Download Overview of the Operations Research Modeling Approach

Network Optimization Models
Chapter 10: Hillier and Lieberman
Chapter 8: Decision Tools for Agribusiness
Dr. Hurley’s AGB 328 Course
Terms to Know

Nodes, Arcs, Directed Arc, Undirected Arc,
Links, Directed Network, Undirected
Network, Path, Directed Path, Undirected
Path, Cycle, Connected, Connected
Network, Tree, Spanning Tree, Arc Capacity,
Supply Node, Demand Node, Transshipment
Node, Sink, Source, Residual Network,
Residual Capacity, Augmenting Path, Cut, Cut
Value, Max-Flow Min-Cut Theorem, Feasible
Solutions Property, Integer Solutions
Property, Reverse Arc, Basic Arcs, Nonbasic
Arcs
Terms to Know Cont.

Spanning Tree Solution, Feasible Spanning
Tree, Fundamental Theorem for the
Network Simplex Method, Program
Evaluation and Review Technique (PERT),
Critical Path Method (CPM), Immediate
Successor, Immediate Predecessor, Project
Network, Activity-on-Arc, Activity-on-Node,
Project Duration, Critical Path, Crashing an
Activity, Crashing the Project, Normal Point,
Crash Point, Marginal Cost Analysis
The Shortest Path Problem


A shortest path problem usually has a node
known as the origin and a node known as
the destination
The objective of this problem is to find the
shortest path from the origin node to the
destination node
◦ The shortest path could be measured in time,
distance, etc.

Since this problem is a special case of the
linear programming problem, the simplex
method could be used to solve it
Mathematical Model for Seervada
Shortest Path Problem
min
𝑤.𝑟.𝑡. 𝑥𝑂𝐴 ,𝑥𝑂𝐵 ,𝑥𝑂𝐶 ,
𝑥𝐴𝐵 ,𝑥𝐴𝐷 ,𝑥𝐵𝐶 ,𝑥𝐵𝐷 ,𝑥𝐵𝐸 ,
𝑥𝐶𝐵 ,𝑥𝐶𝐸 ,𝑥𝐷𝐸 ,𝑥𝐷𝑇 ,𝑥𝐸𝐷 ,𝑥𝐸𝑇
2 𝑥𝑂𝐴 + 5𝑥𝑂𝐵 + 4𝑥𝑂𝐶 + 2𝑥𝐴𝐵 + 7𝑥𝐴𝐷 + 1𝑥𝐵𝐶 + 4𝑥𝐵𝐷
+ 3𝑥𝐵𝐸 + 1𝑥𝐶𝐵 + 4𝑥𝐶𝐸 + 1𝑥𝐷𝐸 + 1𝑥𝐸𝐷 + 5𝑥𝐷𝑇 + 7𝑥𝐸𝑇
Subject To:
𝑥𝑂𝐴 + 𝑥𝑂𝐶 + 𝑥𝑂𝐵 = 1
𝑥𝑂𝐴 − 𝑥𝐴𝐷 − 𝑥𝐴𝐵 = 0
𝑥𝑂𝐵 + 𝑥𝐴𝐵 + 𝑥𝐶𝐵 − 𝑥𝐵𝐶 − 𝑥𝐵𝐷 − 𝑥𝐵𝐸 = 0
𝑥𝑂𝐶 + 𝑥𝐵𝐶 − 𝑥𝐶𝐵 − 𝑥𝐶𝐸 = 0
𝑥𝐴𝐷 + 𝑥𝐵𝐷 + 𝑥𝐸𝐷 − 𝑥𝐷𝐸 − 𝑥𝐷𝑇 = 0
𝑥𝐵𝐸 + 𝑥𝐶𝐸 + 𝑥𝐷𝐸 − 𝑥𝐸𝐷 − 𝑥𝐸𝑇 = 0
𝑥𝐷𝑇 + 𝑥𝐸𝑇 = 1
𝑥𝑂𝐴 , 𝑥𝑂𝐵 , 𝑥𝑂𝐶 , 𝑥𝐴𝐵 , 𝑥𝐴𝐷 , 𝑥𝐵𝐶 , 𝑥𝐵𝐷 , 𝑥𝐵𝐸 , 𝑥𝐶𝐵 , 𝑥𝐶𝐸 , 𝑥𝐷𝐸 , 𝑥𝐷𝑇 , 𝑥𝐸𝐷 , 𝑥𝐸𝑇 ∈ (0,1)
Seervada Spreadsheet Model

Examine in Class
Minimum Spanning Tree Problems
The goal of the minimum spanning tree
problem is to connect all the nodes either
directly or indirectly at the lowest cost
 The minimum spanning tree problem will
have one less link than the number of
nodes in the optimal solution
 The solution to the minimum spanning
tree problem can be accomplished using
the greedy algorithm

Greedy Algorithm
Choose any two nodes initially and
connect them
 Identify the closest unconnected node
and then connect it

◦ Continue until all nodes have been connected
to the tree
◦ Ties can be broken arbitrarily
The Maximum Flow Problem

The purpose of the maximum flow
problem is to get as much flow through
the network based on the capacity
constraints of the network
◦ This can be measured by the amount leaving
the source or by the amount entering the sink

It has a source where supply originates
from and a sink which absorbs the supply
that makes it through the network
Augmenting Path Algorithm for
Maximum Flow Problems

Identify an augmenting path that takes flow
from the source to the sink in the residual
network such that every arc on this path has
strictly positive residual
◦ If this path does not exist, you have the optimal

Identify the residual capacity c* by finding
the minimum of the residual capacities of the
arcs on the path
◦ Increase the flow in this path by c*

Decrease the residual capacities by c* for
each arc on the augmenting path
Max-Flow Min-Cut Theorem

Another way of figuring out the maximum
flow is by using the Max-Flow Min-Cut
Theorem
◦ The theorem states that if you have a single
source and sink, then the maximum flow
through the network is equal to the smallest
cut value for all the cuts of the network
 The cut value is found by summing up all the arcs
which are directly affected by the cut of a network
 A cut is defined as a set of directed arcs that separate the
source from the sink
Mathematical Model for the
Seervada Max Flow Problem
max
𝑤.𝑟.𝑡. 𝑥𝑂𝐴 ,𝑥𝑂𝐵 ,𝑥𝑂𝐶 ,𝑥𝐴𝐵 ,𝑥𝐴𝐷 ,𝑥𝐵𝐶,
𝑥𝐵𝐷 ,𝑥𝐵𝐸 ,𝑥𝐶𝐸 ,𝑥𝐷𝑇 ,𝑥𝐸𝐷 ,𝑥𝐸𝑇
𝑥𝑂𝐴 +𝑥𝑂𝐵 + 𝑥𝑂𝐶
𝑥𝑂𝐴 −𝑥𝐴𝐵 − 𝑥𝐴𝐷 = 0
𝑥𝑂𝐵 + 𝑥𝐴𝐵 − 𝑥𝐵𝐶 − 𝑥𝐵𝐷 − 𝑥𝐵𝐸 = 0
𝑥𝑂𝐶 + 𝑥𝐵𝐶 − 𝑥𝐶𝐸 = 0
𝑥𝐴𝐷 + 𝑥𝐵𝐷 − 𝑥𝐷𝑇 + 𝑥𝐸𝐷 = 0
𝑥𝐵𝐸 + 𝑥𝐶𝐸 − 𝑥𝐸𝐷 − 𝑥𝐸𝑇 = 0
0 ≤ 𝑥𝑂𝐴 ≤ 5, 0 ≤ 𝑥𝑂𝐵 ≤ 7, 0 ≤ 𝑥𝑂𝐶 ≤ 4, 0 ≤ 𝑥𝐴𝐵 ≤ 1,
0 ≤ 𝑥𝐴𝐷 ≤ 3, 0 ≤ 𝑥𝐵𝐶 ≤ 2, 0 ≤ 𝑥𝐵𝐷 ≤ 4, 0 ≤ 𝑥𝐵𝐸 ≤ 5,
0 ≤ 𝑥𝐶𝐸 ≤ 4, 0 ≤ 𝑥𝐷𝑇 ≤ 9, 0 ≤ 𝑥𝐸𝐷 ≤ 1,0 ≤ 𝑥𝐸𝑇 ≤ 6
Note: This is based on Figure 10.11 in the text.
Excel Formulation of Seervada Max
Flow Problem

Examined in Class
In Class Max Flow Activity (Not
Graded)
0
7
B
8
7
A
0
6
C
6
0
0
4
D
3
E
0
7
10
0
0
F
0
0 G 6
5
0
I
0
0 H 7
Minimum Cost Flow Problem
Requirements
At least one supply node
At least one demand node
The network is directed and connected
If the node is not a supply or demand node,
then it is a transshipment node
 Flow through an arc is directed
 There is enough arc capacity to get the total
supply to the total demand
 Costs are proportional to the amount of
flow
 The objective is to minimize cost




Minimum Cost Flow General
Mathematical Model
xij = the arc representing the flow from nodes i to
j
 cij = the cost of flow through xij
 uij = the arc capacity for xij
 bi = net flow generated at node i

◦ Supply node (bi > 0), demand node (bi < 0),
transshipment node (bi = 0)
𝑀𝑖𝑛𝑖𝑚𝑖𝑚𝑧𝑒 𝑍 = 𝑛𝑖=1 𝑛𝑗=1 𝑐𝑖𝑗 𝑥𝑖𝑗
 Subject to:
𝑛
𝑛
 𝑗=1 𝑥𝑖𝑗 − 𝑗=1 𝑥𝑗𝑖 = 𝑏𝑖 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑛𝑜𝑑𝑒 𝑖
 0 ≤ 𝑥𝑖𝑗 ≤ 𝑢𝑖𝑗 for each arc

Minimum Cost Flow Problem in
Excel

We will examine the Distribution
Unlimited Co. in class
What is Project Management
Project management can be defined as the
coordination of activities with the potential use
of many organizations, both internal and
external to the business, in order to conduct a
large scale project from beginning to end.
 There are two management science techniques
that are used for project management:

◦ Program and Evaluation Review Technique (PERT)
◦ Critical Path Method (CPM)
PERT/CPM

PERT
◦ PERT was designed to examine projects from the
standpoint of uncertainty.

CPM
◦ CPM was designed to examine projects from the
standpoint of costs.
PERT and CPM techniques have been combined
over time.
 PERT and CPM both rely heavily on the use of
networks to help plan and display the
coordination of all the activities for a project.

The Reliable Construction
Company
Reliable has just secured a contract to
construct a new plant for a major manufacturer.
 The contract is for $5.4 million to cover all
costs and any profits.
 The plant must be finished in a year.

◦ A penalty of $300,000 will be assessed if Reliable
does not complete the project within 47 weeks.
◦ A bonus of $150,000 will be paid to Reliable if the
plant is completed within 40 weeks.
Needed Terminology

Activity
◦ A distinct task that needs to be performed as
part of the project.

Start Node
◦ This is a node that represents the beginning of
the project.

Finish Node
◦ This node represents the end of the project.
Needed Terminology Cont.

Immediate Predecessor
◦ These are activities that must be completed by no
later than the start time of the given activity.

Immediate Successor
◦ Given the immediate predecessor of an activity, this
activity becomes the immediate successor of each of
these immediate predecessors.
◦ If an immediate successor has multiple immediate
predecessors, then all must be finished before an
activity can begin.
Activity List for Reliable Construction
Activity
Activity Description
Immediate
Predecessors
Estimated
Duration (Weeks)
A
Excavate
—
2
B
Lay the foundation
A
4
C
Put up the rough wall
B
10
D
Put up the roof
C
6
E
Install the exterior plumbing
C
4
F
Install the interior plumbing
E
5
G
Put up the exterior siding
D
7
H
Do the exterior painting
E, G
9
I
Do the electrical work
C
7
J
Put up the wallboard
F, I
8
K
Install the flooring
J
4
L
Do the interior painting
J
5
M
Install the exterior fixtures
H
2
N
Install the interior fixtures
K, L
6
Questions Needed to be Answered
How can the project be displayed graphically?
 How much time is required to finish the project
if no delays occur?
 When is earliest start and finish times for each
activity if no delays occur?
 What activities are critical bottleneck activities
where delays must be avoided to finish the
project on time?

Questions Needed to be Answered
Cont.
For non bottleneck activities, how much can an
activity be delayed and yet still keep the project
on time?
 What is the probability of completing the
project by the deadline?
 What is the least amount of money needed to
expedite the project to obtain the bonus?
 How should costs be monitored to keep the
project within budget?

Project Network
A project network is a network diagram
that uses nodes and arcs to represent the
progression of the activities for a project
from start to finish.
 Three pieces of information needed:

◦ Activity information
◦ Precedence relationship
◦ Time information
Project Network Cont.

Two types of project networks
◦ Activity-on-Arc (AOA)
 On this diagram, the activity is represented on an
arc, while a node is used to separate an activity
from its immediate predecessors.
◦ Activity-on-Node (AON)
 On this diagram, the activity is represented by the
node, while the arc is used to showed the
precedence relationship between the activities.
START
A
Activity C ode
0
A. Excavate
2
B. Foundation
C. Rough wall
B
D. Roof
4
E. Exterior plumbing
C
F. Interior plumbing
10
G. Exterior siding
H. Exterior painting
D
E
6
4
I
I. Electrical work
7
J. Wallboard
K. Flooring
L. Interior painting
G
F
7
5
M. Exterior fixtures
N. Interior fixtures
J
H
8
9
K
M
4
L
2
N
FINISH
0
6
5
Scheduling Using PERT/CPM
A path through a project network is a
route that follows a set of arcs from the
start node to the finish node.
 The length of a path is defined as the sum
of the durations of the activities of the
path.

◦ What are the paths and their corresponding
lengths for Reliable?
Critical Path
This is the path that has the longest
length through the project.
 The shortest time that a project can
conceivably be finished is the critical path.

◦ Why?
More Terminology

Earliest start time of an activity (ES)
◦ The time at which an activity will begin if there are no
delays in a project.

Earliest finish time of an activity (EF)
◦ The time at which an activity will finish if there are no
delays in a project.

Latest start time of an activity (LS)
◦ The latest possible time that an activity can start
without delaying the project.
More Terminology Cont.

Latest finish time of an activity (LF)
◦ The latest possible time that an activity can be
completed without delaying the project.

Forward pass
◦ The process of moving through a project
from start to finish to determine the earliest
start and finish times for the activities in the
project.
More Terminology Cont.

Backward pass
◦ The process of moving through a project from finish
to start to determine the latest start and finish times
for the activities in the project.

Slack for an activity
◦ The amount of time that a particular activity can be
delayed without delaying the whole project.
◦ It is calculated by taking the difference between the
latest finish time with the earliest finish time.
More Terminology Cont.

Earliest start time rule
◦ The earliest start time for an activity is equal
to the largest of the earliest finish times of its
immediate predecessors.

Latest finish time rule
◦ The latest finish time is equal to the smallest
of the latest start times of its immediate
successors.
Procedure for Obtaining Earliest
Times
Step 1: For the activity that starts the project,
assign an earliest start time of zero, i.e., ES=0.
 Step 2: For each activity whose ES has just been
obtained, calculate its earliest finish time as ES
plus duration of the activity.
 Step 3: For each new activity whose immediate
predecessors have EF values, obtain its ES by
using the earliest start time rule.

Procedure for Obtaining Earliest
Times Cont.
Step 4: Apply step 2 to calculate EF.
 Step 5: Repeat step 3 until ES and EF have
been obtained for all activities including
the finish node.

Procedure for Obtaining Latest
Times
Step 1: For each of the activities that
together complete the project, set its
latest finish time equal to the earliest
finish time of the finish node.
 Step 2: For each activity whose LF value
has just been obtained, calculate its latest
start time as LS equals LF minus the
duration of the activity.

Procedure for Obtaining Latest
Times Cont.
Step 3: For each new activity whose
immediate successors now have LS values,
obtain its LF by applying the latest finish
time rule.
 Step 4: Apply step 2 to calculate its LS.
 Step 5: Repeat step 3 until LF and LS have
been obtained for all activities.

START
D
G
6 S = (16, 20)
F = (22, 26)
9
S = (0, 0)
F = (0, 0)
S = (0, 0)
F = (2, 2)
A
2
B
4
S = (2, 2)
F = (6, 6)
C
10
S = (6, 6)
F = (16, 16)
E
4
S = (22, 26)
7 F = (29, 33)
H
0
S = (16, 16)
F = (20, 20)
F
5
I
7
J
8
S = (16, 18)
F = (23, 25)
S = (20, 20)
F = (25, 25)
S = (25, 25)
F = (33, 33)
S = (29, 33)
F = (38, 42)
K
M
4 S = (33, 34)
F = (37, 38)
2 S = (38, 42)
F = (40, 44)
N
FINISH
0 S = (44, 44)
F = (44, 44)
6
L
5 S = (33, 33)
F = (38, 38)
S = (38, 38)
F = (44, 44)
Ways of Finding the Critical Path
Examine all the paths and find the path
with the maximum length.
 Calculate the slack for an activity.

◦ If the slack is zero, it is on the critical path.
◦ If the slack is positive, it is not on the critical
path.
Time-Cost Trade-Offs

Reliable had an incentive bonus of
$150,000 to finish the project in 40
weeks.
◦ Is it worth while for Reliable to speed-up the
project?
Crashing
Crashing an activity refers to taking on
extra expenditures in order to reduce the
duration of an activity below its expected
value.
 Crashing a project refers to crashing a
number of activities to reduce the
duration of the project.

CPM Method of Time-Cost TradeOffs
This is a method concerned with whether
it is worthwhile to crash activities to
reduce the anticipated duration of the
project to a desired value.
 This assumes that there is a trade-off
between time and cost that has an inverse
relationship.

More Terminology
Normal Point is the time and cost of an
activity when it is performed in a normal
way.
 Crash point show the time and cost when
the activity is fully crashed.

Graph of Normal and Crash Points
Activity
cost
Crash cost
Crash
Normal
Normal cost
Crash time
Normal time
Activity duration
Marginal Cost Analysis
It is a method of using the marginal cost
of crashing individual activities on the
current critical path to determine the
least expensive way of reducing the
project duration to an acceptable level.
 This method requires you to calculate the
cost per desired time unit and compare
each cost with the other costs.

Activity
Normal
Crash
Normal
Crash
Maximum
Reduction
in Time (weeks)
A
2
1
$180,000
$280,000
1
$100,000
B
4
2
320,000
420,000
2
50,000
C
10
7
620,000
860,000
3
80,000
D
6
4
260,000
340,000
2
40,000
E
4
3
410,000
570,000
1
160,000
F
5
3
180,000
260,000
2
40,000
G
7
4
900,000
1,020,000
3
40,000
H
9
6
200,000
380,000
3
60,000
I
7
5
210,000
270,000
2
30,000
J
8
6
430,000
490,000
2
30,000
K
4
3
160,000
200,000
1
40,000
L
5
3
250,000
350,000
2
50,000
M
2
1
100,000
200,000
1
100,000
N
6
3
330,000
510,000
3
60,000
Time (weeks)
Cost
Crash Cost
per Week
Saved
Marginal Cost Analysis Cont.
Once the marginal cost for crashing each
activity has been conducted, you next
want to choose the crashing that has the
smallest marginal cost.
 Next, calculate the effect that the crash
has on each path.

◦ Note: Crashing can potentially cause another
path to become a critical path.
Solving Crashing Problems Using LP

There are three decisions to be made:
◦ The start time of each activity
◦ The reduction in each activity due to crashing
◦ The finish time of the project

LP model will be examined in class.