linear programming: transportation problem j. loucks , st. edward's university (austin, tx,...

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Linear Programming: Transportation Linear Programming: Transportation ProblemProblem

J. Loucks, St. Edward's University (Austin, TX, USA)

Chapter 2BChapter 2B

http://www.stedwards.edu/academics/schools/business/jloucks

http://www.stedwards.edu/academics/schools/business/jloucks

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

Transportation problem (TP), as well as many Transportation problem (TP), as well as many other problems, are all examples of network other problems, are all examples of network problems.problems.

Efficient solution algorithms exist to solve Efficient solution algorithms exist to solve network problems. network problems.

Transportation ProblemTransportation Problem

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TP can be formulated as linear programs and TP can be formulated as linear programs and solved by general purpose linear programming solved by general purpose linear programming codes. codes.

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

However, there are many computer packages, However, there are many computer packages, which contain separate computer codes for the which contain separate computer codes for the TP which take advantage of its network TP which take advantage of its network structure.structure.


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Transportation ProblemTransportation Problem The TP problem has the following characteristics:The TP problem has the following characteristics:

• mm sources and sources and nn destinations destinations• number of variables is m x nnumber of variables is m x n• number of constraints is m + n (constraints number of constraints is m + n (constraints

are for source capacity and destination are for source capacity and destination demand)demand)

• costs appear only in objective function costs appear only in objective function (objective is to minimize total cost of shipping)(objective is to minimize total cost of shipping)

• coefficients of decision variables in the coefficients of decision variables in the constraints are either 0 or 1constraints are either 0 or 1

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The The transportation problemtransportation problem seeks to minimize seeks to minimize the total shipping costs of transporting goods the total shipping costs of transporting goods from from mm origins (each with a supply origins (each with a supply ssii) to ) to nn destinations (each with a demand destinations (each with a demand ddjj), when ), when the unit shipping cost from an origin, the unit shipping cost from an origin, ii, to a , to a destination, destination, jj, is , is ccijij..

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


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Transportation ProblemTransportation Problem Network RepresentationNetwork Representation

1

2

3

1

2

cc1111

cc1212

cc1313

cc2121 cc2222

cc2323

dd11

dd22

dd33

ss11

s2

SOURCESSOURCES DESTINATIONSDESTINATIONS

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The Relationship of TP to LPThe Relationship of TP to LP TP is a special case of LPTP is a special case of LP How do we formulate TP as an LP?How do we formulate TP as an LP?

• Let xLet xijij = quantity of product shipped from = quantity of product shipped from source i to destination jsource i to destination j

• Let cLet cijij = per unit shipping cost from source i to = per unit shipping cost from source i to destination jdestination j

• Let sLet sii be the row i total supply be the row i total supply• Let dLet djj be the column j total demand be the column j total demand

The LP formulation of the TP problem is:The LP formulation of the TP problem is:

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LP FormulationLP FormulationThe linear programming formulation in terms of The linear programming formulation in terms of

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

Min Min ccijijxxijij

i ji j

s.t. s.t. xxijij << ssii for each origin for each origin ii jj

xxijij >> ddjj for each destination for each destination jj ii

xxijij >> 0 for all 0 for all ii and and jj

The Relationship of TP to LPThe Relationship of TP to LP

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Transportation ProblemTransportation Problem LP Formulation Special CasesLP Formulation Special Cases

The following special-case modifications to the The following special-case modifications to the linear programming formulation can be made:linear programming formulation can be made:• Minimum shipping guarantees from Minimum shipping guarantees from ii to to jj: :

xxijij >> LLijij

• Maximum route capacity from Maximum route capacity from ii to to jj:: xxijij << LLijij

• Unacceptable routes: Unacceptable routes: delete the variable (or put a very high cost, delete the variable (or put a very high cost, called the Big-M method, on a selected pair of called the Big-M method, on a selected pair of i and j)i and j)

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To solve the transportation problem by its special To solve the transportation problem by its special purpose algorithm, it is required that the sum of purpose algorithm, it is required that the sum of the supplies at the origins equal the sum of the the supplies at the origins equal the sum of the demands at the destinations. If the total supply is demands at the destinations. If the total supply is greater than the total demand, a dummy greater than the total demand, a dummy destination is added with demand equal to the destination is added with demand equal to the excess supply, and shipping costs from all origins excess supply, and shipping costs from all origins are zero. Similarly, if total supply is less than are zero. Similarly, if total supply is less than total demand, a dummy origin is added. total demand, a dummy origin is added.

When solving a transportation problem by its When solving a transportation problem by its special purpose algorithm, unacceptable shipping special purpose algorithm, unacceptable shipping routes are given a cost of +routes are given a cost of +MM (a large number). (a large number).


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The transportation problem is solved in two The transportation problem is solved in two phases: phases: • Phase I — Obtaining an initial feasible solutionPhase I — Obtaining an initial feasible solution• Phase II — Moving toward optimalityPhase II — Moving toward optimality

In Phase I, the Minimum-Cost Procedure can be In Phase I, the Minimum-Cost Procedure can be used to establish an initial basic feasible solution used to establish an initial basic feasible solution without doing numerous iterations of the simplex without doing numerous iterations of the simplex method.method.

In Phase II, the Stepping Stone Method, using the In Phase II, the Stepping Stone Method, using the MODI method for evaluating the reduced costs MODI method for evaluating the reduced costs may be used to move from the initial feasible may be used to move from the initial feasible solution to the optimal one.solution to the optimal one.


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Phase I - Minimum-Cost Method (simple and Phase I - Minimum-Cost Method (simple and intuitive)intuitive)• Step 1: Step 1: Select the cell with the least cost. Select the cell with the least cost.

Assign to this cell the minimum of its remaining Assign to this cell the minimum of its remaining row supply or remaining column demand.row supply or remaining column demand.

• Step 2: Step 2: Decrease the row and column Decrease the row and column availabilities by this amount and remove from availabilities by this amount and remove from consideration all other cells in the row or consideration all other cells in the row or column with zero availability/demand. (If both column with zero availability/demand. (If both are simultaneously reduced to 0, assign an are simultaneously reduced to 0, assign an allocation of 0 to any other unoccupied cell in allocation of 0 to any other unoccupied cell in the row or column before deleting both.) GO TO the row or column before deleting both.) GO TO STEP 1.STEP 1.

Transportation AlgorithmTransportation Algorithm

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Transportation AlgorithmTransportation Algorithm Phase II - Stepping Stone MethodPhase II - Stepping Stone Method

• Step 1: Step 1: For each unoccupied cell, calculate the For each unoccupied cell, calculate the reduced cost by the MODI method described reduced cost by the MODI method described below. below. Select the unoccupied cell with the Select the unoccupied cell with the most negative reduced cost. (For maximization most negative reduced cost. (For maximization problems select the unoccupied cell with the problems select the unoccupied cell with the largest reduced cost.) If none, STOP.largest reduced cost.) If none, STOP.

• Step 2: Step 2: For this unoccupied cell generate a For this unoccupied cell generate a stepping stone path by forming a closed loop stepping stone path by forming a closed loop with this cell and occupied cells by drawing with this cell and occupied cells by drawing connecting alternating horizontal and vertical connecting alternating horizontal and vertical lines between them. lines between them. Determine the minimum allocation where a Determine the minimum allocation where a subtraction is to be made along this path. subtraction is to be made along this path.

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Transportation AlgorithmTransportation Algorithm Phase II - Stepping Stone Method (continued)Phase II - Stepping Stone Method (continued)

• Step 3: Step 3: Add this allocation to all cells where Add this allocation to all cells where additions are to be made, and subtract this additions are to be made, and subtract this allocation to all cells where subtractions are to allocation to all cells where subtractions are to be made along the stepping stone path. be made along the stepping stone path. (Note: An occupied cell on the stepping stone (Note: An occupied cell on the stepping stone path now becomes 0 (unoccupied). path now becomes 0 (unoccupied).

If more than one cell becomes 0, If more than one cell becomes 0, make only one unoccupied; make the others make only one unoccupied; make the others occupied with 0's.) GO TO STEP 1.occupied with 0's.) GO TO STEP 1.

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Transportation AlgorithmTransportation Algorithm MODI Method (for obtaining reduced costs)MODI Method (for obtaining reduced costs)

Associate a number, Associate a number, uuii, with each row and , with each row and vvjj with each column. with each column.• Step 1: Step 1: Set Set uu11 = 0. = 0.• Step 2: Step 2: Calculate the remaining Calculate the remaining uuii's and 's and vvjj's 's

by solving the relationship by solving the relationship ccijij = = uuii + + vvjj for for occupied cells.occupied cells.

• Step 3: Step 3: For unoccupied cells (For unoccupied cells (ii,,jj), the reduced ), the reduced cost = cost = ccijij - - uuii - - vvjj..

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A A transportation tableautransportation tableau is given below. Each is given below. Each cell represents a shipping route (which is an cell represents a shipping route (which is an arc on the network and a decision variable in arc on the network and a decision variable in the LP formulation), and the unit shipping the LP formulation), and the unit shipping costs are given in an upper right hand box in costs are given in an upper right hand box in the cell. the cell.

SupplySupply

3030

5050

3535

2020

4040

3030

3030

DemandDemand 101045452525

S1S1

S2S2

D3D3D2D2D1D1

1515

Example 1: TPExample 1: TP

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Building Brick Company (BBC) has orders for 80 Building Brick Company (BBC) has orders for 80 tons of bricks at three suburban locations as tons of bricks at three suburban locations as follows: Northwood — 25 tons, Westwood — 45 follows: Northwood — 25 tons, Westwood — 45 tons, and Eastwood — 10 tons. BBC has two plants, tons, and Eastwood — 10 tons. BBC has two plants, each of which can produce 50 tons per week. each of which can produce 50 tons per week. How should end of week shipments be made to fill How should end of week shipments be made to fill the above orders given the following delivery cost the above orders given the following delivery cost per ton:per ton:

NorthwoodNorthwood WestwoodWestwood EastwoodEastwood Plant 1 24 Plant 1 24 30 30 40 40

Plant 2 Plant 2 30 30 40 40 42 42

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Example 1: TPExample 1: TP Initial Transportation TableauInitial Transportation Tableau

Since total supply = 100 and total demand Since total supply = 100 and total demand = 80, a dummy destination is created with = 80, a dummy destination is created with demand of 20 and 0 unit costs.demand of 20 and 0 unit costs.

4242

4040 00

004040

3030

3030

DemandDemand

SupplySupply

5050

5050

2020101045452525

DummyDummy

Plant 1Plant 1

Plant 2Plant 2

EastwoodEastwoodWestwoodWestwoodNorthwoodNorthwood

2424

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Example 1: TPExample 1: TP Least Cost Starting ProcedureLeast Cost Starting Procedure

• Iteration 1: Iteration 1: Tie for least cost (0), arbitrarily Tie for least cost (0), arbitrarily select select xx1414. Allocate 20. Reduce . Allocate 20. Reduce ss11 by 20 to 30 by 20 to 30 and delete the Dummy column.and delete the Dummy column.

• Iteration 2: Iteration 2: Of the remaining cells the least cost Of the remaining cells the least cost is 24 for is 24 for xx1111. Allocate 25. Reduce . Allocate 25. Reduce ss11 by 25 to 5 by 25 to 5 and eliminate the Northwood column.and eliminate the Northwood column.

• Iteration 3: Iteration 3: Of the remaining cells the least cost Of the remaining cells the least cost is 30 for is 30 for xx1212. Allocate 5. Reduce the Westwood . Allocate 5. Reduce the Westwood column to 40 and eliminate the Plant 1 row.column to 40 and eliminate the Plant 1 row.

• Iteration 4: Iteration 4: Since there is only one row with Since there is only one row with two cells left, make the final allocations of 40 two cells left, make the final allocations of 40 and 10 to and 10 to xx2222 and and xx2323, respectively., respectively.

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Example 1: TPExample 1: TP Iteration 1Iteration 1

• MODI MethodMODI Method1. Set 1. Set uu11 = 0 = 02. Since 2. Since uu11 + + vvjj = = cc11jj for occupied cells in for occupied cells in

row 1, thenrow 1, then vv11 = 24, = 24, vv22 = 30, = 30, vv44 = 0. = 0.

3. Since 3. Since uuii + + vv22 = = ccii22 for occupied cells in for occupied cells in column 2, column 2, then then uu22 + 30 = 40, hence + 30 = 40, hence uu22 = 10. = 10.

4. Since 4. Since uu22 + + vvjj = = cc22jj for occupied cells in for occupied cells in row 2, thenrow 2, then

10 + 10 + vv33 = 42, hence = 42, hence vv33 = 32. = 32.

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• MODI Method (continued)MODI Method (continued)Calculate the reduced costs (circled numbers on Calculate the reduced costs (circled numbers on the the next slide) by next slide) by ccijij - - uuii + + vvjj..

Unoccupied CellUnoccupied Cell Reduced CostReduced Cost (1,3) 40 - 0 - 32 = 8(1,3) 40 - 0 - 32 = 8 (2,1) 30 - 24 -10 = -4(2,1) 30 - 24 -10 = -4 (2,4) 0 - 10 - 0 = -10(2,4) 0 - 10 - 0 = -10

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Example 1: TPExample 1: TP Iteration 1 TableauIteration 1 Tableau

2525 5 5

-4 -4

+8 +8 20 20

40 40 10 10 -10 -10 4242

4040 00

004040

3030

3030

vvjj

uuii

1010

00

00323230302424

DummyDummy

Plant 1Plant 1

Plant 2Plant 2


2424

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• Stepping Stone MethodStepping Stone MethodThe stepping stone path for cell (2,4) is (2,4), (1,4), The stepping stone path for cell (2,4) is (2,4), (1,4), (1,2), (2,2). The allocations in the subtraction cells (1,2), (2,2). The allocations in the subtraction cells are 20 and 40, respectively. The minimum is 20, and are 20 and 40, respectively. The minimum is 20, and hence reallocate 20 along this path. Thus for the next hence reallocate 20 along this path. Thus for the next tableau:tableau:xx2424 = 0 + 20 = 20 (0 is its current allocation) = 0 + 20 = 20 (0 is its current allocation)

xx1414 = 20 - 20 = 0 (blank for the next tableau) = 20 - 20 = 0 (blank for the next tableau) xx1212 = 5 + 20 = 25 = 5 + 20 = 25 xx2222 = 40 - 20 = 20 = 40 - 20 = 20 The other occupied cells remain the same.The other occupied cells remain the same.

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• MODI MethodMODI MethodThe reduced costs are found by calculating The reduced costs are found by calculating the the uuii's 's and and vvjj's for this tableau.'s for this tableau.1. Set 1. Set uu11 = 0. = 0.2. Since 2. Since uu11 + + vvjj = = ccijij for occupied cells in row 1, then for occupied cells in row 1, then

vv11 = 24, = 24, vv22 = 30. = 30.3. Since 3. Since uuii + + vv22 = = ccii22 for occupied cells in column 2, for occupied cells in column 2, then then uu22 + 30 = 40, or + 30 = 40, or uu22 = 10. = 10.4. Since 4. Since uu22 + + vvjj = = cc22jj for occupied cells in row 2, then for occupied cells in row 2, then

10 + 10 + vv33 = 42 or = 42 or vv33 = 32; and, 10 + = 32; and, 10 + vv44 = 0 or = 0 or vv44 = -10.= -10.

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• MODI Method (continued)MODI Method (continued)Calculate the reduced costs (circled Calculate the reduced costs (circled

numbers on the numbers on the next slide) by next slide) by ccijij - u- uii + + vvjj..

Unoccupied CellUnoccupied Cell Reduced CostReduced Cost (1,3) (1,3) 40 - 0 - 32 = 840 - 0 - 32 = 8 (1,4) (1,4) 0 - 0 - (-10) 0 - 0 - (-10)

= 10= 10 (2,1) (2,1) 30 - 10 - 24 = -430 - 10 - 24 = -4

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2525 25 25

-4 -4

+8 +8 +10 +10

20 20 10 10 20 20 4242

4040 00

004040

3030

3030

vvjj

uuii

1010

00

-6-6363630302424

DummyDummy

Plant 1Plant 1

Plant 2Plant 2


2424

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• Stepping Stone MethodStepping Stone MethodThe most negative reduced cost is = -4 determined The most negative reduced cost is = -4 determined by by xx2121. The stepping stone path for this cell is (2,1),. The stepping stone path for this cell is (2,1),(1,1),(1,2),(2,2). The allocations in the subtraction (1,1),(1,2),(2,2). The allocations in the subtraction cells are 25 and 20 respectively. Thus the new cells are 25 and 20 respectively. Thus the new solution is obtained by reallocating 20 on the solution is obtained by reallocating 20 on the stepping stone path. Thus for the next tableau:stepping stone path. Thus for the next tableau:

xx2121 = 0 + 20 = 20 (0 is its current allocation) = 0 + 20 = 20 (0 is its current allocation) xx1111 = 25 - 20 = 5 = 25 - 20 = 5 xx1212 = 25 + 20 = 45 = 25 + 20 = 45 xx2222 = 20 - 20 = 0 (blank for the next tableau) = 20 - 20 = 0 (blank for the next tableau) The other occupied cells remain the same.The other occupied cells remain the same.

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• MODI MethodMODI MethodThe reduced costs are found by calculating The reduced costs are found by calculating the the uuii's 's and and vvjj's for this tableau.'s for this tableau.1. Set 1. Set uu11 = 0 = 02. Since 2. Since uu11 + + vvjj = = cc11jj for occupied cells in row 1, then for occupied cells in row 1, then

vv11 = 24 and = 24 and vv22 = 30. = 30.3. Since 3. Since uuii + + vv11 = = ccii11 for occupied cells in column 2, for occupied cells in column 2, then then uu22 + 24 = 30 or + 24 = 30 or uu22 = 6. = 6. 4. Since 4. Since uu22 + + vvjj = = cc22jj for occupied cells in row 2, then for occupied cells in row 2, then

6 + 6 + vv33 = 42 or = 42 or vv33 = 36, and 6 + = 36, and 6 + vv44 = 0 or = 0 or vv44 = - = -6. 6.

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• MODI Method (continued)MODI Method (continued)Calculate the reduced costs (circled Calculate the reduced costs (circled

numbers on the numbers on the next slide) by next slide) by ccijij - - uuii + + vvjj..

Unoccupied CellUnoccupied Cell Reduced CostReduced Cost (1,3) (1,3) 40 - 0 - 36 = 4 40 - 0 - 36 = 4 (1,4) (1,4) 0 - 0 - (-6) 0 - 0 - (-6)

= 6= 6 (2,2) (2,2) 40 - 6 - 30 = 4 40 - 6 - 30 = 4

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Since all the reduced costs are non-Since all the reduced costs are non-negative, this is the optimal tableau.negative, this is the optimal tableau.

55 45 45

20 20

+4 +4 +6 +6

+4 +4 10 10 20 20 4242

4040 00

004040

3030

3030

vvjj

uuii

66

00

-6-6363630302424

DummyDummy

Plant 1Plant 1

Plant 2Plant 2


2424

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Example 1: TPExample 1: TP Optimal SolutionOptimal Solution

FromFrom ToTo AmountAmount CostCostPlant 1 Northwood 5 120Plant 1 Northwood 5 120

Plant 1 Westwood 45 1,350Plant 1 Westwood 45 1,350 Plant 2 Northwood 20 600Plant 2 Northwood 20 600 Plant 2 Eastwood 10 Plant 2 Eastwood 10 420420

Total Cost = $2,490Total Cost = $2,490

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Des MoinesDes Moines(100 units (100 units capacity)capacity)

ClevelandCleveland(200 units(200 unitsReq.)Req.)

Boston (200Boston (200units req.)units req.)

Fort LauderdaleFort Lauderdale(300 units capacity)(300 units capacity)

Albuquerque Albuquerque (300 units(300 unitsReq.)Req.) Evansville (300Evansville (300

units capacity)units capacity)

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To (Destination)

From(Sources)

Albuquerque Boston Cleveland

Des Moines $5 $4 $3

Evansville $8 $4 $3

FortLauderdale

$9 $7 $5

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To Factory

From Albu-querque

Boston Cleve-land

capacity

Des 5 4 3 100MoinesEvansville 8 4 3 300

Fort 9 7 5 300LauderdaleWarehouserequirement

300 200 200 700


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Example 2: TPExample 2: TP What is the initial solution?What is the initial solution?

• Since we want to minimize the cost, it is Since we want to minimize the cost, it is reasonable to “be greedy” and start at the reasonable to “be greedy” and start at the cheapest cell ($3/unit) while shipping as much cheapest cell ($3/unit) while shipping as much as possible. This method of getting the initial as possible. This method of getting the initial feasible solution is called the least-cost rulefeasible solution is called the least-cost rule

• Any ties are broken arbitrarily. Any ties are broken arbitrarily. (Des Moines to Cleveland - $3)(Des Moines to Cleveland - $3)(Evansville to Cleveland - $3)(Evansville to Cleveland - $3)Let’s start ad cell 1-3 (Des Moines to Cleveland)Let’s start ad cell 1-3 (Des Moines to Cleveland)

• In making the flow assignments make sure not In making the flow assignments make sure not to exceed the capacities and demands on the to exceed the capacities and demands on the margins!margins!

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


StartStart +100+100 00

100100

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


+100+100

+100+100

00

100100

200200

00

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


+100+100

+200 +100+200 +100

00

100100

200200

0000

00

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


+100+100

+200 +100+200 +100

+300+300

00

100100

200200

0000

00

00

00

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Example 2: TPExample 2: TP The initial feasible solution via the least-cost rule The initial feasible solution via the least-cost rule

is:is:• Ship 100 units from Des Moines to ClevelandShip 100 units from Des Moines to Cleveland• Ship 100 units from Evansville to ClevelandShip 100 units from Evansville to Cleveland• Ship 200 units from Evansville to BostonShip 200 units from Evansville to Boston• Ship 300 units from Fort Lauderdale to Ship 300 units from Fort Lauderdale to

AlbuquerqueAlbuquerque The total cost of this solution is:The total cost of this solution is:

• 100(3) + 100(3) + 200(4) + 300(9) = $4,100100(3) + 100(3) + 200(4) + 300(9) = $4,100 Is the current solution optimal?Is the current solution optimal?

• Perhaps, but we are not sure. Hence, some Perhaps, but we are not sure. Hence, some more work is needed.more work is needed.

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Example 2: TPExample 2: TP The cells that have a number in them are The cells that have a number in them are

referred to as basic cells (or referred to as basic cells (or basic variablesbasic variables))• In general, all basic variables are strictly In general, all basic variables are strictly

positivepositive• When at least one basic variable is equal to When at least one basic variable is equal to

zero, the corresponding solution is called zero, the corresponding solution is called degeneratedegenerate

• When a solution is degenerate, exercise When a solution is degenerate, exercise caution in using sensitivity reportscaution in using sensitivity reports

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Example 2: TPExample 2: TP How many basic variables (cells) should there be How many basic variables (cells) should there be

in the transportation tableau?in the transportation tableau?• Answer: (R+C-1), where R = number of rows Answer: (R+C-1), where R = number of rows

and C = number of columns. Here, (R+C-1) = and C = number of columns. Here, (R+C-1) = 3+3-1 = 53+3-1 = 5

What are they (there are 4 of them)?What are they (there are 4 of them)?• Ship 100 units from Des Moines to ClevelandShip 100 units from Des Moines to Cleveland• Ship 100 units from Evansville to ClevelandShip 100 units from Evansville to Cleveland• Ship 200 units from Evansville to BostonShip 200 units from Evansville to Boston• Ship 300 units from Fort Lauderdale to Ship 300 units from Fort Lauderdale to

AlbuquerqueAlbuquerque

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Example 2: TPExample 2: TP Is this solution degenerate?Is this solution degenerate?

• Answer: Yes, because (R+C-1) = 3+3-1 = 5 > Answer: Yes, because (R+C-1) = 3+3-1 = 5 > 44

• Hence, we can arbitrarily make (5-4) = 1 Hence, we can arbitrarily make (5-4) = 1 variable which is at level zero basic. variable which is at level zero basic.

• Let us make Des Moines to Albuquerque a Let us make Des Moines to Albuquerque a basic variable at level zerobasic variable at level zero

How do we determine if the current solution is How do we determine if the current solution is optimal?optimal?• Let us price currently nonbasic variables Let us price currently nonbasic variables

(cells), which are all at level zero using the (cells), which are all at level zero using the stepping stonestepping stone algorithm algorithm

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


0 0 +100 +100

+200 +100 +200 +100

+300+300

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Example 2: TPExample 2: TP Evansville – Albuquerque:Evansville – Albuquerque:

• 8 – 3 + 3 – 5 = 3 > 08 – 3 + 3 – 5 = 3 > 0

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


0 0 +100 +100

+200 +100 +200 +100

+300+300

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Example 2: TPExample 2: TP Des Moines - Boston:Des Moines - Boston:

• 4 – 3 + 3 - 4 = 04 – 3 + 3 - 4 = 0

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


0 0 +100 +100

+200 +100 +200 +100

+300+300

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Example 2: TPExample 2: TP Fort Lauderdale - Boston:Fort Lauderdale - Boston:

• 7 – 9 + 5 - 3 + 3 – 4 = -1 < 07 – 9 + 5 - 3 + 3 – 4 = -1 < 0

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700


0 0 +100 +100

+200 +100 +200 +100

+300+300

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Example 2: TPExample 2: TP Fort Lauderdale - Cleveland:Fort Lauderdale - Cleveland:

• 5 –29 + 5 - 3 = -2 < 05 –29 + 5 - 3 = -2 < 0

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Example 1: TPExample 1: TP The objective function improvement potentials The objective function improvement potentials

for the nonbasic variables are:for the nonbasic variables are:• Evansville – Albuquerque: 3 > 0Evansville – Albuquerque: 3 > 0• Des Moines - Boston: 0 = 0Des Moines - Boston: 0 = 0• Fort Lauderdale - Boston: -1 < 0Fort Lauderdale - Boston: -1 < 0• Fort Lauderdale - Cleveland: Fort Lauderdale - Cleveland: -2 < 0-2 < 0

The largest improvement (here: reduction in The largest improvement (here: reduction in cost) in the objective function can be achieved cost) in the objective function can be achieved by increasing shipments from Fort Lauderdale to by increasing shipments from Fort Lauderdale to ClevelandCleveland

What i2 the limit on the increase?What i2 the limit on the increase?

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To Factory

From Albu-querque

Boston Cleve-land

capacity



300 200 200 700

Example 2 TPExample 2 TP

0 0 +K+K +100 +100 -K-K

+200 +100 +200 +100

+300 +300 -K-K +K+K

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Example 2: TPExample 2: TP If we increase shipments from Fort Lauderdale to If we increase shipments from Fort Lauderdale to

Cleveland by K, we must reduce shipments in the Cleveland by K, we must reduce shipments in the basic cells in the same row and column so that basic cells in the same row and column so that the totals on the margins remain the samethe totals on the margins remain the same

Clearly, because negative quantities cannot be Clearly, because negative quantities cannot be shipped, we must impose that:shipped, we must impose that:• K >= 0 (no problem)K >= 0 (no problem)• 300 – K >= 0 (K <= 300)300 – K >= 0 (K <= 300)• 0 + K >= 0 (no problem)0 + K >= 0 (no problem)• 100 – K >=0 (K <=100)100 – K >=0 (K <=100)Hence, let K = 100Hence, let K = 100

The new basis becomes:The new basis becomes: