Slides by John Loucks St. Edward’s University

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Slides by

JohnLoucks

St. Edward’sUniversity

Modifications byA. Asef-Vaziri

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Chapter 6, Part ADistribution and Network Models

Transportation Problem• Network Representation• General LP Formulation

Assignment Problem• Network Representation• General LP Formulation

Transshipment Problem• Network Representation• General LP Formulation

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Transportation, Assignment, and Transshipment Problems

A network model is one which can be represented by a set of nodes, a set of arcs, and functions (e.g. costs, supplies, demands, etc.) associated with the arcs and/or nodes.

Transportation, assignment, transshipment, shortest-route, and maximal flow problems of this chapter as well as the minimal spanning tree and PERT/CPM problems (in others chapter) are all examples of network problems.

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Transportation, Assignment, and Transshipment Problems

Each of the five problems of this chapter can be formulated as linear programs and solved by general purpose linear programming codes.

For each of the five problems, if the right-hand side of the linear programming formulations are all integers, the optimal solution will be in terms of integer values for the decision variables.

However, there are many computer packages that contain separate computer codes for these problems which take advantage of their network structure.

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Transportation Problem

The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply si) to n destinations (each with a demand dj), when the unit shipping cost from an origin, i, to a destination, j, is cij.

The network representation for a transportation problem with two sources and three destinations is given on the next slide.

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Network Representation

2

c1

1 c12

c13c21

c22c23

d1

d2

d3

s1

s2

Sources Destinations

3

2

1

1

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Transportation Problem: Example #1

Acme Block Company has orders for 80 tons ofconcrete blocks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, andEastwood -- 10 tons. Acme has two plants, each of which can produce 50 tons per week. Delivery cost per ton from each plant to each suburban location is shown on the next slide.

How should end of week shipments be made to fillthe above orders?

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Delivery Cost Per Ton

Northwood Westwood Eastwood Plant 1 24 30 40 Plant 2 30 40 42


Decision Variables. the tons of concrete blocks, xij , to be shipped from source i to destination j.

Northwood Westwood Eastwood Plant 1 x11 x12 x13

Plant 1 x21 x22 x23

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Define the Objective Function Minimize the total shipping cost. Min: (shipping cost per ton for each origin

to destination) × (number of pounds shipped from each origin to each destination).

Min: 24x11 + 30x12 + 40x13 + 30x21 + 40x22 + 42x23


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Define the Constraints Supply constraints: (1) x11 + x12 + x13 = 50 (2) x21 + x22 + x23 = 50 Demand constraints: (4) x11 + x21 = 25 (5) x12 + x22 = 45 (6) x13 + x23 = 10 Non-negativity of variables: xij > 0, i = 1, 2 and j = 1, 2, 3


= Constraints

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Define the Constraints Supply constraints: (1) x11 + x12 + x13 ≤ 50 (2) x21 + x22 + x23 ≤ 50 Demand constraints: (4) x11 + x21 ≥ 25 (5) x12 + x22 ≥ 45 (6) x13 + x23 ≥ 10 Non-negativity of variables: xij > 0, i = 1, 2 and j = 1, 2, 3


≤ and ≥ Constraints

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Partial Spreadsheet Showing Problem Data


A B C D E F G H12 Constraint X11 X12 X13 X21 X22 X23 RHS3 #1 1 1 1 504 #2 1 1 1 505 #3 1 1 256 #4 1 1 457 #5 1 1 108 Obj.Coefficients 24 30 40 30 40 42 30

LHS Coefficients

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Transportation Problem: Example #1Cost Table

Northwood Westwood EastwoodPlant 1 24 30 40Plant 2 30 40 42

`

Decision Variable TableNorthwood Westwood Eastwood LHS RHS

Plant 1 5 45 0 50 ≤ 50Plant 2 20 0 10 30 ≤ 50LHS 25 45 10 2490

≥ ≥ ≥RHS 25 45 10

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Transportation Problem: Example #1 Sensitivity Report

Variable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$9 Plant 1 Northwood 5 0 24 4 4$C$9 Plant 1 Westwood 45 0 30 4 36$D$9 Plant 1 Eastwood 0 4 40 1E+30 4$B$10 Plant 2 Northwood 20 0 30 4 4$C$10 Plant 2 Westwood 0 4 40 1E+30 4$D$10 Plant 2 Eastwood 10 0 42 4 42

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$11 LHS Northwood 25 30 25 20 20$C$11 LHS Westwood 45 36 45 5 20$D$11 LHS Eastwood 10 42 10 20 10$E$9 Plant 1 LHS 50 -6 50 20 5$E$10 Plant 2 LHS 30 0 50 1E+30 20

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The Navy has 9,000 pounds of material in Albany,Georgia that it wishes to ship to three installations:San Diego, Norfolk, and Pensacola. They require 4,000, 2,500, and 2,500 pounds, respectively. Government regulations require equal distribution of shippingamong the three carriers.

The shipping costs per pound for truck, railroad, and airplane transit are shown on the next slide. Formulate and solve a linear program to determine the shipping arrangements (mode, destination, and quantity) that will minimize the total shipping cost.

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DestinationMode San Diego Norfolk

PensacolaTruck $12 $ 6

$ 5Railroad 20 11

9Airplane 30 26

28


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Define the Decision VariablesWe want to determine the pounds of material, xij , to be shipped by mode i to destination j. The following table summarizes the decision variables:

San Diego Norfolk Pensacola

Truck x11 x12 x13

Railroad x21 x22 x23

Airplane x31 x32 x33


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Define the Objective Function Minimize the total shipping cost. Min: (shipping cost per pound for each

mode per destination pairing) x (number of pounds shipped by mode per destination pairing).

Min: 12x11 + 6x12 + 5x13 + 20x21 + 11x22 + 9x23

+ 30x31 + 26x32 + 28x33


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Define the Constraints Equal use of transportation modes: (1) x11 + x12 + x13 = 3000 (2) x21 + x22 + x23 = 3000 (3) x31 + x32 + x33 = 3000 Destination material requirements: (4) x11 + x21 + x31 = 4000 (5) x12 + x22 + x32 = 2500 (6) x13 + x23 + x33 = 2500 Non-negativity of variables: xij > 0, i = 1, 2, 3 and j = 1, 2,

3


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Cost TableSan Siego Norfolk Pensacola

Truck 12 6 5Railroad 20 11 9Airplane 30 26 28

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Decision Variable TableSan Siego Norfolk Pensacola LHS RHS

Truck 1000 2000 0 3000 ≤ 3000Railroad 0 500 2500 3000 ≤ 3000Airplane 3000 0 0 3000 ≤ 3000LHS 4000 2500 2500 142000

≥ ≥ ≥RHS 4000 2500 2500

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Linear Programming Formulation Using the notation: xij = number of units shipped from origin i to destination j cij = cost per unit of shipping from origin i to destination j si = supply or capacity in units at origin i dj = demand in units at destination j

continued

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Linear Programming Formulation (continued)

1 1Min

m n

ij iji j

c x

1 1,2, , Supply

n

ij ij

x s i m

1 1,2, , Demand

m

ij ji

x d j n

xij > 0 for all i and j

>

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LP Formulation Special Cases• Total supply exceeds total demand:

• Total demand exceeds total supply: Add a dummy origin with supply equal to the shortage amount. Assign a zero shipping cost per unit. The amount “shipped” from the dummy origin (in the solution) will not actually be shipped.


No modification of LP formulation is necessary.

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LP Formulation Special Cases (continued)• The objective is maximizing profit or

revenue:

• Minimum shipping guarantee from i to j: xij > Lij

• Maximum route capacity from i to j: xij < Lij

• Unacceptable route: Remove the corresponding

decision variable.


Solve as a maximization problem.

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Assignment Problem

An assignment problem seeks to minimize the total cost assignment of m workers to m jobs, given that the cost of worker i performing job j is cij.

It assumes all workers are assigned and each job is performed.

An assignment problem is a special case of a transportation problem in which all supplies and all demands are equal to 1; hence assignment problems may be solved as linear programs.

The network representation of an assignment problem with three workers and three jobs is shown on the next slide.

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Assignment Problem


2

3

1

2

3

1 c11c12

c13

c21 c22

c23

c31 c32

c33

Agents Tasks

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An electrical contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects.

ProjectsSubcontractor A B C Westside 50 36 16

Federated 28 30 18 Goliath 35 32 20

Universal 25 25 14How should the contractors be assigned so that totalmileage is minimized?

Assignment Problem: Example

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Network Representation50

361628

301835 32

2025 25

14

West.

C

B

A

Univ.

Gol.

Fed.

ProjectsSubcontractors


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Linear Programming Formulation

Min 50x11+36x12+16x13+28x21+30x22+18x23

+35x31+32x32+20x33+25x41+25x42+14x43 s.t. x11+x12+x13 < 1

x21+x22+x23 < 1 x31+x32+x33 < 1 x41+x42+x43 < 1 x11+x21+x31+x41 = 1 x12+x22+x32+x42 = 1 x13+x23+x33+x43 = 1 xij = 0 or 1 for all i and j

Agents

Tasks


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Min 50x11+36x12+16x13+28x21+30x22+18x23

+35x31+32x32+20x33+25x41+25x42+14x43 s.t. x11+x12+x13 < 1

x21+x22+x23 < 1 x31+x32+x33 < 1 x41+x42+x43 < 1 x11+x21+x31+x41 ≥ 1 x12+x22+x32+x42 ≥ 1 x13+x23+x33+x43 ≥ 1 xij = 0 or 1 for all i and j

Agents

Tasks


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Cost TableProject A Project B Project C

Westside 50 36 16Federated 28 30 18Goliath 35 32 20Universal 25 25 14

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Decision Variable TableSan Siego Norfolk Pensacola LHS RHS

Westside 0 0 1 1 ≤ 1Federated 1 0 0 1 ≤ 1Goliath 0 0 0 0 ≤ 1Universal 0 1 0 1 ≤ 1LHS 1 1 1 69

≥ ≥ ≥RHS 1 1 1

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The optimal assignment is:

Subcontractor Project Distance Westside C 16

Federated A 28Goliath (unassigned) Universal B 25

Total Distance = 69 miles


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Linear Programming Formulation Using the notation: xij = 1 if agent i is assigned

to task j 0 otherwise cij = cost of assigning agent i to

task j

Assignment Problem

continued

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Assignment Problem

1 1Min

m n

ij iji j

c x

11 1,2, , Agents

n

ijj

x i m

11 1,2, , Tasks

m

iji

x j n


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LP Formulation Special Cases• Number of agents exceeds the number of

tasks:

• Number of tasks exceeds the number of agents: Add enough dummy agents to equalize the number of agents and the number of tasks. The objective function coefficients for these new variable would be zero.

Assignment Problem

Extra agents simply remain unassigned.

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Assignment Problem

LP Formulation Special Cases (continued)• The assignment alternatives are evaluated in terms of revenue or profit:

Solve as a maximization problem.• An assignment is unacceptable:

Remove the corresponding decision variable.

• An agent is permitted to work t tasks:

1 1,2, , Agents

n

ijj

x t i m

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Transshipment Problem

Transshipment problems are transportation problems in which a shipment may move through intermediate nodes (transshipment nodes)before reaching a particular destination node.

Transshipment problems can be converted to larger transportation problems and solved by a special transportation program.

Transshipment problems can also be solved by general purpose linear programming codes.

The network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations is shown on the next slide.

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2

3

4

5

6

7

1c13

c14

c23

c24c25

c15

s1

c36

c37

c46c47

c56

c57

d1

d2

Intermediate NodesSources Destinationss2

DemandSupply

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Using the notation: xij = number of units shipped from node i to node j

cij = cost per unit of shipping from node i to node j

si = supply at origin node i dj = demand at destination node j

continued

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all arcsMin ij ijc x

arcs out arcs ins.t. Origin nodes ij ij ix x s i


arcs out arcs in0 Transhipment nodesij ijx x

arcs in arcs out Destination nodes ij ij jx x d j


continued

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LP Formulation Special Cases• Total supply not equal to total demand• Maximization objective function• Route capacities or route minimums• Unacceptable routesThe LP model modifications required here areidentical to those required for the special

cases inthe transportation problem.

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The Northside and Southside facilities of Zeron Industries supply three firms (Zrox, Hewes, Rockrite) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc.

Currently weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockrite. Both Arnold and Supershelf can supply at most 75 units to its customers.

Additional data is shown on the next slide.

Transshipment Problem: Example

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Because of long standing contracts based on past orders, unit costs from the manufacturers to the suppliers are:

Zeron N Zeron S Arnold 5 8 Supershelf 7 4

The costs to install the shelving at the various locations are:

Zrox Hewes Rockrite Zeron N 1 5 8

Zeron S 3 4 4


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ARNOLD

WASHBURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

15

8

34

4

Arnold

SuperShelf

Hewes

Zrox

ZeronN

ZeronS

Rock-Rite


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Linear Programming Formulation• Decision Variables Defined

xij = amount shipped from manufacturer i to supplier j

xjk = amount shipped from supplier j to customer k

where i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7

(Rockrite)• Objective Function Defined

Minimize Overall Shipping Costs: Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 +

5x36 + 8x37 + 3x45 + 4x46 + 4x47


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Constraints DefinedAmount Out of Arnold: x13 + x14 < 75Amount Out of Supershelf: x23 + x24 < 75Amount Through Zeron N: x13 + x23 - x35 - x36 - x37 = 0Amount Through Zeron S: x14 + x24 - x45 - x46 - x47 = 0Amount Into Zrox: x35 + x45 = 50Amount Into Hewes: x36 + x46 = 60Amount Into Rockrite: x37 + x47 = 40

Non-negativity of Variables: xij > 0, for all i and j.


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Cost TableZeron N Zeron S Zrox Hewes Rockrite

Arnold 5 8 Zeron N 1 5 8Supershelf 7 4 Zeron S 3 4 4

Decision Variable TableZeron N Zeron S Zrox Hewes Rockrite

Arnold 75 0 75 ≤ 75 Zeron N 50 25 0 75Supershelf 0 75 75 ≤ 75 Zeron S 0 35 40 75

75 75 50 60 40 1150≥ ≥ ≥

` 50 60 40

Zeron N 0 = 0Zeron S 0 = 0

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Solution

ARNOLD

WASHBURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

15

8

3 4

4

Arnold

SuperShelf

Hewes

Zrox

ZeronN

ZeronS

Rock-Rite

75

75

50

25

35

40


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Transshipment Problem: ExampleVariable Cells

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$B$8 Arnold Zeron N 75 0 5 2 2$C$8 Arnold Zeron S 0 2 8 1E+30 2$B$9 Supershelf Zeron N 0 4 7 1E+30 4$C$9 Supershelf Zeron S 75 0 4 2 1E+30$I$8 Zeron N Zrox 50 0 1 3 6$J$8 Zeron N Hewes 25 0 5 2 2$K$8 Zeron N Rockrite 0 3 8 1E+30 3$I$9 Zeron S Zrox 0 3 3 1E+30 3$J$9 Zeron S Hewes 35 0 4 2 2$K$9 Zeron S Rockrite 40 0 4 3 10

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$D$14 Zeron N 0 5 0 0 75$D$15 Zeron S 0 6 0 0 25$D$8 Arnold 75 0 75 1E+30 0$D$9 Supershelf 75 -2 75 25 0$I$10 Zrox 50 6 50 0 50$J$10 Hewes 60 10 60 0 25$K$10 Rockrite 40 10 40 0 25

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Computer Output (continued) Constraint Slack/Surplus Dual

Values 1 0.000

0.000 2 0.000

2.000 3 0.000

-5.000 4 0.000

-6.000 5 0.000

-6.000 6 0.000 -

10.000 7 0.000 -

10.000


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Computer Output (continued)

OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper

Limit X13 3.000 5.000 7.000 X14 6.000 8.000

No Limit X23 3.000 7.000

No Limit X24 No Limit 4.000

6.000 X35 No Limit 1.000

4.000 X36 3.000 5.000

7.000 X37 5.000 8.000

No Limit X45 0.000 3.000

No Limit X46 2.000 4.000

6.000 X47 No Limit 4.000

7.000


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Computer Output (continued)

RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper

Limit 1 75.000 75.000

No Limit 2 75.000 75.000

100.000 3 -75.000 0.000

0.000 4 -25.000 0.000

0.000 5 0.000 50.000

50.000 6 35.000 60.000

60.000 7 15.000 40.000

40.000


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Transshipment Transformed into Transportation Problem

Cost TableZeron N Zeron S Zrox Hewes Rockrite

Arnold 5 8 1000 1000 1000Supershelf 7 4 1000 1000 1000Zeron N 0 1000 1 5 8Zeron S 1000 0 3 4 4

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Decision Variable TableZeron N Zeron S Zrox Hewes Rockrite LHS RHS

Arnold 75 0 0 0 0 75 = 75Supershelf 0 75 0 0 0 75 = 75Zeron N 75 0 50 25 0 150 = 150Zeron S 0 75 0 35 40 150 = 150LHS 150 150 50 60 40 1150

= = = = =RHS 150 150 50 60 40

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End of Chapter 6, Part A

Slides by John Loucks St. Edward’s University

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