Transportation lecture

Engineering Management 12/19/2013

M M HASAN, LECTURER, AIE, HSTU, DINAJPUR 1

The Transportation Model

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The Transportation Model Characteristics

• A product is transported from a number of sources to a number of destinations at the minimum possible cost.

• Each source is able to supply a fixed number of units of the product, and each destination has a fixed demand for the product.

• The linear programming model has constraints for supply at each source and demand at each destination.

• All constraints are equalities in a balanced transportation model where supply equals demand.

• Constraints contain inequalities in unbalanced models where supply does not equal demand.



Balanced Transportation Problem

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

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

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Transportation Model Example Problem Definition and Data

- Problem:How many tons of wheat to transport from each grain elevator to each mill on a monthly basis in order to minimize the total cost of transportation ?

- Data: Grain Elevator Supply Mill Demand

1. Kansas City 150 A. Chicago 200

2. Omaha 175 B. St.Louis 100

3. Des Moines 275 C. Cincinnati 300

Total 600 tons Total 600 tons



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Transportation Model Example Model Formulation

minimize Z = $6x1A + 8x1B + 10x1C + 7x2A + 11x2B + 11x2C + 4x3A + 5x3B + 12x3C

subject to x1A + x1B + x1C = 150

x2A + x2B + x2C = 175

x3A + x3B+ x3C = 275

x1A + x2A + x3A = 200

x1B + x2B + x3B = 100

x1C + x2C + x3C = 300

xij 0

Network of transportation routes for wheat shipments

where xij = tons of wheat from each grain elevator, i, i = 1, 2, 3, to each mill j, j = A,B,C

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Solution of the Transportation Model Tableau Format

• Transportation problems are solved manually within a tableau format.

• Each cell in a transportation tableau is analogous to a decision variable that indicates the amount allocated from a source to a destination.

• The supply and demand values along the outside rim of a tableau are called rim values.

The Transportation Tableau



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Solution of the Transportation Model

Solution Methods

• Transportation models do not start at the origin where all decision values are zero; they must instead be given an initial feasible solution.

• Initial feasible solution determination methods include:

- northwest corner method

- minimum cell cost method

- Vogel’s Approximation Method

• Methods for solving the transportation problem itself include:

- stepping-stone method and

- modified distribution method.

The Northwest Corner Method

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To From

A B C Supply

1 6 8 10

150

2 7 11 11

175

3 4 5 12

275

Demand 200 100 300 600

- In the northwest corner method the largest possible allocation is made to the cell in the upper left-hand corner of the tableau , followed by allocations to adjacent feasible cells.

150<200




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To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

175

3 4 5 12

275

Demand 50 100 300


200-150=50

150-150=0


12

To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

175

3 4 5 12

275

Demand 50 100 300


50<175




13

To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

125 50

3 4 5 12

275

Demand 0 100 300


50-50=0

175-50=125


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To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

125 50

3 4 5 12

275

Demand 0 100 300


100<125




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To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

25 50 100

3 4 5 12

275

Demand 0 0 300


100-100=0

125-100=25


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To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

25 50 100

3 4 5 12

275

Demand 0 0 300


25<300




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To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

0 50 100 25

3 4 5 12

275

Demand 0 0 275


300-25=275

25-25=0


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To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

0 50 100 25

3 4 5 12

275

Demand 0 0 275


275=275




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To From

A B C Supply

1 6 8 10

0 150

2 7 11 11

0 50 100 25

3 4 5 12

0 275

Demand 0 0 0


275-275=0

275-275=0

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- The initial solution is complete when all rim requirements are satisfied.

- Transportation cost is computed by evaluating the objective function:

Z = $6x1A + 8x1B + 10x1C + 7x2A + 11x2B + 11x2C + 4x3A + 5x3B + 12x3C

= 6(150) + 8(0) + 10(0) + 7(50) + 11(100) + 11(25) + 4(0) + 5(0) + 12(275)

= $5,925

The Initial NW Corner Solution



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Summary of Steps

1. Allocate as much as possible to the cell in the upper left-hand corner, subject to the supply and demand conditions.

2. Allocate as much as possible to the next adjacent feasible cell.

3. Repeat step 2 until all rim requirements are met.

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The Minimum Cell Cost Method (1 of 3)

- In the minimum cell cost method as much as possible is allocated to the cell with the minimum cost followed by allocation to the feasible cell with minimum cost.

To From

A B C Supply

1 6 8 10

150

2 7 11 11

175

3 4 5 12

275

Demand 200 100 300 600



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To From

A B C Supply

1 6 8 10

150

2 7 11 11

175

3 4 5 12

75 200

Demand 0 100 300

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To From

A B C Supply

1 6 8 10

150

2 7 11 11

175

3 4 5 12

75 200

Demand 0 100 300



25



To From

A B C Supply

1 6 8 10

150

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 25 300

26



To From

A B C Supply

1 6 8 10

150

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 25 300



27



To From

A B C Supply

1 6 8 10

150

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 25 300

28



To From

A B C Supply

1 6 8 10

125 25

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 0 300



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To From

A B C Supply

1 6 8 10

125 25

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 0 300

30



To From

A B C Supply

1 6 8 10

125 25

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 0 300



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To From

A B C Supply

1 6 8 10

0 25 125

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 0 175

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To From

A B C Supply

1 6 8 10

0 25 125

2 7 11 11

175

3 4 5 12

0 200 75

Demand 0 0 175



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To From

A B C Supply

1 6 8 10

0 25 125

2 7 11 11

0 175

3 4 5 12

0 200 75

Demand 0 0 0

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- The complete initial minimum cell cost solution; total cost = $4,550.

- The minimum cell cost method will provide a solution with a lower cost than the northwest corner solution because it considers cost in the allocation process.

The Initial Solution



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The Minimum Cell Cost Method Summary of Steps

(3 of 3)

1. Allocate as much as possible to the feasible cell with the minimum transportation cost, and adjust the rim requirements.

2. Repeat step 1 until all rim requirements have been met.

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Vogel’s Approximation Method (VAM) (1 of 5)

- Method is based on the concept of penalty cost or regret.

- In VAM the first step is to develop a penalty cost for each source and destination.

- Penalty cost is calculated by subtracting the minimum cell cost from the next higher cell cost in each row and column.

The VAM Penalty Costs



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To From

A B C Supply

Row Penalty Costs

1 6 8 10

150 (8-6)=2

2 7 11 11

175 (11-7)=4

3 4 5 12

275 (5-4)=1

Demand 200 100 300 600

Column Penalty Costs

(6-4)=2 (8-5)=3 (11-10)=1

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To From

A B C Supply

Row Penalty Costs

1 6 8 10

150 (8-6)=2

2 7 11 11

0 - 175

3 4 5 12

275 (5-4)=1

Demand 25 100 300


(6-4)=2 (8-5)=3 (12-10)=2



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To From

A B C Supply

Row Penalty Costs

1 6 8 10

150 (10-6)=4

2 7 11 11

0 - 175

3 4 5 12

175 (12-4)=8 100

Demand 25 0 300


(6-4)=2 - (12-10)=2

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To From

A B C Supply

Row Penalty Costs

1 6 8 10

150 -

2 7 11 11

0 - 175

3 4 5 12

150 - 25 100

Demand 0 0 300


- - (12-10)=2



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To From

A B C Supply

Row Penalty Costs

1 6 8 10

0 - 150

2 7 11 11

0 - 175

3 4 5 12

150 - 25 100

Demand 0 0 150


- -

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To From

A B C Supply

Row Penalty Costs

1 6 8 10

0 - 150

2 7 11 11

0 - 175

3 4 5 12

0 - 25 100 150

Demand 0 0 0


- -



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Vogel’s Approximation Method (VAM) To

From

A B C Supply Row Penalty Costs

1 2 3 4

1 6 8 10 150

0 2 2 4 - 150

2 7 11 11 175

0 4 - - - 175

3

4 5 12 275 175 150

0

1 1 8 - 25 100 150

Demand 200 25 0

100 0

300 150

0 600


1 2 3 1

2 2 3 2

3 2 - 2

4 - - 2

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- The initial VAM solution; total cost = $5,125

- VAM and minimum cell cost methods both provide better initial solutions than does the northwest corner method.

The Initial VAM Solution



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Vogel’s Approximation Method (VAM) Summary of Steps

1. Determine the penalty cost for each row and column.

2. Select the row or column with the highest penalty cost.

3. Allocate as much as possible to the feasible cell with the lowest transportation cost in the row or column with the highest penalty cost.

4. Repeat steps 1, 2, and 3 until all rim requirements have been met.

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The Stepping-Stone Solution Method (1 of 12)

- Once an initial solution is derived, the problem must be solved using either the stepping-stone method or the modified distribution method (MODI).

- The initial solution used as a starting point in this problem is the minimum cell cost method solution because it had the minimum total cost of the three methods used.

The Minimum Cell Cost Solution



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The Stepping-Stone Solution Method

To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 200 75

Demand 200 100 300 600

• 1A - 2A-2B-3C

- The stepping-stone method determines if there is a cell with no allocation that would reduce cost if used.

- An empty cell that will reduce cost is a potential entering variable.

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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 200 75

Demand 200 100 300 600

• 1A=6-8+5-4= -1 • 2A= • 2B= • 3B=

+1 -1

+1 -1

• Must subtract one ton from another allocation along that row.

• A requirement of this solution method is that units can only be added to and subtracted from cells that already have allocations, thus one ton must be added to a cell as shown.



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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 200 75

Demand 200 100 300 600

• 1A=6-8+5-4= -1 • 2A=7-11+10-8+5-4= 22-23= -1 • 2B= • 3B=

+1

-1 +1

-1

+1 -1

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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 200 75

Demand 200 100 300 600

• 1A=6-8+5-4= -1 • 2A=7-11+10-8+5-4= 22-23= -1 • 2B= 11-11+10-8= 2 • 3B=

+1

-1 +1

-1



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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 200 75

Demand 200 100 300 600

• 1A=6-8+5-4= -1 • 2A=7-11+10-8+5-4= 22-23= -1 • 2B= 11-11+10-8= +2 • 3B=12-10+8-5= +5

+1

-1 +1

-1

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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 200 75

Demand 200 100 300 600

• 1A=6-8+5-4= -1 • 2A=7-11+10-8+5-4= 22-23= -1 • 2B= 11-11+10-8= +2 • 3B=12-10+8-5= +5

- After all empty cells are evaluated, the one with the greatest cost reduction potential is the entering variable.

- A tie can be broken arbitrarily.



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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 200 75

Demand 200 100 300 600

• 1A=6-8+5-4= -1 • 2A=7-11+10-8+5-4= 22-23= -1 • 2B= 11-11+10-8= +2 • 3B=12-10+8-5= +5

- When reallocating units to the entering variable (cell), the amount is the minimum amount subtracted on the stepping-stone path.

- At each iteration one variable enters and one leaves (just as in the simplex method).

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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 175 100

Demand 200 100 300 600

- Check to see if the solution is optimal.



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To From

A B C Supply

1 6 8 10

150 25 125

2 7 11 11

175 175

3 4 5 12

275 175 100

Demand 200 100 300 600

• 1B=8-5+4-6= +1

• 2A=7-11+10-6=0

• 2B=11-11+10-6+4-5=+3

• 3C=12-10+6-4=+4

• The stepping-stone process is repeated until none of the empty cells will reduce costs (i.e., an optimal solution).

• In example, evaluation of four paths indicates no cost reductions, therefore solution is optimal.

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- Solution and total minimum cost :

x1A = 25 tons, x2C = 175 tons, x3A = 175 tons, x1C = 125 tons, x3B = 100 tons

Z = $6(25) + 8(0) + 10(125) + 7(0) + 11(0) + 11(175) + 4(175) + 5(100) + 12(0)

= $4,525



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- A multiple optimal solution occurs when an empty cell has a cost change of zero and all other empty cells are positive.

- An alternate optimal solution is determined by allocating to the empty cell with a zero cost change.

- Alternate optimal total minimum cost also equals $4,525.

The Alternative Optimal Solution

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The Stepping-Stone Solution Method Summary of Steps

1. Determine the stepping-stone paths and cost changes for each empty cell in the tableau.

2. Allocate as much as possible to the empty cell with the greatest net decrease in cost.

3. Repeat steps 1 and 2 until all empty cells have positive cost changes that indicate an optimal solution.



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The Modified Distribution Method (MODI) (1 of 6)

- MODI is a modified version of the stepping-stone method in which math equations replace the stepping-stone paths.

- In the table, the extra left-hand column with the ui symbols and the extra top row with the vj symbols represent values that must be computed.

- Computed for all cells with allocations :

ui + vj = cij = unit transportation cost for cell ij.

The Minimum Cell Cost Initial Solution

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The Modified Distribution Method (MODI) (2 of 6) - Formulas for cells containing allocations:

x1B: u1 + vB = 8

x1C: u1 + vC = 10

x2C: u2 + vC = 11

x3A: u3 + vA = 4

x3B: u3 + vB = 5

- Five equations with 6 unknowns, therefore let u1 = 0 and solve to obtain:

vB = 8, vC = 10, u2 = 1, u3 = -3, vA= 7

The Initial Solution with All ui and vj Values



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- Each MODI allocation replicates the stepping-stone allocation.

- Use following to evaluate all empty cells:

cij - ui - vj = kij

where kij equals the cost increase or decrease that would occur by allocating to a cell.

- For the empty cells in Table 26:

x1A: k1A = c1A - u1 - vA = 6 - 0 - 7 = -1

x2A: k2A = c2A - u2 - vA = 7 - 1 - 7 = -1

x2B: k2B = c2B - u2 - vB = 11- 1 - 8 = +2

x3C: k3C = c3C - u3 -vC = 12 - (-3) - 10 = +5

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- After each allocation to an empty cell, the ui and vj values must be recomputed.

The Second Iteration of the MODI Solution Method



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The Modified Distribution Method (MODI) (5 of 6) - Recomputing ui and vj values:

x1A: u1 + vA = 6, vA = 6 x1C: u1 + vC = 10, vC = 10 x2C: u2 + vC = 11, u2 = 1

x3A: u3 + vA = 4, u3 = -2 x3B: u3 + vB = 5, vB = 7

The New ui and vj Values for the Second Iteration

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- Cost changes for the empty cells, cij - ui - vj = kij;

x1B: k1B = c1B - u1 - vB = 8 - 0 - 7 = +1

x2A: k2A = c2A - u2 - vA = 7 - 1 - 6 = 0

x2B: k2B = c2B - u2 - vB = 11 - 1 -7 = +3

x3C: k2B = c2B - u3 - vC = 12 - (-2) - 10 = +4

- Since none of the values are negative, solution obtained is optimal.

- Cell 2A with a zero cost change indicates a multiple optimal solution.



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The Modified Distribution Method (MODI) Summary of Steps

1. Develop an initial solution.

2. Compute the ui and vj values for each row and column.

3. Compute the cost change, kij, for each empty cell.

4. Allocate as much as possible to the empty cell that will result in the greatest net decrease in cost (most negative kij)

5. Repeat steps 2 through 4 until all kij values are positive or zero.

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The Unbalanced Transportation Model (1 of 2)

- When demand exceeds supply a dummy row is added to the tableau.

An Unbalanced Model (Demand . Supply)



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The Unbalanced Transportation Model (2 of 2)

- When supply exceeds demand, a dummy column is added to the tableau.

- The dummy column (or dummy row) has no effect on the initial solution methods or the optimal solution methods.

An Unbalanced Model (Supply . Demand)

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Degeneracy (1 of 3)

- In a transportation tableau with m rows and n columns, there must be m + n - 1 cells with allocations; if not, it is degenerate.

- The tableau in the figure does not meet the condition since 3 + 3 -1 = 5 cells and there are only 4 cells with allocations.

The Minimum Cell Cost Initial Solution



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Degeneracy (2 of 3)

- In a degenerate tableau, all the stepping-stone paths or MODI equations cannot be developed.

-To rectify a degenerate tableau, an empty cell must artificially be treated as an occupied cell.

The Initial Solution

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Degeneracy (3 of 3)

- The stepping-stone path s and cost changes for this tableau:

2A 2C 1C 1A

x2A: 7 - 11 + 10 - 6 = 0

2B 2C 1C 1B

x2B: 11 - 11 + 10 - 8 = + 2

3B 1B 1A 3A

x3B: 5 - 8 + 6 - 4 = - 1

3C 1C 1A 3A

x3C: 12 - 10 + 6 - 4 = + 4

The Second Stepping-Stone Iteration



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Prohibited Routes

- A prohibited route is assigned a large cost such as M.

- When the prohibited cell is evaluated, it will always contain the cost M, which will keep it from being selected as an entering variable.

Limitations of Transportation Problem

• One commodity ONLY: any one product supplied and demanded at multiple locations – Merchandise – Electricity, water

• Invalid for multiple commodities: (UNLESS transporting any one of the multiple commodities is completely independent of transporting any other commodity and hence can be treated by itself alone) – Example: transporting product 1 and product 2 from the

supply points to the demand points where the total amount (of the two products) transported on a link is subject to a capacity constraint

– Example: where economy of scale can be achieved by transporting the two products on the same link at a larger total volume and at a lower unit cost of transportation



Limitations of Transportation Problem

– Difficult to generalize the technique to accommodate (these are generic difficulty for “mathematical programming,” including linear and non-linear programming

• Economy of scale the per-unit cost of transportation on a link decreasing with the volume (nonlinear and concave; there is a trick to convert a “non-linear program with a piecewise linear but convex objective function to a linear program; no such tricks exists for a piecewise linear but concave objective function)

• Fixed-cost: transportation usually involves fixed charges. For example, the cost of truck rental (or cost of trucking in general) consists of a fixed charge that is independent of the mileage and a mileage charge that is proportional to the total mileage driven. Such fixed charges render the objective function NON-LINEAR and CONCAVE and make the problem much more difficult to solve

Assignment:

• Consider a transportation problem with three ware houses and four markets. The ware house capacities are W1=3, W2=7 and W3=5. The market demands are M1=4, M2=3, M3=4 and M4=4. The unit cost is given by the following table –

a) Design an initial shipping program by northwest corner method

b) Design an initial shipping program by minimum cell cost method

c) Design an initial shipping program by Vogel’s Approximation Method

d) Find the optimum shipping program

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M1 M2 M3 M4

W1 2 2 2 1

W2 10 8 5 4

W3 7 6 6 8

Education

Transportation lecture