (x + y –2xy)y
= 2 [xy +y2/2 + xy2]01
= 2 [x + ½ – x]
= 2[½]
=1
40
Question 1:
If f(X1, X2) =(X21X2 + X31X22 + X1) given the intervals 0<= X1<=2, 1<=X2 <=3.
Find the marginal distribution of x = f(x).
Find the marginal distribution of y = f(x).
Find the Expected value of X1 (E(X1)).
Find the variance of X1 (Var (X1)).
Solution:
Pr {0<=X1<=2} = 02 (X21X2 + X31X22 + X1)x
= [X31X2 /3+ X41X22 /4+ X21/2]02
= [8X2 /3+ 4X22 + 2] – [0]
= 8X2 /3+ 4X22 + 2
Pr {1<=X2<=3} = 13 (X21X2 + X31X22 + X1)y
= [X21X22 /2+ X31X32 /3+ X1X2]13
= [9X21 /2+ 27X31 /3+ 3X2]13 –[X21 /2+ X32 /3+ X1]
= 9X21 /2+ 27X31 /3+ 3X2 – X21 /2- X32 /3 - X1
= 8X21 /2+ 26X31 /3+ 2X2
Expected value of E(X1) =02 X. f(X1)x
= 02 X(X21X2 + X31X22 + X1)x
= 02 (X31X2 + X41X22 + X21)x
= [X41X2 /4+ X51X22 /5+ X31/3]02
= [4X2 + 32X22 /5 + 8 /3] – [0]
= 4X2 + 32X22 /5 + 8 /3
Variance of X1 = Var (X1) = 02 ([X1 - E(X1)]2 . f(X1)x
= 02 ([X1 - 4X2 + 32X22 /5 + 8 /3]2 * (X21X2 + X31X22 + X1)x.
Question 2:
A manufacturing company produces two products bicycles and roller skates. Its fixed costs
production is: $1200 per week. Its variables costs of production are: $40 for each bicycle
produced and $15 for each pair of roller skates. Its total weekly costs in producing x
bicycles and y pairs of roller skates are therefore c= cost.
C(x, y) = 1200 + 40x + 15y for example; in producing x = 20 bicycles and y = 30 pairs of
roller skates/ week.
The manufacture experiences total cost of:
C(20, 30) = 1200 + 40(20) + 15(30)
= 1200 + 800 + 450
= 2450.
41
Question 3:
A manufacturing of Automobile tyres produces 3 different types: regular, green and blue
tyres. If the regular tyres sell for $60 each, the green tyres for $50 each and the blue tyres
for $100 each. Find a function giving the manufacture’s total receipts or revenue from the
of x regular tyres and y green tyres and z blue tyres.
R(x, y, z) = 60x + 50y +100z.
Solution:
Since the receipts of the sale of any tyre type is the price per tyre times the number of tyres
sold:
The total receipts are:
R(x, y, z) = 60x + 50y + 100z
For example receipts from the sell of 10 tyres of each type would be:
R(10,10,10) = 60(10) + 50(10) + 100(10)
= 600 + 500 + 1000
= $2100
 PARTIAL DIFFERENTIATION:
For a function “f” of a single variable, the derivative f measures the rate at which the
values of f(x) change as the independent variable x change.
A partial derivative of a function i.e. f(X1, X2, X3.. Xn) of several variables is just the rate at
which the values of the function change as one of the independent variable changes and all
others are held constant.
Question 3:
For the function f(x, y) = X3 + 4X2Y3 + Y2
Find
 f / x
 f / y
 f(-2; 3)
Solution:
f
/ x = 3X2 + 8XY3
f
/ y = 12X2Y2 + 2Y
f(-2; 3) = X3 + 4X2Y3 + Y2
= (-2)3 + 4(-2)2(3)3 + (3)2
= -8 + 16(27) +9
= 433
42
Question 4:
A company produces electronic typewriters and word processors, it sells the electronic
typewriters for $100 each and word processors for $300 each. The company has
determined that its weekly sales in producing x electronic writers and y word processors
are given by the following joint cost function.
C(x, y) = 200 + 50x +8y + X2 + 2Y2
Find the numbers of x and y of machines that the company should manufacture and sell
weekly in order to maximize profits.
Solution:
Revenue function is given by:
R(x, y) = 100x + 300y
Profit = Revenue – Cost.
Then Profit function is given by:
P(x, y) = R(x, y) – C(x, y)
= (100x + 300y) – (200 + 50x +8y + X2 + 2Y2)
= 50x + 292y – 200 - X2 - 2Y2
To find the critical points of turning points of x and y. We set the partial derivative = 0.
Thus
p /
x => 50 – 2x = 0
 50 = 2x
 x = 25
p
/ y => 292 – 4y =0
 292 = 4y
 y = 73
The production schedule for maximum profit is therefore x = 25 type writers and y = 73
word processors which yields a profit of
P = 50(25) + 292(73) – 200 – 625 – 2(73)2
= 1250 + 21316 – 200 – 625 – 1065
= 22566 – 11493
= $11083
43
 NECESSARY AND SUFFICIENT CONDITIONS FOR EXTREMA:
The necessary condition or the GRADIENT VECTOR of the extrema determines the
turning points or critical points of a function.
Let X0 be a variable representing the turning point and represented mathematically as:
X0 = (A0, B0, … N0).
In general form; a necessary condition or gradient vector for X0 to be an extrema point of
f(x) is that the gradient ()  f (X0) = 0.
Question 1:
Given f(X1, X2, X3) =(X1 + 2X3 + X2X3 – X21 - X22 - X23)
Find the gradient vector for X0 i.e. f (X0) = 0.
Solution:
The necessary condition (gradient vector) f (X0) = 0 is given by:
f
/ x1 => 1 - 2X1 = 0.
1 - 2X1 = 0. [1]
f
/ x2 => X3 - 2X2 = 0.
X3 - 2X2 = 0. [2]
f
/ x3 => 2 + X2 - 2X3 = 0.
2 + X2 - 2X3 = 0. [3]
(a) Finding X1 is given by 1 = 2X1
X1 = ½
(b) Equation 2 is given by X3 - 2X2 = 0.
X3 = 2X2.
(c) On equation 3 where therefore substitute X3 with 2X2.
Thus 2 + X2 - 2X3 = 0.
 2 + X2 – 2(2X2) = 0.
 2 + X2 – 4X2 = 0.
 2 – 3X2 = 0.
 X2 = 2/3.
Therefore X3 = 2X2.
X3 = 2(2 / 3)
X3 = 4/3.
Therefore
X0 = (½, 2/3, 4/3)
44
A sufficient condition for X0 a point to be extremism is that the HESHIAN matrix (denoted
by H) evaluate at X0 is:
i. Positive definite when X0 is a Minimum point.
ii. Negative definite when X0 is a Maximum point.
The Hessian matrix is achieved by finding the 2nd Partial derivation of the first Partial
derivative of each equation with respect to all variables defined.
Thus the Hessian matrix is evaluated at the point X0
H
/X0 =
2f/
2
2f/
2
2f/
2
X 1,
X1 X 2,
X1 X 3
2f/
2
2f/
2
2f/
2
X2 X 1,
X 2,
X2 X 3
2f/
2
2f/
2 2f/
2
X3 X 1,
X3 X 2,
X 3
f
/ x1 => 1 - 2X1 = 0.
f
/ x2 => X3 - 2X2 = 0.
f
/ x3 => 2 + X2 - 2X3 = 0.
To establish the sufficiency the function has to have:
H
/X0 = -2
0
0
0
-2
1
0
1
-2
Since the Hessian matrix is 3 by 3 matrix then:
 Find the 1st Principal Minor determinant of 1 by 1 matrix in the Hessian matrix.
 Find the 2nd Principal Minor determinant of 2 by 2 matrix in the Hessian matrix.
 Find the 3rd Principal Minor determinant of 3 by 3 matrix in the Hessian matrix.
The Positive definite when X0 is a Minimum point is evaluated as:
 When 1st PMD = + ve.
 When 2nd PMD = +ve.
 When 3rd PMD = +ve.
Or
 When 1st PMD = - ve.
 When 2nd PMD = +ve.
 When 3rd PMD = +ve.
Thus 3 by 3 Hessian matrix the number of positive number should be greater than one.
(Should be two or more).
45
The Negative definite when X0 is a Maximum point is evaluated as:
 When 1st PMD = - ve.
 When 2nd PMD = - ve.
 When 3rd PMD = - ve.
Or
 When 1st PMD = + ve.
 When 2nd PMD = - ve.
 When 3rd PMD = - ve.
Thus 3 by 3 Hessian matrix the number of negative number should be greater than two.
(Should be two or more).
H
/X0 = -2
0
0
0
-2
1
0
1
-2
Thus the 1st PMD of (-2) = -2
Thus the 2nd PMD of
–2
0
0
-2
0
-2
1
0
1
-2
= (-2 * -2) – (0 * 0)
=4
The 3rd PMD =
=
-2 –2
1
-2
0
0
1 -0 0
-2
0
1 +0 0
-2
0
-2
1
= -2 {(-2 * -2) – (1 * 1)} – 0 (0 – 0) + 0 (0 – 0)
= - 2 (3)
=-6
Thus the PMD is equal to –2, 4 and –6 and H/X0 is negative definite and X0 = (½, 2/3, 4/3)
represents a Maximum point.
46
Question 2:
Given f(X1, X2, X3) =(-X1 + 2X3 - X2X3 + X21+ X22 - X23)
i. Find the gradient vector for X0 i.e. f (X0) = 0.
ii. Determine the nature of the turning points using Hessian Matrix.
Solution:
The necessary condition (gradient vector) f (X0) = 0 is given by:
f
/ x1 => -1 + 2X1 = 0.
-1 + 2X1 = 0. [1]
f
/ x2 => -X3 + 2X2 = 0.
-X3 + 2X2 = 0. [2]
f
/ x3 => 2 - X2 - 2X3 = 0.
2 - X2 - 2X3 = 0. [3]
(b) Finding X1 is given by -1 = 2X1
X1 = ½
(b) Equation 2 is given by -X3 + 2X2 = 0.
X3 = 2X2.
(c) On equation 3 where therefore substitute X3 with 2X2.
Thus 2 - X2 - 2X3 = 0.
 2 - X2 – 2(2X2) = 0.
 2 - X2 – 4X2 = 0.
 2 – 5X2 = 0.
 X2 = 2/5.
Therefore X3 = 2X2.
X3 = 2(2 / 5)
X3 = 4/5.
Therefore
H
/X0 =
X0 = (½, 2/5, 4/5)
2f/
2
2f/
2
2f/
2
X 1,
X1 X 2,
X1 X 3
2f/
2
2f/
2
2f/
2
X2 X 1,
X 2,
X2 X 3
2f/
2
2f/
2 2f/
2
X3 X 1,
X3 X 2,
X 3
47
H
/X0 = 2
0
0
0
2
-1
0
-1
-2
Thus the 1st PMD of (2) = 2
Thus the 2nd PMD of
2
0
0
2
0
2
-1
0
-1
-2
= (2 * 2) – (0 * 0)
=4
The 3rd PMD =
=
-2 2
-1
2
0
0
-1 - 0 0
-2
0
-1 + 0 0
-2
0
2
-1
= 2 {(2 * -2) – (-1 * -1)} – 0 (0 – 0) + 0 (0 – 0)
= 2 (-4) - (1)
= 2 (-5)
= - 10
Thus the PMD is equal to 2, 4 and –10 and H/X0 is positive definite and X0 = (½, 2/5, 4/5)
represents a Minimum point.
 NON LINEAR ALGORITHMS: (COMPUTATIONS)

THE GRADIENT METHODS:
The general idea is to generate successive iterative points, starting from a given initial
point, in the direction of the fast and increase (maximization of the function).
The method is based on solving the simultaneous equations representing the necessary
conditions for optimality namely f (X0) = 0.
Termination of the gradient method occurs at the point where the gradient vector becomes
null. This is only a necessary condition for optimality suppose that f(x) is maximized.
Let X0 be the initial point from which the procedure starts and define f (Xk) as the
gradient of f at Kth point Xk.
This result is achieved if successive point Xk and Xk+1 are selected such that
Xk+1 = Xk + rk f (Xk) where rk is a parameter called Optimal Step Size.
The parameter rk is determined such that Xk+1 results in the largest improvement in f. In
other words, if a function h(r) is defined such that h(r) = f(Xk )+ rk f (Xk). This function is
then differentiated and equate zero to the differentiatable function to obtain the value of rk.
48
Question 3:
Consider maximizing f(X1, X2) =(4X1 + 6X2 - 2X21- 2X2X1 - 2X22)
And let the initial point be given by X0(1, 1).
Hint in X0(1, 1) X1 =1 and X2 = 1
Solution:
Find f (X0) = (f/x1, f/x2)
= (4 – 4X1 – 2X2; 6 – 2X1 - 4X2)
1st iteration
Step 1: Find f (X0) = (4 – 4 – 2; 6 – 2 - 4)
= (-2; 0)
Step 2: Find Xk+1 = Xk + rk f (Xk)
 X0+1 = X0 + rk f (X0)
 X1 = (1, 1) + r(-2; 0)
 (1, 1) + (-2r, 0)
 (1 + -2r, 1)
 (1 – 2r, 1)
Thus h(r) = f(Xk )+ rk f (Xk)
= f(X0 )+ r f (X0)
= f(1 – 2r; 1)
= 4(1 – 2r) + 6(1) – 2(1 – 2r)2 – 2(1)(1 – 2r) - 2(1)2.
= 4(1 – 2r) + 6 – 2(1 – 2r) 2 – 2(1 – 2r) - 2.
= 4(1 – 2r) – 2(1 – 2r) + 6 – 2 – 2(1 – 2r)2.
= (4 – 2)(1 – 2r) + 4 – 2(1 – 2r)2.
= 2(1 – 2r) – 2(1 – 2r)2 + 4.
= – 2(1 – 2r)2 +2(1 – 2r) + 4.
= – 2(1 – 2r)2 + 2 – 4r + 4.
h1(r) = 0






– 2(1 – 2r)2 + 2 – 4r + 4 = 0.
- 4 * -2(1 – 2r) – 4 = 0.
8(1 – 2r) + - 4 = 0.
8 – 16r – 4 = 0
4 –16r = 0.
r=¼
The optimum step size yielding the maximum value of h(r) is h1 = ¼.
This gives X1= (1 –2(¼); 1)
= (1 - ½; 1)
= (½; 1)
49
UNIT 4
PROJECT MANAGEMENT WITH PERT/CPM
HOURS:
20
 PROJECT MANAGEMENT:
TERMS USED IN PROJECT MANAGEMENT:

PROJECT:
Is a combination of interrelated activities that must be executed in a certain
order before the entire task can be completed.

ACTIVITY:
Is a job requiring time and resource for its completion.

ARROW:
Represents a point in time signifying the completion of some activities and the
beginning of others.

NETWORK:
Is a graphic representation of a project’s operation and is composed of
activities and nodes.
RULES FOR CONSTRUCTING NETWORK DIAGRAM:
 Each activity is represented by one and only one arrow in the network.
 No two activities can be identified by the same head and tail events. If activities
A and B can be executed simultaneously, then a dummy activity is introduced
either between A and one end event or between B and one end event. Dummy
activities do not consume time or resources. Another use of the dummy activity:
suppose activities A and B must precede C while activity E is preceded by B
only.
 To ensure the correct precedence relationships in the network diagram, the
following questions must be answered as every activity is added to the network:
 What activities must be completed immediately before this activity
can start.
 What activities must follow this activity?
 What activities must occur concurrently with this activity?
50
Question 1:
ACTIVITY
A
B
C
D
E
F
G
H
I
J
K
L
PRECEDED BY
Initial activity
A
A
B
B
C
C
F
D
G, H
E
I
DURATION (Weeks)
10
9
7
6
12
6
8
8
4
11
5
7
Find the critical path and the time for completing the project.
Solution:
2
19 25
B 9
0
0 0
A
10
D
6
4
25 31
I
4
9
29 35
E 12
L 7
5
31 37
1
10 10
7
25 31
K
5
10
42 42
dummy
J 11
C 7
G 8
3
17 17
F
6
6
23 23
H
8
8
31 31
 EARLISET START TIME:
Represents all the activities emanating from i. Thus ESi represent the earliest occurrence
time of event i.
Earliest finish time is given by:
EF = Max {ESi + D}
51
 LATESET COMPLETION TIME:
It initiates the backward pass. Where calculations from the “end” node and moves to
the “start” node.
Latest start time is given by:
LSi = Min {LF – D}
 DETERMINATION OF THE CRITICAL PATH:
A Critical path defines a chain of critical that connects the start and end of the arrow
diagram. An activity is said to be critical if the delay in its start will cause a delay in the
completion date of the entire project. Or it is the longest route, which the project should
follow until its completion date of the entire project.
The critical path calculations include two phases:
 FORWARD PASS:
Is where calculations begin from the “start” node and move to the “end” node. At
each node a number is computed representing the earliest occurrence time of the
corresponding event.
 BACKWARD PASS:
Begins calculations from the “end” node and moves to the “start” node. The
number computed at each node represents the latest occurrence time of the
corresponding event.
 DETERMINITION OF THE FLOATS:
A Float or Spare time can only be associated with activities which are non critical. By
definition activities on the critical path cannot have floats.
There are 3 types of floats.

TOTAL FLOAT:
This is the amount of time a path of activities could be delayed without affecting the
overall project duration.
Total Float = Latest Head Time – Earliest Tail time – duration.
= LS – ES.
= LF – ES – D
= LF – EF or EC.
52
FREE FLOAT:
This is the amount of time an activity can be delayed without affecting the
commencement of a subsequent activity at its earliest start time.
Free Float = Earliest Head Time – Earliest Tail Time – Duration.
= LF – ES – D
= ESj – ESi – D.

INDEPENDENT FLOAT:
This is the amount of time an activity can be delayed when all preceding activities
are completed as late as possible and all succeeding activities completed as early as
possible.
Independent Float = EF – LS – D.

NORMAL
ACTIVITY
A
B
C
D
E
F
G
H
I
J
K
L
EARLIEST TIME
TIME ES
10
9
7
6
12
6
8
8
4
11
5
7
LATEST TIME
EF = ES + D LS = LF – D LF
0
10
10
19
19
17
17
23
25
31
31
29
10
19
17
25
31
23
25
31
29
42
36
36
0
16
10
25
25
17
23
23
31
31
37
35
10
25
17
31
37
23
31
31
35
42
42
42
TOTAL FLOAT
=LS – ES
0
6
0
6
6
0
6
0
6
0
6
6
Question 2:
Draw the network for the data given below then find the critical path as well total float
and free float.
ACTIVITY (I, J)
DURATION
(0, 1)
(0, 2)
(1, 3)
(2, 3)
(2, 4)
(3, 4)
(3,5)
(3, 6)
(4, 5)
(4, 6)
(5, 6)
2
3
2
3
2
0
3
2
7
5
6
53
Solution:
2
1
4
2
3
6
6
2
2
3
5
13 13
0
0 0
3
6
6
19 19
Dummy
3
7
2
3
2
3
ACTIVITY
D
ES
(0, 1)
(0, 2)
(1, 3)
(2, 3)
(2, 4)
(3, 4)
(3,5)
(3, 6)
(4, 5)
(4, 6)
(5, 6)
2
3
2
3
2
0
3
2
7
5
6
0
0
2
3
3
6
6
6
6
6
13
6
5
4
6
EF=ES+D
2
3
4
6
5
6
9
8
13
11
19
LS=LF-D
LF
TOTAL
Float
FREE
Float
2
0
4
3
4
6
10
17
6
14
13
4
3
6
6
6
6
13
19
13
19
19
2
0
2
0
1
0
4
11
0
8
0
2
0
2
0
1
0
4
11
0
8
0
54
 PERT ALGORITHM:
 PROBABILISTIC TIME DURATION OF ACTIVITIES.
The following are steps involved in the development of probabilistic time duration of
activities.
 Make a list of activities that make up the project including immediate
predecessors.
 Make use of step 1 sketch the required network.
 Denote the Most Likely Time by Tm, the Optimistic Time by To and
Pessimistic time by Tp.
 Using beta distribution for the activity duration the Expected Time Te
for each activity is computed by using the formula:
Te = (To + 4Tm + Tp) / 6.
 Tabulate various times i.e. Expected activity times, Earliest and Latest
times and the EST and LFT on the arrow diagram.
 Determine the total float for each activity by taking the difference
between EST and LFT.
 Identify the critical activities and the expected date of completion of the
project.
 Using the values of Tp and To compute the variance (2) of each
activity’s time estimates by using the formula: 2 = {{Tp – To} / 6}2.
 Compute the standard normal deviate by:
Zo = (Due date – Expected date of Completion) / Project variance.
 Use Standard normal tables to find the probability P (Z <= Zo) of
completing the project within the scheduled time, where Z ~ N(0,1).
Question 3:
A project schedule has the following characteristics:
Activity
1–2
2–3
2–4
3–5
4–5
4–6
5–7
6–7
7–8
7–9
8 – 10
9 – 10
Most Likely Time
2
2
3
4
3
5
5
7
4
6
2
5
Optimistic Time
Pessimistic Time
1
1
1
3
2
3
4
6
2
4
1
3
3
3
5
5
4
7
6
8
6
8
3
7
55
I.
II.
III.
IV.
Construct the project network.
Find expected duration and variance for each activity.
Find the critical path and expected project length.
What is the probability of completing the project in 30 days.
Solution:
3
4 8
5
8 12
4
5
7
17 17
6
9
23 23
2
3
1
0 0
2
7
4
5
2
2 2
10
28 28
3
2
4
5 5
6
10 10
5
Expected job Time
8
21 26
Te = (To + 4Tm + Tp) / 6.
2 = {{Tp – To} / 6}2.
Variance
Activity
Tm
To
Tp
Te
2
1–2
2–3
2–4
3–5
4–5
4–6
5–7
6–7
7–8
7–9
8 – 10
9 – 10
2
2
3
4
3
5
5
7
4
6
2
5
1
1
1
3
2
3
4
6
2
4
1
3
3
3
5
5
4
7
6
8
6
8
3
7
2
2
3
4
3
5
5
7
4
6
2
5
0.111
0.111
0.445
0.111
0.111
0.445
0.111
0.111
0.445
0.445
0.111
0.445
Critical path (*) comprises of activities (1 –2), (2 - 4), (4 –6), (6 –7), (7 –9) and (9 –10)
Expected project length is = 28 days.
56
Variance 2 = 0.111 + 0.445 + 0.445 + 0.111 + 0.445 + 0.445
= 2.00 (on critical path only)
(iv) Probability of completing the project in 30 days is obtained by:
Zo = (Due date – Expected date of Completion) / Project variance.
= (30 – 28) / 2.
= 1.414 (Look this from Normal tables)
Now from Standard Normal tables Z= 0.4207.
P (t <= 30) = P (Z <= 1.414)
= 0.5 + 0.4207
= 0.9207
0
1.414
This shows that the probability of meeting the scheduled time will be 0.9207
 COST CONSIDERATIONS IN PERT / CPM:
The cost of a project includes direct costs and indirect costs. The direct costs are associated
with the individual activities and the indirect costs are associated with the overhead costs
such as administration or supervision cost. The direct cost increase if the job duration is to
be reduced whereas the indirect costs increase if the job duration is to be increased.

TIME COST OPTIMIZATION PROCEDURE:
The process of shortening a project is called Crashing and is usually achieved by adding
extra resources to an activity. Project crashing involves the following steps:




Critical Path: Find the normal critical path and identify the critical activities.
Cost Slope: Calculate the cost slope for the different activities by using the
Formula: COST SLOPE = Crash cost – Normal cost.
Normal Time – Crash Time.
Ranking: Rank the activities in the ascending order of cost slope.
Crashing: Crash the activities in the critical path as per the ranking i.e.
activities having lower cost slope would be crashed first to the maximum
extent possible. Calculate the new direct cost by cumulatively adding the cost
of crashing to the normal cost.
57


Parallel Crashing: As the critical path duration is reduced by the crash in step
3 other paths become critical i.e. we get parallel critical paths. This means
that project duration can be reduced by simultaneous crashing of activities in
the parallel critical paths.
Optimal Duration. Crashing as per Step 3 and step 4 an optimal project is
determined. It would be the time duration corresponding to which the total
cost (i.e. Direct cost plus Indirect cost) is a minimum.
Question 4:
For the network given below find the optimum cost schedule for the completion of the
project:
JOB
1–2
2–3
2–4
3–4
3–5
3–6
4–5
5–6
NORMAL
CRASH
TIME
COST $
TIME
COST $
10
9
7
6
9
10
6
7
60
75
90
100
50
40
50
70
8
6
4
5
7
8
4
5
120
150
150
140
80
70
70
110
Solution:
JOB
*1 – 2
*2 – 3
2–4
3–4
3–5
3–6
 4–5
 5–6
COST SLOPE
30 --->(4)
25 ---> (3)
20
40 ---> (5)
15
15
10 ---> (1)
20 ---> (2)
=(120 –60) / (10 – 8)
58
6
38 38
10
3
19 19
9
1
0 0
10
6
2
10 10
9
7
7
4
25 25
6
5
31 31
The critical path = 1 –2, 2 –3, 3 –4, 4 –5 and 5 –7.
Expected project Length = 38 days.
Associated with 38 days the minimum direct project cost
= 60 + 75 + 90 + 100 + 50 + 40 + 50 + 70
= $535
In order to reduce the project duration we have to crash at least one of the jobs on the
critical path. This is being done because crashing of the job not on the critical path does not
reduce the project length.
1st Crashing:
On critical path the minimum cost slope is job 4 –5 and is to be crashed at extra cost of $10
per day.
3
19 19
6
36 36
10
9
1
0 0
10
6
2
10 10
9
7
7
4
25 25
4
5
29 29
Duration of project = 36 days and Total cost = $535 + $10 * 2 = $555.
59
2nd Crashing:
Now crash job 5 –6 and is to be crashed at extra cost of $20 per day.
3
19 19
6
34 34
10
9
1
0 0
10
6
2
10 10
9
5
7
4
25 25
4
5
29 29
Duration of project = 34 days and Total cost = $555 + $20 * 2 = $595.
3rd Crashing:
Now crash job 2 –3 for 3 days and is to be crashed at extra cost of $25.
3
16 16
6
31 31
10
6
1
0 0
10
6
2
10 10
9
5
7
4
22 22
4
5
26 26
Duration of project = 31 days and Total cost = $595 + $25 * 3 = $670.
4th Crashing:
60
Now crash job 1 –2 for 2 days and is to be crashed at extra cost of $30.
3
14 14
6
29 29
10
6
1
0 0
8
6
2
8 8
9
5
7
4
20 20
5
24 24
4
Duration of project = 29 days and Total cost = $670 + $30 * 2 = $730.
5th Crashing:
Final crash job 3 –4 for 1 day and it is to be crashed at extra cost of $40 and two critical
paths occurs.
3
14 14
6
28 28
10
6
1
0 0
8
5
2
8 8
9
5
7
4
19 19
4
5
23 23
Duration of project = 28 days and Total cost = $730+ $40 * 1 = $770.
Optimum Duration of project = 28 days and Total Cost = $770.
UNIT 5:
RANDOM VARIABLES AND THEIR PROBABILITY:
HOURS:
20.
61
 RANDOM VARIABLES AND PROBABILITY FUNCTIONS (DISCRETE):
Given a Sample space S = {1,2,3,4,5,6} we may therefore use the variable such as X to
represent an outcome in the sample space such a variable is called Random variable.
When the outcome in a sample space are represented by values in a random variable the
assignment of probabilities to the outcome can be thought of as a function for which the
domain is the sample space, we refer to this as the probability function written as Pr.
We use the following notation with probability function Pr {X = a} which means the
probability associated with the outcome a while Pr {X in E} means the probability
associated with event E.
Given S = {1,2,3,4,5,6}
a. Find the probability of S = 3.
b. Find the probability of S = 5.
c. Find the probability of X in E when E = 1,2.3.
Solution:
i. P (S = 3) = 1/6.
ii. P (S = 5) = 1/6.
iii. P (X in E) = ½.

PROBABILITY DENSITY FUNCTIONS:
 PROPERTIES OF DISCRETE RANDOM VARIABLE:
a. It is a discrete variable.
b. It can only assume values x1, x2. …xn.
c. The probabilities associated with these values are p1, p2. …pn.
Where P(X = x1) = p1.
P(X = x2) = p2.
.
.
P(X = xn) = pn.
Then X is a discrete random variable if p1 + p2. …pn = 1.
This can be written as ∑ P(X = x) =1.
all x
Question 1:
The P.d.f. of a discrete random variable Y is given by P (Y=y) = cy2, for y = 0,1,2,3,4.
Given that c is a constant, find the value of c.
Solution:
62
Y
P(Y =y)
0
0
1
c
2
4c
3
9c
4
16c
= ∑ P(X = x) =1.
all x
 1 = c + 4c + 9c + 16c
 1 = 30c
 c = 1/30.
Question 2:
The Pdf. of a discrete random variable X is given by P (X=x) = a(¾)x , for x = 0,1,2,3...
Find the value of the constant a.
Solution:
= ∑P(X = x) =1.
all x
P(X = 0) = a(¾)0.
P(X = 1) = a(¾)1.
P(X = 2) = a(¾)2.
P(X = 3) = a(¾)3 and so on.
So ∑P(X = x) =a + a(¾) + a(¾)2 + a(¾)3 + …
all x
= a( 1 + ¾ + (¾)2 + (¾)3 + …)
= a ( 1/1- ¾) -> (sum of an infinite G.P with first term 1 and common ratio ¾)
= a(4)
4a = 1
a=¼
 EXPECTED VALUE/ MEAN / AVERAGE:
For a random variable X associated with a sample space {x1, x2. …xn} the concept of
expected value is the generalization of the average of numbers {x1, x2. …xn}.
 EXPECTED VALUE WITH SAME PROBABILITIES:
63
The expected value of X with same probabilities is given by:
E(x) = X1 + X2 + …Xn/n.
Question 3:
Given that an die is thrown 6 times and the recordings are as follows then calculate the
expected mean or mean score
Score x
P(X = x)
1
1/
6
2
1/
6
3
1/
6
4
1/
6
5
1/
6
6
1/
6
Solution:
E(x) = X1 + X2 + …Xn/n.
= 1 + 2 + 3 + 4 + 5 + 6/6
= 21/6
= 7/2
= 3.5
 EXPECTED VALUE WITH DIFFERENT PROBABILITIES:
The expected value of X with different probabilities is given by:
E(x) = P1 * X1 + P2 * X2 + …Pn * Xn.
Question 4:
Given a random variable X which has a Pdf shown below. Calculate the expected mean.
X
P(X = x)
-2
0.3
-1
0.1
0
1
0.15 0.4
2
0.05
Solution:
E(x) = P1 * X1 + P2 * X2 + …Pn * Xn
= -2 * 0.3 + -1 * 0.1 + 0 * 0.15 + 1 * 0.4 + 2 * 0.05
= -0.2
Question 5:
A venture capital firm is determined based on the past experience that for each $100
invested in a high technology startup company; a return of $400 is experienced 20% of
time. A return of $100 is experienced 40% of the time and zero (0) total loss is
experienced 40% of the time.
What is the firm’s expected return based on this data?.
Solution:
S = {400, 100, 0}
E(x) = P1 * X1 + P2 * X2 + …Pn * Xn
64
= 400 * 0.2 + 100 * 0.4 + 0 * 0.4.
= 80 + 40 + 0
= $120
Question 6:
Given that an unbiased die was thrown 120 times and the recordings are as follows then
calculate the expected mean or mean score.
Score x
Frequency f
1
15
2
22
3
23
4
19
5
23
6
18
Total = 120
Solution:
E(x) = ∑fx/∑f.
= (15 + 44 + 69 + 76 + 115 +108)/120.
= 3.558
Question 7:
The random variable X has Pdf P(X=x) for x = 1,2,3.
X
P(X = x)
1
0.1
2
0.6
3
0.3
Calculate:
a. E(3).
b. E(x).
c. E(5x).
d. E(5x + 3).
e. 5E(x) + 3.
f. E(x2).
g. E(4x2 - 3).
h. 4E(x2) – 3.
Solution:
X
5x
5x + 3
x2
4x2 – 3
P(X = x)
1
5
8
1
1
0.1
2
10
13
4
13
0.6
3
15
18
9
33
0.3
a. E(3) = ∑P(X = x).
all x
65
= ∑3P(X = x).
all x
= 3(0.1) + 3(0.6) + 3(0.3).
= 3.
b. E(x) =∑xP(X = x).
all x
=1(0.1) +2(0.6) +3(0.3)
= 2.2
c. E(5x) = ∑5xP(X = x).
all x
= 5(0.1) + 10(0.6) +15(0.3)
= 11
d. E(5x + 3) = ∑(5x + 3)P(X = x).
all x
= 8(0.1) + 13(0.6) + 18(0.3)
= 14
e. 5E(x) + 3 = 5(2.2) + 3
= 14
f. E(x2) = ∑ x2P(X = x).
all x
= 1(0.1) + 4(0.6) + 9(0.3)
= 5.2
g. E(4x2 - 3) = ∑(4x2 – 3)P(X = x).
all x
= 1(0.3) + 13(0.6) + 33(0.3)
= 17.8
66
h. 4E(x2) – 3 = 4(5.2) – 3
=20.8 – 3
= 17.8
 VARIANCE:
The expected value of a random variable is a measure of central tendency i.e. what
values are mostly likely to occur while Variance is a measure of how far apart the
possible values are spread again weighted by their respective probabilities.
The formula for variance is given by:
Var (x) = E(x – μ)2 this can be reduced to
Var (x) = E(x2) - μ2
Question 8:
The random variable X has probability distribution shown below.
x
P(X =x)
1
0.1
2
0.3
3
0.2
4
0.3
5
0.1
Find:
μ = E(x).
Var(x) using the formula E(x – μ)2
E(x2)
Var(x) using the formula E(x2) - μ2
i.
ii.
iii.
iv.
Solution:
i.
E(x) =μ = ∑xP(X = x).
all x
=1(0.1) + 2(0.3) + 3(0.2) + 4(0.3) + 5(0.1)
=3
ii.
Var (x) = E(x – μ)2
=∑(x – 3)2P(X = x).
all x
X
(x – 3)
(x – 3)2
P(X = x)
1
-2
4
0.1
2
-1
1
0.3
3
0
0
0.2
4
1
1
0.3
5
2
4
0.1
67
= 4(0.1) + 1(0.3) + 0(0.2) + 1(0.3) + 4(0.1)
= 1.4
iii.
E(x2) = ∑x2P(X = x).
all x
= 1(0.1) + 4(0.3) + 9(0.2) + 16(0.3) + 25(0.1)
= 10.4
iv.
Var(x) = E(x2) - μ2
= 10.4 – 9
= 1.4
 STANDARD DEVIATION:
Is the square root of its variance given by the following formula:
δ = √Var (x).
Question 9:
From the question given above find the standard deviation for part (iv).
δ = √Var (x).
δ = √1.4
= 1.183215957
= 1.18
 CUMULATIVE DISTRIBUTION FUNCTION:
When we had a frequency distribution, the corresponding Cumulative frequencies were
obtained by summing all the frequencies up to a particular value.
In the same way if X is a discrete random variable, the corresponding Cumulative
Probabilities are obtained by summing all the probabilities up to a particular value.
If X is a discrete random variable with Pdf P(X = x) for x = x1, x2. …xn then the
Cumulative distribution function is given by:
F(t) = P(X <= t)
= ∑t P(X = x).
x = x1
The Cumulative Distribution is sometimes called Distribution function.
Question 10:
68
The probability distribution for the random variable X is given below then constructs
the Cumulative distribution table.
X
0
P(X =x) 0.03
1
0.04
2
0.06
3
0.12
4
0.4
5
0.15
6
0.2
Solution:
F(t) = ∑t P(X = x).
x = x1
So
F(0) = P(X <= 0) = 0.03
F(1) = P(X <= 1) = 0.03 + 0.04 = 0.07
F(2) = P(X <= 2) = 0.03 + 0.04 + 0.06 = 0.13 and so on.
The Cumulative Distribution table will be as follows:
X
0
1
F(x) 0.03 0.07
2
0.13
3
0.25
4
0.65
5
0.8
6
1
Question 11:
For a discrete random variable X the Cumulative distribution function F(x) is given below:
X
F(x)
1
0.2
2
0.32
3
0.67
4
0.9
5
1
Find:
a) P(x = 3).
b) P(x > 2).
Solution:
a. F(3) = P(x = 3) => P(x = 1) + P(x = 2) + P(x = 3)
= 0.67
F(2) = P(x <= 2) => P(x = 1) + P(x = 2)
= 0.32
Therefore P(x = 3) = 0.67 – 0.32
= 0.35
b. P(x > 2) = 1 – P(x <= 2)
= 1 – F(2)
= 1 – 0.32
= 0.68
 PROBABILITY DISTRIBUTION (CONTINUOUS RANDOM VARIABLES):
69
A random variable X that can be equal to any number in an interval, which can be
either finite or infinite length, is called a Continuous Random Variable.

PROBABILITY DENSITY FUNCTIONS:
There are 2 essential properties of Pdf:
 Because probabilities cannot be negative. The integral of a function must be nonnegative for all choices of interval [a, b] i.e. f(x) >= 0 for all values in the sample
space for the random variable X.
 Since the probability associated with the entire sample space is always 1. The
integral of f(x) of the entire sample space = 1.
Question 1:
A continuous random variable has Pdf f(x) where f(x) = kx, 0<= x <= 4.
i. Find the value of constant k.
ii. Sketch y = f(x).
iii. Find P(1 <= X <= 2½).
Solution:
i.
∫ f(x) ∂x = 1.
all x
∫04 kx ∂x = 1.
[kx2/2]04 = 1.
8k = 1
k=⅛
ii.
Sketch of y = f(x).
½
y = ⅛x
70
0
4
P(1<= x <= 2½) = ∫1
2½
[⅛x]∂x.
= [x2/16]1
2½
= 0.328
Question 2:
A continuous random variable has Pdf f(x) where
Kx
k(4 – x)
0
f(x)=
0<= x <= 4.
2<= x <= 4
otherwise
a) Find the value of constant k.
b) Sketch y = f(x).
Solution:
∫
b
D => a P ∂x +
∫ab Q ∂x = 1.
∫02 kx ∂x + ∫24 k(4 – x) ∂x = 1.
=>
=> [kx2/2]02 + [4xk - kx2/2]24 = 1.
=> [4k/2] – [0] + {[16k - 16k/2] – [8k - 4k/2]} = 1.
 [2k] + {[8k] – [6k]} = 1.
 4k = 1
 k=¼
c. Sketch y = f(x).
X
Y
0
0
1
2
¼
½
F(x) = kx.
X
Y
2
½
3
4
¼
0
F(x) = k(4 – x)
3
¾
4
1
71
1
¾
½
¼
0
1
2
3
4
 EXPECTED VALUE OR MEAN (CONTINUOUS RANDOM VARIABLE):
For a continuous random variable X defined within finite interval [a, b] with
continuous Pdf f(x) then the expected value or mean is given by:
∫ab x. f(x) ∂x
E(x) =
Question 3:
A continuous random variable has Pdf f(x) where
6/ x
7
6/ x(2
7
f(x)=
– x)
0
i.
ii.
0<= x <= 1.
1<= x <= 2
otherwise
Find E(x).
Find E(x2).
Solution:
∫
b
D => a P ∂x +
∫ab Q ∂x = 1.
E(x) =
∫ab x. f(x) ∂x
E(x) =
∫01 6/7 x2 ∂x + ∫12 6/7 x2(2 – x)∂x
= 6/7[x3/3]01 + 6/7[2/3x3 – x4/4]12
= 6/7[⅓] + 6/7{16/3 – 4 – (⅔ - ¼)}
= 6/7[5/4]
= 15/14
72
E(x2) =
∫ab x2. f(x) ∂x
E(x2) =
∫01 6/7 x3 ∂x + ∫12 6/7 x3(2 – x)∂x
= 6/7[x4/4]01 + 6/7[x4/2 – x5/5]12
= 6/7[¼] + 6/7{8 - 32/5 – (½ - 1/5)}
= 6/7[31/20]
= 93/70
 VARIANCE AND STANDARD DEVIATION:
The variance and Standard Deviation associated with a continuous random variable X
on the sample space [a, b] is given by:
∫
b
Var (x) = a [x - E(x)]2 . f(x) ∂x or Var(x) = E(x2) - μ2 ∂x
and Standard deviation = √Var (x).
E(x) = μ.
Question 4:
A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4.
Find:
a) E(x).
b) E(x2).
c) Var (x).
d) The standard deviation of x.
e) Var(3x +2).
Solution:
i.
E(x) =
∫ab x. f(x) ∂x
E(x) =
∫04 ⅛x2 ∂x
= ⅛[x3/3]04
= 8/3
73
ii.
E(x2) =
∫ab x2. f(x) ∂x
∫04 ⅛x3 ∂x
=
= ⅛[x4/4]04
= ⅛(64)
=8
iii.
Var(x) = E(x2) - μ2 ∂x
= E(x2) - E2(x) ∂x
= 8 – (8/3)2
= 8/9
iv.
Standard Deviation =>  = √Var (x).
= √8/9
= 2√2/3
v.
Var(3x + 2) = 9 Var(x) this has been obtained form the concept
Var (ax)= Var a2(x).
= 9 (8/9)
=8
 MODE:
The Mode is the value of X for which f(x) is greatest in the given range of X. It is usually
to draw a sketch of y = f(x) and this will give an idea of the location of the Mode.
For some Probability Density functions it is possible to determine the mode by finding
the maximum point of the curve y = f(x) from the relationship f1(x) = 0.
f1(x) = d/∂x * f(x).
Question 5:
A continuous random variable has Pdf f(x) where f(x) = 3/80(2 + x)(4 – x), 0<= x<= 4.
a) Sketch y = f(x).
b) Find the mode.
Solution:
X
0
1
2
3
4
74
Y
/80
/80
24
27
/80
24
/80
15
0
a)
/80
Mode
/80
f(x) = 3/80(2 + x)(4 – x)
27
24
/80
15
0
1
2
3
4
b) The mode => f(x) = 3/80(2 + x)(4 – x)
= 3/80(8 + 2x – x2)
f1(x) = (2+ 2x)
f1(x) = 0.
0 = 2+ 2x
2x = 2
x=1
 MEDIAN:
The median splits the area under the curve y = f(x) into 2 halves so if the value of the
Median is m. Therefore the formula for the median is given by:
∫am f(x) ∂x = 0.5.
F(m) = 0.5
Question 6:
A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4.
Find:
a. The median m.
Solution:
m =>
∫am f(x) ∂x = 0.5.
F(m) = 0.5
f(x) = ⅛x ∂x
75
0.5 = m2/16
m2 = 8
m = 2.83
 CUMULATIVE DISTRIBUTION FUNCTION: F(x)
When considering a frequency distribution the corresponding cumulative frequencies
were obtained by summing all the frequencies up to a particular value.
In the same way if X is a continuous random variable with Pdf f(x) defined for a<=x<=b
then the Cumulative Distribution Function is given by F(t):
F(t) = P(X <= t) =

∫at f(x) ∂x
PROPERTIES OF CDF:
 F(b) =
∫ab f(x) ∂x = 1.
∫
t
 If f(x) is valid for - <= x <=  then F(t) = - f(x) ∂x where the interval is
taken over all values of x <= t.
 The Cumulative distribution function is sometimes known as just as the
distribution function.
Question 6:
A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4.
Find:
i. The Cumulative distribution function F(x).
ii. Sketch y = F(x).
iii. Find P(0.3 <=x<= 1.8).
Solution:
i.
∫at f(x) ∂x
t
F(t) = ∫0 ⅛x ∂x
F(t) =
= ⅛[x2/2]0t
= t2/16
F(t) = t2/16 0<=t<=4
76
NB: (1) F(4) = 42/16 = 1
F(x) =
ii.
0
x2/16
1
x <= 0.
0<= x <= 4
x >= 4
Sketch y = F(x).
X
Y
0
0
1
1
2
/16
1
3
/2
9
4
1
/16
1
F(x) = 1
9
/16
1
/2
1
/16
F(x)= x2/16
0 1
2
3
4
iii. P(0.3 <= x <= 1.8) = F(1.8) – F(0.3)
F(1.8) = (1.8)2/16
= 0.2025
F(0.3) = (0.3)2/16
= 0.005625
Therefore P(0.3 <= x <= 1.8) = F(1.8) – F(0.3)
= 0.2025 – 0.005625
= 0.196875
= 0.197
Question 7:
A continuous random variable has Pdf f(x) where
x/
f(x)=
3
-2x/
0
3
+2
0<= x <= 2.
2<= x <= 3
otherwise
a. Sketch y = f(x).
b. Find the Cumulative distribution function F(x).
c. Sketch y = F(x).
77
d. Find P(1 <= X <= 2.5)
e. Find the median m.
Solution:
i. Sketch y = f(x).
X
Y
0
0
X
Y
1
⅓
2
⅔
2
⅔
3
0
⅔
y = x/3
⅓
y =-2x/3 +2
0
ii.
1
2
CDF = F(t) =
3
∫0t x/3∂x
= [x2/6]0t
= t2/6
F(t) = x2/6
0<=x<=2
NB: F(2) = 22/6 = ⅔
F(t) = F(2) + (Area under the curve y = -2x/3 +2 between 2 and t)
So
F(t) = F(2) +
∫2t (-2x/3 +2) ∂x
= F(2) + [-x2/3 + 2x]2t
= ⅔ + {-t2/3 +2t – ( -4/3 + 4)}
= -t2/3 +2t – 2
2<= t <= 3
78
NB: F(2) = -9/3 + 6 – 2 = 1
Therefore CDF =
x2/6
- x2/ +2x -2
3
1
f(x)=
iii.
0<= x <= 2.
2<= x <= 3.
x >= 3.
Sketch of y = F(x).
y=1
1
y = - x2/3 +2x -2
2
/3
y = x2/6
1
/3
0
iv.
1
2
3
P(1 <= X <= 2.5) = F(2.5) – F(1) as 2.5 is in the range 2<= x <=3.
 F(2.5) = - x2/3 +2x –2
 F(2.5) = - (2.5)2/3 +2(2.5) –2
 = 11/12
F(1) = x2/6 as 1 is in the range 0 <= x <= 2.
F(1) = x2/6
F(1) = 12/6

 = 1/6

Therefore P(1 <= X <= 2.5) = F(2.5) – F(1)
= 11/12 - 1/6
= 0.75
v.
m =>
∫am f(x) ∂x = 0.5 where m is the median.
F(2) = ⅔ so the median must lie in the range 0 <= x <= 2.
F(m) = m2/6
m2/6 = 0.5
79
m2 = 3.
m = 1.73
 OBTAINING THE PDF FROM THE CDF:
The Probability Density Function can be obtained from the Cumulative Distribution
function as follows:
∫
t
Now F(t) = a f(x) ∂x
So
f(x) = d/∂x * F(x).
= F1(x).
a<= t <= b.
NB: The gradient of the F(x) curve gives the value of f(x).
Question 8:
A continuous random variable has Pdf f(x) where
F(x)=
0
x3/27
1
x <= 0.
0<= x <= 3
x >= 3.
Find the Pdf of X, f(x) and sketch y = f(x).
Solution:
a. f(x) = d/∂x * F(x).
= d/∂x(x3/27).
= 3x2/27
= x2/9
Therefore the Pdf is equal to:
x2/9
0<=x<=3
0
otherwise.
f(x) =
b. Sketch of y = f(x).
y = x2/9
1
0
1
2
3
80
Question 9:
A continuous random variable X takes values in the interval 0 to 3.
It is given that P(X > x) = a + bx3, 0 <= x <= 3.
i. Find the values of the constants a and b.
ii. Find the Cumulative distribution function F(x).
iii. Find the Probability density function f(x).
iv.
Show that E(x) = 2.25.
v. Find the Standard deviation.
Solution:
a. P(X > x) = a + bx3, 0 <= x <= 3.
So P(X > 0) = 1 and P(X > 3) = 0.
i.e. a + b(0) = 1 and a + b(27) = 0
Therefore a = 1 and 1 + 27b = 0.
B = -1/27.
So P(X > x) = 1 - x3/27,
0 <= x <= 3.
P(X <= x) = x3/27 (CDF)
b. Now
X3/27 0<=x<=3
F(x) =
1
x > 3.
c. f(x) = d/∂x * F(x).
= d/∂x(x3/27).
= 3x2/27
= x2/9
d. E(x) =
∫ab x. f(x) ∂x
=
∫03 x. x2/9∂x
=
∫03 x3/27∂x
= [x4/36]03
= 2.25
81
e. Var(x) =
=
∫ab [x - E(x)]2 . f(x) ∂x = ∫ab x2.f(x)∂x - E2(X)
∫03 x4/9∂x – 2.252.
=[x5/45]03 - 5.0625.
= 0.3375
f. δ = √ Var (x).
= √ 0.3375
= 0.581
 RELATIONSHIPS AMONG PROBABILITY DISTRIBUTIONS:
 JOINT PROBABILITY DISTRIBUTION:
Question 1:
2(X + Y - 2XY)
Given f(X, Y) = 0
i.
ii.
iii.
0<= X<=1, 0<= Y<=1
Otherwise
Show that this is a PDF.
Find P(0 <= X <=½), (0 <= Y <=¼).
Find CDF.
Solution:
b
b
a) =∫a ∂X∫a ∂Y
1
1
=∫0 ∂X∫0 ∂Y [2(X + Y - 2XY)]
1
= 2∫0 ∂X [(XY + Y2/2 - XY2)]01
1
= 2∫0 ∂X [(X + ½ - X)]
= 2[(X2/2 + ½X - X2/2)]01
= 2[½ + ½ - ½]
=2*½
=1
82
b
b
b) =∫a ∂X∫a ∂Y
½
¼
=∫0 ∂X∫0 ∂Y [2(X + Y - 2XY)]
½
= 2∫0 ∂X[(XY + Y2/2 - XY2)]0¼
½
= 2∫0 ∂X[(¼X + 1/32 - 1/16X)]
½
= 2∫0 ∂X[(3/16X + 1/32)]
= 2[(3/32X2 + 1/32X)] 0½
= 2[(3/128 + 1/64)]
= 2 * 5/128
= 5/64
b
b
c) F(u, v) =∫a ∂u∫a ∂v
x
y
=∫0 ∂U∫0 ∂V[2(U + V – 2UV)]
= 2∫0x∂U[(UV + V2/2 - UV2)]0y
x
= 2∫0 ∂U[(UY + Y2/2 - UY2)]
= 2[(YU2/2 + UY2/2 - U2X2/2)]0X
= 2[(X2Y/2 + XY2/2 - X2X2/2)]
= X2Y + XY2 - X2X2
NB: Given CDF to get PDF we differentiate the function.
Given PDF to get CDF we integrate the function.
Given a PDF to prove that it is a PDF integrate until the answer is 1.
Given PDF the marginal distribution is found by integrating.
Given CDF the marginal distribution is found by differentiating.
Question 1:
Given the CDF = x2 + 3xyz + z.
Find the corresponding PDF.
Solution:
To get the corresponding PDF we differentiate with respect to x, y, z.
f(x, y, z) =x2 + 3xyz + z
= d/∂x(2x + 3yz).
= d/∂y(3z).
= d/∂z (3).
=3
83
UNIT 6
DECISION THEORY
HOURS:
20
 DECISION THEORY:
 DECISION UNDER RISK:
A. THE EXPECTED VALUE CRITERION:
Muchadura buys and sell tomatoes at $3 and $8 a case respectively. Because tomatoes are
perishable, they must be bought a day and sold after that, they become valuable.
A 90 days observation of a business reveals the following information in cases.
Daily Sales
Number Of Days Sold
Probability Of Demand
10 cases
11 cases
12 cases
13 cases
18
36
27
9
(18 / 90) =
(36 / 90) =
(27 / 90) =
(9 / 90) =
Total number of days
90
=
0.2
0.4
0.3
0.1
1
The Probability of demand = Number of days sold / Total number of days.
There are only 4 sales volumes but their sequence in UNKNOWN. The problem is how
many cases must Muchadura stock for selling the following day. If he stocks more or less
than the demand on any day he will suffer some losses.
Example:
If he stocks 13 cases on a particular day and the customer demand on that particular day is
10 cases, then he gets a profit of 3 cases.
Let buying price = $3.
Let Selling price = $8.
NB: This method if getting Maximum profit = Selling Price – Buying Price per case = $5.
Then 10 * 5 = $50 and perishable ones = 3 * 3 = $9 and then the Maximum profit = $50 - $9
= $41.
For the 3 unsold cases he suffer a loss for each equivalent to the Buying price. All such
losses are calculated in a conditional profit table. I.e. a table of profit on a given supply and
demand levels as illustrated below.
84
STOCK ACTION
EXPECTED PROFIT
Possible
Demand
10
11
12
13
Probability
of Demand
10
11
12
13
50
50
50
50
47
55
55
55
44
52
60
60
41
49
57
65
Total
10
Profits
11
12
13
0.2
0.4
0.3
0.1
10
20
15
5
9.4
22
16.5
5.5
8.8
20.8
18
6
8.2
19.6
17.1
6.5
1
50
53.4
53.6
51.4
Muchadura must stock 12 cases which give the highest Expected profit of $53.60 on a daily
bases to get the maximum profit over a long time.
B. EXPECTED PROFIT WITH PERFECT INFORMATION:
Suppose Muchadura had perfect information i.e. complete and accurate information about
the future. Sales demand would still vary from 10 to 13 cases in their respective
probabilities. But Muchadura would not know in advance how many cases would be
needed each day.
The table below shows the conditional profits value in such situation.
STOCK ACTION
EXPECTED PROFIT
Possible
Demand
Probability
of Demand
10
10
11
12
13
Total
11
12
13
50
55
60
65
10
0.2
0.4
0.3
0.1
10
1
10
Profits
11
12
13
22
18
6.5
22
18
6.5
Maximum possible profit = 10 + 22 + 18 + 6.5.
= $56.50
85
C. MINIMISING EXPECTED LOSS:
There are two types of Losses:
 Overstocking
 Under stocking.
OVERSTOCKING:
Means you will loss the buying amount.
UNDER STOCKING:
It means you will loss the profit.
STOCK ACTION
EXPECTED LOSS
Possible
Demand
10
11
12
13
Total
Probability
of Demand
10
11
12
13
0
5
10
15
3
0
5
10
6
3
0
5
9
6
3
0
10
Loss
11
12
13
0.2
0.4
0.3
0.1
0
2
3
1.5
0.6
0
1.5
1
1.2
1.2
0
0.5
1.8
2.4
0.9
0
1
6.5
3.1
2.9
5.1
The Expected Minimum Loss is $2.9. The values above the zero diagonal are due to
overstocking and those below are due to under stocking.
D. THE EXPECTED VALUE OF PERFECT INFORMATION:
The expected value of perfect information is calculated as:
The expected Profit with perfect information – The expected value Criterion.
= 56.50 – 53.60
= 2.9
In general the expected value of perfect information = Minimum Expected Loss.
86
E. ITEMS WITH A SALVAGE VALUE:
Main items do not become completely worthless after their prime deaths. They will have
reduced values (Salvage values). The Salvage values must be considered when computing
conditional profits or losses.
Consider the previous case when the cost price was $3 and selling price was $8. Any unsold
cases will be disposed / salvage of at $4 the profit will be $4 - $3 = $1.
STOCK ACTION
EXPECTED PROFIT
Possible
Demand
Probability
of Demand
10
11
12
13
10
11
12
13
50
50
50
50
51
55
55
55
52
56
60
60
53
57
61
65
Total
10
Profits
11
12
13
0.2
0.4
0.3
0.1
10
20
15
5
10.2
22
16.5
5.5
10.4
22.4
18
6
10.6
22.8
18.3
6.8
1
50
54.2
56.8
58.2
Muchadura must stock 13 cases, which give the maximum profit of $58.20.
 DECISION MAKING UNDER UNCERTAINITY:
Under conditions of uncertainty, only payoffs are known and nothing is known about
the likelihood of each state of nature.
Different persons have suggested several decisions rules for making decisions under
such situations:
A. MAXIMIN (PESSIMISTIC):
It is based upon the consecutive approach to assume that the worst possible is going
to happen. The decision maker considers each alternative and locates the minimum
payoff for each and then selects that alternative which maximizes the minimum
payoffs.
STEPS INVOLVED:
i.
ii.
Determine the minimum assured payoffs for each alternative.
Choose that alternative which corresponds to the Maximum of the above
minimum payoff.
87
B. MAXIMAX CRITERIA (OPTIMISTIC):
Is based upon extreme optimistic, the decision maker selects that particular strategy
which corresponds to the maximum of the maximum payoffs of each strategy.
STEPS INVOLVED:
i.
ii.
Determine the maximum possible payoff for each alternative.
Select that alternative which corresponds to the maximum of the above
maximum payoff.
C. HURWICZ:
In order to overcome the disadvantage of extreme pessimistic (Maximin) and
extreme Optimism (Maximax) criteria. Hurwicz introduced the concept of coefficient of optimism or pessimism as .
Therefore the Hurwicz concept is given by:
 Maximin j {V (ai, j)} + (1 - ) Maximax j {V(ai, j)}
D. LAPLACE CRITERION:
This criterion is based on what is known as the principal if insufficient reason. Since
the probabilities associated with the occurrences of 1, 2….n are unknown, we do
not have enough information to conclude that these probabilities will be different.
For if this is not the case, we should be able to determine these probabilities and the
situation will no longer be a decision under uncertainty. Thus because of insufficient
reason to believe otherwise; the states of 1, 2….n are equally likely to occur.
The Laplace principal assumes that of 1, 2….n are likely to occur therefore the
probability = 1/n where n = number of occurrences.
E. MAXIMIN REGRET CRITERION (SALVAGE):
This is a less conservative criterion. In this case a new loss matrix is created such
that V(Ai, Qj) is replaced by R(Ai, Qj) which is defined by:
Max {V (Ai, Qj) – V(Ai, Qj)} if V is profit
R(Ai, Qj)=
V(Ai, Qj) – Min {V(Ai, Qj)} if V is loss
R(Ai, Qj) is the difference between the best choice in column Qj and the values of
V(Ai, Qj) in the same column.
R(Ai, Qj) is a regret matrix.
88
Question 1:
Given the following data:
Supplier
Level
A1
A2
A3
A4
Q1
Customers Category
Q2
Q3
Q4
5
8
21
30
10
7
18
22
18
8
12
9
25
23
21
25
Using the above data:
Find
i. Maximin Criterion.
ii. Maximax Criterion.
iii. Hurwicz Criterion.
iv. Laplace Criterion.
v. (Salvage) Maximin Regret Criterion.
Solution:
a) Maximin Criterion.
Supplier
Level
A1
A2
A3
A4
Q1
Customers Category
Q2
Q3
Q4
5
8
21
30
10
7
18
22
18
8
12
9
25
23
21
25
Maximin
5
7
12
9
Maximin = 12
The optimal decision is to take supply level A3
b) Maximax Criterion.
Supplier
Level
A1
A2
A3
A4
Q1
Customers Category
Q2
Q3
Q4
5
8
21
30
10
7
18
22
18
8
12
9
25
23
21
25
Maximax
25
23
21
30
Maximax = 30
The optimal decision is to take supply level A4
89
c) Minimax Criterion.
Supplier
Level
A1
A2
A3
A4
Q1
Customers Category
Q2
Q3
Q4
5
8
21
30
10
7
18
22
18
8
12
19
25
23
21
25
Minimax
25
23
21
30
Minimax = 21
The optimal decision is to take supply level A3
d) Hurwicz Criterion.
Supplier
Level
A1
A2
A3
A4
Q1
Customers Category
Q2
Q3
Q4
5
8
21
30
10
7
18
22
18
8
12
19
25
23
21
25
Hurwicz: =  Maximin j {V (ai, j)} + (1 - ) Maximax j {V(ai, j)}
Let assume that α = ½.
Maximin
5
7
12
9
Maximax
25
23
21
30
 Maximin + (1 - ) Maximax
15
15
16.5
19.5
The optimal decision is to take supply level A4
e) Laplace Criterion.
Supplier
Level
A1
A2
A3
A4
Q1
Customers Category
Q2
Q3
Q4
5
8
21
30
10
7
18
22
18
8
12
19
25
23
21
25
90
Laplace: = 1/n
Expected costs for different actions A1, A2, A3 and A4 are:
E(A1) = ¼(5 + 10 + 18 + 25) = 14.5
E(A2) = ¼(8 + 7 + 8 + 23) = 11.5
E(A3) = ¼(21 + 18 + 12 + 21) = 18
E(A1) = ¼(30 + 22 + 19 + 25) = 24
The optimal decision is to take supply level A2
f) Savage Minimax Regret Criterion.
Customers Category
Q1
Q2
Q3
Q4
Supplier
Level
A1
A2
A3
A4
A1
A2
A3
A4
5
8
21
30
10
7
18
22
18
8
12
19
25
23
21
25
Q1
Q2
Q3
Q4
0
3
16
25
3
0
11
15
10
0
4
11
4
2
0
4
Max{ R(Ai, Qj)}
10
3
16
25
Minimax
The optimal decision is to take supply level A2 which gives Minimax.
Question 2:
A firm has to decide on its advertising campaign. It has a choice between Tv,
Newspaper, Poster and Radio advertising. The return on the advertising medium is
measured by the number of potential customers who have the opportunity to see
each medium. This will depend on the type of the weather. The figures in the table
are the numbers of potential customers in thousands. The firm has funds for using
only to one medium.
Poor Moderate
Tv
Newspaper
Poster
Radio
200
180
110
210
190
160
140
190
Weather Good
170
150
140
160
Excellent
130
130
190
110
91
Decide which medium the firm should use based on the following information:
i. Maximin Criterion.
ii. Maximax Criterion.
iii. Hurwicz Criterion.
iv. Laplace Criterion.
v. (Salvage) Maximin Regret Criterion.
Solution:
a) Maximin Criterion.
Tv
Newspaper
Poster
Radio
Poor Mod
Good Excellent
200
180
110
210
170
150
140
160
190
160
140
190
130
130
190
110
Maximin
130
130
110
110
Maximin = 130
The best medium to use for advertising is Tv or Newspaper since they have the
highest number of (possible) potential customers of 130
b) Maximax Criterion.
Tv
Newspaper
Poster
Radio
Poor Mod
Good Excellent
200
180
110
210
170
150
140
160
190
160
140
190
130
130
190
110
Maximax
200
180
190
210
Maximax = 210
The best medium to use for advertising is the Radio since it has the highest number
of (possible) potential customers of 210.
92
c) Minimax Criterion.
Tv
Newspaper
Poster
Radio
Poor Mod
Good Excellent
200
180
110
210
170
150
140
160
190
160
140
190
130
130
190
110
Minimax
200
180
190
210
Minimax = 180
The best medium to use for advertising is the Newspaper with 180 potential
customers.
d) Hurwicz Criterion.
Poor
Tv
Newspaper
Poster
Radio
200
180
110
210
Mod Good
190
160
140
190
170
150
140
160
Excellent
130
130
190
110
Hurwicz: =  Maximin j {V (ai, j)} + (1 - ) Maximax j {V(ai, j)}
Let assume that α = 0.7
Maximin
130
130
110
110
Maximax
200
180
190
210
 Maximin + (1 - ) Maximax
151
145
109.7
140
The best medium to use for advertising is the Tv since it has the highest number of
(possible) potential customers of 151
93
g) Laplace Criterion.
Tv
Newspaper
Poster
Radio
Poor Mod
Good Excellent
200
180
110
210
170
150
140
160
190
160
140
190
130
130
190
110
Laplace: = 1/n
Expected costs for different actions A1, A2, A3 and A4 are:
E(Tv) = ¼(200 + 190 + 170 + 130) = 172 500
E(Newspaper) = ¼(180 +160 + 150 + 130) = 155 000
E(Poster) = ¼(110 + 140 + 140 + 190) = 145 000
E(Radio) = ¼(210 + 190 + 160 + 110) = 167 500
The best medium to use for advertising is Tv since it has the highest number of
(possible) potential customers who amount to 172 500
h) Savage Minimax Regret Criterion.
Tv
Newspaper
Poster
Radio
Tv
Newspaper
Poster
Radio
Poor Mod
Good Excellent
200
180
110
210
170
150
140
160
190
160
140
190
130
130
190
110
Poor Mod
Good Excellent
90
70
0
100
30
10
0
20
50
20
0
50
20
20
80
0
Max {R(Ai, Qj)}
90
70
80
100
Minimax
The best medium to use for advertising is Newspaper, which minimize maximum
view ship.
94
UNIT 7:
THEORY OF GAMES:
HOURS:
20
 GAME THEORY:
It deals with Decision under Uncertainty which involves 2 or more intelligent opponents in
which each opponent aim to optimize his / her own decision at the expense of the other
opponent. Typical examples include launching an advertisement campaign. (2) Competing
for products or planning war techniques for opposing enemy.
In game theory an opponent is referred to as a Player. Each player has the number of
choices, limited or unlimited called Strategies. The outcomes payoffs of a game are
summarized as functions of different strategies for each player.
A game with 2 players were a gain of one player is equal to the loss of another player is
called a Two Person Zero Sum game.
To illustrate two person zero sum game considers a coin-matching situation in which each
of the players A and B select a Head H or Tail T.
If the outcome match Head or Tail Player A wins $1 from Player B otherwise Player A
loses $1 to Player B.
In this game each player has 2 strategies Head or Tail which yields the following 2*2
game matrix expressed in terms of payoff to Player A.
Player B
H
T
Player A
H 1
T
-1
-1
1
 OPTIMUM SOLUTION OF TWO PERSON ZERO SUM GAME:
Is obtained by using minimax Maximin criterion according to which Player A (whose
strategies represents rows) select a strategy (mixed or Pure) which maximize his minimum
gains, the minimum being taken over all the strategies of Player B.
In similar way Player B selects his strategy that minimize his maximum loses.

VALUE OF THE GAME:
Is the maximum guaranteed gain to Player A or the minimum possible loss to Player B
denoted by V.
When Maximin possible value is equal to Minimax value of the corresponding pure
strategies, the game is said to have the Optimum Strategies and has a Saddle point.
95
Player B
2
3
1
Player A
1
2
3
8
6
7
Minimax
2
5
3
8
9
7
-4
5
4
5
18
10
9
Maximin
2
5
-4
18
The game has a Saddle point which exists at row 2 column 2 and the value of the game = 5.
Player B
2
3
1
Player A
1
2
3
Minimax
1
2
6
3
1
2
6
Maximin
1
1
1
6
3
1
3
6
NB: The optimum solution which shows that Maximin <> Minimax means it is a mixed
strategy i.e. there is no agreement (poor strategy). Below and above show that there is no
Saddle point which means Maximin <> Minimax.
1
Player A
1
2
3
4
Minimax
5
6
8
3
Player B
2
3
-10
7
7
4
8
7
9
8
15
-1
15
4
Maximin
-10
1
2
-1
0
1
2
4
4
Consider the following game G.
B1
Player A
A1
A2
Minimax
Player B
B2
2
-2
2
6
-1
Maximin
2
-1
6
Maximin = Minimax which is a condition strictly determinable game hence the game is
strictly determinable whatever  maybe. The value of the game = 2 with the best strategy
for Player A from A1 and the best strategy for Player B from B1.
96
Question 1:
Find the range of values of P and Q, which will render the entry (2,2) a Saddle point of the
game.
Player B
B1
B2
B3
Maximin
4
5
Player A
A1 2
2
7
q
A2 10
7
p
8
A3 4
4
Minimax
10
7
8
Ignoring values of P and Q determine the Maximin and Minimax value of the payoff
matrix. The Maximin =f and Minimax = f thus there exists a Saddle point at position (2,2).
This impose a condition on P as P<= f and on Q as Q >= f.
Hence the required range for P and Q is = 7<= Q, P<= 7.
Question 2:
For what values of  is the game with the following payoff. Payoff is strictly determinable.
B1
Player A
Minimax
A1
A2
A3
Player B
B2
B3

-1
-2
-1
6

4
2
-7

6
Maximin
2
-7
-2
2
This shows that the value of the game (V) lies between –1 and 2.
That is –1 <= V <= 2. For a strictly determinable we have –1 <=  <= 2.
97
 GRAPHICAL SOLUTION OF 2 BY N & M BY 2 GAMES:
The graphical solution of 2 by N and M by 2 is only applicable to games in which at least on
of the Players have 2 strategies only.
Consider the following 2 by N games.
Player B
Y1
Y2
Player A
X1
X2
A11
A21
A12
A22
Y3
A1n
A2n
The expected payoff of the corresponding to the pure strategies of B are given below:
B’s Pure Strategies
1
2
N
A’s Expected Payoff
(A11 –A12) X1 + A21
(A21 – A22) X1 + A22
(A2n –A2n) X1 + A2n
This shows that A’s average payoff varies linearly with X1.
According to the Minimax criterion for mixed strategies games Player A should select the
value V that minimize his maximum expected playoffs. This may be done by plotting a
straight line as function of X1.
98
Question 3:
Consider the following 2 by 4 games.
1
Player A
1
2
2
4
Player B
2
2
3
3
3
2
4
-1
6
Solution:
B’s Pure Strategies
A’s Expected Payoff
(2 – 4) X1 + 4
(2 – 3) X1 + 3
(3 – 2) X1 + 2
(-1 –7) X1 + 6
1
2
3
4
B’s Pure Strategies
A’s Expected Payoff
1
2
3
4
6
5
4
3
2
1
X1 0
-1
-2
- 2X1 + 4
-X1 + 3
X1 + 2
- 7X1 + 6
1/2
6
5
4
3
2
1
0 X1
-1
-2
This is a point of integration of any 2 of the lines or more i.e. lines 2, 3 and 4.
A’s optimum strategy is X1 = ½ and X2 = 5/2 and the value of the game is obtained by
substituting X1 in the equations of any of the lines passing the intersection of the point.
The value of the game = 5/2.
99
Question 4:
Consider the following 2 by 3 game, find B’s pure strategies and A’s expected payoff using
the graphically method.
Player B
1
2
3
Player A
1
2
1
8
3
5
11
2
Solution:
B’s Pure Strategies
1
2
3
11
10
9
8
7
6
5
4
3
2
1
X1 0
-1
-2
A’s Expected Payoff
- 7X1 + 8
- 2X1 + 5
9X1 + 2
H
Maximin
11
10
9
8
7
6
5
4
3
2
1
0 X1
-1
-2
Since Player A wishes to Maximize his Minimum expected payoff we consider the highest
point of intersection on the lower elevation of A’s expected payoff equations.
This point it represents the maximum value of the game for A.
Lines 2 and 3 passes through it defines the relative moves B2 and B3 it along B needs to
play.
100
The solution of the original 2 * 3 game reduces to a 2 by 2 payoff matrix.
Player A
Player B
B1
B2
A1 3
11
X1+ X2 = 1
A2 5
Y1 + Y2 = 1
2
Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained
by:
X1 = 3 – 11 / (3 + 2) – (5 + 11)
= 8 / 11
X2 = 1 - 8
/ 11 = 3 / 11
The value of the game V = (2 * 3) – (5 * 11)
= 49
/ (3 + 2) – (5 + 11)
/ 11
101
Question 4:
Consider the following 2 by 4 game, find B’s pure strategies and A’s expected payoff using
the graphically method.
Player B
1
2
3
4
Player A
1
2
1
2
3
5
-3
4
7
-6
Solution:
B’s Pure Strategies
A’s Expected Payoff
1
2
3
4
11
10
9
8
7
6
5
4
3
2
1
X1 0
-1
-2
-3
-4
-5
-6
- X1 + 2
- 2X1 + 5
- 7X1 + 4
13X1 - 6
H
Maximin
11
10
9
8
7
6
5
4
3
2
1
0 X1
-1
-2
-3
-4
-5
-6
102
The solution of the original 2 * 4 game reduces to a 2 by 2 payoff matrix.
Player A
Player B
B1
B2
A1 -3
7
X1+ X2 = 1
A2
Y1 + Y2 = 1
4
-6
Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained
by:
X1 = -3 – 7 / (-3 + -6) – (4 + 7)
= 11 / 20
X2 = 1 - 11
Y1
/ 20 = 9 / 20
= -6 – 7 / (-3 + -6) – (4 + 7)
Y2 = 1 - 13
= 13 / 20
/ 20 = 7 / 20
The value of the game V = (-2 * -6) – (7 * 4)
= 40
/ (3 + 2) – (5 + 11)
/ 20 = 2
 MINIMAX PROBLEMS:
Obtain the optimum strategy for both persons and the value of the game for zero person
sum game whose payoff matrix is as follows:
Player B
1
2
Player A
1
2
3
4
5
6
1
3
-1
4
2
-5
-3
5
6
1
2
0
103
Solution:
A’s Pure Strategies
B’s Expected Payoff
1
2
3
4
5
6
- 4Y1 - 3
- 2Y1 + 5
- 6Y1 + 6
5Y1 + 1
2Y1 + 2
-5Y1 + 0
11
10
9
8
7
6
5
4
3
2
1
Y1 0
-1
-2
-3
-4
-5
-6
-7
Minimax
H
11
10
9
8
7
6
5
4
3
2
1
0 Y1
-1
-2
-3
-4
-5
-6
-7
The solution of the original 6 * 2 game reduces to a 2 by 2 payoff matrix.
Player B
B1
B2
Player A
A1
A2
3
5
4
1
X1+ X2 = 1
Y1 + Y2 = 1
104
Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained
by:
X2 = 3 – 5 / (3 + 1) – (4 + 5)
=2/5
X1 = 1 - 2
Y1
/5 =3/5
= 1 – 5 / (3 + 1) – (4 + 5)
Y2 = 1 - 4
=4/5
/5 =1/5
The value of the game V =
(1 * 3) – (4 * 5)
= 17
/ (3 + 1) – (4 + 5)
/5
 GENERAL SOLUTION OF M * N RECTANGULAR GAMES:
In a rectangular game with M * N payoff matrix if there does exists any saddle point and is
not possible to reduce the size of the game to 2 by 2 payoff matrix.
The following methods are generally used to solve the game:
 Linear programming method.
 Iterative Method.
Question 5:
Solve the following game; consider the game problem in which B’s linear programming
problem.
Player B
1
2
3
Maximin
8
4
2
Player A
1
2
2
8
4
2
2
1
2
8
3
1
Minimax
8
8
8
Player B
Y1
Y2
Y3
Player A
X1
X2
X3
8
2
2
4
8
2
2
4
8
105
B’s linear programming problem is as follows:
Maximize Z = Y1 + Y2 + Y3
Subject to:
8Y1 + 4Y2 + 2Y3 <= 1 {1}
2Y1 + 8Y2 + 4Y3 <= 1 {2}
Y1 + 2Y2 + 8Y3 < = 1 {3}
Y1 , Y2 , Y3 >= 0
TABLEAU 1:
Y1
Y2
Y3
S1
S2
S3
Solution
S1
8
4
2
1
0
0
1
S2
2
8
4
0
1
0
1
S3
1
2
8
0
0
1
1
Z
1
1
1
0
0
0
0
TABLEAU 2:
Y1
Y2
Y3
S1
S2
S3
Solution
Y1
1
0.5
0.25
0.125 0
0
0.125
S2
0
7
3.5
-0.25 1
0
0.75
S3
0
1.5
7.75
-0.125 0
1
0.875
Z
0
0.5
0.75
-0.125 0
0
-0.125
S3
Solution
PC
PC
TABLEAU 3:
Y1
Y2
Y3
S1
Y1
1
0
0
0.107 -0.072 0
0.072
Y2
0
0.5
0.036 0.143 0
0.107
S3
0
0
7
-0.179 -0.215 1
0.715
0
0.5
-0.143 -0.072 0
-0.1785
Z
0
1
S2
106
TABLEAU 3:
Y1
Y2
Y3
S1
Y1
1
0
0
0.107 -0.072
Y2
0
1
0
0.049 0.159 -0.072 0.057
Y3
0
0
1
-0.026 -0.031 0.143 0.102
0
0
-0.156 -0.087 -0.072 -0.2295
Z
0
S2
S3
0
Solution
0.072
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum
Z, Y1= 0.072, Y2= 0.057 and Y1= 0.102 producing Z = $0.2295.
UNIT 8:
INVENTORY MODELLING:
HOURS:
20
 INVENTORY SYSTEM MODEL:
Inventory deals with manufacturing sufficient stock of goods (parts and raw materials)
to ensure a smooth operation of a business activity.

GENERALIZED INVENTORY MODEL:
The ultimate objective of inventory model is to answer the following two questions:
 How much to order.
 When to order.
The answer to the 1st question is expressed in terms of economic order quantity (EOQ),
which is the optimum amount that should be ordered every time an order is placed and
may vary with time depending on the situation under consideration.
The 2nd part depends on the type of the inventory type. If the inventory system requires
periodic review i.e. equal time intervals (e.g. every week or month). The time for acquiring
a new order usually coincides with the beginning of each time interval.
If a system is of continuous review type a reorder point is usually specified by an inventory
level at which a new order must be placed.
We can express the solution of the general inventory problems as follows:
107
 PERIODIC REVIEW CASE:
Receive a new order of the amount specified by the order quantity at equal intervals
of order time, which can be weekly or monthly.
 CONTINUOUS REVIEW CASE:
When the inventory level reaches the re-order point places an order whose size is
equal to the quantity.
The order quantity and re-order point are normally determined by minimizing the
total inventory cost, the point which can be expressed as a function of its principle
components in the following manner.
Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.

PURCHASING COST:
Purchasing cost is the price of the quantity of goods to be supplied. The purchasing
cost becomes an important factor when the commodity unit prices become
dependant on the size of the order. This situation is normally expressed in terms of
quantity discount or price break where the unit price of an item decrease with the
increase of order quantity.

SETUP COST / ORDERING COST:
Represents a fixed charge incurred when an order is placed. Thus to satisfy the
demand for a given time period, ordering of smaller quantities will result in a higher
set up cost during the period than if the demand is satisfied by placing large orders.

HOLDING COST / CARRYING COST:
Represents the cost of carrying inventory or represents the cost of maintaining
inventory e.g. warehouse rental, security, handling, depreciation, stock
maintenance, insurance and loses of interest on capital.

SHORTAGE COST:
Is a penalty incurred when we run out of stock of the needed commodity. It
generally includes costs due to the loss of customer’s goodwill as well as potential
loss in income. We need to minimize the total cost involved in stocking operations by
determining the optimum quantity for the replenishment order Q.
108
 INVENTORY NOTATION:
 Tc – Total inventory cost per unit time.
 D – Total demand per unit time.
 Ci – Cost of production or purchase per unit expressed as $ per unit.
 Q – Quantity per order.
 Q* - Optimum quantity = EOQ.
 H – Holding cost or Carrying cost per unit of inventory per time.
 T – Time between orders or length of inventory cycle Q/D.
 K – Cost of processing an order or Ordering cost / Setup cost.
 L – Leading time.
 Cs – Shortage cost.
Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.
Tc = CiD + QH/2 + DK/Q + Cs.
We need to minimize Tc depending on the quantity Q hence by differentiation
we get
(Tc)/Q = (QH/2)/ Q + (DK/Q)/Q + (CiD)/Q + (Cs)/Q
0 = H/2 – DKQ2
0 = H/2 – DK/Q2
Multiply both sides by 2Q2 we get 0 = HQ2 – 2DK
0 = HQ2 – 2DK
HQ2 = 2DK
2DK
Q2 =
/H
2DK
Q* = 
/H (This is the formula for EOQ)
109
Economic Order Quantity is a deterministic model referred to as Wilson Economic Lot
Size.
 EOQ ASSUMPTIONS:
Tc = CiD + QH/2 + DK/Q + Cs.
The above model will hold for the following assumptions:
1)
2)
3)
4)
Demand should be constant.
Replenishment should be instantaneous.
Shortages are not allowed.
Fixed quantity order per inventory order level should be maintained.
Question 1:
The demand for an item is 18000 units per month and the holding cost per unit is $14.40
per year and the cost of ordering is $400. No shortages are allowed and the replenishment
rate is instantaneous.
a) Determine the optimum order quantity.
b) The total cost per year of the inventory if the cost of one unit is $1.
c) The number of orders per year.
d) The time between orders.
Solution:
Since all the information asked for are years. Convert all information in months to years.
D = 18000 * 12 = 216000 per year.
H = $14.40 per year.
K = $400.
EOQ => Q* = 
2DK
/H


2* 21600 * 400/14.40
172800000/14.40
 12000000
 3464.101615
 3464
110
Tc = CiD +
QH
/2 + DK/Q.
= (1 * 216000) + (3464 * 14.40)/2 + (216000 * 400)/3464
= 216000 + 24940.80 + 24942.26
= 265883.06
The number of orders per year = D/Q
=216000/3464
= 62.36 orders
The time between orders =
Q
/D
= 3464/216000
= 0.016
= 0.02
Question 2:
The demand for a particular item is 18000 units per year. Holding cost per unit is $1.20 per
year and the cost of procurement is $400. No shortages are allowed and the replenishment
rate is instantaneous.
a. Determine the optimum order quantity.
b.The number of orders per year.
c. The time between orders.
d.The total cost per year of the inventory if the cost of one unit is $1.
Solution:
D = 18000 per year.
H = $1.20 per year.
K = $400.
EOQ => Q* = 
2DK
/H


2* 18000 * 400/1.20
14400000/1.20
 12000000
 3464.101615
 3464
111
The number of orders per year =
D
/Q
=18000/3464
= 5.196 orders
The time between orders = Q/D
or 365/5.196 = 70.24 days
= 3464/18000
= 0.1924
= 0.19 days
Tc = CiD + QH/2 + DK/Q.
= (1 * 18000) + (3464 * 1.20)/2 + (18000 * 400)/3461
= 18000 + 206.40 + 2078.52
= 22156.92
Question 3:
Suppose a company has soft drinks products. It has a constant annual demand rate of 3600
cases. A case of soft drinks cost the Harare drinks company $3. Ordering cost is $20 per
order and holding cost is charged at 25% of the cost per unit. There are 250 working days
per year and the lead-time is 5 days.
Identify the following aspects of inventory policy:
i.
Economic order quantity.
ii. Re-order point.
iii. Cycle time (Length) = Q/D.
iv.
The total annual cost.
Solution:
D = 3600.
H = 25% of $3.
K = $20 per order.
Ci = $3 per unit.
i.
EOQ => Q* = 


2DK
/H
2* 3600 * 20

/0.75
144000

/0.75
 192000
 438.178
 438
112
ii.
Re-order point = Daily Demand * Lead-time.
= 3600/250 * 5
= 72 crates
iii.
Cycle time = Q/Daily demand.
Daily Demand is given by = 3600/250 = 14.40
Cycle Time = 438/14.40
= 30.42 days
Tc = CiD + QH/2 + DK/Q.
iv.
= (3 * 3600) + (438 * 0.75)/2 + (20 * 3600)/438
= 10800 + 164.25 + 164.38356
= 11128.633
= 11128.63
`
 RE-ORDER POINT:
Is a point in which a new order should be placed.

i.
ii.
iii.
GENERAL RULES TO DETERMINE THE RE-ORDER POINT:
Determine the number of days the order should be placed (Cycle length) = Q/D.
Determine the number of days which is the minimum level to re-order new entities
and is given by Lead-time – Cycle Length (Q/D) where Lead-time is the time between
the placement of an order and its receipt.
Determine the level of inventory (stock) to re-order new items and is given by:
Number of days which gives minimum level to order new items * Demand.
113
Question 4:
The daily demand for a commodity is approximately 100 units. Every time an order is
placed a fixed cost of $100 is incurred and daily holding cost per unit inventory $0.02.
If the lead-time is 12 days.
Determine:
i. Economic order quantity.
ii. Re-order point.
iii. Cycle time.
iv.
Minimum inventory level.
v. Level of inventory stock.
Solution:
D = 100.
H = $0.02.
K = $100.
i.
2DK
EOQ => Q* = 
/H
2* 100 * 100



/0.02
20000/0.02
 1000000
 1000
ii.
Re-order point = Demand * Lead-time
= 100 * 12
= 1200
iii.
Cycle time = Q/D
= 1000/100
= 10.
vi.
Minimum inventory level = Lead-time – Cycle Length.
= 12 – 10
= 2 days
vii.
Level of inventory stock = Minimum inventory level * Demand.
= 2 * 100
= 200 orders.
114
 EOQ WITH GRADUAL REPLENISHMENT / ECONOMIC PRODUCTION LOT
SIZE:
Suppose a demand of an item in a company is D units per unit time and the company
can produce them at a rate of R units per unit time where R>D. The cost of set-up is $K
and the holding cost per unit time is $H. We wish to determine the Optimum
manufacturing quantity and the total cost per year assuming the cost of one unit is $Ci.
It is assumed that shortages are not allowed.
EOQ => Q* = 
2DK
/H(1 – D/R)
Question 5:
A company uses 100000 units per year that cost $3 each. The carrying costs are 1% per
month and the ordering cost are $2.50 per year.
What is the economic order quantity made the items themselves on a machine with a
potential capacity of 600000 units per year.
Solution:
D = 100000.
K = $2.50.
H = 1% = 0.01 * 12 * 3 = 0.36
EOQ => Q* = 
=
2DK
/H(1 – D/R)
2* 100000 * 2.50
/0.36(1 – 100000/600000)
1
= 500000/0.36(1 – /6)
=1666666.667
= 1290.994449
=1291.
115
Question 6:
A company manufactures an item, which is also used in the company. The demand of this
item is 18000 units per year and the production rate is 3000 per month. The cost of set up is
$500. The holding cost of 1 unit per month is $0.15.
Determine the optimum manufacturing quantity and the total cost per year assuming the
cost of 1 unit is $2 and shortages are not allowed.
Solution:
D = 18000 per year.
H = $0.15 * 12 = $1.8 per year.
K = 500
Ci = $2
R = 3000 * 12 = 36000 per year.
EOQ => Q* = 
=
2DK
/H(1 – D/R)
2* 18000 * 500
/1.8 *(1 – 18000/36000)
= 18000000/0.36(1 – 0.5)
== 18000000/0.18
=3162.27766
= 3162
Tc = CiD + QH/2 + DK/Q.
= (2 * 18000) + (3162 * 1.8)/2 + (18000 * 500)/3162
= 36000 + 2845.8 + 2846.29981
= 41692.09981
= 41692
116
 INVENTORY CONTROL (TYPES OF CONTROL SYSTEMS):
There are 2 two basic inventory control systems:
 Reorder Level
 Periodic Review system.

PERIODIC REVIEW SYSTEM:
This is the reviewing of all stocks at fixed intervals periodically. Therefore
absolute stocks can be eliminated. Variable quantities can be ordered.

THE REORDER LEVEL SYSTEM:
The reorder level system results in fixed quantities being ordered at variable
intervals dependent upon demand. The reorder level system usually has
three (3) control levels namely:
 The reorder level.
 The minimum level.
 The maximum level.
NB:
 The Reorder Level is calculated so that if the worst
anticipated position occurs, Stock would be replenished in
time.
 The Minimum Level is calculated so that management will
be warned when demand is above average. There may be
no danger but the situation needs watching.
 The Maximum Level is calculated so that management will
be warned when demand is the minimum anticipated and
quencequently the stock level is likely to rise.
Question 7:
The following is an illustration of a company with a simple manual reorder level system.
Normal Usage
= 110 per day.
Minimum Usage
= 50 per day.
Maximum Usage
= 140 per day.
Lead time
= 25 – 30 days.
EOQ
= 5000
Using the above data calculate the various control levels i.e.
 Reorder level.
 Minimum Usage.
 Maximum Usage. [20 marks].
117
Solution:
a) Reorder Level = Maximum Usage * Maximum Lead time.
= 140 * 30
= 4200 units
b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage)
= 4200 – (27.5 * 110)
= 4200 – 3025
= 1175 units
c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage)
= 4200 + 5000 – (25 * 50)
= 7950
Question 8:
The following is an illustration of a company with a simple manual reorder level system.
Normal Usage
= 560 per day.
Minimum Usage
= 250 per day.
Maximum Usage
= 750 per day.
Lead time
= 15 – 20 days.
EOQ
= 10000
Using the above data calculate the various control levels i.e.
 Reorder level.
 Minimum Usage.
 Maximum Usage. [20 marks].
Solution:
a) Reorder Level = Maximum Usage * Maximum Lead time.
= 750 * 20
= 15000 units
b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage)
= 15000 – (17.5 * 560)
= 15000 – 9800
= 5200 units
c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage)
= 15000 + 10000 – (15 * 250)
= 21250
118
UNIT 8:
REGRESSION ANALYSIS AND CORRELATION:
HOURS:
20

CORRELATION:
Correlation is a measure of strength of a linear relationship between 2 sets of numbers.
There are 2 measures of correlation namely:
 The Pearson Product Moment Correlation Coefficient represented by r.
 The Spearman’s Rank Correlation Coefficient represented by R.
A coefficient of +1 indicates a perfect Positive relationship whilst a coefficient of – 1 shows
perfect negative correlation.
Hence the correlation coefficient will have values ranging between – 1 and + 1 inclusive.
A coefficient of zero implies no correlation.
SCATTER DIAGRAM:
It is a way of representing a set of data by a scatter of plots or dots. One variable is
plotted on the X-axis and another on the Y-axis. Normally variable X is the one
controlled over (independent variable) and Y variable is the variable that you are
interested in (dependent variable).
POSITIVE CORRELATION:
*
*
* *
Delivery
*
*
Time
*
* *
* *
*
Number of deliveries
NEGATIVE CORRELATION:
*
* *
* *
* *
* *
* *
*
Temperature
NO CORRELATION:
*
*
*
*
* * *
*
*
*
* *
*
*
*
*
* * *
*
*
* * *
*
*
Age
NON-LINEAR CORRELATION:
**
* *
**
* *
**
* *
**
* *
**
* *
**
*
*
Quantity produced
Salary
The first diagram is a positive because number of deliveries increase so apparently does the
delivery time.
The second is a negative correlation because as air temperature increases the heating cost
falls.
For the third correlation shows that there is no correlation existing between salary and age of
an employee.
119
For non-linear correlation it suggest that quantity produced and efficiency are correlated but
not linearly.
Given two sets of data represented by the variables X and Y the Product Moment Correlation
Coefficient is given by:
r =  (X - ) (Y – Ý)

 (X - )2 * (Y – Ý)2
OR
n ΣXY
ΣX
-

ΣY
r =
2
2
[n Σ X - (ΣX)]
2
2
[n Σ Y - (ΣY)]
Question 1:
Calculate r for the data given below:
X
Y
15
60
24
45
25
50
30
35
35
42
40
46
45
28
65
20
70
22
75
15
Solution:
X
15
24
25
30
35
40
45
65
70
75
424
Y
60
45
50
35
42
46
28
20
22
15
363
XY
900
1080
1250
1050
1470
1840
1260
1300
1540
1125
12835
X2
225
576
625
900
1225
1600
2025
4225
4900
5625
21926
Y2
3600
2025
2500
1225
1764
2116
784
400
484
225
15123
120
n ΣXY
ΣX
-

ΣY
r =
2
2
2
[n Σ X - (ΣX)]
10 * 12835
2
[n Σ Y - (ΣY)]
- 424 * 363
r =
10* 21926 - 179776 * 10 * 15123 - 131769
128350 - 153912
r =
39484 * 19461
- 25562
r =
27719.9959
r = - 0.922150238
r = - 0.92
Question 2:
Calculate r for the data given below:
X
Y
1
5
2
3
6
10.5 15.5 25
5
16
6
22.5
Solution:
X
1
2
3
4
5
6
21
Y
5
10.5
15.5
25
16
22.5
94.5
X2
XY
5
21
46.5
100
80
135
387.5
1
4
9
16
25
36
91
Y2
25
110.25
240.25
625
256
506.25
1762.75
121
n ΣXY
ΣX
-

ΣY
r =
2
2
[n Σ X - (ΣX)]
6 * 387.5
2
2
[n Σ Y - (ΣY)]
- 21 * 94.5
r =
6* 91 - 441
* 6 * 1762.75 – 8930.25
2325 – 1984.5
r =
546 - 441 * 10576 – 8930.25
340.5
r =
415.7598467
r = 0.819
(B) THE SPEARMAN’S RANK CORRELATION COEFFICIENT:
When two sets of data are given positions or ranked, we use Spearman’s Rank Correlation
Coefficient represented by R. It is given by:
R = 1 - 6 Σd2
n(n2 -1)
Where d = difference between pairs of ranked values.
n = number of pairs of ranking.
1 = is an independent number.
122
Question 2:
A group of eight students were tested in Maths and English and the ranking in the two
subjects are given below:
Student
Maths
English
A
2
3
B
7
6
C
6
4
D
1
2
E
4
5
F
3
1
G
5
8
H
8
7
Solution:
Students Maths
A
2
B
7
C
6
D
1
E
4
F
3
G
5
H
8
TOTAL
English
3
6
4
2
5
1
8
7
D
-1
1
2
-1
-1
2
-3
1
D2
1
1
4
1
1
4
9
1
= 22
R = 1 - 6 Σd2
n(n2 -1)
R = 1 – 6 * 22
8(82 -1)
R = 1 - 132
8* 63
R = 1 – 0.2619
= 0.738
 REGRESSION:
This is a statistical technique that can be used for short to medium term forecasting. It
seeks to establish the line of “best fit” for some observed data.
These are 2 ways of obtaining the trend of a set of data:
 The graphical method.
 The method of least squares.
123
 THE GRAPHICAL METHOD:
In this method a scatter diagram is plotted and a straight line is produced through the
middle of the points representing the trend line.
This line can then be used (when extended) to predict a future value.
y
* Trend line or line of best fit.
* *
*
*
*
*
*
*
*
*
x
 METHOD OF LEAST SQUARES:
This method gives an equation of the trend line or regression line. The equation is
given by:
Y = A + Bx
Where B = n ΣXY - ΣX  ΣY
n ΣX2 - (ΣX)2
and A = Y - BX
x is an independent variable.
n is the number of items in the list.
Question 3:
Consider the following data:
X
Y
i.
ii.
1
6
2
4
3
3
4
5
5
4
6
2
Find the equation of the regression line.
Find Y when X = 4 using the equation.
124
Solution:
X
1
2
3
4
4
6
21
Y
6
4
3
5
4
2
24
XY
6
8
9
20
20
21
75
X2
1
4
9
16
25
36
91
B = n ΣXY - ΣX  ΣY
n ΣX2 - (ΣX)2
B = 6(75) – 21 * 24
6 (91) – (21)2
= 450 – 504
546 – 441
= -18/35
X = X/n
21/
6
= 3.5
Y = Y/n
24/
6
=4
A = Y - BX
= 4 – (7/2 * - 54/105)
= 4 + 9/5
=29/5
Y = A + Bx
= 29/5 – 18/35X
(ii) When X = 4
Y = A + Bx
= 29/5 – 18/35* 4
= 29/5 – 72/35
= 131/35 or 326/35
125
Question 4:
For data below calculate the equation of the regression Y upon X.
X
Y
98
14
78
9
74
10
80
11
80
10
83
11
95
12
100
13
97
11
75
9
Solution:
X
98
78
74
80
80
83
95
100
97
75
860
Y
14
9
10
11
10
11
12
13
11
9
110
XY
1372
720
740
880
800
913
1140
1300
1067
675
9589
X2
9604
6084
5476
6400
6400
6889
9025
10000
9409
5625
74912
B = n ΣXY - ΣX  ΣY
n ΣX2 - (ΣX)2
B = 10(9589) – 860 * 110
10 (74912) – (860)2
= 95890 – 94600
749120 – 739600
= 0.1355
X = X/n
Y = Y/n
860/
10
110/
10
= 86
= 11
A = Y - BX
= 11 – (0.1355 * 86)
= 11 – 11.653
= - 0.6533
Y = A + Bx
= - 0.6533 + 0.1355X
126
UNIT 10
INTRODUCTION TO QUEUEING THEORY
HOURS:
20

QUEUEING MODELS:
It involves a customer’s standpoint where a waiting line is created. Customers arrive at
the facility and they join a waiting line (or Queue). The server chooses a customer from
the waiting line to begin service. Upon the completion of a service, the process of choosing
a new (waiting) customer is repeated.
It is assumed that no time is lost between the completion of a service and the admission of
a new customer into the facility.
The major actors in queuing situation are Customer and the Server. The interaction
between the Customer and the Serve are of interest only in as far as it relates to the
period of time the customer needs to complete a service. From the standpoint of customer
arrivals we will be interested in the time intervals that separate successive arrivals.
The customer arrivals and service times are summarized in terms of probability
distributions normally referred to as: arrivals and service time distribution.
These distributions may represent situations where customers arrive and are served
individually {banks or supermarkets}.
Customers may arrive and or be served in groups are referred to as Bulk queues.
In queue model the manner of choosing customer from the waiting line to start service is
referred to as Service discipline.
The Service discipline occurs as FCFS rule (first come first service), LCFS rule (last come
first served) and SIRO (service in random order).
Customers arriving at a facility may be put in Priority queues such that those with higher
priority will receive preference to start service first.
The facility may include more than one server thus allowing as many customers as the
number of servers to be serviced simultaneously this is referred as Parallel service.
Facilities which comprise a number of series stations through which the customer may
pass before service is completed (e.g. processing of a product on a sequence of machines)
is referred to as Queues in series or Tandem queues.
A combination of Parallel and Series is called Network queues.
Calling source is the source from which calls for service (arrivals of customer) are
generated.
Queue size the number of limited customers that may be allowed at a facility may be
because of space limitation (e.g. car spaces in a drive-in bank).
127

PURE BIRTH MODEL / PROCESS:
Pure birth model is process situation where customer arrive and never leave. E.g. issuing
of birth certificates for newborn babies. These certificates are normally kept as
permanent records in a Central office administered by the state health department.
The birth of new babies and issuing of birth certificates is a completely random process
that can be described ay a Poisson distribution.
Assuming that  is the rate at which birth certificates are issued, the pure birth process of
having n arrivals (birth certificates) during the time period t is given as:
Pn(t) = (t)n e-t
n!
Where n = 0, 1, 2, 3… and  is the rate of arrival per unit time with the expected number of
arrivals during t being equal to t.
Question 1:
Suppose that the birth in a state are spaced over time according to a an exponential
distribution with one birth occurring every 7 minutes on the average.
Solution:
Since the average inter arrival (inter birth) time = 7 minutes the birth rate in the state is
computed as:
 = 24{hours} * 60{minutes}
7
= 205.7 births per day.
The number of birth in the state per year =:
t = 205.7 * 365
= 75.080 births per year
The probability of no births in any one-day is computed as:
Pn(t) = (t)n e-t
n!
P0(1) = (205.7 * 1)0 e-205.7 * 1  0.
0!
Suppose that we are interest in the probability of issuing 45 birth certificates by the end of
a period of 3 hours given that 35 certificates were issued in the first 2 hours.
We observe that since birth occur according to a Poisson process the required probability
reduces to having 45 –35 = 10 births in one (3 – 2) hours.
Given  = 60/7 = 8.57 births per hour.
128
P10(1) = (8.57 * 1)10 e-8.57 * 1
10!
P10(1) = (2137018192) * (0.000189712)
3628800
 0.11172
Question 1:
Suppose that the clerk who enters the information from the birth certificate into a
computer normally wait at least 5 certificates to be accumulated. What it the probability
that the clerk will be waiting a new batch every hour.
Solution:
 = / = births per hour.
P5(1) = (* 1)5 e- * 1
5!
 0.92868
129
 PURE DEATH MODEL / PROCESS:
Pure death model is process situation where customers are withdrawals. Consider the
situation of stocking units of an item at start of the week to meet customer’s demand
during the week. If we assume that customer demand occurs at the rate  units per
week and that the demand process is completely random, the associated probability of
having n items remaining in stock after time t is given by the following Truncated
Poisson distribution.
Pn(t) = (t)N - n e-t
where n = 1, 2, 3 … N.
(N – n)!
Pn(t) = 1 – N n-1 * Pn(t)
Question 2:
At the beginning of each week, 15 units of an inventory item are stocked for use during
the week. Withdrawals from stock occur only during the first 6 days (business is closed
on Sundays) and follow a Poisson distribution with 3 units per day.
When the stock level reaches 5 units, a new order of 15 units is placed for delivery at
the beginning of next week. Because of the nature of the item all units left at the end of
the week are discarded.
Solution:
We can recognize that the consumption rate is  = 3 units per day.
Suppose that we are interested in computing the probability of having 5 units (reorder
level) on day t then we compute it as:
Pn(t) = (t)N - n e-t
(N – n)!
P5(t) = (3t)15 - 5 e-3t
(15 – 5)!
NB: P5(t) represents the probability of reordering on day t. This probability peaks at t =
3 and then declines as we advance through the week. If we are interested in the
probability of reordering by day t we must compute the cumulative probability of
having 5 units or less on day t.
i.e. Pn <= 5(t) = P0(t) + P1(t) + …..+ P5(t).
130

STEADY STATE MEASURES OF PERFORMANCE:
The steady state measures of performance is used to analyze the operation of the queuing
situation for the purpose of making recommendations about the design of the system.
Among these measures of performance are:
 The expected number of customers waiting.
 The expected waiting time per customer.
 The expected utilization of the service facility.
The system comprises both the queue and the service facility.
Let:
Ls = expected number of customers in system.
Lq = expected number of customers in Queue.
Ws = expected waiting time in system.
Wq = expected waiting time in Queue.
Suppose that we are considering a Service Facility with c parallel servers then from the
definition of Pn we get:
Ls = n=0 * npn
Lq = n=c + 1 * (n – c)pn
NB: A strong relationship exists between Ls and Ws (also Lq and Wq) so that either
measure is automatically determined from the other.
Let eff be the effective average arrival rate (independent of the number in the system n)
then:
Ls = eff * Ws
Lq = eff * Wq
The value of eff is determined from the state dependent n and the probabilities Pn is given
as:
eff = n=0 * npn
NB: A direct relationship exists between Ws and Wq, which is equal to:
Expected waiting time in system = Expected waiting time in queue + expected service time.
Ws = Wq + 1/.
Given that  is the Service rate per busy server, the Expected Service time = 1/.
131
Ws = Wq + 1/.
Multiplying both sides by eff we obtain:
Ls = Lq + eff/ = expected number of customers in system.
The expected utilization of a service is defined as a function of the average number of busy
servers.
Since the difference between Ls and Lq must equal the Expected number of busy servers
we obtain:
Expected number of busy servers = C = Ls – Lq = eff/.
The Percent utilization of a service facility with c parallel servers is given by:
Percent utilization = C/C * 100
NB: In summary
the following order:
Given Pn we can compute the system’s measure of performance in
Pn Ls = n=npn Ws = Ls/eff
Wq C = Ls – Lq.
Wq = Ws - 1/
Lq = eff *
Question 3:
Job orders arriving at a production Facility is divided into three groups. Group1 will take
the highest priority for processing; Group3 will be processed only if there are no waiting
orders from group 1 and 2. It is assumed that a job once admitted to the facility must be
completed before any new job is taken in.
Orders from groups 1,2 and 3 occur according to Poisson distribution with mean 4, 3 and 2
per day respectively. The service times for the three groups are constant with rates 10, 9
and 10 per day respectively.
Find the following:
I.
II.
III.
IV.
The expected waiting time in the system for each of the three queues.
The expected waiting time in the system for any customer.
The expected number of waiting jobs in each of the three groups.
The expected number waiting in the system.
132
Solution:
133
HIGHER NATIONAL DIPLOMA IN INFORMATION TECHNOLOGY
SUBJECT:
OPERATIONS RESEARCH
CODE:
710 / 04 /S02
UNIT 1
APPROXIMATIONS
TRAPEZIUM RULE
Any definite integral may be thought of as an area under which a given curve for a specific
interval. We can evaluate a definite of when an indefinite integral is known.
Some functions may not have a single integral hence an approximation method could be
used to evaluate them.
Trapezium divides the area under the curve into small trapeziums whose sum
approximates the required integral.
Y-axis
Y =f(x)
A
B
X-axis
Area of trapezium = ½ (sum of 2 sides)*height
Consider the function y = f (x) to be the integral over the interval a, b i.e. limits are from a
to b.
We obtain the approximation of this definite integral by dividing the area under f (x) into
any vertical trapezium width with delta x (δx)
 TRAPEZIUM RULE
½ δx {f (x0)+2f(x1)+2f(x2)+2f(x3)+……….f (xn)}
Where δx = b - a
n
X0 = a
X1 = x0 + δx
X2 = x1 + δx
X3 = x2 + δx
Xn = b
134
135