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Transportation Problem

Its is also known as distribution problem

it arises when there are no. of sources each given with quantity of product or

capacity

And there are a no. of destinations to which the product is to be transported

The transportation problem gives an optimum plan which results in least cost

for transportation

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The Transportation Table

The transportation table is set up with the following steps:

Step 1: set up transportation table with m rows (for source) & n column (for

destination)

Step 2: Add an additional row, (m + 1)throw, as demand & an

additional column, (n + 1)thcolumn, as Supply / capacity

Step 3: Enter individual figures of demand in the row titled demand & individual

figures of capacity in the column titled capacity

Step 4: Sum up the capacityfigures and demandfigures (which must tally) and enter

their respective totals at the intersection of capacity column and demand row

Step 5: Insert per unit shipping cost in the lower left hand corners of the cells

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Eg. Daily production of Spice mobilesin terms of units produced

varies from factory to factory is given below:

Factory Capacity: Chandigarh Gurgaon Kanpur

Units / day 30 40 50

Distribution centre: Jaipur Kolkata Delhi Chennai

Demand(units/day) 35 28 32 25

The costs of transportation (Rs.) of each mobile is given as under:

Factory Jaipur Kolkata Delhi Chennai

Gurgaon 5 11 9 7

Chandigarh 6 8 8 5

Kanpur 8 9 7 13

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Step 1: Objective is to minimize the costs in the transportation

Step 2: Set up a Transportation Table

5 11 9 7

6 8 8 5

8 9 7 13

35 28 32 25

30

40

50

Capacity

Demand

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Step 3: Develop an initial basic feasible solution

1. North West Corner Method

Step 1: begin with North west corner cell (the upper left hand corner) of the

transportation table

Step 2: Allocate as many units as possible such that the minimumof

demand or capacitycomes in that cell & total of either is also justified

Step 3: If the demand in the column is satisfied move to the right cell in the

next columnorif the capacity for the row is exhausted , move down to

the cell in the next row.

Step 5: Go to Step 2 and Repeat until the demand in the column or thecapacity in the row are exhausted completely

Step 6: if both demand and capacity are exhausted before , then there is a

tie for the next allocation. Make the next allocation of value in the

cell in either the column or the next row

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5 11 9 7

6 8 8 5

8 9 7 13

35 28 32 25

30

40

50

Capacity

Demand

30

5 28 7

25 25

Hence total cost = 30x 6 + 5x 5 + 28x 11 + 7x 9 + 25x 7 + 25x 13

= 1076

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2. Least Cost Method

Step 1: Determine the lowest (smallest ) cost among all the rows of the

transportation table

Step 2: Identify the row and allocate the maximum feasible quantity (minimum

of the demand and capacity)in the box corresponding to the smallestcost in the row , then eliminate that row (column) where an allocation

is made and demand column or capacity row is satisfied completely

Step 3: Repeat sets 1 & 2 for the reduced transportation table until all the

requirements are satisfied. Whenever the minimum cost is more thanone than make an arbitrary choice among the minimum costs

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5 11 9 7

6 8 8 5

8 9 7 13

35 28 32 25

30

40

50

Capacity

Demand

25

35

32

5

18

5

Hence total cost = 25x 5 + 5x 8 + 35x 5 + 5x 11 + 18x 9 +7x 32

= 781

The least cost is more economical than NWC method as it gives a

lower cost.

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3. Vogels Approximation Method

Step 1: For each row of the transportation table, identify the smallest andnext to smallest costs. Calculate difference between these two for

each row and column

Step 2: select the row or the column with the largest difference (penalty)

Step 3: allocate the maximum feasible quantity(minimum of the demandand capacity)in the cell with the minimum costin the selected row or

column

Step 4: eliminate (cross out) that row or column, where an allocation is made

Step 5: re-calculate row and column difference for each row and each column

of the reduced transportation table

Step 6: go to step 2 and repeat the procedure until all the requirements are

satisfied

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5 11 9 7

6 8 8 5

8 9 7 13

35 28 32 25

30

40

50

Capacity

Demand

1Column Diff 1 1 2

Row

Diff

1

2

1

35

30

5

50

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11 9 7

8 8 5

9 7 13

28 32 25

30

5

50

Capacity

Demand

1Column Diff 1 2

Row

Diff

3

2

2

25 5

5

50

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11 9

8 8

9 7

28 32

5

5

50

Capacity

Demand

1Column Diff 1

Row

Diff

0

2

232

5

5

18

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11

8

9

28

5

5

18

Capacity

Demand

5

5

18

0

0

0

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5 11 9 7

6 8 8 5

8 9 7 13

35 28 32 25

30

40

50

Capacity

Demand

35

5 25

5

18 32

Hence total cost = 25x 5 + 5x 8 + 35x 5 + 5x 11 + 18x 9 +7x 32

= 781

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Step 4: Examine the initial solution for feasibility. The solution is feasible if

it has allocations in m + n - 1 cells. Here 3 + 41 = 6 cells are

occupying the allocationshence the solutions has 6 feasible

solutions, where m = rows & n = columns in transportation table.

Step 5: test the initial basic feasible solution for OPTIMALITY, which can be done

by either of the following:

- Stepping stone Method

- Modified distribution method (or MODI method)

Modified distribution method (or MODI method)

Step 1: For all the occupied cells, i.e where we have made the allocations

during Vogel's Approximation method, determine a set of no.s uifor

each row and vjfor each column by solving the system of equations

ui+ vj= cij. We can assign value for u1= 0 (arbitrary value) and

calculate the other values based on it.

Step 2: calculate the opportunity costs for all the unoccupied cells by

using the relationship :

ij= ui+ vj- cij

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Step 3: check the sign of each opportunity cost. If all the opportunity cost

are non positive, an OPTIMUMsolution has been reached.

Otherwise go to the next step.

Step 4: select the largest positiveopportunity cost calculated in step 4. the

unoccupied (non basic) cell corresponding to this + ve value

becomes occupied in the next iteration

Step 6: Assign alternate + & - signs at the corner points of occupied cells

on the closed path starting with + sign at the current entering cell

(unoccupied cell)

Step 5: Determine a closed path for the current entering cellthat starts and

ends at this unoccupied cell. Take Right angle turns (90o)in this closed

path only at the occupied and current entering cell

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Step 7: Determine the least no. of units that can be allocated to the occupied

cells havingsign.Add this quantity to all the cells on the closed

path marked with a (+) sign and subtract the same from all the cells

marked with (-) sign.

Step 8: Go to step 3 & repeat the procedure until an optimum solution is

reached

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5 11 9 7

6 8 8 5

8 9 7 13

2 8 6 5

0

3

1

ui

vj

35

5 25

5

18 32

1

2

3

1 2 3 4

Taking u1= 0; & putting values (costs) of occupied cells

u1+ v2= c12 ; v2= 8

u1+ v4= c14 ; v4 = 5

u2+v1= c21; v1= 2

u2+v2= c22; u2 = 3

u3+v2= c32 ; u3 = 1

u3+v3= c33 ; v3= 6

Unoccupied cells are: c11, c13, c23, c24, c31, c34

Occupied cells are: c12, c14, c21, c22, c32, c33

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5 11 9 7

6 8 8 5

8 9 7 13

2 8 6 5

0

3

1

ui

vj

35

5 25

5

18 32

1

2

3

1 2 3 4

ij= ui+ vj - cij

Finding opportunity costs for all unoccupied cells

11= u1+ v1 - c11 = 0 + 26 = - 4

13= u1+ v3 - c13 = 0 + 68 = - 2

23= u2+ v3c23 = 3 + 69 = 0

24= u2+ v4c24 = 3 + 57 = 1

31= u3+ v1c31 = 1 + 28 = - 5

34= u3+ v4c34 = 1 + 513 = - 7

+ ve no.

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Assignment Problems

Its is a special case of transportation problems wherein the number of

resources (origins) equal no. of activities (destinations).

The capacity demand value is exactly one unit, i.e. only one unit can be

supplied from each origin and each destination also requires exactly one unit.

The objective is to determine which origin should supply one unit to which

destination so that the total cost is minimum.

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Solution procedure for assignment problem (Hungarian Method)

Step 1: determine the cost matrix from the given problem

i) if the no. of jobs equal to no. of facilities, go to step 3

ii) if the no. of jobs does not equal the no. of facilities go to step 2

Step 2: Add a dummy job or dummy facility so that the cost matrix becomes a

square matrix. The cost entries of the dummy job / facility are always zero.Step 3: select the smallest element in each row of the given cost matrix and then

subtract the same from each element to that row.

Step 4: in the reduced matrix obtained in step 3, locate the smallest value of

each column & then subtract the same from each element of that column.

Each column & row now have at least one zero.

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iii) If a row / column has two or more zeroes & one cant be chosen by inspection

then assign arbitrarily any of these zeroes & cross off all other zeroes of that row

/ column

iv) Repeat step i) to iii) above successively until the chin of assigning orcross X ends

Step 6: if the no. of assignments are equal to n (the order of the cost matrix), anoptimum solution is reached

But if the assignments are less than n (the order of the matrix), go to step 7.

Step 7: draw the minimum no. of horizontals & / or verticals lines to cover all the

zeroes of the reduced matrixStep 8: develop the new revised cost matrix as follows:

i) find the smallest value of the reduced matrix not covered by any of the lines

ii) subtract this value from all the uncovered values and add the same to all the

values lying at the intersection of any two lines.

Step 9: go to step 5 and repeat the procedure until an optimum solution is obtained

Step 5: in the modified matrix obtained in step 4, search for an optimal

assignment as follows:

i) examine the rows successively until a row with a single zero is found. Make an

assignment indicated by to this zero and cross out X all other zeros in the

column.ii) Repeat the procedure for each column of the reduced matrix.

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40 30 40 30

50 40 60 20

60 20 30 20

10 20 10 30

30 30 20 30

Its a matrix of 5 x 4 order so we will add a dummy to it

1 2 3 4ContractorsHangars

A

B

C

D

E

Bids from Contractors (in lacs Rs.)

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40 30 40 30 0

50 40 60 20 0

60 20 30 20 0

10 20 10 30 0

30 30 20 30 0

Dummy

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30 10 30 10 0

40 20 50 0 0

50 0 20 0 0

0 0 0 10 0

20 10 10 10 0

First looking in the rows that have single zero

Now no row left behind which has any single zero in

itself then go for column

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To draw the minimum no. of horizontals & or verticals lines to cover all

the zeroes of the reduced matrix

1. Mark () rows that do not have any assigned zeroes

2. Then Mark () columns that have zeroes in the marked rows

3. And in that marked columnfinally mark () rows that have assigned

zeroes.4. Draw lines through all the marked columns& unmarked rows.

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30 10 30 10 0

40 20 50 0 0

50 0 20 0 0

0 0 0 10 0

20 10 10 10 0

Subtracting the least uncoveredvalue (10) from all the values which

are NOT cross off by lines& adding it at the intersection of the lines

Again going to Step 5 & repeating the procedure

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20 0 20 0 0

40 20 50 0 10

50 0 20 0 10

0 0 0 10 10

10 0 0 0 0

Again going to Step 5 & repeating the procedure

First looking in the rows that have single zero

Now since the no. of assignments are 5 and the order is also 5 , the

optimal solution has been reached

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199

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151

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6,

2007

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2007

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2007

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2007

Januar

y 3,

2008

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ry 2,

2008

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3,

2008

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2008

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2008

June

1,

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Chemical A

Chemical B

Chemical C

idle Time days completed days remaining