Transportation problem

19-Mar-20 Dr. Abdulfatah Salem 2
Transportation Problem

 Deals with transportation of a product available at several sources a number of different
destinations in such a way that the total transportation cost is minimum.
 The origin of this model dates back to 1941 when Fl Hitchcock presented a study on the subject.
Assumptions in the transportation model:-
 Total quantities of an item available at different sources is equal to total requirement at different
destinations.
 Items can be transported conveniently from all sources to destinations.
 The unit transportation cost of the item from all sources to destinations is certainly and precisely
known.
 The transportation cost on a given route is directly proportional to the number of units shipped on
that route.
 The objective is to minimize the total transportation cost for the organization as a whole and not
for individual supply and distribution centers.

The transportation problem seeks to minimize the total shipping costs of
transporting goods from m origins (each with a supply si) to n destinations (each
with a demand dj), when the unit shipping cost from an origin, i, to a destination, j,
is cij.

Each 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 the
cell.

The transportation problem is solved in two phases:
Phase I -- Obtaining an initial feasible solution
Phase II -- Moving toward optimality
In Phase I, the Minimum-Cost Procedure can be used to
establish an initial basic feasible solution without doing
numerous iterations of the simplex method.
In Phase II, the Stepping Stone, by using the MODI method
for evaluating the reduced costs may be used to move from
the initial feasible solution to the optimal one.

There are a number of methods for generating an initial feasible
solution for a transportation problem.
Consider three of the following
(i) North West Corner Method
(ii) Least Cost Method
(iii) Vogel’s Approximation Method

EX.

EX.
0
50
150
450

EX.
0
0
150
50 100 25
0 275
0
275

EX.
0
0
150
275
50 100 25
0 0
0
0
0

EX.
- 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.
- The initial solution is complete when all rim requirements are satisfied.
-Total Transportation Cost = 6(150) + 7(50) + 11(100) + 11(25) + 12(275)
= 5,925

National shipments Co. of laptops to distribution centers
EX.
Distribute to 4 centers,
N, demand 100k
S, demand 140k
E, demand 210k
W, demand 250k
Production at 3 facilities,
A, supply 200k
B, supply 350k
C, supply 150k

Supply
N S E W
A
16 13 22 17
200
100 100
B
14 13 19 15
350
40 210 100
C
9 20 23 10
150
150
Demand 100 140 210 250
Source
Destination
Z = 10770

The allocation according to this method is very useful as it takes into consideration
the lowest cost and therefore, reduce the computation as well as the amount of time
necessary to arrive at the optimal solution.
Step 1
(a) Select the cell with the lowest transportation cost among all the rows or columns of the
transportation table.
(b) If the minimum cost is not unique, then select arbitrarily any cell with this minimum cost.
Step 2
Allocate as many units as possible to the cell determined in Step 1 and eliminate that row
(column) in which either supply is exhausted or demand is satisfied.
Repeat Steps 1 and 2 for the reduced table until the entire supply at different plants is exhausted
to satisfied the demand at different warehouses.

EX.

Sol.
4
200
0
75
400

Sol.
5
200
0
0
25
75
325

Sol.
8
200
0
0
0
75
25 125
300

Sol.
10
200
0
0
0
75
25 0
175
125
175

Sol.
11
200
0
0
0
75
25 0
0
125
175 0
0

Sol.
200
0
0
0
75
25 0
0
125
175 0
0

The total transportation cost = 25(8) + 125(10) + 175(11) + 200(4) + 75(5)
= 4550
Sol.

This method is preferred over the other two methods because the initial feasible
solution obtained is either optimal or very close to the optimal solution.
Step 1:
Compute a penalty for each row and column in the transportation table.
Penalty (difference) = (Second smallest cij in row/col) - (Smallest cij in row/col)
Step 2:
Identify the row or column with the largest penalty.
Step 3:
Repeat steps 1 and 2 for the reduced table until entire supply at plants are
exhausted to satisfy the demand as different warehouses.

Ex.

Sol.
132
2
4
1
175 0
25
425
Cost1 = 175*7
= 1225
425

2
1
232
100
0
175
Cost2 = 100*5
= 500
325

4
8
22
25
0
150
Cost3 = 25*4
= 100
300

150
150 0
0
0
0
Cost4 = 150*10 + 150*12
= 3300
Total cost = 1225 + 500 + 100 + 3300
= 5125
0

Destination
Supply
Source N S E W
A
16 13 22 17
200
B
14 13 19 15
350
C
9 20 23 10
150
Demand 100 140 210 250
Ex.

Destination
Supply diff
Source N S E W
A
16 13 22 17
200 3
B
14 13 19 15
350 1
C
9 20 23 10
150 1
Demand 100 140 210 250
diff 5 0 3 5
Sol.

Destination
Supply diff
Source N S E W
A
16 13 22 17
200 3
B
14 13 19 15
350 1
C
9 20 23 10
0 0
Demand 100 140 210 100
diff 5 0 3 5
150

Destination
Supply diff
Source N S E W
A
16 13 22 17
200 3
B
14 13 19 15
350 1
C
9 20 23 10
0 0
Demand 100 140 210 100
diff 2 0 3 2
150

Destination
Supply diff
Source N S E W
A
16 13 22 17
60 1
B
14 13 19 15
350 1
C
9 20 23 10
0 0
Demand 100 0 210 100
diff 2 0 3 2
150
`140

Destination
Supply diff
Source N S E W
A
16 13 22 17
60 1
B
14 13 19 15
140 1
C
9 20 23 10
0 0
Demand 100 0 0 100
diff 2 0 0 2
150
`140
210

Destination
Supply diff
Source N S E W
A
16 13 22 17
60 1
B
14 13 19 15
40 1
C
9 20 23 10
0 0
Demand 0 0 0 100
diff 0 0 0 2
150
`140
210100

Destination
Supply diff
Source N S E W
A
16 13 22 17
0 0
B
14 13 19 15
0 0
C
9 20 23 10
0 0
Demand 0 0 0 100
diff 0 0 0 0
150
`140
210100
60
40
Z = 10330

3/19/2020 Dr. Abdulfatah Salem 39
The transportation problem is solved in two phases:
Phase I -- Obtaining an initial feasible solution
Phase II -- Moving toward optimality
In Phase I, the Minimum-Cost Procedure can be used to establish an
initial basic feasible solution without doing numerous iterations of the
simplex method.
In Phase II, the Stepping Stone, by using the MODI method for
evaluating the reduced costs may be used to move from the initial
feasible solution to the optimal one.

Once an initial solution has been found, the next step is to test that
solution for optimality. The following two methods are widely used
for testing the solutions:
 Stepping Stone Method
 Modified Distribution Method
The two methods differ in their computational approach but give
exactly the same results and use the same testing procedure.

In this method we calculate the net cost change that can be obtained by
introducing any of the unoccupied cells into the solution.
Step 1:
Make sure that the number of occupied cells is exactly equal to m+n-1, where
m=number of rows and n=number of columns.
Step 2:
Evaluate each unoccupied cells by following its closed path and determine its net
cost change.
Step 3:
Determine the quality to be shipped to the selected unoccupied cell. Trace the
closed path for the unoccupied cell and identify the minimum quality by
considering the minus sign in the closed path.

EX

200 75
25 125
175
Sol.
Using
least
cost
method
Total cost = 25*8 + 125*10 + 175*11 + 200*4 + 75*5 = 4550

Sol.
Total cost = 25*8 + 125*10 + 175*11 + 200*4 + 75*5 = 4550
Unused
cells

Quality of change
(opportunity for improvement)
Path 1 : 6 – 8 + 5 – 4 = -1
Path 2 : 11 – 11 + 10 - 8 = +2
Path 3 : 12 – 10 + 8 - 5 = +5
Path 4 : 7 – 11 + 10 – 8 + 5 - 4 = -1

Quality of change
Path 1 : 6 – 8 + 5 – 4 = -1
Path 4 : 7 – 11 + 10 – 8 + 5 - 4 = -1

200 75
25 125
175

175 100
25 125
175

175 100
25 125
175
8 - 6 + 4 – 5 = + 1
7 - 11 + 10 – 6 = 0
11 - 11 + 10 - 6 + 4 – 5 = + 3
12 – 10 + 6 – 4 = + 4
So the case is optimal
The optimal transportation cost = 25*6 + 125*10 + 175*11 + 175*4 + 100*5 = 4525
Quality of change

EX.
S1 S2 S3
Factory
Capacity
F1 100
F2 300
F3 300
Warehouse
Req.
300 200 200
5
8
9 7 5
4
4
3
3

Sol.
S1 S2 S3
Factory
Capacity
F1 100 100
F2 200 100 300
F3 100 200 300
Warehouse
Req.
300 200 200
5
8
9 7 5
4
4
3
3
North
West
Corner
method
Total cost = 100*5 + 200*8 + 100*4 + 100*7 + 200*5 = 4200

100
200 100
100 200
5
8
9 7 5
4
4
3
3
Path 1 = 9 - 7 + 4 - 8 = -2
Path 2 = 4 - 4 + 8 – 5 = +3
Path 3 = 3 – 5 + 7 - 4 + 8 - 5 = +5
Path 4 = 3 - 5 + 7 - 4 = +1
Quality of change

100
200 100
100 200
5
8
9 7 5
4
4
3
3Opportunity for improvement
Path 1 = 9 - 7 + 4 - 8 = -2
100
100 200
100 200
5
8
9 7 5
4
4
3
3
Total cost
= 200*4 + 200x5 +
100*5 + 100*8 + 100x9
= 4000

100
100 200
100 200
5
8
9 7 5
4
4
3
3
100
200 100
200 100
5
8
9 7 5
4
4
3
3
New path = 3 – 5 + 9 – 8 = -1
The optimal cost = 200*4 + 200x9 +
100*5 + 100*5 + 100x3
= $ 3900
4 – 4 + 3 - 5 + 9 - 5 + 4 = + 2
3 - 5 + 9 – 5 = + 2
8 - 3 – 5 + 9 = + 1
7 – 5 + 3 – 4 = + 1
Quality of change

100
100 200
100 200
5
8
9 7 5
4
4
3
3
100
200 100
200 100
5
8
9 7 5
4
4
3
3
New path = 3 – 5 + 9 – 8 = -1
The optimal cost = 200x4 + 200x9 + 100x5 + 100x5 + 100x3 = $ 3900
4 – 4 + 3 - 5 + 9 - 5 + 4 = + 2
3 - 5 + 9 – 5 = + 2
8 - 3 – 5 + 9 = + 1
7 – 5 + 3 – 4 = + 1
Quality of change

EX.
P1 P1 P1 P1 Supply
W1
4 6 8 8
40
W1
6 8 6 7
60
W1
5 7 6 8
50
Demand 20 30 50 50 150
The following table indicate the unit transportation cost between the
warehouses and the production locations of the steel company.
Calculate the most optimum solution

Sol.
P1 P1 P1 P1 Supply
W1
4 6 8 8
40
W1
6 8 6 7
60
W1
5 7 6 8
50
Demand 20 30 50 50 150
By using VAM we get the initial feasible solution
The transportation cost will be = 4x10 + 6x30 + 6x50 + 7x10 + 5x10 + 8x40 = 960
No. of occupied cells = 6 which is exactly equals to 3 + 4 – 1 (m + n – 1)
10 30
50 10
10 40

4 6 8 8
6 8 6 7
5 7 6 8
10 30
50 10
10 40
8 - 6 + 7 – 8 + 5 – 4 = + 2
8 - 4 + 5 – 8 = + 1
6 - 5 + 8 – 7 = + 2
8 - 7 + 8 – 5 + 4 – 6 = + 2
7 - 5 + 4 – 6 = 0
6 - 8 + 7 - 6 = -1
So, there is an opportunity for
improvement
Quality of change

4 6 8 8
6 8 6 7
5 7 6 8
10 30
10 50
10 40
8 - 6 + 5 – 4 = + 3
8 - 7 + 6 – 6 + 5 - 4 = + 2
6 - 5 + 6 – 6 = + 1
8 – 6 + 6 - 5 + 4 - 6 = + 1
7 - 5 + 4 – 6 = 0
8 -6 + 6 -7 = +1
So, there is no opportunity for
improvement
Quality of change
The optimal cost = 4x10 + 6x30 + 7x50 + 6x10 + 6x40 + 5x10 = 920

P1 P1 P1 P1 Supply
W1
4 6 8 8
40
W1
6 8 6 7
60
W1
5 7 6 8
50
Demand 20 30 50 50 150
10 30
10 50
10 40
The final distribution is as indicated on the table
With total transportation cost = $ 920

In case there are a large number of rows and columns, then Modified distribution (MODI) method
would be more suitable than Stepping Stone method
Step 1
Add a column on the right hand side of the TP table and title it ui. Also add a row at the bottom
of the TP table and title it vj.
Step 2
i) Assign value to ui=0 To any of the variable ui or vj, assign any arbitrary value. Generally the
in the first row i.e. u1 is assigned the value equal to zero.
ii) Determine values of the vj in the first row using the value of u1 = 0 and the cij values of the
occupied cells in the first row by applying the formula ui + vj = cij
iii) Determine ui and vj values for other rows and columns with the help of the formula ui + vj =
cij using the ui and vj values already obtained in steps a), b) above and cij values of each of
the occupied cells one by one.
iv) Check the solution for degeneracy. If the solution is degenerate [ie no. of occupied cells is
less than (m+n-1)], then this method will not be applicable.

Step 3
Calculate the net opportunity cost for each of the unoccupied cells using the formula
δij = cij - (ui + vj) .
If all unoccupied cells have positive δij value, then, the solution is optimal.
Multiple optimality: If, however, one or more unoccupied cells have δij value equal to zero, then
the solution is optimal but not unique.
Non optimal solution: If one or more unoccupied cells have negative δij value, then the solution is
not optimal.
Largest positive dj value: The unoccupied cell with the largest negative δij value is identified.
Step 4
A closed loop is traced for the unoccupied cell with the largest δij value. Appropriate quantity is
shifted to the unoccupied cell and also from and to the other cells in the loop so that the
transportation cost comes down.
Step 5
The resulting solution is once again tested for optimality.
If it is not optimal, then the steps from 1 to 4 are repeated, till an optimal solution is obtained

Dest.1 Dest.2 Dest.3
Source1
Source2
Source3
7
5
4 5
7
8
96
3
20
28
17
21 25 19
EX.

Source1 20
Source2 4 5 19
Source3 17
7
5
4 5
7
8
96
3
Total transportation cost = 6x20 + 5x4 + 7x5 + 3x19 + 4x17 = 300
Sol.
•By using the least
cost method we get
the initial feasible
solution
•No. of occupied
cells = 5 which is
exactly equals to 3 +
3 – 1 (m + n – 1)

Source1 20
Source2 4 5 19
Source3 17
7
5
4 5
7
8
96
3
U1= 0
U2= 1
U3= 0
V1= 4 V2= 6 V3= 2
For each occupied
cell
Cij = ui + vj

Source1 20
Source2 4 5 19
Source3 17
7
5
4 5
7
8
96
3
U1 = 0
U2 = 1
U3 = 0
V1 = 4 V2 = 6 V3 = 2
73
6-1
Calculating the
opportunity cost
δij for each
unoccupied cell
For each occupied
cell
δij = cij - (ui + vj)

Source1 20
Source2 4 5 19
Source3 17
7
5
4 5
7
8
96
3
U1 = 0
U2 = 1
U3 = 0
V1 = 4 V2 = 6 V3 = 2
-1

Source1 20
Source2 9 19
Source3 12 5
7
5
4 5
7
8
96
3
U1= 0
U2= 0
U3= -1
V1= 5 V2= 6 V3= 3

Source1 20
Source2 9 19
Source3 12 5
7
5
4 5
7
8
96
3
U1 = 0
U2 = 0
U3 = -1
V1 = 5 V2 = 6 V3 = 3
6
6
1
2
The optimal transportation cost = 6x20 + 5x9 + 3x19 + 4x12 + 5x5 = 295

Store 1 Store 2 Store 3 Store 4
Plant A
Plant A
Plant A
7
4
2 16
2
5
83
5 10
6
60
100
40
20 50 50 80
EX.

Plant A
20 40
Plant A
10 50 40
Plant A
40
7
4
2 16
2
5
83
5 10
6
60
100
40
20 50 50 80
Total transportation cost = 7x20 + 3x40 + 2x10 + 5x50 + 10x40 + 1x40 = 970
•By using the north
west corner method
we get the initial
feasible solution
•No. of occupied
cells = 6 which is
exactly equals to 3 +
4 – 1 (m + n – 1)

Plant A
20 40
Plant A
10 50 40
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -10
V1 = 7 V2 = 3 V3 = 6 V4 = 11

Plant A
20 40
Plant A
10 50 40
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -10
V1 = 7 V2 = 3 V3 = 6 V4 = 11
-2
2 -5
5 13 9
Calculating the
opportunity cost
δij for each
unoccupied cell

Plant A
20 40
Plant A
10 50 40
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -10
V1 = 7 V2 = 3 V3 = 6 V4 = 11
-5

Plant A
20 40
Plant A
50 50
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -5
V1 = 7 V2 = 3 V3 = 6 V4 = 6
No. of occupied
cells = 5 which is
less than to 6 (m +
n – 1)
So, we are in case
of degeneracy
To solve this case
we have to increase
the No. of occupied
cells by adding a
value 0 to
unoccupied cell and
complete solution

Plant A
20 0 40
Plant A
50 50
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -5
V1 = 7 V2 = 3 V3 = 6 V4 = 6
2
1080
-2 5
Calculating the
opportunity cost
δij for each
unoccupied cell

Plant A
20 0 40
Plant A
50 50
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -5
V1 = 7 V2 = 3 V3 = 6 V4 = 6
-2
Total transportation cost = 7x20 + 3x0 +6x40 +2x50 + 5x50 + 1x40 = 770

Plant A
20 40
Plant A
20 30 50
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -5
V1 = 7 V2 = 3 V3 = 6 V4 = 6

Plant A
20 40
Plant A
20 30 50
Plant A
40
7
4
2 16
2
5
83
5 10
6
U1 = 0
U2 = -1
U3 = -5
V1 = 5 V2 = 3 V3 = 6 V4 = 6
2
Total optimal transportation cost = 3x20 + 6x40 + 4x20 + 2x30 + 5x50 + 1x40 = 730
2
5
2 8 8
Calculating the
opportunity cost
δij for each
unoccupied cell
All are positive

Unbalanced problem
Demand less than supply
Demand greater than supply
 Degeneracy
More than one optimal solution


Demand less than supply
Demand greater than supply
 In real-life problems, total demand is not equal to total supply.
 These unbalanced problems can be handled easily by using dummy sources or
dummy destinations.
 If total supply is greater than total demand, a dummy destination (warehouse),
with demand exactly equal to the surplus, is created.
 If total demand is greater than total supply, introduce a dummy source (factory)
with a supply equal to the excess of demand over supply.
 Regardless of whether demand or supply exceeds the other, shipping cost
coefficients of zero are assigned to each dummy location or route because no
shipments will actually be made from a dummy factory or to a dummy
warehouse.
 Any units assigned to a dummy destination represent excess capacity, and units
assigned to a dummy source represent unmet demand.

Demand less than supply Demand greater than supply
D1 D2 supply
S1
5 2
120
S2
3 7
80
Demand 60 115
D1 D2 dummy supply
S1
5 2 0
120
S2
3 7 0
80
Demand 60 115 S > D25
D1 D2 supply
S1
5 2
150
S2
3 7
75
Demand 60 250
D1 D2 supply
S1
5 2
150
S2
3 7
75
dummy 0 0
Demand 60 250 D < S
85

 Degeneracy occurs when the number of occupied cells or routes in a transportation
table solution is less than the number of rows plus the number of columns minus 1.
# Occupied Squares = Rows + Columns – 1
 Such a situation may arise in the initial solution or in any subsequent solution.
 Degeneracy requires a special procedure to correct the problem.
 Without enough occupied squares to trace a closed path for each unused route, it
would be impossible to apply the stepping-stone method or to calculate δij value
needed for the MODI technique.
 To handle degenerate problems, create an artificially occupied cell by placing a zero
(representing a fake shipment) in one of the unused cells and then treat that cell as if
it were occupied.
 Note that the cell chosen must be in such a position as to allow all stepping-stone
paths to be closed.
 Although there is usually a good deal of flexibility in selecting the unused cell that
will receive the zero.

As with LP problems, it is possible for a Transportation Problem to have multiple
optimal solutions.
Such is the case when one or more of the improvement indices that we calculate
for each unused cell is zero in the optimal solution.
This means that it is possible to design alternative shipping routes with the same
total shipping cost.
The alternate optimal solution can be found by shipping the most to this unused
square using a stepping-stone path.
Practically speaking, multiple optimal solutions provide management with greater
flexibility in selecting and using resources.

Transportation problem

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