Chapter 6, Part A Distribution and Network...

11SlideSlide© 2008 Thomson South-Western. All Rights Reserved© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6, Part ADistribution and Network Models

n Transportation Problem• Network Representation• General LP Formulation

n Assignment Problem• Network Representation• General LP Formulation

n Transshipment Problem• Network Representation• General LP Formulation


Transportation, Assignment, and Transshipment Problems

n 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.

n 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.


Transportation, Assignment, and Transshipment Problems

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

n For each of the five models, 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.

n However, there are many computer packages (including The Management Scientist) that contain separate computer codes for these models which take advantage of their network structure.


Transportation Problem

n 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 shipping cost from an origin, i, to a destination, j, is cij.

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



n Network Representation

22

c11c12

c13

c21

c22c23

d1

d2

d3

s1

s2

Sources Destinations

33

22

11

11

66SlideSlide© 2008 Thomson South-Western. All Rights Reserved© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


n Linear Programming Formulation

Using the notation:xij = number of units shipped from

origin i to destination jcij = cost per unit of shipping from

origin i to destination jsi = supply or capacity in units at origin idj = demand in units at destination j

continued



n Linear Programming Formulation (continued)

1 1Min

m n

ij iji j

c x= =åå

1 1Min

m n

ij iji j

c x= =åå

=

£ =å K1

1,2, , Supplyn

ij ij

x s i m=

£ =å K1

1,2, , Supplyn

ij ij

x s i m

=

= =å K1

1,2, , Demandm

ij ji

x d j n=

= =å K1

1,2, , Demandm

ij ji

x d j n

xij > 0 for all i and j


n LP Formulation Special Cases• The objective is maximizing profit or revenue:

• Minimum shipping guarantee from i to j:

xij > Lij

• Maximum route capacity from i to j:

xij < Uij

• Unacceptable route:

Remove the corresponding decision variable.


Solve as a maximization problem.


Transportation Problem: Example #1

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

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


n Delivery Cost Per Ton

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



n 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


n Partial Spreadsheet Showing Optimal Solution


A B C D E F G10 X11 X12 X13 X21 X22 X2311 Dec.Var.Values 5 45 0 20 0 1012 Minimized Total Shipping Cost 24901314 LHS RHS15 50 <= 5016 30 <= 5017 25 = 2518 45 = 4519 10 = 10Eastwood Demand

Westwood DemandNorthwood Demand

ConstraintsPlant 1 CapacityPlant 2 Capacity


n Optimal Solution

From To Amount CostPlant 1 Northwood 5 120Plant 1 Westwood 45 1,350Plant 2 Northwood 20 600Plant 2 Eastwood 10 420

Total Cost = $2,490



n Partial Sensitivity Report (first half)


Adjustable CellsFinal 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 CellsFinal 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


n Partial Sensitivity Report (second half)


ConstraintsFinal 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

ConstraintsFinal 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



The Navy has 9,000 pounds of material in Albany,Georgia that it wishes to ship to three installations:San Diego, Norfolk, and Pensacola. Theyrequire 4,000, 2,500, and 2,500 pounds,respectively. Government regulationsrequire 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 todetermine the shipping arrangements(mode, destination, and quantity) thatwill minimize the total shipping cost.



DestinationMode San Diego Norfolk Pensacola

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



n Define the Decision Variables

We 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 PensacolaTruck x11 x12 x13

Railroad x21 x22 x23

Airplane x31 x32 x33



n 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



n Define the ConstraintsEqual 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



n The Management Scientist Output

OBJECTIVE FUNCTION VALUE = 142000.000Variable Value Reduced Cost

x11 1000.000 0.000x12 2000.000 0.000x13 0.000 1.000 x21 0.000 3.000 x22 500.000 0.000 x23 2500.000 0.000 x31 3000.000 0.000 x32 0.000 2.000 x33 0.000 6.000



n Solution Summary• San Diego will receive 1000 lbs. by truck

and 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.



Assignment Problem

n 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.

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

n 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.

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


Assignment Problem


22

33

11

22

33

11c11

c12c13

c21 c22

c23

c31 c32

c33

Agents Tasks



Using the notation:

xij = 1 if agent i is assigned to task j0 otherwise

cij = cost of assigning agent i to task j

Assignment Problem

continued



Assignment Problem

1 1Min

m n

ij iji j

c x= =åå

1 1Min

m n

ij iji j

c x= =åå

11 1,2, , Agents

n

ijj

x i m=

£ =å K1

1 1,2, , Agentsn

ijj

x i m=

£ =å K

11 1,2, , Tasks

m

iji

x j n=

= =å K1

1 1,2, , Tasksm

iji

x j n=

= =å K



n 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 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.


Assignment Problem

n 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=

£ =å K1

1,2, , Agentsn

ijj

x t i m=

£ =å K


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

How should the contractors be assigned to minimize total mileage costs?

Assignment Problem: Example


n Network Representation50

361628

30

1835 32

2025 25

14

West.West.

CC

BB

AA

Univ.Univ.

Gol.Gol.

Fed.Fed.

ProjectsSubcontractors




Min 50x11+36x12+16x13+28x21+30x22+18x23+35x31+32x32+20x33+25x41+25x42+14x43

s.t. x11+x12+x13 < 1x21+x22+x23 < 1x31+x32+x33 < 1x41+x42+x43 < 1x11+x21+x31+x41 = 1x12+x22+x32+x42 = 1x13+x23+x33+x43 = 1

xij = 0 or 1 for all i and j

Agents

Tasks



n The optimal assignment is:

Subcontractor Project DistanceWestside C 16Federated A 28Goliath (unassigned) Universal B 25

Total Distance = 69 miles



Transshipment Problem

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

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

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

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




22

33

44

55

66

77

11c13

c14

c23

c24c25

c15

s1

c36

c37

c46

c47

c56

c57

d1

d2

Intermediate NodesSources Destinations

s2

DemandSupply




Using the notation:xij = number of units shipped from node i to node jcij = cost per unit of shipping from node i to node jsi = supply at origin node idj = demand at destination node j

continued



ååMin ij iji j

c xååMin ij iji j

c x

- £å ås.t. Origin nodes ij ji ij j

x x s i- £å ås.t. Origin nodes ij ji ij j

x x s i


- =å å 0 Transhipment nodes ik kji j

x x k- =å å 0 Transhipment nodes ik kji j

x x k

- =å å Destination nodes ij ji ji i

x x d j- =å å Destination nodes ij ji ji i

x x d j


continued



n 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.


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


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

Zeron N Zeron SArnold 5 8

Supershelf 7 4

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

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




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



n Linear Programming Formulation• Decision Variables Defined

xij = amount shipped from manufacturer i to supplier jxjk = 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 DefinedMinimize Overall Shipping Costs: Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37

+ 3x45 + 4x46 + 4x47



n 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.



n The Management Scientist Solution

Objective Function Value = 1150.000Variable Value Reduced Costs

X13 75.000 0.000X14 0.000 2.000X23 0.000 4.000X24 75.000 0.000X35 50.000 0.000X36 25.000 0.000X37 0.000 3.000X45 0.000 3.000X46 35.000 0.000X47 40.000 0.000



n 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



n The Management Scientist Solution (continued)

Constraint Slack/Surplus Dual Prices1 0.000 0.0002 0.000 2.0003 0.000 -5.0004 0.000 -6.0005 0.000 -6.0006 0.000 -10.0007 0.000 -10.000




OBJECTIVE COEFFICIENT RANGESVariable Lower Limit Current Value Upper Limit

X13 3.000 5.000 7.000X14 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 7.000X37 5.000 8.000 No LimitX45 0.000 3.000 No LimitX46 2.000 4.000 6.000X47 No Limit 4.000 7.000




RIGHT HAND SIDE RANGESConstraint Lower Limit Current Value Upper Limit

1 75.000 75.000 No Limit2 75.000 75.000 100.0003 -75.000 0.000 0.0004 -25.000 0.000 0.0005 0.000 50.000 50.0006 35.000 60.000 60.0007 15.000 40.000 40.000



Summary

n Transportation Problemn Assignment Problem

• Special case of Transportation Problem• si = 1, di =1

n Transshipment Problem• can be transformed into Transportation Problem

Chapter 6, Part A Distribution and Network...

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