Network Optimization ModelsChapter 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, NonbasicArcs

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

min𝑤.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 + 7𝑥𝐸𝑇Subject 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

max𝑤.𝑟.𝑡. 𝑥𝑂𝐴,𝑥𝑂𝐵,𝑥𝑂𝐶,𝑥𝐴𝐵,𝑥𝐴𝐷,𝑥𝐵𝐶,

𝑥𝐵𝐷,𝑥𝐵𝐸,𝑥𝐶𝐸,𝑥𝐷𝑇,𝑥𝐸𝐷,𝑥𝐸𝑇

𝑥𝑂𝐴+𝑥𝑂𝐵 + 𝑥𝑂𝐶

𝑥𝑂𝐴−𝑥𝐴𝐵 − 𝑥𝐴𝐷 = 0𝑥𝑂𝐵 + 𝑥𝐴𝐵 − 𝑥𝐵𝐶 − 𝑥𝐵𝐷 − 𝑥𝐵𝐸 = 0

𝑥𝑂𝐶 + 𝑥𝐵𝐶 − 𝑥𝐶𝐸 = 0𝑥𝐴𝐷 + 𝑥𝐵𝐷 − 𝑥𝐷𝑇 + 𝑥𝐸𝐷 = 0𝑥𝐵𝐸 + 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)

A

D

C

B

E

H

G

F

I

8

7

6

7

106 3

5

7

6

47

0

0

0

0

00

0

0

0

0

0

0

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 = 𝑖=1𝑛 𝑗=1

𝑛 𝑐𝑖𝑗𝑥𝑖𝑗 Subject to:

𝑗=1𝑛 𝑥𝑖𝑗 − 𝑗=1

𝑛 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.

A

START

G

H

M

F

J

K L

N

Activity Code

A. Excavate

B. Foundation

C. Rough wall

D. Roof

E. Exterior plumbing

F. Interior plumbing

G. Exterior siding

H. Exterior painting

I. Electrical work

J. Wallboard

K. Flooring

L. Interior painting

M. Exterior fixtures

N. Interior fixtures

2

4

10

746

7

9

5

8

4 5

6

2

0

0FINISH

D IE

C

B

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.

A

START

G

H

M

F

J

FINISH

K L

N

D IE

C

B

2

4

10

746

7

9

5

8

4 5

6

2

S = (0, 0) F = (2, 2)

S = (2, 2) F = (6, 6)

S = (16, 20) F = (22, 26)

S = (16, 16) F = (20, 20)

S = (16, 18) F = (23, 25)

S = (20, 20) F = (25, 25)

S = (22, 26) F = (29, 33)

S = (6, 6) F = (16, 16)

S = (0, 0) F = (0, 0)

S = (25, 25) F = (33, 33)

S = (33, 33) F = (38, 38)

S = (38, 38) F = (44, 44)

S = (33, 34) F = (37, 38)

S = (29, 33) F = (38, 42)

S = (38, 42) F = (40, 44)

S = (44, 44) F = (44, 44)

0

0

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 Trade-

Offs 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 duration

Activity cost

Crash cost

Normal cost Normal

Crash

Crash time Normal time

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.

Time (weeks) Cost MaximumReduction

in Time (weeks)

Crash Costper Week

SavedActivity Normal Crash Normal Crash

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

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.