Management Science Chapter 5 Transportation, Assignment, and Transshipment Problems - A11.4801i Chapter 5 Transportation Problem

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Management Science

Chapter 5

Chapter 5 Transportation, Assignment, and

Transportation, Assignment, and

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Transportation, Assignment, and

Transshipment Problems

 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,

and functions (e.g. costs, supplies, demands,

etc.) associated with the arcs and/or nodes.

 Transportation, assignment, and Transportation, assignment, and

transshipment problems of this chapter, as

well as the shortest route, minimal spanning

tree, and maximal flow problems (Chapter 9)

and PERT/CPM problems (Chapter 10) are all

examples of network problems.

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Transportation, Assignment, and

Transshipment Problems

 Each of the three models of this chapter Each of the three models of this chapter

(transportation, assignment, and transshipment (transportation, assignment, and transshipment

models) can be formulated as linear programs models) can be formulated as linear programs

and solved by general purpose linear and solved by general purpose linear

programming Algorithms (simplex method). programming Algorithms (simplex method).

 For each of the three models, if the right-hand For each of the three models, if the right-hand

side of the linear programming formulations are side of the linear programming formulations are all integers, the optimal solution will be in terms all integers, the optimal solution will be in terms

of integer values for the decision variables. of integer values for the decision variables.

 However, there are many computer packages However, there are many computer packages

(including

(including The Management Scientist, DS, QSBThe Management Scientist, DS, QSB) ) which contain separate computer codes for

which contain separate computer codes for these models which take advantage of their these models which take advantage of their

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Transportation Problem



The

transportation problem

seeks to

minimize the total shipping costs of

transporting goods from

m

origins or

sources (each with a supply

s

i_i

) to

n

destinations (each with a demand

d

j_j

),

when the unit shipping cost from source,

i

, to a destination,

_{, to a destination,}

j

_j

, is

_{, is}

c

_c

ij_ij

.



The

network representation

for a

transportation problem with two sources

and three destinations is given on the

next slide.

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Transportation Problem

 Network RepresentationNetwork Representation

1

2

3

1

2

c

c₁₁₁₁

c

c₁₂₁₂

c

c₁₃₁₃

c

c₂₁₂₁ _c_c

22 22

c

c₂₃₂₃

d

d₁₁

d

d₂₂

d

d₃₃

s

s₁₁

s₂

SOURCES

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Transportation Problem

 LP FormulationLP Formulation

The linear programming formulation in terms of the

amounts shipped from the sources to the destinations,

amounts shipped from the sources to the destinations, xxij ij , ,

can be written as:

Min Min _ccijijxxij ij (total transportation cost)(total transportation cost)

i ji j

s.t. s.t. _xxijij << ssii for each source for each source i i

(supply constraints)

jj

xxijij = d = djj for each destination for each destination j (demand j (demand

constraints)

ii

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Transportation Problem

 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

the supplies at the sources equal the sum of the

demands at the destinations. If the total supply is

greater than the total demand, a dummy

destination is added with demand equal to the

excess supply, and shipping costs from all sources

are zero. Similarly, if total supply is less than

total demand, a dummy source is added.

 When solving a transportation problem by its When solving a transportation problem by its

special purpose algorithm, unacceptable shipping

routes are given a cost of +

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Transportation Problem

 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

the LP formulation), and the unit shipping

costs are given in an upper right hand box in

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Problem formulation

 The LP model for this problem is as follows:The LP model for this problem is as follows:

Min Z = 15 X

Min Z = 15 X1111 + 30 X + 30 X1212 + 20 X + 20 X1313 + 30 X + 30 X2121 + 40X + 40X2222 + 35X + 35X2323

S.t.

X

X1111 + X + X1212 + X + X1313 ≤ 50 ≤ 50

Supply constraintsSupply constraints X

X2121 + X + X2222 + X + X2323 ≤ 30 ≤ 30

X

X1111 + X + X2121 = 25 = 25 X

X1212 + X + X22 22 = 45= 45 X

X1313 + X + X2323 = 10 demand constraints = 10 demand constraints X

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Transportation Problem

 The transportation problem is solved in two The transportation problem is solved in two

phases: phases:

• Phase I -- Obtaining an initial feasible solution_{Phase I -- Obtaining an initial feasible solution} • Phase II -- Moving toward optimality_{Phase 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, In Phase II, the Stepping Stone, by using thethe Stepping Stone, by 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

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Initial Tableau

 There are many method for finding the initial There are many method for finding the initial tableau for the transportation problem which

tableau for the transportation problem which

are:

1.

1. Northwest cornerNorthwest corner

2.

2. Minimum cost of the rowMinimum cost of the row

3.

3. Minimum cost of the columnMinimum cost of the column

4.

4. Least costLeast cost

5.

5. Vogle’s approximation methodVogle’s approximation method

6.

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Northwest corner

 Northwest corner: Begin by selecting XNorthwest corner: Begin by selecting X₁₁₁₁ (that is, (that is, start in the northwest corner of the transportation

start in the northwest corner of the transportation

tableau). Therefore, if X

tableau). Therefore, if Xij_ij was the last basic variable was the last basic variable

(occupied cell) selected, then select X

(occupied cell) selected, then select Xij+1ij+1 (that is, (that is,

move one column to the right) if source I has any

supply remaining. Otherwise, next select X

supply remaining. Otherwise, next select Xi+1 ji+1 j (that (that

is, move one row down).

is, move one row down). SupplySupply

30

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 Phase I - Minimum-Cost MethodPhase I - Minimum-Cost Method

• Step 1: _{Step 1:}Select the cell with the least cost. Assign to Select the cell with the least cost. Assign to this cell the minimum of its remaining row supply or

this cell the minimum of its remaining row supply or

remaining column demand.

• Step 2: _{Step 2:}Decrease the row and column availabilities by Decrease the row and column availabilities by this amount and remove from consideration all other

this amount and remove from consideration all other

cells in the row or column with zero availability/

demand. (If both are simultaneously reduced to 0,

assign an allocation of 0 to any other unoccupied cell in

the row or column before deleting both.) GO TO STEP 1.

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Transportation 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 below.

Select the unoccupied cell with the most negative

reduced cost. (For maximization problems select

the unoccupied cell with the 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 with

this cell and occupied cells by drawing connecting

alternating horizontal and vertical lines between

them.

Determine the minimum allocation where a

subtraction is to be made along this path.

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Transportation 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

be made along the stepping stone path.

(Note: An occupied cell on the

stepping stone path now becomes 0

(unoccupied). If more than one cell becomes

0, make only one unoccupied; make the

others occupied with 0's.)

GO TO STEP 1.

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Transportation 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 =

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Example: BBC

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 Eastwood

Eastwood

Plant 1 24 Plant 1 24 30 30 40

40

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Example: BBC

 Initial Transportation TableauInitial Transportation Tableau

Since total supply = 100 and total demand =

80, a dummy destination is created with demand

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Example: BBC

 Least Cost Starting ProcedureLeast Cost Starting Procedure

• Iteration 1: _{Iteration 1:}Tie for least cost (0), arbitrarily select Tie for least cost (0), arbitrarily select

x

x14₁₄. Allocate 20. Reduce . Allocate 20. Reduce ss11 by 20 to 30 and by 20 to 30 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.

• Iteration 4: _{Iteration 4:}Since there is only one row with two Since there is only one row with two cells left, make the final allocations of 40 and 10

cells left, make the final allocations of 40 and 10

to

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Example: BBC

 Initial tableauInitial tableau

25

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Example: BBC

 Iteration 1Iteration 1

• MODI Method_{MODI Method}

1. Set

1. Set uu1₁ = 0 = 0

2. Since

2. Since uu1₁ + + vvjj = = cc11jj for occupied cells in row 1, for occupied cells in row 1,

then then

vv1₁ = 24, = 24, vv22 = 30, = 30, vv44 = 0. = 0.

3. Since

3. Since uui_i + + vv22 = = ccii22 for occupied cells in column for occupied cells in column

2,

2, then then uu2₂ + 30 = 40, hence + 30 = 40, hence uu22 = 10. = 10.

4. Since

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

then then

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Example: BBC

• MODI Method (continued)_{MODI Method (continued)}

Calculate the reduced costs (circled numbers

on the

on 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 = (2,4) 0 - 10 - 0 = -10-10

• Since some of the reduced cost are negative, the current

solution is not optimal.

• Cell (2,4) has the most negative; therefore, it will be the

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 Iteration 1 TableauIteration 1 Tableau

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Example: BBC

• Stepping Stone Method_{Stepping Stone Method}

The stepping stone path for cell (2,4) is (2,4), (1,4),

(1,2), (2,2). The allocations in the subtraction cells are

20 and 40, respectively. The minimum is 20, and

hence reallocate 20 along this path. Thus for the next

tableau:

tableau: x

x2424 = 0 + 20 = 20 (0 is its current allocation) = 0 + 20 = 20 (0 is its current allocation)

xx1414 = 20 - 20 = 0 (blank for the next = 20 - 20 = 0 (blank for the next tableau)

tableau)

xx12₁₂ = 5 + 20 = 25 = 5 + 20 = 25

xx2222 = 40 - 20 = 20 = 40 - 20 = 20

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25

Total transportation cost is $2570 = 2770 – 10 (20)

Reduced cost of cell (2,4)

New

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Example: BBC

The reduced costs are found by calculating

the

the uui_i's and 's 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 uu_i_i + + vv₂₂ = = cc_i_i₂₂ 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

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Example: BBC

on the

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

(1,3) (1,3) 40 - 0 - 32 = 840 - 0 - 32 = 8

(1,4) (1,4) 0 - 0 - (-10) 0 - 0 - (-10) = 10

= 10

_{Since there is still negative reduced cost for cell (2,1),}

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

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Example: BBC

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Example: BBC

• Stepping Stone Method_{Stepping Stone Method}

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

solution is obtained by reallocating 20 on the 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)

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Example: BBC

The reduced costs are found by calculating

The reduced costs are found by calculating the the uui_i's 's

and

and vvj_j's for this tableau.'s for this tableau.

1. Set

1. Set uu1₁ = 0 = 0

2. Since

2. Since uu₁₁ + + vv_j_j = = cc₁₁_j_j 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 uu_i_i + + vv₁₁ = = cc_i_i₁₁ 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.

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Example: BBC

on the

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

(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-negative,

this is the optimal tableau.

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Example: BBC

 Optimal SolutionOptimal Solution

FromFrom ToTo AmountAmount CostCost Plant 1 Northwood 5 120

Plant 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