Transportation, Assignment,and Transshipment Problems

Pertemuan 7

Matakuliah : K0442-Metode KuantitatifTahun : 2009

Bina Nusantara University 3

Material Outline

• Transportation ProblemNetwork RepresentationGeneral LP Formulation

• Assignment ProblemNetwork RepresentationGeneral LP Formulation

• Transshipment ProblemNetwork RepresentationGeneral LP Formulation

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, and transshipment problems of this chapter as well as the PERT/CPM problems (in another chapter) are all examples of network problems.

Transportation, Assignment, and Transshipment Problems

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

• For each of the three 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.

• 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

• 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.

Transportation Problem• Network Representation

22

cc1111

cc1212

cc1313

cc2121

cc2222

cc2323

dd11

dd22

dd33

ss11

s2

SourcesSources DestinationsDestinations

33

22

11

11

Transportation Problem

• LP FormulationThe LP formulation in terms of the

amounts shipped from the origins to the destinations, xij , can be written as:

Min cijxij i j

s.t. xij < si for each origin i j

xij = dj for each destination j

i

xij > 0 for all i and j

• LP Formulation Special CasesThe following special-case modifications to

the linear programming formulation can be made:– 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.

Transportation Problem

Example: Acme Block Co.

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. 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 fill

the above orders?

AcmAcm

eeAcmAcm

ee

Delivery Cost Per Ton

Northwood Westwood Eastwood

Plant 1 24 30 40

Plant 2 30 40 42

Example: Acme Block Co.

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

Example: Acme Block Co.

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

Example: Acme Block Co.

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

Example: Acme Block Co.

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.

Assignment Problem

• Network Representation

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3333

1111

2222

3333

1111cc1111

cc1212

cc1313

cc2121cc2222

cc2323

cc3131 cc3232

cc3333

AgentsAgents TasksTasks

• LP Formulation

Min cijxij i j

s.t. xij = 1 for each agent i j

xij = 1 for each task j i xij = 0 or 1 for all i and j

– Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than 1 job.

Assignment Problem

LP Formulation Special Cases

• Number of agents exceeds the number of tasks:

xij < 1 for each agent i j

• 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

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 a tasks:

xij < a for each agent i j

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 14

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

Example: Who Does What?

Example: Who Does What?

Network Representation

5050

3636

1616

2828

3030

1818

3535 3232

2020

25252525

1414

WestWest..WestWest..

CCCC

BBBB

AAAA

Univ.Univ.Univ.Univ.

Gol.Gol.Gol.Gol.

Fed.Fed. Fed.Fed.

ProjectsSubcontractors

Example: Who Does What? 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

Example: Who Does What?

• The optimal assignment is:

Subcontractor Project Distance Westside C 16

Federated A 28Goliath (unassigned) Universal B 25

Total Distance = 69 miles

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.

Transshipment Problem

• Network Representation

22 22

3333

4444

5555

6666

77 77

11 11cc1313

cc1414

cc2323

cc2424

cc2525

cc1515

ss11

cc3636

cc3737

cc4646

cc4747

cc5656

cc5757

dd11

dd22

Intermediate NodesIntermediate NodesSourcesSources DestinationsDestinations

ss22

DemandDemandSupplySupply

Transshipment Problem

• Linear Programming Formulation xij represents the shipment from node i to node j

Min cijxij i j

s.t. xij < si for each origin i j

xik - xkj = 0 for each intermediate i j node k

xij = dj for each destination j i xij > 0 for all i and j

Example: Zeron Shelving

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.

Example: Zeron Shelving

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 Thomas 1 5 8

Washburn 3 4 4

Example: Zeron Shelving

• Network Representation

ARNOLD

WASHBURN

ZROX

HEWES

7575

7575

5050

6060

4040

55

88

77

44

1155

88

33

4444

ArnoldArnold

SuperSuperShelfShelf

HewesHewes

ZroxZrox

ZeronZeronNN

ZeronZeronSS

Rock-Rock-RiteRite

Example: Zeron Shelving• 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

Example: Zeron Shelving

• Constraints DefinedAmount Out of Arnold: x13 + x14 < 75

Amount Out of Supershelf: x23 + x24 < 75

Amount Through Zeron N: x13 + x23 - x35 - x36 - x37 = 0Amount Through Zeron S: x14 + x24 - x45 - x46 - x47 = 0Amount 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 and j.

Example: Zeron Shelving Optimal Solution (from The Management Scientist )

Objective Function Value = 1150.000

Variable Value Reduced Costs

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

Example: Zeron ShelvingExample: Zeron Shelving Optimal Solution

ARNOLD

WASHBURN

ZROX

HEWES

7575

7575

5050

6060

4040

55

88

77

44

1155

88

33 44

44

ArnoldArnold

SuperSuperShelfShelf

HewesHewes

ZroxZrox

ZeronZeronNN

ZeronZeronSS

Rock-Rock-RiteRite

7575

7575

5050

2525

3535

4040

Example: Zeron Shelving

Optimal Solution (continued)

Constraint Slack/Surplus Dual Prices

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

Example: Zeron Shelving Optimal Solution (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

Example: Zeron Shelving

Optimal Solution (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