Computational Methods forManagement and Economics

Carla Gomes

Module 8aThe transportation model

The transportation and assignment problems

• Special types of linear programming problems.

• The structure of these problems leads to algorithms – streamlined versions of the simplex method - more efficient than the standard simplex method.

The transportation problem

Prototype example : P&T Company

• Main product of P&T Company – canned peas– 3 canneries

– 4 distribution centers

• Shipping costs – major expense in management

– GOAL – minimize shipping costs

P&T Company Distribution ProblemCANNERY 1 Bellingham

CANNERY 2 Eugene

WAREHOUSE 1 Sacramento

WAREHOUSE 2 Salt Lake City

WAREHOUSE 3 Rapid City

WAREHOUSE 4 Albuquerque

CANNERY 3 Albert Lea

Shipping Data

Cannery Output Warehouse Allocation

Bellingham 75 truckloads Sacramento 80 truckloads

Eugene 125 truckloadsSalt Lake City

65 truckloads

Albert Lea 100 truckloads Rapid City 70 truckloads

Total 300 truckloads Albuquerque 85 truckloads

Total300

truckloads

Current Shipping Plan

Warehouse

From \ To Sacramento Salt Lake City Rapid City Albuquerque

Cannery

Bellingham 75 0 0 0

Eugene 5 65 55 0

Albert Lea 0 0 15 85

Shipping Cost per Truckload

Warehouse

From \ ToSacramento

Salt Lake City

Rapid City Albuquerque

Cannery

Bellingham $464 $513 $654 $867Eugene 352 416 690 791Albert Lea 995 682 388 685

Total shipping cost = 75($464) + 5($352) + 65($416) + 55($690) + 15($388) + 85($685)= $165,595

Terminology for a Transportation Problem

P&T Company Problem

Truckloads of canned peas

Canneries

Warehouses

Output from a cannery

Allocation to a warehouse

Shipping cost per truckload from a cannery to a warehouse

General Model

Units of a commodity

Sources

Destinations

Supply from a source

Demand at a destination

Cost per unit distributed from a source to a destination

Characteristics of Transportation Problems

• The Requirements Assumption

– Each source has a fixed supply of units, where this entire supply must be distributed to the destinations.

– Each destination has a fixed demand for units, where this entire demand must be received from the sources.

• The Feasible Solutions Property

– A transportation problem will have feasible solutions if and only if the sum of its supplies equals the sum of its demands.

• The Cost Assumption

– The cost of distributing units from any particular source to any particular destination is directly proportional to the number of units distributed.

– This cost is just the unit cost of distribution times the number of units distributed.

The Transportation Model

Any problem (whether involving transportation or not) fits the model for a transportation problem if:

1. It can be described completely in terms of a table that identifies all the sources, destinations, supplies, demands, and unit costs, and

2. satisfies both the requirements assumption and the cost assumption.

The objective is to minimize the total cost of distributing the units.

The P&T Co. Transportation ProblemTransportation Tableau

Unit Cost

Destination(Warehouse): Sacramento Salt Lake City Rapid City Albuquerque Supply

Source (Cannery)

Bellingham $464 $513 $654 $867 75

Eugene 352 416 690 791 125

Albert Lea 995 682 388 685 100

Demand 80 65 70 85 300

Network Representation

S1

S2

S3

D4

D2

D1

D3

75

125

100

80

65

70

85

Supplies Demands

SourcesDestinations

(Bellingham)

(Eugene)

(Alber t Lea)

(Sacramento)

(Salt Lake City)

(Rapid City)

(Albuquerque)

464513

654867

352 416690

791

995 682

685

388

This graph is “bipartite.” That is, the nodes are partitioned into two parts and arcs have one endpoint in each part.

The Transportation Problem is an LP

1. Decision Variable:

Since we have to determine how much electricity is sent from each plant to each city;

xij = Amount of commodity produced at source i and sent to destination j

x13 = truckloads produced at cannery 1 and sent to warehouse 3

Network Representation

S1

S2

S3

D4

D2

D1

D3

75

125

100

80

65

70

85

Supplies Demands

SourcesDestinations

(Bellingham)

(Eugene)

(Alber t Lea)

(Sacramento)

(Salt Lake City)

(Rapid City)

(Albuquerque)

464513

654867

352 416690

791

995 682

685

388

This graph is “bipartite.” That is, the nodes are partitioned into two parts and arcs have one endpoint in each part.

x11

x12

x13

x14

x21

x22

x23

x24

x31

x32

x33

x34

Shipping cost – source i destination j

Objective function

Since we want to minimize the total cost of shipping from canneries to wharehouses;

Minimize Z = $464x11 + $513x12 + $654x13 + $867x14 + $352x21 + $416x22

+ $690x23 + $791x24 + $995x31 + $682x32 + $388x33 + $685x34

Supply Constraints

Since each supply point has a limited production capacity;

Cannery 1: x11 + x12 + x13 + x14 = 75Cannery 2: x21 + x22 + x23 + x24 = 125Cannery 3: x31 + x32 + x33 + x34 = 100

Demand Constraints

Since each supply point has a limited production capacity;

Warehouse 1: x11 + x21 + x31 = 80Warehouse 2: x12 + x22 + x32 = 65Warehouse 3: x13 + x23 + x33 = 70Warehouse 4: x14 + x24 + x34 = 85

Sign Constraints

Since a negative amount of truckloads can not be shipped all Xij’s must be non negative;

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

The Transportation Problem is an LPLet xij = the number of truckloads to ship from cannery i to warehouse j

(i = 1, 2, 3; j = 1, 2, 3, 4)

Minimize Cost = $464x11 + $513x12 + $654x13 + $867x14 + $352x21 + $416x22

+ $690x23 + $791x24 + $995x31 + $682x32 + $388x33 + $685x34

subject toCannery 1: x11 + x12 + x13 + x14 = 75Cannery 2: x21 + x22 + x23 + x24 = 125Cannery 3: x31 + x32 + x33 + x34 = 100Warehouse 1: x11 + x21 + x31 = 80Warehouse 2: x12 + x22 + x32 = 65Warehouse 3: x13 + x23 + x33 = 70Warehouse 4: x14 + x24 + x34 = 85

andxij ≥ 0 (i = 1, 2, 3; j = 1, 2, 3, 4)

The Transportation Problem is an LP with special matrix A structure

Let xij = the number of truckloads to ship from cannery i to warehouse j(i = 1, 2, 3; j = 1, 2, 3, 4)

Minimize Cost = c11 x11 + c12 x12 + c13 x13 + c14 x14 + c21 x21 + c22 x22

+ c23 x23 + c24 x24 + c31 x31 + c32 x32 + c33 x33 + c34 x34

subject toSource1: x11 + x12 + x13 + x14 = supply1Source 2: x21 + x22 + x23 + x24 = supply2 Source 3: x31 + x32 + x33 + x34 = supply3 Destination 1: x11 + x21 + x31 = demand1Destination 2: x12 + x22 + x32 = demand2

Destination 3: x13 + x23 + x33 = demand3Destination 4: x14 + x24 + x34 = demand4

andxij ≥ 0 (i = 1, 2, 3; j = 1, 2, 3, 4)

The Node-Arc Incidence Matrix (for directed graphs)

2

34

1

a

b

c

d

e

A Directed Graph

1 1 0 0 0

0 1 1 0 1

0 0 0 1 1

1 0 1 1 0

•Have a row for each node

a b c d e

1

2

3

4

•Have a column for each arc

•Put a 1 in row i- column j if arc j starts at node i.

•Put a -1 in row i- column j if arc j ends at node i.

What would happen if arc (4,2) became arc (2,4)?

On Incidence Matrices• If the constraint matrix for a linear program is a node-arc

incidence matrix (at most one 1 and at most one –1 per column), then the linear program solves in integer optima.

• Node arc incidence matrix shows up in Linear Programs.

The constraint matrix of a transportationproblem is a node-arc incidence matrix

in disguise.

The Transportation Problem is an LPLet xij = the number of truckloads to ship from cannery i to warehouse j

(i = 1, 2, 3; j = 1, 2, 3, 4)

Minimize Cost = $464x11 + $513x12 + $654x13 + $867x14 + $352x21 + $416x22

+ $690x23 + $791x24 + $995x31 + $682x32 + $388x33 + $685x34

subject toCannery 1: x11 + x12 + x13 + x14 = 75Cannery 2: x21 + x22 + x23 + x24 = 125Cannery 3: x31 + x32 + x33 + x34 = 100Warehouse 1: -x11 - x21 - x31 = -80Warehouse 2: - x12 - x22 -x32 = -65Warehouse 3: -x13 - x23 - x33 = -70Warehouse 4: -x14 - x24 - x34 = -85

andxij ≥ 0 (i = 1, 2, 3; j = 1, 2, 3, 4)

The Node-Arc Incidence Matrix:Transportation Problem is an LP

1 1 1 1 1 1 1 1

1 1 1 1

-1 -1 -1 -1 -1 -1

-1 -1 -1 1 -1 -1

Spreadsheet Formulation

34567891011121314151617

B C D E F G H I JUnit Cost Destination (Warehouse)

Sacramento Salt Lake City Rapid City AlbuquerqueSource Bellingham $464 $513 $654 $867

(Cannery) Eugene $352 $416 $690 $791Albert Lea $995 $682 $388 $685

Shipment Quantity Destination (Warehouse)(Truckloads) Sacramento Salt Lake City Rapid City Albuquerque Total Shipped Supply

Source Bellingham 0 20 0 55 75 = 75(Cannery) Eugene 80 45 0 0 125 = 125

Albert Lea 0 0 70 30 100 = 100Total Received 80 65 70 85

= = = = Total CostDemand 80 65 70 85 $152,535

Note: excel does not have a specialized simplexalgorithm to solve transportation problems.

Integer Solutions Property

As long as all its supplies and demands have integer values, any transportation problem with feasible solutions is guaranteed to have an optimal solution with integer values for all its decision variables. Therefore, it is not necessary to add constraints to the model that restrict these variables to only have integer values.

On the integrality Property• The fact that solutions to the transportation problem are

integral is an amazing property.

• In general, solutions to IP are fractional.

• But solutions to the transportation problem are integral.

• Structure of the matrix of technological coefficients (A matrix) and the fact that the RHS are integral – in general, if there is at most one 1 and at most one –1 in any column of the constraint matrix, then every basic feasible solution is integer (so long as RHS is integral.)

• For many applications, we want to restrict variables to be integer valued.

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 most si units.

2. A set of n demand points to which the good is shipped. Demand point j must receive at least di units of the shipped good.

3. Each unit produced at supply point i and shipped to demand point j incurs a variable cost of cij.

Features of this transportation problem

• The constraint matrix is (or can be made to be) the node arc incidence matrix of the network

• If supplies/demands are integral, then the flows are also integral.

• If the total supply is equal to the total demand, then all supply and demand constraints hold with equality

• Very efficient special purpose solution techniques exist

• Applications to shipment of goods

Features of this transportation problem

• The constraint matrix is (or can be made to be) the node arc incidence matrix of the network

• If supplies/demands are integral, then the flows are also integral.

• If the total supply is equal to the total demand, then all supply and demand constraints hold with equality

• Very efficient special purpose solution techniques exist

• Applications to shipment of goods

xij = number of units shipped from supply 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

Balanced Transportation Problem

If Total supply equals to total demand, the problem is said to be a balanced transportation problem:

nj

j

j

mi

i

i ds11

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 the dummy demand point are not real, they are assigned a cost of zero.

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 feasible solution. 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 unmet demand and as one can guess this time the total penalty cost is desired to be minimum

Distribution System at Proctor and Gamble

• Proctor and Gamble needed to consolidate and re-design their North American distribution system in the early 1990’s.– 50 product categories– 60 plants– 15 distribution centers– 1000 customer zones

• Solved many transportation problems (one for each product category).• Goal: find best distribution plan, which plants to keep open, etc.• Closed many plants and distribution centers, and optimized their product sourcing

and distribution location.• Implemented in 1996. Saved $200 million per year.For more details, see 1997 Jan-Feb Interfaces article, “Blending OR/MS, Judgement,

and GIS: Restructuring P&G’s Supply Chain”, downloadable from course web site.

Dummy Destination

Balancing a TP if total supply exceeds total demand

Northern Airplane (Production Scheduling)Northern Airplane Company produces commercial airplanes. The last stage in production is to produce the jet engines and install them.

– The company must meet the delivery deadline indicated in column 2.An option is to produce some engines one month or more before they are scheduled for installation and store them.

– Production and storage costs vary from month to month.

Maximum

Production

Unit Cost of Production ($million)

Unit Costof Storage(*)($thousand)Month

ScheduledInstallations

1 10 25 1.08 15

2 15 35 1.11 15

3 25 30 1.10 15

4 20 10 1.13Question: How many engines should be produced in each of the four months so that the total of the production and storage costs will be minimized?

(*) storage cost is incurred at the end of the month for just those engines that are being held over to the next month;

• Source i - production of jet engines in month i (i = 1,2 ,3 4);

• Destination j – installation of jet engines in month j (j = 1,2 ,3 4)

• xij = number of engines produce in month i to be installed in month j

• cij = cost associated with each unit of xij

Cost per unit for production + storage i < j

i >j???

Northern Airplane (Production Scheduling)

Cost per unit distributed

Destinations

(installation in month j)

Source

(production in month i) 1 2 3 4 SUPLLY

1 1.080 1.095 1.110 1.125 ?

2 ? 1.110 1.125 1.140 ?

3 ? ? 1.100 1.115 ?

4 ? ? ? 1.130 ?

Demand 10 15 25 30

Question: How many engines should be produced in each of the four months so that the total of the production and storage costs will be minimized?

(*) storage cost is incurred at the end of the month for just those engines that are being held over to the next month;

Transportation Tableau Cost per unit distributed

Destinations

(installation in month j)

Source

(production in month i) 1 2 3 4

5(D)(*) SUPLLY

1 1.080 1.095 1.110 1.125 0 25

2 M 1.110 1.125 1.140 0 35

3 M M 1.100 1.115 0 30

4 M M M 1.130 0 10

Demand 10 15 25 20 30

(*) the dummy destination can be seen as a slack variable that represents the unused production capacity. Cost is zero because it is the cost of distributing to a fictional destination. Note that it would be inappropriate to assign M since we do not want to force the corresponding values to be zero. In fact these values need to sum 30.

Dummy Source

Balancing a transportation problem if total supply is less than total demand

Metro Water (Distributing Natural Resources)

Metro Water District is an agency that administers water distribution in a large goegraphic region. The region is arid, so water must be brought in from outside the region.

– Sources of imported water: Colombo, Sacron, and Calorie rivers.– Main customers: Cities of Berdoo, Los Devils, San Go, and Hollyglass.

Cost per Acre Foot

Berdoo Los Devils San Go Hollyglass

Available

(million acre feet)

Colombo River

$160 $130 $220 $170 50

Sacron River 140 130 190 150 60

Calorie River 190 200 230 — 50

Min. Needed 30 70 0 10

Requested 50 70 30 Question: How much water should Metro take from each river, and how much should they

send from each river to each city?

What’s the problem with the previous table to look like a transportation

tableau?• It is not clear what the demands are at the

destinations

The amount to be received at each destination is a decision variable, with an upper bound and a lower bound – in the transportation model it should be a constant.

• Also, here we have excess demand dummy source

What’s the problem with the previous table to look like a transportation

tableau?

Hollyglass upper-bound =

Total supply – min requested =

(50 + 60 +50) – (30 + 70 + 0) = 60

How do we solve the problem of having constant demands?

Metro Water Transportation Tableau

Berdoo

(min)

(1)

Berdoo

(extra)

(2)

Los Devils

(3)

San Go

(4)

Hollyglass

(5)

Available

(million acre feet)

Colombo River (1) $160 $160 $130 $220 $170 50

Sacron River (2) 140 140 130 190 150 60

Calorie River (3) 190 190 200 230 M 50

Dummy (4) M 0 70 M 0 50

Demand 30 20 70 30 60 260