Set2_transport & Networks

7/28/2019 Set2_transport & Networks

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Chapter 7Transportation, Assignment and

Transshipment Problems


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Applications Physical analogof nodes Physical analogof arcs Flow

Communication

systems

phone exchanges,

computers,

transmission

facilities, satellites

Cables, fiber optic

links, microwave

relay links

Voice messages,

Data,

Video transmissions

Hydraulic systemsPumping stations

Reservoirs, LakesPipelines

Water, Gas, Oil,

Hydraulic fluids

Integrated

computer circuits

Gates, registers,

processorsWires Electrical current

Mechanical systems

JointsRods, Beams,

SpringsHeat, Energy

Transportation

systems

Intersections,

Airports,

Rail yards

Highways,

Airline routes

Railbeds

Passengers,

freight,

vehicles,

operators

Applications of Network

Optimization


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Description

A transportation problem basically deals with the

problem, which aims to find the best way to fulfill

the demand of n demand points using the

capacities of m supply points. While trying to findthe best way, generally a variable cost of shipping

the product from one supply point to a demand

point or a similar constraint should be taken into

consideration.


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7.1 Formulating Transportation

ProblemsExample 1: Powerco has three electric

power plants that supply the electric needs

of four cities.

The associated supply of each plant anddemand of each city is given in the table 1.

The cost of sending 1 million kwh ofelectricity from a plant to a city depends on

the distance the electricity must travel.


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Transportation tableau

A transportation problem is specified by

the supply, the demand, and the shipping

costs. So the relevant data can be

summarized in a transportation tableau.The transportation tableau implicitly

expresses the supply and demand

constraints and the shipping cost between

each demand and supply point.


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Table 1. Shipping costs, Supply, and Demand

for Powerco Example

From To

City 1 City 2 City 3 City 4 Supply

(Million kwh)

Plant 1 $8 $6 $10 $9 35

Plant 2 $9 $12 $13 $7 50

Plant 3 $14 $9 $16 $5 40Demand

(Million kwh)45 20 30 30

Transportation Tableau


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Solution

1. Decision Variable:

Since we have to determine how much electricity

is sent from each plant to each city;

Xij = Amount of electricity produced at plant i

and sent to city j

X14 = Amount of electricity produced at plant 1

and sent to city 4


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2. Objective function

Since we want to minimize the total cost of shipping

from plants to cities;

Minimize Z = 8X11+6X12+10X13+9X14

+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34


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3. Supply Constraints

Since each supply point has a limited productioncapacity;

X11+X12+X13+X14


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4. Demand Constraints

Since each supply point has a limited productioncapacity;

X11+X21+X31 >= 45X12+X22+X32 >= 20

X13+X23+X33 >= 30

X14+X24+X34 >= 30


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5. Sign Constraints

Since a negative amount of electricity can not be

shipped all Xijs must be non negative;

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)


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LP Formulation of Powercos Problem

Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34

S.T.: X11+X12+X13+X14 = 30

X14+X24+X34 >= 30

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)


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General Description of a Transportation

Problem

1. A set of m supply points from which a good is

shipped. Supply point i can supply at mostsi

units.

2. A set of n demand points to which the good is

shipped. Demand pointj must receive at least di

units of the shipped good.

3. Each unit produced at supply point i and shippedto demand pointj incurs a variable costofcij.


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Xij = number of units shipped fromsupply point i to

demand point j

),...,2,1;,...,2,1(0

),...,2,1(

),...,2,1(..

min

1

1

1 1

njmiX

njdX

misXts

Xc

ij

mi

i

jij

nj

j

iij

mi

i

nj

j

ijij


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Balanced Transportation Problem

If Total supply equals to total demand, theproblem is said to be a balanced

transportation problem:

nj

j

j

mi

i

i ds11


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Balancing a TP if total supply exceeds total

demand

If total supply exceeds total demand, we

can balance the problem by adding dummy

demand point. Since shipments to thedummy demand point are not real, they are

assigned a cost of zero.


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Balancing a transportation problem if total

supply is less than total demand

If a transportation problem has a total

supply that is strictly less than total

demand the problem has no feasiblesolution. There is no doubt that in such a

case one or more of the demand will be left

unmet. Generally in such situations a

penalty cost is often associated with unmetdemand and as one can guess this time the

total penalty cost is desired to be minimum


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7.2 Finding Basic Feasible

Solution for TPUnlike other Linear Programming

problems, a balancedTP with m supply

points and n demand points is easier to

solve, although it has m + n equality

constraints. The reason for that is, if a set

of decision variables (xijs) satisfy all but

one constraint, the values for xijs willsatisfy that remaining constraintautomatically.


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Methods to find the bfs for a balanced TP

There are three basic methods:

1. Northwest Corner Method

2. Minimum Cost Method

3. Vogels Method


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1. Northwest Corner Method

To find the bfs by the NWC method:

Begin in the upper left (northwest) corner of the

transportation tableau and set x11 as large as

possible (here the limitations for setting x11 to a

larger number, will be the demand of demandpoint 1 and the supply of supply point 1. Your

x11 value can not be greater than minimum of

this 2 values).


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According to the explanations in the previous slide

we can set x11=3 (meaning demand of demand

point 1 is satisfied by supply point 1).5

6

2

3 5 2 3

3 2

6

2

X 5 2 3


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After we check the east and south cells, we saw that

we can go east (meaning supply point 1 still has

capacity to fulfill some demand).

3 2 X

6

2

X 3 2 3

3 2 X

3 3

2

X X 2 3


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After applying the same procedure, we saw that we

can go south this time (meaning demand point 2

needs more supply by supply point 2).

3 2 X

3 2 1

2

X X X 3

3 2 X

3 2 1 X

2

X X X 2


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Finally, we will have the following bfs, which is:

x11

=3, x12

=2, x22

=3, x23

=2, x24

=1, x34

=2

3 2 X

3 2 1 X

2 X

X X X X


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2. Minimum Cost Method

The Northwest Corner Method dos not utilize shippingcosts. It can yield an initial bfs easily but the total

shipping cost may be very high. The minimum cost

method uses shipping costs in order come up with a

bfs that has a lower cost. To begin the minimum costmethod, first we find the decision variable with the

smallest shipping cost (Xij). Then assignXijits largest

possible value, which is the minimum ofsi and dj


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After that, as in the Northwest Corner Method we

should cross out row i and column j and reduce the

supply or demand of the noncrossed-out row orcolumn by the value of Xij. Then we will choose the

cell with the minimum cost of shipping from the

cells that do not lie in a crossed-out row or column

and we will repeat the procedure.


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An example for Minimum Cost Method

Step 1: Select the cell with minimum cost.

2 3 5 6

2 1 3 5

3 8 4 6

5

10

15

12 8 4 6


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Step 2: Cross-out column 2

2 3 5 6

2 1 3 5

8

3 8 4 6

12 X 4 6

5

2

15


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Step 3: Find the new cell with minimum shipping

cost and cross-out row 2

2 3 5 6

2 1 3 5

2 8

3 8 4 6

5

X

15

10 X 4 6


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cost and cross-out row 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

X

X

15

5 X 4 6


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cost and cross-out column 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5

X

X

10

X X 4 6


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cost and cross-out column 3

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4

X

X

6

X X X 6


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Step 7: Finally assign 6 to last cell. The bfs is found

as: X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4 6

X

X

X

X X X X


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3. Vogels Method

Begin with computing each row and column a penalty.The penalty will be equal to the difference between

the two smallest shipping costs in the row or column.

Identify the row or column with the largest penalty.

Find the first basic variable which has the smallestshipping cost in that row or column. Then assign the

highest possible value to that variable, and cross-out

the row or column as in the previous methods.

Compute new penalties and use the same procedure.


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An example for Vogels MethodStep 1: Compute the penalties.

Supply Row Penalty

6 7 8

15 80 78

Demand

Column Penalty 15-6=9 80-7=73 78-8=70

7-6=1

78-15=63

15 5 5

10

15


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Step 2: Identify the largest penalty and assign the

highest possible value to the variable.

Supply Row Penalty

6 7 8

5

15 80 78

Demand

Column Penalty 15-6=9 _ 78-8=70

8-6=2

78-15=63

15 X 5

5

15


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Supply Row Penalty

6 7 8

5 5

15 80 78

Demand

Column Penalty 15-6=9 _ _

_

_

15 X X

0

15


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Supply Row Penalty

6 7 8

0 5 5

15 80 78

Demand

Column Penalty _ _ _

_

_

15 X X

X

15


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Step 5: Finally the bfs is found as X11=0, X12=5,

X13=5, and X21=15

Supply Row Penalty

6 7 8

0 5 5

15 80 78

15

Demand

Column Penalty _ _ _

_

_

X X X

X

X


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7.3 The Transportation Simplex

Method

In this section we will explain how the simplex

algorithm is used to solve a transportation problem.


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How to Pivot a Transportation Problem

Based on the transportation tableau, the followingsteps should be performed.

Step 1. Determine (by a criterion to be developedshortly, for example northwest corner method) the

variable that should enter the basis.

Step 2. Find the loop (it can be shown that there isonly one loop) involving the entering variable and

some of the basic variables.Step 3. Counting the cells in the loop, label them aseven cells or odd cells.


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Step 4. Find the odd cells whose variable assumes thesmallest value. Call this value . The variablecorresponding to this odd cell will leave the basis. To

perform the pivot, decrease the value of each odd cell

by and increase the value of each even cell by . The

variables that are not in the loop remain unchanged.The pivot is now complete. If =0, the enteringvariable will equal 0, and an odd variable that has a

current value of 0 will leave the basis. In this case a

degenerate bfs existed before and will result after thepivot. If more than one odd cell in the loop equals ,you may arbitrarily choose one of these odd cells to

leave the basis; again a degenerate bfs will result


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7.5. Assignment Problems

Example: Machineco has four jobs to be completed.

Each machine must be assigned to complete one job.

The time required to setup each machine for completing

each job is shown in the table below. Machinco wants tominimize the total setup time needed to complete the

four jobs.


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Setup times

(Also called the cost matrix)Time (Hours)

Job1 Job2 Job3 Job4

Machine 1 14 5 8 7

Machine 2 2 12 6 5

Machine 3 7 8 3 9

Machine 4 2 4 6 10


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The Model

According to the setup table Machincos problem can beformulated as follows (fori,j=1,2,3,4):

10

1

1

1

1

1

1

11..

10629387

5612278514min

44342414

43332313

42322212

41312111

44434241

34333231

24232221

14131211

4443424134333231

2423222114131211

ijij orXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXXXXXXts

XXXXXXXX

XXXXXXXXZ


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For the model on the previous page note that:

Xij=1 if machine i is assigned to meet the demands of

jobj

Xij=0 if machine i is not assigned to meet the demands

of jobj

In general an assignment problem is balancedtransportation problem in which all supplies and

demands are equal to 1.


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The Assignment Problem

In general the LP formulation is given as

Minimize 1 1

1

1

1 1

1 1

0

, , ,

, , ,

or 1,

n n

ij ij

i j

n

ijj

n

iji

ij

c x

x i n

x j n

x ij

Each supply is 1

Each demand is 1

Comments on the Assignment


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Comments on the Assignment

Problem The Air Force has used this for assigning

thousands of people to jobs.

This is a classical problem. Research on the

assignment problem predates research on LPs.

Very efficient special purpose solution techniquesexist.

10 years ago, Yusin Lee and J. Orlin solved a problemwith 2 million nodes and 40 million arcs in hour.

Al h h h i i l b


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Although the transportation simplex appears to be very

efficient, there is a certain class of transportation

problems, called assignment problems, for which the

transportation simplex is often very inefficient. For thatreason there is an other method called The Hungarian

Method. The steps of The Hungarian Method are as

listed below:

Step1. Find a bfs. Find the minimum element in each rowof the mxm cost matrix. Construct a new matrix by

subtracting from each cost the minimum cost in its row.

For this new matrix, find the minimum cost in eachcolumn. Construct a new matrix (reduced cost matrix) by

subtracting from each cost the minimum cost in its

column.


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Step2. Draw the minimum number of lines (horizontaland/or vertical) that are needed to cover all zeros in the

reduced cost matrix. If m lines are required , an optimalsolution is available among the covered zeros in the

matrix. If fewer than m lines are required, proceed to step

3.

Step3. Find the smallest nonzero element (call its valuek) in the reduced cost matrix that is uncovered by the

lines drawn in step 2. Now subtract k from eachuncovered element of the reduced cost matrix and add k

to each element that is covered by two lines. Return to

step2.


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7.6 Transshipment Problems

A transportation problem allows only shipments that godirectly from supply points to demand points. In many

situations, shipments are allowed between supply points

or between demand points. Sometimes there may also

be points (called transshipment points) through whichgoods can be transshipped on their journey from a

supply point to a demand point. Fortunately, the optimal

solution to a transshipment problem can be found by

solving a transportation problem.

Transshipment Problem


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Transshipment Problem

An extension of a transportation problem

More general than the transportation problem in that in thisproblem there are intermediate transshipment points. Inaddition, shipments may be allowed between supply points

and/or between demand points

LP Formulation Supply point: it can send goods to another point but cannot

receive goods from any other point

Demand point It can receive goods from other points but

cannot send goods to any other point Transshipment point: It can both receive goods from other

points send goods to other points


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The following steps describe how the optimal solution to

a transshipment problem can be found by solving a

transportation problem.

Step1. If necessary, add a dummy demand point (with a

supply of 0 and a demand equal to the problems excess

supply) to balance the problem. Shipments to the dummyand from a point to itself will be zero. Let s= total

available supply.

Step2. Construct a transportation tableau as follows: A

row in the tableau will be needed for each supply point

and transshipment point, and a column will be needed for

each demand point and transshipment point.


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Each supply point will have a supply equal to itsoriginal supply, and each demand point will have a

demand to its original demand. Let s= total available

supply. Then each transshipment point will have a supply

equal to (points original supply)+s and a demand equal

to (points original demand)+s. This ensures that anytransshipment point that is a net supplier will have a netoutflow equal to points original supply and a netdemander will have a net inflow equal to points original

demand. Although we dont know how much will beshipped through each transshipment point, we can be

sure that the total amount will not exceed s.

Transshipment Example


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Transshipment Example Example 5: Widgetco manufactures widgets at two

factories, one in Memphis and one in Denver. The

Memphis factory can produce as 150 widgets, and theDenver factory can produce as many as 200 widgetsper day. Widgets are shipped by air to customers inLA and Boston. The customers in each city require130 widgets per day. Because of the deregulation of

airfares, Widgetco believes that it may be cheaperfirst fly some widgets to NY or Chicago and then flythem to their final destinations. The cost of flying awidget are shown next. Widgetco wants to minimizethe total cost of shipping the required widgets tocustomers.

Transportation Tableau Associated


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Transportation Tableau Associatedwith the Transshipment Example

NY Chicago LA Boston Dummy Supply Memphis $8 $13 $25 $28 $0 150 Denver $15 $12 $26 $25 $0 200

NY $0 $6 $16 $17 $0 350

Chicago $6 $0 $14 $16 $0 350 Demand 350 350 130 130 90

Supply points: Memphis, Denver

Demand Points: LA Boston

Transshipment Points: NY, Chicago

The problem can be solved using the transportation simplexmethod

Limitations of Transportation Problem


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One commodity ONLY: any one product supplied

and demanded at multiple locations Merchandise Electricity, water

Invalid for multiple commodities: (UNLESStransporting any one of the multiple commodities iscompletely independent of transporting any othercommodity and hence can be treated by itself alone) Example: transporting product 1 and product 2 from the

supply points to the demand points where the total amount(of the two products) transported on a link is subject to acapacity constraint

Example: where economy of scale can be achieved bytransporting the two products on the same link at a largertotal volume and at a lower unit cost of transportation



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Difficult to generalize the technique to accommodate(these are generic difficulty for mathematical

programming, including linear and non-linearprogramming

Economy of scale the per-unit cost of transportation on a linkdecreasing with the volume (nonlinear and concave; there is a trick

to convert a non-linear program with a piecewise linear butconvex objective function to a linear program; no such tricks exists

for a piecewise linear but concave objective function)

Fixed-cost: transportation usually involves fixed charges. Forexample, the cost of truck rental (or cost of trucking in general)

consists of a fixed charge that is independent of the mileage and a

mileage charge that is proportional to the total mileage driven.

Such fixed charges render the objective function NON-LINEAR

and CONCAVE and make the problem much more difficult to

solve


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Chapter 8

Network Models


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Networks are Everywhere

Physical NetworksRoad NetworksRailway Networks

Airline traffic NetworksElectrical networks, e.g., the power grid

Abstract networksorganizational charts

precedence relationships in projects

Others?


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Overview:

Networks and graphs are powerfulmodeling tools.

Most OR models have networks or graphs

as a major aspect Each representation has its advantages

Major purpose of a representation

efficiency in algorithms ease of use


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Description

Many important optimization problems can be analyzed

by means of graphical or network representation. In this

chapter the following network models will be discussed:

1. Shortest path problems2. Maximum flow problems

3. CPM-PERT project scheduling models

4. Minimum Cost Network Flow Problems5. Minimum spanning tree problems

/


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WHAT IS CPM/PERT FOR?

CPM/PERT are fundamental tools of project

management and are used for one of a kind, oftenlarge and expensive, decisions such as building

docks, airports and starting a new factory. Such

decisions can be described via mathematicalmodels, but this is not essential. Some would

argue that CPM/PERT is not a pure OR topic.

CPM/PERT really falls into gray area that can beclaimed by fields other than OR also.


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General Comments on CPM/PERT vs. ALB

Assembly Line Balancing (ALB) are naturally not

discussed in this text, but it is important to be aware of thehuge difference between the ALB and CPM/PERT

concepts because the precedence diagrams look so similar.

Activity on node (AON) method of network precedence

diagram drawing and the ALB diagram are identical

looking at first. The ALB deals with small repetitive

items such as TVs while CPM/PERT deals with large one

of a kind projects.

Network Analysis and Their LP


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y

Connections

Several Major Basic Classes of Network Problems How to recognize and formulate them? What are the

features they can be used to model? What are their

limitations?

All can be formulated as an LP

Several important ad-hoc algorithms and their rationaleswill be provided

Most efficient transportation of goods, information

etc. through a network Transportation problem (already discussed)

Transshipment problem



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yConnections

Most efficient way to go from one point to another in adistance network or networks representing non-distance

phenomenon, e.g., the cost network representingproduction, inventory, and other costs

Shortest path problem:

Find the shortest path between two points in a network Dijkstra algorithm Limitations: Breakdown of the Dijkstras algorithm when side

constraints are added LP formulation

LP formulation to accommodate some side constraints, e.g.,disallowing use of two particular links in the shortest path

An example application to non-distance contexts:Minimum production and inventory cost in the context ofdynamic programming.



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yConnections

Maximum amount of flow from one point toanother in a capacitated network

Maximum flow problem

The flow-augmenting algorithm

Limitations: breakdown of the flow-augmentingalgorithm when side-constraints are added

LP formulation

LP formulation to accommodate some side-

constraints, e.g., no link carrying more than 30% of thewhole flow so as to avoid drastic reduction of flow afterfailure of any one flow-carrying link


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8.1 Basic Definitions

A graph or network is defined by two sets of symbols: Nodes: A set of points or vertices(call it V) are callednodes of a graph or network.

Arcs: An arc consists of an ordered pair of vertices andrepresents a possible direction of motion that may occur

between vertices.

1 2

Nodes

1 2

Arc

Chain: A sequence of arcs such that every arc has


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Chain: A sequence of arcs such that every arc hasexactly one vertex in common with the previous arc is

called a chain.

1 2

Common vertex

between two arcs


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Path: A path is a chain in which the terminal node ofeach arc is identical to the initial node of next arc.

For example in the figure below (1,2)-(2,3)-(4,3) is a

chain but not a path; (1,2)-(2,3)-(3,4) is a chain and a

path, which represents a way to travel from node 1 to

node 4.

2 3

1 4

Essence of Dijkstras Shortest- Path


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jAlgorithm

Key Points regarding the nature of thealgorithm

In each iteration, the shortest path from the originto one of the rest of the nodes is found. That is, we

obtain one new solved node in each iteration.(More than one such path and node may be foundin one iteration when there is a tie. There may alsoexist multiple shortest paths from the origin tosome nodes.)

The algorithm stops when the shortest path to thedestination is found



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jAlgorithm

General thought process involved in eachiteration

Let S be the current set of solved nodes (the setof nodes whose shortest paths from the origin been

found), N be the set of all nodes, and NS be theset of unsolved nodes

1. The next solved node should be reachable directlyfrom one of the solved nodes via one direct link or arc

(these nodes can be called neighboring nodes of thecurrent solved nodes). Therefore, we consider only such

nodes and all the links providing the access from the

current solved nodes to these neighboring nodes (but no

other links).



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jAlgorithm

2. For each of these neighboring nodes, find theshortest path from the origin via only currentsolved nodes and the corresponding distance from

the origin

3. In general, there exist multiple such neighboringnodes.The shortest path to one of these nodes is

claimed to have been found. This node is the one

that has the shortest distance from the origin

among these neighboring nodes being considered.

Call this new node solved node.

Algorithm for the Shortest Path Problem


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g

Objective of the nth iteration: Find the nth nearest node to theorigin (to be repeated for n = 1, 2, until the nth nearest node is

the destination) Input for the nth Iteration: (n1) nearest nodes to the origin(solved for at the previous iterations), including their shortestpath and distance from the origin. (These nodes plus the originwill be called solved nodes; the others are unsolved nodes)

Candidates for the nth nearest node: Each solved node that isdirectly connected by a link to one or more unsolved nodesprovides one candidate the unsolved node with the shortestconnecting link (ties provide additional candidates)

Calculation of nth nearest node: For each solved node and itscandidate, add the distance between them and the distance of the

shortest path from the origin to this solved node. The candidatewith the smallest such total distance is the nth nearest node (tiesprovide additional solved nodes), and its shortest path is the onegenerating this distance.

The Road System for Seervada Park


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y Cars are not allowed into the park There is a narrow winding road system for trams and

for jeeps driven by the park rangers The road system is shown without curves in the next slide Location O is the entrance into the park Other letters designate the locations of the ranger stations

The scenic wonder is at location T The numbers give the distance of these winding roads inmiles

The park management wishes to determine whichroute from the park entrance to station T has the

smallest total distance for the operation of the trams

The Road System for Seervada Park


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y

O

A

B

C

D

E

T

2

1

2

3

4

7

41

7

5

5

4

Dijkstras Algorithm for Shortest Path on a


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j gNetwork with Positive Arc Lengths

Oth iteration: Shortest distance from node O toNode O. S = {O}. Ist iteration:

Step 1: Neighboring Nodes = {A, B, C}

Step 2: Shortest path from O to neighboring nodesthat traverse through the current set of solvednodes S. Min {2, 5, 4} = 2 (corresponding to nodeA).

Step 3: The shortest path from O to A has beenfound with a distance of 2. S = {O, A}

Dijkstras Algorithm for Shortest Path on aN k i h P i i A L h


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j gNetwork with Positive Arc Lengths

O

A

B

C

Solved Nodes2

5

4

Dijkstras Algorithm for Shortest Path on ak i h i i h


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Network with Positive Arc Lengths

2nd Iteration:Step 1: Neighboring nodes = {B, C, D}

Step 2: Min (Min (2 + 2, 5), 4, (2 + 7)) = 4.

Step 3: Shortest path from B and C has been

found. S = {O, A, B, C}

O

A D

C

B

7

2

5

4

Current Solved Nodes

(2)

(0)

Dijkstras Algorithm for Shortest Path on ak i h i i A h


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3rd Iteration:

Step 1: Neighboring nodes = {D, E}. Only AD, BD, BE,and CE

Step 2: Min(Min(2 + 7, 4+4), Min(4 + 3, 4+4)) = 7

Step 3: The shortest path to E has been found S = {O, A, B,C, E}

O

A

B

C

D

E

7

4

3

4

Current Solved

Nodes (2)

(4)

(4)

Dijkstras Algorithm for Shortest Path on aN k i h P i i A L h


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Network with Positive Arc Lengths Iteration 4

Step 1: Include (only) Nodes D and T. Includeonly arcs AD, BD, ED, & ET

Step 2: Min((min(2+7, 4+4, 7+1), (7+7))) = 8

Step 3: Shortest path from node O to Node D hasbeen found. S = {O, A, B, C, D, E}

O

A

B

C

E

D

T

(2)

(4)

(4) (7)

7

4

1

7Current solved nodes

Dijkstras Algorithm for Shortest Path on aN t k ith P iti A L th


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Iteration 5

Step 1: Include only node T and include arcs DT and ET Step 2: Min(8+5, 7+7) = 13 (no other competing nodes)

Step 3: The shortest path from the origin to T, thedestination, has been found, with a distance of 13

O

A

B

C

D

E

T(0)

(2)

(4)

(4)

(8)

(7)

5

7

Current solved nodes

Dijkstras Algorithm for Shortest Path on a


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Network with Positive Arc Lengths Final Solution

Incidentally, we have also found the nthnearest node from the origin sequentially

O

A

B

C

D

E

T

2

4(0)

2

4

3 1

5

(2)

(4)

(4)

(7)

(8)

(13)

Shortest-Path Algorithm Applied toSeervada Park Problem


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Seervada Park Problemn Solved Nodes

Directly Connected

to Unsolved Nodes

Closest

Connected

UnsolvedNode

Total

Distance

Involved

Nth

Nearest

Node

Minimum

Distance

Last

Connection

1 O A 2 A 2 OA

2, 3 O

A

C

B

4

2+2=4

C

B

4

4

OC

AB

4 A

B

C

D

E

E

2+7=9

4+3=7

4+4=8

E 7 BE

5 A

B

E

D

D

D

2+7=9

4+4=8

7+1=8

D

D

8

8

BD

ED

6 D

E

T

T

8+5=13

7+7=14

T 13 DT

LP Formu at on o t e ortest PatP bl


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Problem Consider the following shortest path problem

from node 1 to node 6 : denotes a link

Send one unit of flow from node 1 to node 6

1

2

3

4

5

6

4

3

2

3

3

2

1

2

5

4

3

6

7

2

LP Formulation of the Shortest PathP bl


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Problem Use flow conservation constraints

(Outflow from any nodeinflow to that node) = 0For origin = 1For destination = -1

For all other nodes = 0Let xj denote the flow along link j, j = 1, 2, .., 7, xj

= 0 or 1

It turns out that this 0-1 constraints can be replacedby 0 xj 1, which can in turn be replaced by xj 0

LP Formulation of the Shortest PathP bl


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Problem

Min 4x1 + 3x2 + 3x3 + 2x4 + 3x5 + 2x6 + 2x7

S.t. x1 + x2 = 1

-x1 + x3 + x4 = 0

- x2 + x5 = 0 - x3 + x6 = 0

- x4x5 + x7 = 0

- x6x7 = -1 Xj 0, j = 1, 2, , 7 (xj integers)

Maximum Flow Problem


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Maximum flow problem description All flow through a directed and connected network

originates at one node (source) and terminates at one

another node (sink)

All the remaining nodes are transshipment nodes

Flow through an arc is allowed only in the directionindicated by the arrowhead, where the maximum amount of

flow is given by the capacity of that arc. At the source, all

arcs point away from the node. At the sink, all arcs point

into the node

The objective is to maximize the total amount of flow fromthe source to the sink (measured as the amount leaving the

source or the amount entering the sink)

Maximum Flow Problem


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Typical applications

Maximize the flow through a companysdistribution network from its factories to itscustomers

Maximize the flow through a companys supply

network from its vendors to its factoriesMaximize the flow of oil through a system ofpipelines

Maximize the flow of water through a system of

aqueductsMaximize the flow of vehicles through a

transportation network

Maximum Flow Algorithm


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Some Terminology

The residual networkshows the remaining arccapacities for assigning additional flows after some

flows have been assigned to the arcs

O B

25

The residual capacity for flow from node O to Node B

The residual capacity for assigning some flow from node B to node O



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An augmenting path is a directed path from the source tothe sink in the residual network such that every arc on this

path has strictly positive residual capacity

The residual capacity of the augmenting path is theminimum of these residual capacities (the amount of flow

that can feasibly be added to the entire path)

Basic idea Repeatedly select some augmenting path and add a flow

equal to its residual capacity to that path in the original

network. This process continues until there are no more

augmenting paths, so that the flow from the source to thesink cannot be increased further



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The Augmenting Path Algorithm Assume that the arc capacities are either integers or rational

numbers

1. identify an augmenting path by finding somedirected path from the source to the sink in the

residual network such that every arc on this path has

strictly positive residual capacity. If no such path

exists, the net flows already assigned constitute an

optimal flow pattern

2. Identify the residual capacity c* of this augmentingpath by finding the minimum of the residualcapacities of the arcs on this path. Increase the flow in

this path by c*

Maximum Flow Example


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During the peak season the park management of theSeervada park would like to determine how to route

the various tram trips from the park entrance (StationO) to the scenic (Station T) to maximize the numberof trips per day. Each tram will return by the sameroute it took on the outgoing trip so the analysisfocuses on outgoing trips only. To avoid undulydisturbing the ecology and wildlife of the region,strict upper limits have been imposed on the numberof outgoing trips allowed per day in the outbounddirection on each individual road. For each road the

direction of travel for outgoing trips is indicated byan arrow in the next slide. The number at the base ofthe arrow gives the upper limit on the number ofoutgoing trips allowed per day.



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Consider the problem of sending as many units from node O tonode T for the following network (current flow, capacity):

O

A

B

C

D

E

T

(0,5)

(0,2)

(0,1)

(0,5)

(0,4)

(0,6)

(0,4)

(0,1)

(0,3)

(0,9)

(0,7)

(0,4)



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Iteration 1: one of the several augmenting paths is OBET, which has aresidual capacity of min{7, 5, 6} = 5. By assigning the flow of 5 to this path, the

resulting network is shown above

O

A

B

C

D

E

T

(0,5)

(0,2)

(0,1)

(5,5)

(0,4)

(5,6)

(0,4)

(0,1)

(0,3)

(0,9)

(5,7)

(0,4)



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Iteration 2: Assign a flow of 3 to the augmenting path

OADT. The resulting residual network is

O

A

B

C

D

E

T

(3,5)

(0,2)

(0,1)

(5,5)

(0,4)

(5,6)

(0,4)

(0,1)

(3,3)

(3,9)

(5,7)

(0,4)



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OABDT. The resulting residual network is

O

A

B

C

D

E

T

(4,5)

(0,2)

(1,1)

(5,5)

(0,4)

(5,6)

(1,4)

(0,1)

(3,3)

(4,9)

(5,7)

(0,4)



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Iteration 4: Assign a flow of 2 to the augmenting path OBDT. Theresulting residual network is

O

A

B

C

D

E

T

(4,5)

(0,2)

(1,1)

(5,5)

(0,4)

(5,6)

(3,4)

(0,1)

(3,3)

(6,9)

(7,7)

(0,4)



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OCEDT. The resulting residual network is

O

A

B

C

D

E

T

(4,5)

(0,2)

(1,1)

(5,5)

(1,4)

(5,6)

(3,4)

(1,1)

(3,3)

(7,9)

(7,7)

(1,4)



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OCET. The resulting residual network is

O

A

B

C

D

E

T

(4,5)

(0,2)

(1,1)

(5,5)

(2,4)

(6,6)

(3,4)

(1,1)

(3,3)

(7,9)

(7,7)

(2,4)



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There are no more flow augmenting paths, so the currentflow pattern is optimal

13

O

C

A

B

E

D

T

13

4

72

1

3

3

5

6

1

7

2



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Recognizing optimality Max-flow min-cut theorem can be useful A cut is defined as any set of directed arcs

containing at least one arc from every directedpath from the source to the sink

For any particular cut, the cut value is the sumof the arc capacities of the arcs of the cut

The theorem states that, for any network with a

single source and sink, the maximum feasibleflow from the source to the sink equals theminimum cut value for all cuts of the network

8 3 Maximum Flow Problems


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8.3 Maximum Flow Problems

Many situations can be modeled by a network in whichthe arcs may be thought of as having a capacity that

limits the quantity of a product that may be shipped

through the arc. In these situations, it is often desired to

transport the maximum amount of flow from a startingpoint (called the source) to a terminal point (called the

sink). Such problems are called maximum flow

problems.

An example for maximum flow problem


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p pSunco Oil wants to ship the maximum possible amount

of oil (per hour) via pipeline from nodeso to nodesi as

shown in the figure below.

so si21(2)2 (1)3 (2)2

(1)3

a0

3(1)4 (1)1

The various arcs represent pipelines of different diameters. The

maximum number of barrels of oil that can be pumped through

each arc is shown in the table above (also called arc capacity).

Arc Capacity

(so,1) 2

(so,2) 3(1,2) 3

(1,3) 4

(3,si) 1

(2,si) 2

For reasons that will become clear soon, an artificial arc

called a0 is added from the sink to the source. To


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called a0 is added from the sink to the source. To

formulate an LP about this problem first we should

determine the decision variable.

Xij = Millions of barrels of oil per hour that will pass

through arc(i,j) of pipeline.

For a flow to be feasible it needs to be in the following

range:

0


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Suncos goal is to maximizeX0.

Max Z=X0

S.t. Xso,1


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One optimal solution to this LP is Z=3,Xso,1=2, X13=1,

X12=1, Xso,2=1, X3,si=1, X2,si=2, Xo=3.

8.6 Minimum Spanning Tree


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108

p g

ProblemsSuppose that each arc (i,j) in a network has a length

associated with it and that arc (i,j) represents a way of

connecting node i to node j. For example, if each node

in a network represents a computer in a computernetwork, arc(i,j) might represent an underground cable

that connects computer i to computer j. In many

applications, we want to determine the set of arcs in a

network that connect all nodes such that the sum of thelength of the arcs is minimized. Clearly, such a group of

arcs contain no loop.

Minimum Spanning Tree Problem

A di d d d k i b i


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An undirected and connected network is beingconsidered, where the given information includes

some measure of the positive length (distance, cost,time, etc.) associated with each link

Both the shortest path and minimum spanning treeproblems involve choosing a set of links that have theshortest total length among all sets of links thatsatisfy a certain property For the shortest-path problem this property is that the

chosen links must provide a path between the origin andthe destination

For the minimum spanning tree problem, the requiredproperty is that the chosen links must provide a pathbetween each pair of nodes

Some Applications

D i f l i i k (fib i


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Design of telecommunication networks (fiber-opticnetworks, computer networks, leased-line telephone

networks, cable television networks, etc.) Design of lightly used transportation network tominimize the total cost of providing the links (raillines, roads, etc.)

Design of a network of high-voltage electricaltransmission lines Design of a network of wiring on electrical

equipment (e.g., a digital computer system) tominimize the total length of the wire

Design of a network of pipelines to connect a numberof locations

For a network with n nodes, a spanning tree is a group


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111

of n-1 arcs that connects all nodes of the network and

contains no loops.

1

3

212

7

4(1,2)-(2,3)-(3,1) is a loop

(1,3)-(2,3) is the minimum spanning tree

n mum pann ng ree ro emDescription


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112

p

You are given the nodes of the network but not the

links. Instead you are given the potential links and thepositive length for each if it is inserted into the

network (alternative measures for length of a link

include distance, cost, and time)

You wish to design the network by inserting enoughlinks to satisfy the requirement that there be a path

between every pair of nodes

The objective is to satisfy this requirement in a waythat minimizes the total length of links inserted intothe network

Minimum Spanning Tree Algorithm


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Greedy Algorithm 1. Select any node arbitrarily, and then connect (i.e., add a

link) to the nearest distinct node

2. Identify the unconnected node that is closest to aconnected node, and then connect these two nodes (i.e., add

a link between them). Repeat the step until all nodes have

been connected

3. Tie breaking: Ties for the nearest distinct node (step 1)or the closest unconnected node (step 2) may be broken

arbitrarily, and the algorithm will still yield an optimal

solution.

Fastest way of executing algorithm manually is thegraphical approach illustrated next

Example: The State University campus has five

Th di b i


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computers. The distances between computers are given

in the figure below. What is the minimum length of

cable required to interconnect the computers? Note thatif two computers are not connected this is because of

underground rock formations.

4

2

5

3

1

6

4

5

1

3

2

2

2

4

Solution: We want to find the minimum spanning tree.


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Iteration 1: Following the MST algorithm discussedbefore, we arbitrarily choose node 1 to begin. The

closest node is node 2. Now C={1,2}, ={3,4,5}, andarc(1,2) will be in the minimum spanning tree.

4

2

5

3

1

6

4

5

1

3

2

2

2

4

Iteration 2: Node 5 is closest to C. since node 5 is two


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blocks from node 1 and node 2, we may include either

arc(2,5) or arc(1,5) in the minimum spanning tree. Wearbitrarily choose to include arc(2,5). Then C={1,2,5}

and ={3,4}.

4

2

5

3

1

6

4

5

1

3

2

2

2

4

Iteration 3: Since node 3 is two blocks from node 5,i l d (5 3) i th i i i t


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we may include arc(5,3) in the minimum spanning tree.

Now C={1,2,5,3} and ={4}.

4

2

5

3

1

6

4

5

1

3

2

2

2

4

Iteration 4: Node 5 is the closest node to node 4. Thus,


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we add arc(5,4) to the minimum spanning tree.

We now have a minimum spanning tree consisting ofarcs(1,2), (2,5), (5,3), and (5,4). The length of the

minimum spanning tree is 1+2+2+4=9 blocks.

4

2

5

3

1

6

4

5

1

3

2

2

2

4

8 4 CPM and PERT


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8.4 CPM and PERTNetwork models can be used as an aid in the scheduling of large

complex projects that consist of many activities.

CPM: If the duration of each activity is known with certainty, the

Critical Path Method (CPM) can be used to determine the length

of time required to complete a project.

PERT: If the duration of activities is not known with certainty,

the Program Evaluation and Review Technique (PERT) can be

used to estimate the probability that the project will be completed

by a given deadline.

CPM and PERT are used in many applications

i l di th f ll i


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including the following:

Scheduling construction projects such as officebuildings, highways and swimming poolsDeveloping countdown and hold procedure for thelaunching of space crafts

Installing new computer systems

Designing and marketing new products

Completing corporate mergers

Building ships

Project Planning, Scheduling and

C t l


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Control

Planning: organized approach to accomplish the goal ofminimizing elapsed time of project defines objectives and tasks; represents tasks interactions on

a network; estimates time and resources

Scheduling: a time-phased commitment of resources identifies critical tasks which, if delayed, will delay the

projects completion time.

Control: means of monitoring and revising the progressof a project

N k R i


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Network Representation

Tasks (or activities) are represented by arcs Each task has a duration denoted by tj Node 0 represents the start and node n denotes the finish of

the project

Precedence relations are shown by arcs specify what other tasks must be completed before the task in

question can begin.

A path is a sequence of linked tasks going from beginningto end

Critical path is the longest path

To apply CPM and PERT, we need a list of activities

that make up the project The project is considered to be


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that make up the project. The project is considered to be

completed when all activities have been completed. For

each activity there is a set of activities (called thepredecessors of the activity) that must be completed

before the activity begins. A project network is used to

represent the precedence relationships between

activities. In the following discussions the activities willbe represented by arcs and the nodes will be used to

represent completion of a set of activities (Activity on

arc (AOA) type of network).

1 32A B

Activity A must be completed before activity B starts

While constructing an AOA type of project diagram one

should use the following rules:


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should use the following rules:

Node 1 represents the start of the project. An arc should lead

from node 1 to represent each activity that has no predecessors. A node (called the finish node) representing the completion ofthe project should be included in the network.

Number the nodes in the network so that the node representing

the completion time of an activity always has a larger number thanthe node representing the beginning of an activity.

An activity should not be represented by more than one arc in thenetwork

Two nodes can be connected by at most one arc.To avoid violating rules 4 and 5, it can be sometimes necessary to

utilize a dummy activity that takes zero time.

Formulating the CPM Problem


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Formulating the CPM Problem

Input Data:

Precedence relationships and durations

Decision Variable:ESi : Earliest starting times for each of the tasks

Objective:

Minimize the elapsed time of the project

where node n is the last node in the graph

C t i t


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Constraints

If tj is the earliest starting time of a task, ESi isthe earliest starting time of an immediate

predecessor and ti is the duration of theimmediate predecessor, then we have

ESj ESi + ti for every arc (i, j)

Critical Path Definitions


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Earliest Start Time (ES) is the earliest time a taskcan feasibly start

Earliest Finish Time (EF) is the earliest time atask can feasibly end

Latest Start Time (LS) is the latest time a task canfeasibly start, without delaying the project at all.

Latest Finish Time (LF) is the latest time a task

can feasibly end, without delaying the project atall

Critical Path Method


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Critical Path Method

Forward Pass Go through the jobs in order Start each job at the earliest time while satisfying the

precedence constraints

It finds the earliest start and finish times EFi = ESi + ti Earliest start time for an activity leaving a particular

node is equal to the largest of the earliest finish times

for all activities entering the node.

CPM Th B k d P


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CPM: The Backward Pass

Fix the finishing time Look at tasks in reverse order Lay out tasks one at a time on the Gantt chart

starting at the finish and working backwards to thestart

Start the task at its latest starting time LSi = LFi - ti Latest finish time for an activity entering a particular

node is equal to the smallest of the latest start times forall activities leaving the node.

CPM d C iti l P th


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CPM and Critical Path

Theorem: The minimum length of the schedule isthe length of the longest path.

The longest path is called the critical path

Look for tasks whose earliest start time and latest start time are

the same. These tasks are critical, and are on a critical path.

CPM d C iti l P th


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CPM and Critical Path

Critical path are found by identifying those tasks whereES=LS (equivalently, EF=LF)

No flexibility in scheduling tasks on the critical path

The makespan of the critical path equals the LF of thefinal task

Slack is the difference between LS and ES, or LF and EF.

An activity with a slack of zero is on the critical path

An example for CPM


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Widgetco is about to introduce a new product. A list of

activities and the precedence relationships are given in

the table below. Draw a project diagram for this project.

Activity Predecessors Duration(days)

A:train workers - 6

B:purchase raw materials - 9

C:produce product 1 A, B 8

D:produce product 2 A, B 7E:test product 2 D 10

F:assemble products 1&2 C, E 12

Project Diagram for Widgetco


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133

1

65

42

3

A 6

B 9

Dummy

C 8

D 7

E 10

F 12

Node 1 = starting nodeNode 6 = finish node

Project Diagram for Widgetco Forward Pass (ES,EF)


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1

65

42

3

A 6

B 9

Dummy

C 8

D 7

E 10

F 12


(0,6)

(0,9)

(9,17)

(9,16)(16,26)

(26,38)

Project Diagram for Widgetco Backward Pass (LS,LF)


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1

65

42

3

A 6

B 9

Dummy

C 8

D 7

E 10

F 12


(3,9)

(0,9)

(18,26)

(9,16)(16,26)

(26,38)

For Widgetco example ES(i)s and LS(i)s are asfollows:


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follows:

Activity ES(i) LS(i)

A 0 3

B 0 0

C 9 18

D 9 9

E 16 16

F 26 26

According to the table on the previous slide the slacks

are computed as follows:


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are computed as follows:

Activity B: 0

Activity A: 3

Activity D: 0

Activity C: 9

Activity E: 0

Activity F: 0

Critical path


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An activity with a slack of zero is a critical activity A path from node 1 to the finish node that consistsentirely of critical activities is called a critical path.

For Widgetco example B-D-E-F is a critical path.

The Makespan is equal to 38

Using LP to find a critical pathD i i i bl


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139

Decision variable:

Xij:the time that the event corresponding to node j occurs

Since our goal is to minimize the time required tocomplete the project, we use an objective function of:

Z=XF-X1

Note that for each activity (i,j), before j occurs , i must

occur and activity (i,j) must be completed.

Min Z =X6-X1

S T X >=X +6 (Arc (1 3) constraint)


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S.T. X3>=X1+6 (Arc (1,3) constraint)

X2>=X

1+9 (Arc (1,2) constraint)

X5>=X3+8 (Arc (3,5) constraint)




X3>=X2 (Arc (2,3) constraint)

The optimal solution to this LP is Z=38, X1=0, X2=9,

X3=9, X4=16, X5=26, X6=38

On variability of tasksC id 10 i d d k


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Consider 10 independent tasks

each takes 1 unit of time on average the time it takes is uniformly distributed between 0and 2.

CPM (uses expected value) Modeling Randomness

The random schedule takes longer

On Incorporating Variability


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Program Evaluation and Review Technique(PERT)

Attempts to incorporate variability in the durations

Assume mean, m, and variance, s2, of the durationscan be estimated

Simulation

Model variability using any distributionsimulate to see how long a schedule will take

PERTCPM assumes that the duration of each activity is


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CPM assumes that the duration of each activity is

known with certainty. For many projects, this is clearly

not applicable. PERT is an attempt to correct thisshortcoming of CPM by modeling the duration of each

activity as a random variable. For each activity, PERT

requires that the project manager estimate the following

three quantities:

a : estimate of the activitys duration under the mostfavorable conditions

b : estimate of the activitys duration under the leastfavorable conditions

m : most likely value for the activitys duration

Let Tij be the duration of activity (i,j). PERT requires


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Let Tij be the duration of activity (i,j). PERT requires

the assumption that Tij follows a beta distribution.

According to this assumption, it can be shown that themean and variance ofTij may be approximated by

36

)(

var

64)(

2ab

T

bmaTE

ij

ij

PERT requires the assumption that the durations of all


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activities are independent. Thus,

pathji

ijTE),(

)(

pathji

ijT),(

var

: expected duration of activities on any path

: variance of duration of activities on any path

Let CP be the random variable denoting the total


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duration of the activities on a critical path found by

CPM. PERT assumes that the critical path found byCPM contains enough activities to allow us to invoke

the Central Limit Theorem and conclude that the

following is normally distributed:

thcriticalpaji

ijTCP),(

a, b and m for activities in Widgetco


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Activity a b m

(1,2) 5 13 9

(1,3) 2 10 6(3,5) 3 13 8

(3,4) 1 13 7

(4,5) 8 12 10

(5,6) 9 15 12

According to the table on the previous slide:

)513(36135 2


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.

1481

36

)915(var,12

6

48159)(

44.036

)812(var,106

40128)(

436

)113(var,7

6

28131)(

78.236

)313(

var,86

32133

)(

78.136

)210(var,6

6

24102)(

78.1

36

)513(var,9

6

36135)(

2

5656

2

4545

2

3434

2

3535

2

1313

2

1212

TTE

TTE

TTE

TTE

TTE

TTE

Of course, the fact that arc (2,3) is a dummy arc yields

E(T23)=varT23=0


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E(T23) varT23 0

The critical path was B-D-E-F. Thus,

E(CP)=9+0+7+10+12=38

varCP=1.78+0+4+0.44+1=7.22

Then the standard deviation for CP is (7.22)1/2=2.69

And

13.0)12.1()69.2

3835

69.2

38()35(

ZP

CPPCPP

PERT implies that there is a 13% chance that the


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project will be completed within 35 days.

More on Project Management CPM


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CPM

Advantage of ease of useLays out the Gantt chart (nicely visual)

Identifies the critical path

Used in practice on large projects e.g., used for the big dig

Other issuesTasks take resources, which are limited

Task times are really random variables

Unpredictable things happen

Incorporating Resource


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Constraints

Each task can have resources that it needs3 construction workers

1 craneetc

In scheduling, do not use more resourcesthan are available at any time

Makes the problem much more difficult tosolve exactly. Heuristics are used.

Dealing with the unknown


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Dealing with the unknown

VERY hard to model

How does one model totally unforeseen

events?In the Big Dig, there is a leak in digging a

tunnel despite assurances it would not happen

In the Hubble telescope, a firm assures that themirrors are ground properly. By the time the

mistake is discovered, the telescope is in outer

space.

Project Management Software


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Project Management Software

Explosive growth in software packages usingthese techniques

Cost and capabilities vary greatly Yearly survey inPM Network

Microsoft Project is most commonly used

package todayFree 60 day trial versions:

http://www.microsoft.com/office/98/project/

trial/info.htm

Summary


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Project management Simple model: we use estimates of the timefor each task, and we use the precedence

constraints

Set2_transport & Networks

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