Distn & Network Models

7/29/2019 Distn & Network Models

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Slide 2008 Thomson South-Western. All Rights Reserved

Slides by

JOHNLOUCKSSt. EdwardsUniversity


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Chapter 6, Part ADistribution and Network Models

Transportation ProblemNetwork RepresentationGeneral LP Formulation

Assignment Problem

Network Representation

General LP Formulation

Transshipment Problem

Network RepresentationGeneral LP Formulation


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Transportation, Assignment, andTransshipment 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 thearcs and/or nodes.

Transportation, assignment, transshipment

problems of this chapter as well as shortest-route,and maximal flow , minimal spanning tree andPERT/CPM problems (in others chapter) are allexamples of network problems.


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Transportation, Assignment, andTransshipment Problems

Each of the three models of this chapter can be

formulated as linear programs and solved bygeneral purpose linear programming codes.

For each of the three models, if the right-hand sideof the linear programming formulations are all

integers, the optimal solution will be in terms ofinteger values for the decision variables.

However, there are many computer packages(including The Management Scientist) that contain

separate computer codes for these models whichtake 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 morigins (each with a supply si) to n destinations(each with a demand dj), when the unit shippingcost from an origin, i, to a destination,j, is cij.

The network representation for a transportationproblem with two sources and three destinations isgiven on the next slide.


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2

c11

c12

c13

c21

c22

c23

d1

d2

d3

s1

s2

Sources Destinations

3

2

1

1


7/517Slide 2008 Thomson South-Western. All Rights Reserved


Linear Programming Formulation

Using the notation:

xij = number of units shipped from

origin i to destinationj

cij= cost per unit of shipping fromorigin i to destinationj

si = supply or capacity in units at origin i

dj = demand in units at destinationj

continued




Linear Programming Formulation (continued)

1 1

Min

m n

ij iji j

c x

1

1,2, , Supplyn

ij ij

x s i m

1

1,2, , Demandm

ij ji

x d j n

xij > 0 for all i andj



LP Formulation Special Cases

The objective is maximizing profit or revenue:

Minimum shipping guarantee from i toj:

xij > Lij

Maximum route capacity from i toj:

xij < LijUnacceptable route:

Remove the corresponding decision variable.


Solve as a maximization problem.



Transportation Problem: Example #1

Acme Block Company has orders for 80 tons of

concrete blocks at three suburban locations

as follows: Northwood -- 25 tons,

Westwood -- 45 tons, and

Eastwood -- 10 tons. Acmehas 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 fill

the above orders?



Delivery Cost Per Ton

Northwood Westwood Eastwood

Plant 1 24 30 40

Plant 2 30 40 42




Partial Spreadsheet Showing Problem Data


A B C D E F G H

1

2 Constraint X11 X12 X13 X21 X22 X23 RHS

3 #1 1 1 1 50

4 #2 1 1 1 505 #3 1 1 25

6 #4 1 1 45

7 #5 1 1 10

8 Obj.Coefficients 24 30 40 30 40 42 30

LHS Coefficients


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14Slide 2008 Thomson South-Western. All Rights Reserved

Partial Spreadsheet Showing Optimal Solution


A B C D E F G

10 X11 X12 X13 X21 X22 X23

11 Dec.Var.Values 5 45 0 20 0 10

12 Minimized Total Shipping Cost 2490

1314 LHS RHS

15 50


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Optimal Solution

From To Amount Cost

Plant 1 Northwood 5 120

Plant 1 Westwood 45 1,350

Plant 2 Northwood 20 600

Plant 2 Eastwood 10 420

Total Cost = $2,490



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Partial Sensitivity Report (first half)


Adjustable Cells

Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$C$12 X11 5 0 24 4 4$D$12 X12 45 0 30 4 1E+30

$E$12 X13 0 4 40 1E+30 4

$F$12 X21 20 0 30 4 4

$G$12 X22 0 4 40 1E+30 4

$H$12 X23 10.000 0.000 42 4 1E+30

Adjustable Cells

Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$C$12 X11 5 0 24 4 4$D$12 X12 45 0 30 4 1E+30

$E$12 X13 0 4 40 1E+30 4

$F$12 X21 20 0 30 4 4

$G$12 X22 0 4 40 1E+30 4

$H$12 X23 10.000 0.000 42 4 1E+30


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Partial Sensitivity Report (second half)


Constraints

Final Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$E$17 P2.Cap 30.0 0.0 50 1E+30 20

$E$18 N.Dem 25.0 30.0 25 20 20

$E$19 W.Dem 45.0 36.0 45 5 20

$E$20 E.Dem 10.0 42.0 10 20 10

$E$16 P1.Cap 50.0 -6.0 50 20 5

Constraints

Final Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$E$17 P2.Cap 30.0 0.0 50 1E+30 20

$E$18 N.Dem 25.0 30.0 25 20 20

$E$19 W.Dem 45.0 36.0 45 5 20

$E$20 E.Dem 10.0 42.0 10 20 10

$E$16 P1.Cap 50.0 -6.0 50 20 5


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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 shipping

among the three carriers.


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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) thatwill minimize the total shipping cost.



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Destination

Mode San Diego Norfolk Pensacola

Truck $12 $ 6 $ 5

Railroad 20 11 9

Airplane 30 26 28



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Define the Decision Variables

We want to determine the pounds of material, xij ,to be shipped by mode i to destinationj. Thefollowing table summarizes the decision variables:

San Diego Norfolk PensacolaTruck 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 perdestination 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 = 2500Non-negativity of variables:

xij > 0, i = 1,2,3 and j = 1,2,3



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The Management Scientist Output

OBJECTIVE FUNCTION VALUE = 142000.000

Variable Value Reduced Costx11 1000.000 0.000

x12 2000.000 0.000x13 0.000 1.000x21 0.000 3.000x22 500.000 0.000x23 2500.000 0.000

x31 3000.000 0.000x32 0.000 2.000x33 0.000 6.000



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Solution Summary

San Diego will receive 1000 lbs. by truckand 3000 lbs. by airplane.

Norfolk will receive 2000 lbs. by truck

and 500 lbs. by railroad.

Pensacola will receive 2500 lbs. by railroad.

The total shipping cost will be $142,000.



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Assignment Problem

An assignment problem seeks to minimize the total

cost assignment of m workers to m jobs, given thatthe cost of worker i performing jobj is cij.

It assumes all workers are assigned and each job isperformed.

An assignment problem is a special case of atransportation problem in which all supplies and alldemands are equal to 1; hence assignment problemsmay be solved as linear programs.

The network representation of an assignment

problem with three workers and three jobs is shownon the next slide.


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Assignment Problem


2

3

1

2

3

1c11

c12c13

c21c22

c23

c31 c32

c33

Agents Tasks


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Using the notation:

xij = 1 if agent i is assigned to taskj

0 otherwise

cij= cost of assigning agent i to taskj

Assignment Problem

continued


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Assignment Problem

1 1

Min

m n

ij iji j

c x

1

1 1,2, , Agentsn

ij

j

x i m

1

1 1,2, , Tasksm

iji

x j n



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Number of agents exceeds the number of tasks:

Number of tasks exceeds the number of agents:

Add enough dummy agents to equalize thenumber of agents and the number of tasks.The objective function coefficients for thesenew 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 termsof 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, , Agentsn

ijj

x t i m


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An electrical contractor pays his subcontractors a

fixed fee plus mileage for work performed. On a givenday the contractor is faced with three electrical jobsassociated with various projects. Given below are thedistances between the subcontractors and the projects.

ProjectsSubcontractor A B CWestside 50 36 16Federated 28 30 18Goliath 35 32 20Universal 25 25 14

How should the contractors be assigned to minimizetotal mileage costs?

Assignment Problem: Example


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50

36

16

28

30

18

35 32

2025 25

14

West.

C

B

A

Univ.

Gol.

Fed.

ProjectsSubcontractors



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

+35x31+32x32+20x33+25x41+25x42+14x43s.t. x11+x12+x13 < 1

x21+x22+x23 < 1x31+x32+x33 < 1

x41+x42+x43 < 1

x11+x21+x31+x41 = 1

x12+x22+x32+x42 = 1x13+x23+x33+x43 = 1

xij = 0 or 1 for all i andj

Agents

Tasks



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The optimal assignment is:

Subcontractor Project Distance

Westside C 16

Federated A 28

Goliath (unassigned)

Universal B 25

Total Distance = 69 miles



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Transshipment problems are transportation problems

in which a shipment may move through intermediatenodes (transshipment nodes)before reaching aparticular destination node.

Transshipment problems can be converted to largertransportation problems and solved by a specialtransportation program.

Transshipment problems can also be solved bygeneral 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

1

c13

c14

c23

c24c25

c15

s1

c36

c37

c46c47

c56

c57

d1

d2

Intermediate Nodes

Sources Destinationss2

DemandSupply


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Using the notation:

xij = number of units shipped from node i to nodej

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

si= supply at origin node idj= demand at destination node j

continued


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all arcs

Min ij ijc x

arcs out arcs in

s.t. Origin nodesij ij ix x s i


arcs out arcs in

0 Transhipment nodesij ijx x

arcs in arcs out

Destination nodesij ij jx x d j


continued

all arcs

Min ij ijc x

h bl


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Total supply not equal to total demandMaximization objective function

Route capacities or route minimums

Unacceptable routes

The LP model modifications required here are

identical to those required for the special cases in

the transportation problem.

h P bl l


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The Northside and Southside facilities

of Zeron Industries supply three firms(Zrox, Hewes, Rockrite) with customizedshelving for its offices. They both ordershelving 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 forRockrite. Both Arnold and Supershelf cansupply at most 75 units to its customers.

Additional data is shown on the nextslide.

Transshipment Problem: Example

T hi P bl E l


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Because of long standing contracts based on

past orders, unit costs from the manufacturers to thesuppliers are:

Zeron N Zeron SArnold 5 8

Supershelf 7 4

The costs to install the shelving at the variouslocations are:

Zrox Hewes RockriteZeron N 1 5 8Zeron S 3 4 4


T hi P bl E l


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ARNOLD

WASH

BURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

15

8

3

44

Arnold

SuperShelf

Hewes

Zrox

ZeronN

ZeronS

Rock-Rite


T hi P bl E l


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Decision Variables Definedxij = amount shipped from manufacturer i to supplierj

xjk = amount shipped from supplierj 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


T hi t P bl E l


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Constraints Defined

Amount Out of Arnold: x13 + x14 < 75Amount Out of Supershelf: x23 + x24 < 75

Amount Through Zeron N: x13 + x23 - x35 - x36 - x37 = 0

Amount Through Zeron S: x14 + x24 - x45 - x46 - x47 = 0

Amount Into Zrox: x35 + x45 = 50

Amount Into Hewes: x36 + x46 = 60

Amount Into Rockrite: x37 + x47 = 40

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


T hi t P bl E l


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The Management Scientist Solution

Objective Function Value = 1150.000

Variable Value Reduced CostsX13 75.000 0.000

X14 0.000 2.000X23 0.000 4.000X24 75.000 0.000X35 50.000 0.000X36 25.000 0.000

X37 0.000 3.000X45 0.000 3.000X46 35.000 0.000X47 40.000 0.000


T hi t P bl E l


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Solution

ARNOLD

WASH

BURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

15

8

3 4

4

Arnold

SuperShelf

Hewes

Zrox

Zeron

N

ZeronS

Rock-Rite

75


T hi t P bl E l


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The Management Scientist Solution (continued)

Constraint Slack/Surplus Dual Prices

1 0.000 0.000

2 0.000 2.000

3 0.000 -5.0004 0.000 -6.000

5 0.000 -6.000

6 0.000 -10.000

7 0.000 -10.000


T hi t P bl E l


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OBJECTIVE COEFFICIENT RANGES

Variable Lower Limit Current Value Upper LimitX13 3.000 5.000 7.000

X14 6.000 8.000 No LimitX23 3.000 7.000 No LimitX24 No Limit 4.000 6.000X35 No Limit 1.000 4.000X36 3.000 5.000 7.000

X37 5.000 8.000 No LimitX45 0.000 3.000 No LimitX46 2.000 4.000 6.000X47 No Limit 4.000 7.000


T hi t P bl E l


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RIGHT HAND SIDE RANGES

Constraint Lower Limit Current Value Upper Limit

1 75.000 75.000 No Limit

2 75.000 75.000 100.0003 -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 60.0007 15.000 40.000 40.000


End of Chapter 6 Part A


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End of Chapter 6, Part A

Distn & Network Models

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