Solved Example Transportation Problem Help for Transportation Model, Management, Homework Help -...

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EXAMPLE FOR SOLUTION OF TRANSPORTATION PROBLEM

An organization has four destinations and three sources for supply of goods. The transportation cost per unit is

given below. The entire availability is 700 units which exceeds the cumulative demand of 600 units. Decide the

optimal transportation scheme for this case.

Solution

Step 1: Check for balance of supply and demand

S Supply = 250 + 200 + 250 = 700 units

S Demand = 100 + 150 + 250 + 100 = 600 units

Decision Rule

(i) If S Supply = S Demand

then go to next step.

or submit a new assignment

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(ii) Else; if S Supply > S Demand

then, add a “dummy destination” with zero transportation cost.

(iii) Or else; if S supply < S Demand

then, add a “dummy source” with zero transportation cost.

Since, in this problem

S supply > S Demand

Hence; add a “dummy destination” (say D5) with zero transportation cost and balance demand which is difference

in supply and demand (= 100 units).

The initial transportation matrix is now formulated with transportation cost in the small box of each route. Note

that each cell of the transportation matrix represents a potential route.

Introducing dummy column for balancing the supply and demand

Step 2:

(i) Decide the nature of problem : Minimization of transportation-cost

(ii) Make initial assignment

Initial assignment may be done by using any of the following approaches :

(i) Least-cost method

(ii) North-West corner method

(iii) Vogel's approximation method

We would demonstrate all the three methods.



(i) Initial Solution by Least Cost method

Select the lowest transportation (or shipping) cost cell (or route) in the initial matrix. For example: it is route

S1D5, S2D5 and S3D5 in our problem with zero shipping cost.

Allocate the minimum of remaining balance of supply (in last column) and demand (in last row).

Let us select S1D5 route. One can also select other route (S2D5 or S3D5) in case of tie. For S1D5, available supply

is 250 and available demand is 100 units. The lower is 100 units. Hence, allocate 100 units-through this route

(i.e, S1D5).

With this allocation, entire demand of route S1D5 is consumed but supply of corresponding source, S1, is still

(250-100) or 150 units left. This is marked in last column of supply. The entire demand of destination, D5, is

consumed. We get the following matrix (Fig. 12.6) by crossing out the consumed destination (D5):

Now, we leave the consumed routes (i.e., column D5) and work for allocation of other routes.

Next, least cost route is S1D1, with 13 per unit of shipping cost. For this route, the demand is 100 units and

remaining supply is 150 units. We allocate minimum of the two, i.e., 100 units in this route. With this destination,

D1 is consumed but source S1 is still left with (150-100) = 50 units of supply. So, now leave the destination D1

and we get the following matrix.

With 100 units allocation in route S1D5

Assignment for destination D1 and D5 consumed



Now, we work on remaining matrix, which excludes first column (D1) and last column (D5). Next assignment is

due in the least cost route, which is route S2D4. For this route, we can allocate 100 units which is lesser of the

corresponding demand (100 units) and (200 units). By this allocation in route S2D4, the demand of destination D4

is consumed. So, this column is now crossed out.

Assignment with destination D1, D4 and D5 consumed

Now, we work on the remaining matrix which excludes, column, D1, D4 and D5. Next assignment is due in the

least cost route of the remaining routes. Note that we have two potential routes: S1D2 and S2D3. Both have 16

units of transportation cost. In case of any tie (such as this), we select any of the routes. Let us select route,

S1D2, and allocate 50 units (minimum of demand of 150 and supply of remaining 50 units). With this, all supply of

source S1 is consumed. Therefore, cross out row of S1. We get the following matrix:

Destination D1, D4 and D5 source S1 are consumed

Now, remaining allocation is done in route S2D3 (as 100 units). With this source, S2 is consumed. Next allocation

of 100 units is done in route S3D2 and 150 units in route S3D3. Final initial assignment is as follows:

Total cost in this assignment is (13 × 100 + 16 × 50 + 100 × 0 + 16 × 100 + 15 × 100 + 17 × 150) or Rs.

9450.



Initial assignment by least cost method

Step 3: Count the number of filled (or allocated) routes.

Decision rule

(i) If filled route = (m + n – 1) then go for optimality check (i.e. step 5).

(ii) If filled route < (m + n – 1) then the solution is degenerate. Hence, remove degeneracy and go to

step 4.

Here, m = number of destinations, including dummy column, if any

n = number of source, including dummy, row, if any

For our problem (m + n- l) = 5 + 3-1 = 7.

The number of filled route is equal to 7. Hence, problem is not degenerate. Therefore, proceed to step 5.

Initial Assignment by North-West Corner Method (an alternative to least cost method)

This approach is also for making initial assignment, as we have done in the least cost method. Therefore, this

approach should not be applied if initial assignment has already been made by any other method. In the North-

West Corner (NWC) method, we start with the top-left (corner-most) route, which is S:DrIrrespective of cost,

allocation is made in this route for the minimum of supply or demand. In our case, demand for this route is 100

and supply is 250. Therefore, allocate 100 units in this route. With this, column corresponding to D1 is consumed.

Now, work on the remaining matrix, which excludes column Dr Again, select the top-left route. Now, it is cell

S1D2. Allocate in the same way. Thus, 150 units are allocated in this route. Note that, with this, both D2 and S,

are consumed.



Remaining matrix excludes S1, D1 and D2. Hence, allocation in the top-left cell is due in route S2D3. Here, 200

units may be allocated and S2 is now consumed.

Remaining allocations are done in S2D3, S3D4 and S3D5 in sequential order. We get the initial solution by north-

west corner method as follows (Fig. 12.11):

Initial assignment by North-West corner method

For this assignment, the total cost is (13 × 100 + 16 × 150 + 16 × 100 + 0 × 100) or Rs. 9350.

Step 3: Check for degeneracy

(m + n – 1) = 5 + 3 – 1 = 7

Number of filled cells = 6, which is one less than (m + n – 1). Hence, go to step 4 for removing degeneracy.

Step 4: In case of degeneracy, allocate a very-very small quality, (which is zero for all calculation purposes), in

the least cost of un-filled cells. In the above figures of North-West corner method allocation, the least cost un-

filled cells are S1D5 and S2D5. Let us select S1D5 and allocate in this. We get the following allocation after

removing degeneracy.

Initial assignment by North-West corner method after removing degeneracy

Initial Assignment by Vogel’s Approximation Method (VAM)

This is the third alternative method for doing initial assignment of a transportation problem.

In this method, we calculate the difference between the two least-cost routes for each row and column. The

difference is called as penalty cost for not using the least-cost route.



First calculation of Penalty cost in VAM

Highest of all calculated penalty costs is for S3 and (S2). Therefore, allocation is to made in row of source S3. The

route (or cell), which one must select, should be the lowest cost of this row. This route S3D5. Hence, first

allocation is as follows.

First calculation in Vogel’s method

Now, with the first allocation, destination D5 is consumed. We exclude this column and work on the remaining

matrix for calculating the penalty cost. We get the following matrix.

Now for this, source S1 has highest penalty cost. For this row, the least cost route is S1D1. Hence, next

assignment is due in this route:

Second calculation of Penalty cost in VAM

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Second allocation in Vogel’s method

After second allocation, since destination D1 is consumed, we leave this column and proceed for calculation of

next penalty cost. Allocation is done in route S1D2. Since there is tie between all routes, we break the tie by

arbitrarily selecting any route (S1D2 in this case.)

Third calculation of Penalty cost

Third allocation in Vogel’s method



Fourth calculation of Penalty cost in VAM

Fourth allocation in Vogel’s method

With the fourth allocation, column D4 is consumed. In the only left column D3, the allocations of 100 units and

150 units are done in route S2D3 and S4D3 respectively. Thus, we get the following allocations in the Vogel’s

approximation method.

Final allocation through Vogel’s method

The initial cost for this allocation is (13 × 100 + 16 × 150 + 16 × 100 + 15 × 100 + 17 × 150 + 0 × 100) or

equal to Rs. 9350:

Step 3: Check for degeneracy

(m + n – 1) = 7

Number of filled cell = 6, which is one less than (m + n + 1). Hence, go to step 4 for removing the degeneracy.



Step 4: We allocate in the least-cost un-filled cell. This cell is route S1D5 or S2D5. Let us select route

S1D5. Thus, we get following matrix after removing degeneracy.

Final allocation after removing degeneracy in Vogel’s method

Optimization of Initial Assignment

The initial feasible assignment is done by using least-cost method or North-West corner method or Vogel's

approximation method. However, none of these methods guarantees optimal solution. Hence, next step is to

check the optimality of the initial solution.

Step 5: Check the optimality of the initial solution

For this, we have to calculate the opportunity cost of un-occupied routes.

First, we start with any row (or column). Let us select row 1, i.e., source S1; For this row, let us define row

value, u1 = 0. Now consider all filled routes of this row. For these routes, calculate column values v. using

following equation:

u1 + v1 = Cij (For any filled route)

where u1 = row value

vj = column value

Cij = unit cost of assigned route

Once first set of column values (vj is known, locate other routes of filled cells in these columns. Calculate next of

ui (or vj values using above equation. In this way, for all rows and columns, ui and vj values are determined for a

non- degenerate initial solution.

Step 6: Check the optimality

Calculate the opportunity of non-allocated orunfilled routes. For this, use the following equation:



Opportunity unassigned route = ui + vj – Cij

where ui = row value

vj = column value

Cij = unit cost of unassigned route

If the opportunity cost is negative for all unassigned routes, the initial solution is optimal. If in case any of the

opportunity costs is positive, then go to next step.

Step 7: Make a loop of horizontal and vertical lines which joins some filled routes with the unfilled route,

which has a positive opportunity cost. Note that all the corner points of the loop are either filled cells or positive

opportunity cost un-assigned cells.

Now, transfer the minimal of all allocations at the filled cells to the positive opportunity cost cell. ¥or this,

successive corner points from unfilled cell are subtracted with this value. Corresponding addition is done at

alternate cells. In this way, the row and column addition of demand and supply is maintained. We show the

algorithm with our previous problem.

Let us consider the initial allocation of least-cost method (Fig. 12.10) :

For this, we start with row, S1and take u1 = 0. Now S1DpS1D2,and S1D5are filled cells. Hence, for filled cells; (vj =

Cij – ui).

v1 = 13 – 0 = 13

v2 = 16 – 0 = 16

v5 = 0 – 0 = 0



Calculation for ui and vj in least cost initial assignment

Now, cell S3D2 is taken, as this has a vj value. For this cell u3 = 17 – 16 = 1

Now, cell S3D3 is selected, as this has a ui value. For this cell v3 = 17 – 1 = 16

Now, cell S2D3 is selected, as it has a vj value. For this cell u2 = 16 – 16 = 0

Now, cell S2D4 is selected, as it has a ui value. For this cell v4 = 15 – 0 = 0

Thus, all ui and vj are known.

Step 6: Calculate opportunity cost of un-assigned routes.

Unassigned route Opportunity cost (ui + vj – Cij)

S1D3

S1D4

S2D1

S2D2

S2D5

S3D1

S3D4

S3D5

0 + 16 – 19 = –3

0 + 15 – 17 = –2

0 + 13 – 17 = –4

0 + 16 – 19 = –3

0 + 0 – 0 = 0

1 + 13 – 15 = –2

1 + 15 – 16 = 0

1 + 0 – 0 = +1

Since route S3D5 has positive opportunity cost, the solution is non-optimal; hence, we go to next step and make

a loop as follows.



Closed loop for cell S3D5

The revised allocation involves 100 units transfer from cells S1D5 and S3D2 to cells S3D5 and S1D2.

Thus, revised allocation is as follows:

Revised allocation in least-cost assignment

Since above solution is degenerate now, we allocate to the least-cost un-filled cell S1D5. Fresh calculation of ui

and vj is also done in the similar way as explained in Step 5.

For this assignment, the opportunity cost of unassigned cells is as follows.

Now, since un-allocated routes have negative (or zero) opportunity cost, the present assignment is the optimal

one. Thus, optimal allocation of route is given in Figure.

Note that total cost is less than the initial assignment cost of least-cost method (= Rs. 9450).

Similarly, optimality of North-West corner method solution is done.


S1D3

S1D4

S2D1

S2D2

S2D5

S3D1

S3D2

S3D4

0 + 17 – 19 = –2

0 + 16 – 17 = –1

–1 + 13 – 17 = –5

–1 + 16 – 19 = –4

–1 + 0 – 0 = –1

0 + 13 – 15 = –0

0 + 16 – 17 = –1

0 + 16 – 16 = 0

Opportunity cost

Route Unit Cost in this route



S1D1

S1D2

S2D3

S2D4

S3D3

S3D5

100

150

100

100

150

100

13 × 100 = 1300

16 × 150 = 2400

16 × 100 = 1600

15 × 100 = 1500

17 × 150 = 2550

0 × 100 = 0

Total cost = Rs.9350

Optimal allocation in different routes

Calculation of ui and vj for N-W corner method’s initial solutions

Opportunity cost of above assignment is as follows:

Since all opportunity costs are negative or zero, the initial assignment is optimal one with total cost of Rs. 9350.

The optimal assignment of routes is 100 units is S1D1, 150 units in S1D2, 200 units in S2D3, 50 units in S3D3, 100

units in S3D4.

Similarly, the optimality of Vogel’s method’s initial solution is done.

Opportunity cost of above N-W corner assignment is as follows


S1D3

S1D4

S2D1

S2D2

S2D4

S2D5

S3D1

S3D2

0 + 17 – 19 = –2

0 + 16 – 17 = –1

–1 + 13 – 17 = –5

–1 + 16 – 19 = –4

–1 + 16 – 15 = 0

–1 + 0 – 0 = –1

0 + 13 – 15 = –2

0 + 16 – 17 = –1



Calculation of ui and vj for Vogel method’s initial solutions

Opportunity cost of above assignment is as follows:


S1D3

S1D4

S2D1

S2D2

S2D5

S3D1

S3D2

S3D4

0 + 17 – 19 = –2

0 + 16 – 17 = –1

–1 + 13 – 17 = –5

–1 + 16 – 19 = –4

–1 + 0 – 0 = –1

0 + 13 – 15 = –2

0 + 16 – 17 = –1

0 + 16 – 16 = 0

Since all opportunity costs are negative or zero, the initial assignment of Vogel’s solution is optimal with total cost

of Rs. 9350.

The optimal assignment of routes is 100 units in S1D2, 100 units in S2D3, 100 units in S2D4, and 150 units in S3D3.

Note that this solution is different from North-West corner solution but total cost is same and minimum.



The transportation problem solution approach

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