Download page 92 problem 5

Extra
Credit
Project
IE 416: Operations Research I
Robert Delgado
Chris Mui
Amanda Smith
Presented to: Dr. Sima Parisay
California State Polytechnic
University, Pomona
Due: October 20th, 2011
Table of Contents…………………………………………………………………………………………………………….……………………2
Problem Statement ....................................................................................................................................................... 3
Summary of Problem .................................................................................................................................................... 3
Table 1: Chandler Oil Company- Oil Information ............................................................................................ 3
Table 2: Chandler Oil Company- Products Made from Oil .............................................................................. 3
Formulation of Problem.............................................................................................................................................3-6
1.) Define the decision variables ............................................................................................................................. 4
Table 3: Summary of Problem ............................................................................................................................ 4
2.) Provide explanatory information and assumptions ........................................................................................... 4
3.) Formulate Objective Function (O.F) .................................................................................................................. 5
4.) Formulate Constraints .....................................................................................................................................5-6
5.) List All Equations .............................................................................................................................................6-7
Table 4: Constraint Equation List..................................................................................................................... 6
Table 5: LP Form of Constraint Equation List .................................................................................................. 6
Solution using WinQSB ..............................................................................................................................................7-8
WinQSB Screen Shot 1: Original Input....................................................................................................................... 7
Table 6: WinQSB Variable and Name .............................................................................................................. 7
Table 7: WinQSB Constraint and Name ........................................................................................................... 7
WinQSB Screen Shot 2: Output ................................................................................................................................. 8
Report to Manager...................................................................................................................................................9-10
Sensitivity Analysis.................................................................................................................................................10-15
WinQSB Screen Shot 3: Objective Function Graph for Coefficient of Oil 1 for Gas SA ............................................ 11
WinQSB Screen Shot 4: Table for Objective Function SA of Oil 1 for Gas ............................................................... 11
WinQSB Screen Shot 5: RHS Sensitivity Analysis Graph for Oil 1 Availability .......................................................... 13
WinQSB Screen Shot 6: RHS Sensitivity Analysis Table for Oil 1 Availability ........................................................... 13
WinQSB Screen Shot 7: RHS Sensitivity Analysis Table for Demand Gas ................................................................ 15
WinQSB Screen Shot 8: RHS Sensitivity Analysis Table for Demand Gas ................................................................ 15
Simplex Tableau .....................................................................................................................................................16-19
WinQSB Screen Shot 8: Simplex Tableau for Iteration #s1-8 .............................................................................16-17
Table 8: Original Simplex Tableau Table........................................................................................................ 17
Table 9: Simplex Tableau for Iteration #1 .................................................................................................... 18
Table 10: Simplex Tableau for Iteration #2 .................................................................................................. 18
Acknowledgments ...................................................................................................................................................... 19
PowerPoint Slide Printouts ......................................................................................................................................... 20
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Problem Statement
Chandler Oil Company has 5,000 barrels of oil 1 and 10,000 barrels of oil 2. The company sells
two products: gasoline and heating oil. Both products are produced by combining oil 1 and oil 2.
The quality level of each oil is as follows: oil 1- 10; oil 2- 5. Gasoline must have an average
quality level of at least 8, and heating oil at least 6. Demand for each product must be created
by advertising. Each dollar spent advertising gasoline creates 5 barrels of demand and each
spent on heating oil creates 10 barrels of demand. Gasoline is sold for $25 per barrel, heating
oil for $20. Formulate an LP to help Chandler maximize profit. Assume that no oil of either type
can be purchased.
Summary of Problem
•Table 1: Chandler Oil Company – Oil Information
Chandler Oil Company - Oil Information
Oil
# of barrels
Oil Quality
Oil 1
5000
10
Oil 2
10000
5
•Table 2: Chandler Oil Company – Products Made from Oil
Chandler Oil Company - Products Made from Oil
Demand Created
Selling
Product (Blend)
Avg. Quality Level
per $1 spent on
Price per
Advertising
Barrel
Gas
8
5
$25
Heating Oil
6
10
$20
We will condense the tables above further in the report. Explanation will be given.
Dr. Parisay’s comments are in red.
Formulation of Problem
Chandler Oil Company must make two types of decisions: first how much money should be
spent in advertising each of their products (blends): gas and heating oil, and second, how to
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blend each product from the available types of oil. For example, Chandler Oil Company must
decide how many barrels of oil 1 should be used to produce gas.
1.) Define the decision variables
ai = dollars spent daily on advertising blend i (i = 1,2)
xij = barrels of oil i used daily to produce blend j (i = 1,2 ; j = 1,2)
Sign Restrictions:
ai > 0
xij > 0
The definition of the decision variable implies:
x11 + x12 = barrels of oil 1 used daily
x11 + x21 = barrels of gas produced daily
x21 + x22 = barrels of oil 2 used daily
x12 + x22 = barrels of heating oil produced daily
Now that we have defined our decision variables, we can combine Table 1 and Table 2 to
create a more efficient summary of our problem:
•Table 3: Summary of Problem
Demand of Barrels
Avg.
Created per $1
Quality
spent on
Level
Advertising
Selling Price
per Barrel
8
5
$25
6
10
$20
Product
(Blend)
Gas
Heating Oil
Decision Variables
OIL
# of barrels
Oil Quality
x11
x21
x12
x22
1
2
5000
10
10000
5
2.) Provide explanatory information and assumptions
To simplify, let’s assume that gas and heating oil cannot be stored, so it must be sold on the day
it is produced. This implies that for j = 1,2, the amount of blend produced daily should equal the
daily demand for blend j.
If the amount of blend produced daily exceeds daily demand, we would incur unnecessary
production and purchase cost. Even though the problem did not provide us with the production
and purchase cost, we know in a real world environment these costs are real.
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If the amount of blend produced daily is less than daily demand, we fail to meet mandatory
demand and incur unnecessary advertising cost.
Therefore, the goal of this problem is to maximize profit by using the right amounts of oil to
produce the right number of blends.
3.) Formulate Objective Function (O.F)
The basic formula for profit is revenues minus cost.
Profit = Revenue – Cost
Profit = Zmax
Therefore, we must define both revenue and cost in relation to this problem. After defining
revenue and cost, we can determine profit (Zmax) by using the formula above.
 Daily Revenues from Blend Sales (Sales of Gas and Heating Oil)
= 25(x11 + x21) + 20 (x12 + x22)
 Daily Advertising Cost
= a1 + a2
 Daily Profit = Daily Revenues from Blend Sales - Daily Advertising Cost
Daily Profit = [25(x11 + x21) + 20 (x12 + x22)] – [a1 + a2]
 Simplify
Zmax = 25x11 + 25x21 + 20x12 + 20x22 –a1 – a2
4.) Formulate Constraints
Constraint 1: Maximum of 5,000 barrels of oil 1 are available for production
x11 + x12 < 5000
Constraint 2: Maximum of 10,000 barrels of oil 2 are available for production
x21 + x22 < 10,000
Constraint 3: Gasoline must have an average quality level of at least 8.
10x11 +5x12
x11 +x21
≥8
⇒
Simplify
2x11 – 3x21 > 0
Constraint 4: Heating oil must have an average quality level of at least 6.
10x12 +5x22
x12 +x22
≥6
⇒
Simplify
4x12 – x22 > 0
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Constraint 51: Demand of gas is increased by 5 barrels for every dollar spent on
advertising.
x11 + x21 = 5a1
Constraint 61: Demand of heating oil is increased by 10 barrels for every dollar spent on
advertising.
x12 + x22 = 10a2
5.) List all Equations
•Table 4: Constraint Equation List
Description
Equation
Type
Max Profit
Zmax = 25x11 + 25x21 + 20x12 + 20x22 –a1 – a2
Objective Function
Oil 1 Avail.
x11 + x12 < 5000
Constraint
Oil 2 Avail.
x21 + x22 < 10,000
Constraint
Gas Quality
2x11 – 3x21 > 0
Constraint
H. Quality
4x12 – x22 > 0
Constraint
Demand Gas
x11 + x21 = 5a1
Constraint
Demand H.
x12 + x22 = 10a2
Constraint
•Table 5: LP Form of Constraint Equation List
The following table needs modification. Do not define a1 and a2 in constraints as you
have them as ad cost in OF. Replace with a3 and a4. The OF is not standard form.
Refer to the format in your book.
Description
Standard LP Form Equation
Type
Max Profit
Zmax = 25x11 + 25x21 + 20x12 + 20x22 –a1 – a2
Objective Function
Oil 1 Avail.
x11 + x12 + S1 = 5000
Constraint
Oil 2 Avail.
x21 + x22 + S2 =10,000
Constraint
Gas Quality
2x11 – 3x21 - e1 + a1 = 0
Constraint
H. Quality
4x12 – x22 - e2 + a2 = 0
Constraint
Demand Gas
x11 + x21 - 5a1 = 0
Constraint
Demand H.
x12 + x22 - 10a2 = 0
Constraint
1
Remember, the problem stated that demand for each problem must be created by advertising. Also remember in 2.) we
assumed that amount of blend produced daily should equal daily demand. So, here supply equals demand. In layman’s
terms, the amount we produce will be equal to the demand we create through advertising.
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We define slack variables in Oil 1 Avail. and Oil 2 Avail. Slack variable si (si=slack variable for
the ith constraint), which is the amount of resource unused in the ith constraint. If constraint i of
an LP is a < constraint we convert it to an equality constraint by adding a slack variable si.
We define excess variables and artificial variables in Gas Quality and H. Quality. If the ith
constraint of an LP is a > constraint, then it can be converted to an equality by subtracting an
excess variable ei, from the ith constraint. We add an artificial variable too so we have a basic
variable for initial simplex tableau.
Solution using WinQSB
Input into WINQSB:
• WinQSB Screen Shot 1: Original Input
Explanation of WinQSB Variables and Constraints:
•Table 6: WinQSB Variable and Name
•Table 7: WinQSB Constraint and Name
Variable
Name Given
Constraint
Name Given
x11
Oil 1 for Gas
x11 + x12 < 5000
Oil 1 Avail.
x12
Oil 1 for H.
x21 + x22 < 10,000
Oil 2 Avail.
x21
Oil 2 for Gas
2x11 – 3x21 > 0
Gas Quality
x22
Oil 2 for H.
4x12 – x22 > 0
H. Quality
a1
Adv. $ Gas
x11 + x21 - 5a1 = 0
Demand Gas
a2
Adv. $ H.
x12 + x22 - 10a2 = 0
Demand H.7 | P a g e
WinQSB Output:
• WINQSB Screen Shot 2: Output
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Report to Manager
If Chandler Oil Company wants to maximize its profit to $323,000 for the current production of
gasoline and heating oil it should:


Produce 5,000 barrels of gasoline by mixing 3,000 barrels of oil 1 with 2,000 barrels of
oil 2
Produce 10,000 barrels of heating oil by mixing 2,000 barrels of oil 1 with 8,000 barrels
of oil 2
By these combinations we are able to meet the exact quality requirement for gasoline which is 8
and the exact quality requirement for heating oil which is 6. (You concluded this from the
related constraints to be binding. If constraints were not binding you need to calculate the
quality value.)
The production value of 5,000 barrels of gasoline will only remain if the profit per barrel of gas
stays between $18.88-$83.17. (good job) A change in unit profit within this range will cause
maximum profit to change while barrels of gasoline produced remains at 5,000 barrels. Anything
outside this profit range will cause barrels of gas produced to change and maximum profit to
change.
(good explanation) We must take into account both allowable ranges of profit for oil 1 for gas
and oil 2 for gas since gas is a mixture of both. Although oil 1 for gas has a profit range of
$16.83-$83.17 and oil 2 for gas has a profit range of $18.88-$112.25, we consider the
correlation between both oil profit ranges. When taking this into consideration, we determine the
$18.88-$83.17 range mentioned.
Using the same concept, the production value of 10,000 barrels of heating will only remain if the
profit per barrel of heating oil stays between $5.46-$26.13. A change in unit profit within this
range will cause maximum profit to change while barrels of heating oil produced remains at
10,000 barrels. Anything outside this profit range will cause barrels of heating oil produced to
change and maximum profit to change.
We must take into account both allowable ranges of profit for oil 1 for heating oil and oil 2 for
heating oil since heating oil is a mixture of both. Although oil 1 for heating oil has a profit range
of $0-$28.17 and oil 2 for heating oil has a profit range of $5.46-$26.13, we consider the
correlation between both oil profit ranges. When taking this into consideration, we determine the
$5.46-$26.13 range mentioned.
To reach the profit maximization mentioned, we must pay $1000 in advertisement for gas and
$1000 in advertisement for heating oil to generate the demand for the 5,000 barrels of gasoline
and 10,000 barrels of heating oil.
Our solution remains optimal if the range of oil 1 usage is from 2,500-15,000 barrels. For each
additional barrel of oil 1 made available for use, we can increase our profit by $29.70. This in
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turn means that we are willing to buy extra barrels of oil 1, up to 15,000 barrels, for an
extra
cost of up to $29.70 for each barrel (which means we would break even) (This is managerial
application of shadow price. We will cover it after midterm. Good Job) . However, as of now
we only have room to store 5,000 barrels of oil 1 and we have used all this storage.
Our solution remains optimal if the range of oil 2 usage is from 3,333-20,000 barrels. For each
additional barrel of oil 2 made available for use, we can increase our profit by $17.45. This in
turn means that we are willing to buy extra barrels of oil 2, up to 20,000 barrels, for an extra
cost of up to $17.45 for each barrel (which means we would break even). However, as of now
we only have room to store 10,000 barrels of oil 2 and we have used all this storage.
Sensitivity Analysis
 1 Sensitivity Analysis for a Coefficient in the Objective Function that is a Basic
Variable
Choice:
Variable
Name Given
x11
Oil 1 for Gas
We chose to do a sensitivity analysis on x11 (oil 1 for gas) in the objective function.
Why didn’t we choose advertising cost?
We knew we wanted to do an analysis on oil instead of advertising cost. We know that our
supply of oil is fixed. We have assumed that supply equals demand, therefore, if all demand is
met, then demand is also fixed. To maximize profit, all demand will be met. Demand is only
generated through advertising. Because demand is only generated through advertising and our
demand is fixed, then our advertising cost is also fixed. For example, each dollar spent
advertising gasoline creates five barrels of demand for gas, meaning advertising cost per barrel
of gasoline is $0.20. If you look at WinQSB Output 3 you will see this value in the shadow price
column for Demand Gas. Thus, the problem would have to change completely for a sensitivity
analysis on advertising cost to be valid.
Motivation for SA on x11 (oil 1 for gas)
Currently, oil 1 for gas is tied for the highest unit profit of $25 c(j) with oil 2 for gas. However, it
has the highest allowable max c(j) compared with oil 2 for gas. Like previously mentioned, we
are looking at these variables being correlated. Although technically oil 2 for gas has a allowable
max c(j) of $112.25, we have already established that the range for the oils for gas is $18.88$83.17. Therefore this variable x11 contains the $83.17 allowable max c(j) we are looking for. Of
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course, if we do change the unit cost of x11 from $16.83-$83.17 the solution value will remain at
3,000 barrels, but the value of the objective function will change. (good observation)
(This paragraph is a good effort, but not expected for this level of class) To describe the
entering and leaving variables WinQSB says the following:
In a simplex iteration, a decision variable will enter the basis depending on a particular rule.
The typical decision rule is to choose the decision variable with the best Cj-Zj to enter the
basis. For the maximization problem, it is the variable with the most positive Cj-Zj. The
decision rule to choose the leaving basic variable is the one with the minimum ratio of
RHS/A (i,j) in the updated simplex tableau.”
With this definition in mind, we can see the entering and leaving variables in range 1 and range
3.
• WINQSB Screen Shot 3: Objective Function Graph for Coefficient of Oil 1 for Gas SA
This point shows a
unit cost value
outside the
allowable max c(j)
range.
This point shows
that when unit
profit is increased
to $83.17 our
max profit will be
$497,500.
This flat line shows that
the coefficients for x11 on
this line will yield the
same max profit. (x11=0)
This is the current solution. Unit
profit is $25 and our max profit is
$323,000.
• WINQSB Screen Shot 4: Table for Objective Function SA of Oil 1 for Gas
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 1 Sensitivity Analysis for a Coefficient in the Objective Function that is a NonBasic Variable
o There are no non basic variables in the objective function.

1 Sensitivity Analysis for RHS in Constraint that is Binding
Choice:
Constraint
Name Given
x11 + x12 < 5000
Oil 1 Avail.
Motivation for SA on x11 + x12 < 5000 (oil 1 availability)
We chose to do a sensitivity analysis on oil 1 availability because it has the highest shadow
price of $29.70. From the optimal solution we also chose the sensitivity analysis because it had
the highest room for profit.
Oil 1 availability has an allowable max RHS of 15,000 barrels and a shadow price of $29.70. If
we max out our RHS we get 15,000 x $29.70 = $445,500 increase in profit.
Oil 2 Availability has an allowable max RHS of 20,000 barrels and is the only other constraint
with a relatively high shadow price of $17.45. If we max out our RHS on this constraint we get
20,000 x $17.45 = $349,000 increase in profit.
Because oil 1 availability has the higher potential of making more profit, we went with this
choice. In reality, if we were to max out both RHS of oil 1 and oil 2 we would need to get more
storage area for oil and generate more demand by advertisement. (good note)
Obviously, an infinite supply of either oil 1 or oil 2 is infeasible.
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• WINQSB Screen Shot 5: RHS Sensitivity Analysis Graph for Oil 1 Availability
If we increase
barrels of oil 1 to
15,000 our max
profit will be
$620,000.
This is the current
solution. Barrels of
oil 1 used is 5,000
and our max profit
is $323,000.
If we can only obtain 2,500
barrels of oil 1, our max profit
will be $248,750.
• WINQSB Screen Shot 6: RHS Sensitivity Analysis Table for Oil 1 Availability
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 1 Sensitivity Analysis for a RHS in Constraint that is Binding
(This is a good effort; however, try to avoid performing SA on those RHS that
does not have specific/clear meaning, such as here. RHS is zero which does not
have a practical meaning. You got zero through some math operation for simplex
method.)
Choice:
Constraint
Name Given
x11 + x21 - 5a1 = 0
Demand Gas
Motivation for SA on x11 + x21 - 5a1 = 0 (Demand Gas)
We chose to do a sensitivity analysis on Demand Gas because it has the highest shadow price
of $.20 between the two binding constraints available. In terms of profit, it wouldn’t matter if we
chose to increase allowable RHS for Demand Gas or Demand H because the same profit will be
acquired:
Shadow Price of Demand Gas x Max RHS = Amount of Increased Profit Due to Demand Gas
$.20 x 5,000 = $1000
Shadow Price of Demand H. x Max RHS = Amount of Increased Profit Due to Demand H.
$.10 x 10,000 = $1000
However, according to the solution both demand for gas and demand for heating oil is at their
max point. Gas is better because it pays a higher shadow price per barrel than heating oil.
Theoretically this in turn is better for a company because having to produce, move, store, and
sell 5,000 barrels would be easier than having to produce, move, store, and sell 10,000 barrels.
So, you are putting in half the effort but making the same in profit.
Notice that the max allowable for “Demand Gas” is 5000. That is if you increase zero to
5000. Here zero does not have a specific meaning, just equates gas and ad $. Therefore,
you cannot make the above conclusion. It is better not to perform SA.
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• WINQSB Screen Shot 7: RHS Sensitivity Analysis Graph for Demand Gas
• WINQSB Screen Shot 8: RHS Sensitivity Analysis Table for Demand Gas
(As mentioned above, avoid this conclusion. There will be a new set of solution.) From
the Sensitivity Analysis, if our demand increases up to 5000 barrels, so will our profit. If our
demand for gas is equal to 5000 we will have a profit maximization of $323,000 (as shown in
WINQSB Screen Shot 2). However, once our demand goes over 5000 our profit will reduce
because we cannot meet demand. When gas demand equals 8333 barrels our profit will reduce
to $208,333 because more money has to be spent in advertising to create that demand. Any
demand above 8333 barrels is infeasible.
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Simplex Tableau
WinQSB Output:
• WINQSB Screen Shot 7: Simplex Tableau Iteration #s1-8
These shots are from WinQSB
iterations of Simplex method and were not required.
WinQSB’s presentation of simplex tableau is different from your textbook. You need to
know your book’s style for quiz and exam.
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Simplex Tableau Hand Calculation explanation for “Oil 1 for Gas” column:
This table needs modification to follow your book’s style. For example, all coefficients in
Row 0 should be negative. Though, the next step will not change. The term “iteration”
and “step” should be used with care. There are some mistakes below on proper term to
be used. Please update this after discussion of simplex method in class.
•Table 7:Original Simplex Tableau Table
1: Entering
Variable
2: Ratio
Testing
4: Pivot Term
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3: Pivot Row
Note: WinQSB performs simplex method different than what is taught in the supplementary
notes. Although it does use the Gauss-Jordan method of elementary row operation, its
procedure is slightly different. You can search “Simplex Method” on the WinQSB software to
obtain a 6 step procedure as to how WinQSB performs its iterations.
Iteration #1: If we look at the WinQSB output for iteration 1, we can see how the software used
the Gauss-Jordan method of elementary row operation. The goal is to get every number in the
“Oil 1 for Gas” column zero and the pivot term 1. Luckily, the pivot term (highlighted in blue) is
already 1. Now we concentrate on making the other numbers zero. WinQSB achieved this in
two iterations. In the first iteration, row 3 and row 4 were both multiplied by negative one in the
original tableau. This is the result of those operations.
•Table 8: Simplex Tableau for Iteration #1
Iteration #2 this is a step: If you look at iteration #1, the only two numbers in the “Oil 1 for Gas”
column left to make zero are in row 1 and row 3. In the second iteration, algebra was performed
on row 1 and row 3 from the iteration #1 tableau. The following is the result of those operations.
Now, we are completely done with this column.
•Table 9: Simplex Tableau for Iteration #2
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We repeat the method of choosing an entering variable, ratio testing, finding the pivot row, and
finding the pivot term until every column has a number 1 value for its pivot terms and the
remaining column numbers numbers are zero (not including the max row).
Acknowledgements
Bibliography:
Chang, Yih-Long, and Kiran Desai. WinQSB Version 2.0. New York: Wiley, 2003. Print
Winston, Wayne L. Operations Research Application and Algorithms. 4th Edition. New York;
Duxbury, 2003. Print
Software Used:

Microsoft Word 2007 (Windows)

Microsoft Excel 2007 (Windows)

Win QSB, Version 2.0 (Windows)
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