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Understanding and Optimization of Transportation Problem

Dr. Ramesh Kumar Chaturvedi

BBAU, Lucknow

What is Transportation Problem?

• A special case of Linear Programming. • In 1941 by FL Hitchcock, then in 1947 by TC Koopmans

in a study titled “Optimum utilization of Transportation system”

• RIM conditions • Destination demand and Supply • Minimization of transportation cost • Unit cost of transportation is known and fixed • Cost is proportional to number of units transported • Important terms: Initial Basic Feasible solution,

Optimal solution, Degeneracy, Balanced problem

A Transportation Problem

Project A B C Availability

1 2 7 4 50

2 3 3 1 80

3 5 4 7 70

4 1 6 2 140

Demand 70 90 180

Conditions to be fulfilled for IBFS

• Number of allocation should be m+n-1

• All allocations must be independent

Methods

• NWS

• LCM

• VAM

A Transportation Problem solution by NWC

Project A B C Availability

1 2 (50) 7 4 50

2 3 (20) 3(60) 1 80

3 5 4(30) 7(40) 70

4 1 6 2(140) 140

Demand 70 90 180 340

Total cost=1020

A Transportation Problem solution by LCM

Project A B C Availability

1 2 7(20) 4(30) 50

2 3 3 1(80) 80

3 5 4(70) 7 70

4 1(70) 6 2(70) 140

Demand 70 90 180 340

Total cost=830

Steps in VAM

• Construct Transportation Table

• For each row calculate the difference between two lowest cost entries.

• For each column calculate the difference between two lowest cost entries.

• Select the row or column that has largest difference

• In row or column identified in above step select the cell that has lowest cost entry.

• Assign the maximum possible number of units to the cell selected in step 5.

• Reapply the steps 2 to 6 iteratively till total demand is met by supply

A Transportation Problem solution by VAM

Project A B C Availability

1 2 7(20) 4(30) 50

2 3 3 1(80) 80

3 5 4(70) 7 70

4 1(70) 6 2(70) 140

Demand 70 90 180 340

Total cost=830

Variations of Transportation problem

• Degeneracy in transportation problem when No. of allocations is less than m+n-1

• Unbalanced problem

• Converting a maximization problem into minimization problem

Dealing Degeneracy

A B c d E F

1 9 12 9 6 9 10 5

2 7 3 7 7 5 5 6

3 6 5 9 11 3 11 2

4 6 8 11 2 2 10 9

4 4 6 2 4 2 22

Solve the problem with VAM

Dealing Degeneracy

A B c d E F

1 9 12 9(5) 6 9 10 5

2 7 3(4) 7 7 5 5(2) 6

3 6 (1) 5 9(1) 11 3 11 2

4 6(3) 8 11 2(2) 2(4) 10 9

4 4 6 2 4 2 22

Is Number of allocations is less than m+n-1?

Dealing Degeneracy

A B c d E F

1 9 12 9(5) 6 9 10 5

2 7 3(4) 7 7 5 (€) 5(2) 6

3 6 (1) 5 9(1) 11 3 11 2

4 6(3) 8 11 2(2) 2(4) 10 9

4 4 6 2 4 2 22

Solve Degeneracy by assigning € in unoccupied cell with smallest cost, yet closed loop is not

formed.

Problem with degeneracy

Project A B C Availability

1 2 7 4 50

2 3 8 1 80

3 6 4 7 70

4 1 6 2 120

Demand 50 90 180

Solution of Problem with degeneracy (By NWC)

Project A B C Availability

1 2(50) 7 4 50

2 3(€) 8(80) 1 80

3 6 4(10) 7(60) 70

4 1 6 2(120) 120

Demand 50 90 180

Note: Degeneracy may occur in subsequent iterations while optimizing the IBFS

Unbalanced Problem

Total Demand ≠ Total Supply

No feasible solutions can be obtained for such problems

1. In case total demand is more than supply than dummy source is added to fulfil the Balance requirement.

2. In case total supply is more than total demand than dummy destination is added to fulfil the Balance requirement.

An Unbalanced Transportation Problem

Is this Problem Unbalanced?

Project A B C Availability

1 2 7 4 50

2 3 8 1 80

3 6 4 7 70

4 1 6 2 140

Demand 90 90 180

Using Dummy for an Unbalanced Transportation Problem

Project A B C Availability

1 2 (50) 7 4 50

2 3 3 1 80

3 5 4 7 70

4 1 6 2 140

Dummy 0 0 0 20

Demand 90 90 180

Converting a maximization problem into minimization problem

1. Subtract all the costs in transportation table in from highest cost.

2. The Table so obtained is Now transformed into minimization problem and can be solve with methods discussed earlier.

Is the solution (IBFS) best?

• Decision making

• Concept of Bounded rationality in decision making

Types of decision situation

1. Hobson's Choice

2. Certainty

3. Uncertainty

4. Risk

Decision Making

Simon’s Model of decision Making

Is it possible to execute Simons model practically?

Concept of Bounded Rationality

and

Optimality

Optimality in Transportation Problem

Stepping Stone Method

MODI Method

Optimality in Transportation Problem - Stepping Stone Method

•This method involves evaluation of each unoccupied cell for opportunity cost. •If opportunity cost of any unoccupied cell is positive (if more than one unoccupied cell has positive opportunity cost, then the cell with maximum positive OC), then this cell is bought in the solution. Note: Since Each unoccupied cell is to be evaluated for OC then on with maximum OC is selected and then re allocation is to be done, this method becomes cumbersome, hence not much in use. (M-1)(N-1) unoccupied cells are there that has to be evaluated. i.e. mn – (m+n-1) cells

MODI Method Modified Distribution

•MODI method is more apt in optimizing Transportation problem.

•Opportunity cost for unoccupied cells can be calculated arithmetically.

•Only one cell is evaluated by forming the close loop path to improve the solution

Steps in MODI Method 1. Setup IBFS by NWS/LCM/VAM

2. Introduce dual variables corresponding to supply and demand constraints. If there are m supply and n demand points then m+n dual variables will be used. Let for row ui (i=1,2,…m) and for column vj be (j=1,2,…n) be dual variables corresponding to supply and demand constraints. Cij Is the cost in a cell in transportation table.

3. Variables ui and vj are such that, Opportunity cost(OC) = ui + vj - cij .

4. Now, ui + vj = cij for all occupied cells because opportunity cost for occupied cells are Zero.

5. Since the number of dual variables is m+n and number of occupied cells is m+n-1, one variable is assumed arbitrarily, say u1=0

6. After assigning u1=0; Calculate the remaining ui and vj values from the relationship ui + vj = cij

Steps cont…

7. Calculate the opportunity cost for all unoccupied cells from the relationship ui + vj - cij .

8. If opportunity cost of all unoccupied cells is zero or negative than an optimal solution is achieved.

9. In case cells have positive opportunity cost, select the cell with maximum OC. Starting from the selected cell and moving only along horizontal or vertical lines trace a closed path back to the selected cell such a way all corners of this path are occupied cells. Beginning with this cell assign positive (+) and negative (-) signs alternatively to all corners (the occupied cells) of the closed path.

10. v Determine the smallest quantity in a negative position on the closed path. Add this quantity to all corners with positive sign and subtract it from all corners with negative sign. This gives an improved solution.

11. Check this solution for optimality by going through with step 7.

Solving for optimality an IBFS obtained through NWC

Transportation cost is Rs. 1020. Is it Optimal?

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 4 50

u2 2 3 (20) 3(60) 1 80

u3 3 5 4(30) 7(40) 70

u4 4 1 6 2(140) 140

Demand 70 90 180 340

Solving for Optimality

• From all occupied cells calculate the values of u1 and v1 from ui + vj = cij ; assuming u1=0, we get

as follows:

For cell 1A; 2 = u1 + v1; Since u1 = 0, v1=2

For cell 2A; 3= u2 +v1; since v1=2, u2=1

For cell 2B; 3= u2 +v2; since u2=1, v2=2

For cell 3B; 4= u3 +v2; since v2=2, u3=2

For cell 3c; 7= u3 +v3; since u3=2, v3=5

For cell 4c; 2= u4 +v3; since v3=5, u4=-3

Solving for Optimality cont…

• Let us now calculate OC from all unoccupied cells as follows:

For cell 1B; u1 + v2 – C12 = 0+2-7 = -5

For cell 1C; u1 + v3 – C13 = 0+5-4 = 1

For cell 2C; u2 + v3 – C23 = 1+5-1 = 5

For cell 3A; u3 + v1 – C31 = 2+2-5 = -1

For cell 4A; u4 + v1 – C41 = -3+2-1 = -2

For cell 4B; u4 + v2 – C42 = -3+2-6 = -7

Solving for optimality Solution 1

Cell 2C has +5 OC, This solution is not optimal. Hence a closed path is traced from this cell to get improved solution

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 (-5) 4 (+1) 50

u2 2 3 (20) 3(60) 1(+5) 80

u3 3 5 (-1) 4(30) 7(40) 70

u4 4 1(-2) 6 (-7) 2(140) 140

Demand 70 90 180 340

Solving for optimality

40 at cell 3c is the smallest quantity at –ve corner, this will be added and subtracted to +ve

and –ve corners to get new solution.

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 (-5) 4 (+1) 50

u2 2 3 (20) 3(60) _ 1(+5) + 80

u3 3 5 (-1) 4(30) + 7(40) _ 70

u4 4 1(-2) 6 (-7) 2(140) 140

Demand 70 90 180 340

Improved solution 2

Now check this solution for optimality

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 4 50

u2 2 3 (20) 3(20) 1(40) 80

u3 3 5 4(70) 7 70

u4 4 1 6 2(140) 140

Demand 70 90 180 340

Solving for Optimality cont…

• Let us now calculate OC from all unoccupied cells as follows:

• Recalculate the values of u1 and V1 from relationship ui + vj = cij ; assuming u1=0, we get v1=2, v2=2 and v3=0 similarly u2=1, u3=2, u4=2.

• Opportunity cost as follows: • For cell 1B = -5 • For cell 1C; = -4 • For cell 3C; = -5 • For cell 3A; = -1 • For cell 4A; = +3 • For cell 4B; = -2

Calculating opportunity cost for Improved solution 2 we get…..

Cell 4A is still have +3 as OC, hence this needs to be improved, by tracing a closed path from this cell and

re-allocating the supplies to get nest improved solution.

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 (-5) 4 (-4) 50

u2 2 3 (20) 3(20) 1(40) 80

u3 3 5 (-1) 4(70) 7(-5) 70

u4 4 1 (+3) 6 (-2) 2(140) 140

Demand 70 90 180 340

Calculating opportunity cost for Improved solution 2 we get…..

20 at 2A cell is the smallest quantity at –ve corner, this will be added and subtracted to +ve

and –ve corners to get new solution.

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 (-5) 4 (-4) 50

u2 2 3 (20) _ 3(20) 1(40) + 80

u3 3 5 (-1) 4(70) 7(-5) 70

u4 4 1 (+3) + 6 (-2) 2(140) _ 140

Demand 70 90 180 340

Improved solution 3

Check this for optimality

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 4 50

u2 2 3 3(20) 1(60) + 80

u3 3 5 4(70) 7 70

u4 4 1 (20) 6 2(120) 140

Demand 70 90 180 340

Solving for Optimality cont…

• Let us now calculate OC from all unoccupied cells as follows:

• Recalculate the values of u1 and V1 from relationship ui + vj = cij ; assuming u1=0, we get v1=2, v2=5 and v3=3 similarly u2=-2, u3=-1, u4=-1.

• Opportunity cost as follows: • For cell 1B = -2 • For cell 1C; = -1 • For cell 2A; = -3 • For cell 3A; = -4 • For cell 3C; = -5 • For cell 4B; = -2

Optimality test for Improved solution 3

All opportunity costs are Negative, hence this is optimal solution with transportation cost of RS. 760.

v1 v2 v3 v4

Project A B C Availability

u1 1 2 (50) 7 (-2) 4 (-1) 50

u2 2 3 (-3) 3(20) 1(60) 80

u3 3 5 (-4) 4(70) 7(-5) 70

u4 4 1 (20) 6 (-2) 2(120) 140

Demand 70 90 180 340

Thank you

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