Transportation problem

Faculty of Economics and Business Administration
Lebanese University
Chapter 4: Transportation problem
Dr. Kamel ATTAR
attar.kamel@gmail.com
! 2020 !

1
Deﬁnition and basic notation
Transportation Model
Methods for obtaining initial feasible basic solution
Optimality Test: (Method for optimal solution)
1 Deﬁnition and basic notation
2 Transportation Model
Balanced transportation problem
Un-balanced transportation problem
3 Methods for obtaining initial feasible basic solution
North-west corner method
Least cost entry method
Vogel’s Approximation Method (VAM) or penalty method
4 Optimality Test: (Method for optimal solution)
Stepping stone method of optimality test
Dr. Kamel ATTAR | Chapter 4: Transportation problem |

2
Deﬁnition
The transportation problem is a special type of linear programming problem where the objective
consists in minimizing transportation cost of a given commodity from a number of sources or
origins to a number of destinations.
In the transportation problem, we have
• m sources (warehouses, factories) producing
items, and
• n destinations (shops, businesses) requiring these
items.
The items need to be transported from sources to desti-
nations which has associated cost.
• The i-th source denotes by ai has available items
to transfer, and
• The j-th destination denotes by bj has demands
items to be delivered.
It costs cij to deliver one item from i-th source to j-th
destination.

3
Assumption: destinations do not care from which source the items come, and sources do not care
to which destinations they deliver.
Decision variables: xij = the number of items transported from the i-th source to j-th destination.
Objective: The goal is to minimize the total cost of transportation.
Minimize
m
i=1
n
j=1
cij xij
Subject to



n
j=1
xij = ui for i = 1, · · · , m
m
i=1
xij = vi for j = 1, · · · , n
with xij ≥ 0 for i = 1, · · · , m and j = 1, · · · , n

4
TABLE
b1 b2 b3 Supply
a1
c11
x11
c12
x12
c13
x13 u1
a2
c21
x21
c22
x22
c23
x23 u2
a3
c31
x31
c32
x32
c33
x33 u3
Demand v1 v2 v3 z =?

5
# Transportation Model #
There are two different types of transportation problems based on the initial
given information balanced and un-balanced transportation problems.
# Balanced transportation problems #
Cases where the total supply is equal to the total amount demanded.
Factory Transportation Supply
X Y
A 6 8 24
B 4 3 40
Demand 40 24
64
64

6
# Un-balanced transportation problems #
Cases where the total supply is less than or greater than to the total amount demanded.
• When the supply or availability is higher than the demand, a dummy destination is introduced in
the equation to make it equal to the supply (with shipping costs of 0$); the excess supply is
assumed to go to inventory.
X Y
A 6 8 20
B 4 3 50
Demand 30 60
90
70
=⇒
X Y
A 6 8 20
B 4 3 50
dummy 0 0 20
Demand 30 60
90
90
• On the other hand, when the demand is higher than the supply, a dummy source is introduced
in the equation to make it equal to the demand (in these cases there is usually a penalty cost
associated for not fulﬁlling the demand).
X Y
A 6 8 30
B 4 3 60
Demand 20 50
70
90
=⇒
X Y dummy
A 6 8 0 30
B 4 3 0 60
Demand 20 50 20
90
90

7
# Methods for obtaining initial feasible basic solution #
To find the optimal solution we use two steps.
• First we need to find the initial basic feasible solution by using one of the
following three methods
North-west corner , Least cost entry or Vogel’s Approximation Method
• Then we obtain an optimal solution by making successive improvements in
the initial basic feasible solution until no further decrease in transportation
cost is possible. We can use on of the following two methods:
Stepping Stone or Modified Distribution Method

8
# North-West Corner Method #
This method is the most systematic and easiest method for obtaining initial basic feasible solution.
To understand this method more clearly, let us take an example and discuss the rationale of
transportation problem.
Example
Transportation Availability in tons
(Cost per ton in $)
Factory W X Y sugar
A 4 16 8 72
B 8 24 16 102
C 8 16 24 41
Demand in Tons 56 82 77 215
Step‚ This problem is balanced since
m
i=1
ui
total supply
=
n
i=1
vi
total demand
= 215. If the problem is not
balanced, then we open a dummy column or row as the case may be and balance the problem.

9
Stepƒ Construct an empty m × n matrix, completed with rows & columns.
Step„ Indicate the rows and column totals at the end.
W X Y Quantity Supplied
A
4 16 8
72
B
8 24 16
102
C
8 16 24
41
Demand 56 82 77 215

9
Step… Starting with the first cell at the north west corner of the matrix :
• If u1 < v1 , then set x11 = u1 , remove the first row, and decrease v1 to v1 − u1.
• If u1 > v1 , then set x11 = v1, remove the first column, and decrease u1 to u1 − v1.
A
4
56
16 8
72 16
B
8
−
24 16
102
C
8
−
16 24
41
Demand 56 0 82 77 215

9
Step† Repeat:
• If ui < vj , then set xij = ui , remove the i-th row, and decrease vj to vj − ui .
• If ui > vj , then set xij = vj , remove the j-th column, and decrease ui to ui − vj .
A
4
56
16
16
8
− 72 16 0
B
8
−
24 16
102
C
8
−
16 24
41
Demand 56 0 8266 77 215

9
Step† Repeat:
A
4
56
16
16
8
− 72 16 0
B
8
−
24
66
16
102 36
C
8
−
16
−
24
41
Demand 56 0 8266 0 77 215

9
Step† Repeat:
A
4
56
16
16
8
− 72 16 0
B
8
−
24
66
16
36 102 36 0
C
8
−
16
−
24
41
Demand 56 0 8266 0 77 41 215

9
Step‡ Once all the allocations are over, i.e., both rim requirement (column and row i.e., availability
and requirement constraints) are satisﬁed, write allocations and calculate the cost of transportation.
A
4
56
16
16
8
− 72 16 0
B
8
−
24
66
16
36 102 36 0
C
8
−
16
−
24
41 41 0
Demand 56 0 8266 0 77 41 0 215

10
and requirement constraints) are satisﬁed, write allocations and calculate the cost of transportation
Rim Requirements
Stone squares must be equal to the nb. of columns+nb. of rows−1. In this
table we have
3columns + 3rows − 1 = 6 − 1 = 5Stones
Transportation cost
From To Units in tons Cost in $
A W 56 56 × 4 = 224
A X 16 16 × 16 = 256
B X 66 66 × 24 = 1584
B Y 36 36 × 16 = 576
C Y 41 41 × 24 = 984
Total 3624

11
# Least Cost entry Method #
This method takes into consideration the lowest cost and therefore takes less time to solve problem.
To understand this method more clearly, let us resolve the previous example.
Example
(Cost per ton in $)
Factory W X Y sugar
A 4 16 8 72
B 8 24 16 102
C 8 16 24 41
Step‚ This problem is balanced since
m
i=1
ui
total supply
=
n
i=1
vi
total demand
= 215.
If the problem is not balanced, then we open a dummy column or dummy row as the case may be
and balance the problem.

12
Stepƒ Construct an empty m × n matrix, completed with rows & columns.
Step„ Indicate the rows and column totals at the end.
A
4 16 8
72
B
8 24 16
102
C
8 16 24
41
Demand 56 82 77 215

12
Step… Select the cell with the lowest transportation cost cij among all the rows and columns of the
transportation table. If the minimum cost is not unique then select arbitrarily any cell with the lowest
cost. Then allocate as many units as possible to the cell determined and eliminate that row in which
either capacity or requirement is exhausted.
A
4
56
16 8
72 16
B
8
−
24 16
102
C
8
−
16 24
41
Demand 56 0 82 77 215

12
Step† Repeat the steps… for the reduced table until the entire capacities are exhausted to ﬁll the
requirements at the different destination.
A
4
56
16
−
8
16 72 16 0
B
8
−
24 16
102
C
8
−
16 24
41
Demand 56 0 82 77 61 215

12
A
4
56
16
−
8
16 72 16 0
B
8
−
24 16
102
C
8
−
16
41
24
− 41 0
Demand 56 0 8241 77 61 215

12
A
4
56
16
−
8
16 72 16 0
B
8
−
24 16
61 102 41
C
8
−
16
41
24
− 41 0
Demand 56 0 8241 0 77 61 0 215

12
A
4
56
16
−
8
16 72 16 0
B
8
−
24
41
16
61 102 41 0
C
8
−
16
41
24
− 41 0
Demand 56 0 8241 0 77 61 0 215

13
and requirement constraints) are satisﬁed, write allocations and calculate the cost of transportation
Rim Requirements
table we have
Transportation cost
A W 56 56 × 4 = 224
A Y 16 16 × 8 = 128
B X 41 41 × 24 = 984
B Y 61 61 × 16 = 976
C X 41 41 × 16 = 656
Total 2968

14
# Vogel’s Approximation Method #
This method is preferred over the (NWCM) and (LCM), because the initial basic feasible solution
obtained by this method is either optimal solution or very nearer to the optimal solution. So the
amount of time required to calculate the optimum solution is reduced.
To understand this method more clearly, let us resolve the previous example.
Example
(Cost per ton in $)
Factory W X Y sugar
A 4 16 8 72
B 8 24 16 102
C 8 16 24 41

15
Step‚ Find the cells having smallest and next to smallest cost in each row (resp. column) and write
the difference (called penalty) along the side of the table in row penalty (resp. column penalty).
W X Y Supply Penalty
A
4 16 8
72 4 = 8 − 4
B
8 24 16
102 8 = 16 − 8
C
8 16 24
41 8 = 16 − 8
Demand 56 82 77 215
Penalty 4 = 8 − 4 0 = 16 − 16 8 = 16 − 8

15
Stepƒ Select the cell with the lowest transportation cost cij among all the rows and columns of the
transportation table. If the minimum cost is not unique then select arbitrarily any cell with the lowest
cost. Then allocate as many units as possible to the cell determined and eliminate that row in which
either capacity or requirement is exhausted. If there is a tie in the values of penalties then select
the cell where maximum allocation can be possible
A
4
−
16
−
8
72 72 0 4 = 8 − 4
B
8 24 16
102 8 = 16 − 8
C
8 16 24
41 8 = 16 − 8
Demand 56 82 77 5 215
Penalty 4 = 8 − 4 0 = 16 − 16 8 = 16 − 8
The maximum penalty is 8, occurs in column Y. The minimum cij in this column is c13 = 8.
The maximum allocation in this cell is min(72, 77) = 72.
It satisfy supply of A and adjust the demand of Y from 77 to 5 (77 − 72 = 5).

15
Step„ Repeat this steps until all supply and demand values are 0.
A
4
−
16
−
8
72 72 0
B
8 24 16
102 8 = 16 − 8
C
8 16 24
41 8 = 16 − 8
Demand 56 82 77 5 215
Penalty 4 = 8 − 4 8 = 24 − 16 8 = 24 − 16
The maximum penalty is 8, occurs in row B.
The minimum cij in this row is c21 = 8.
It satisfy demand of W and adjust the supply of B from 102 to 46 (102 − 56 = 46).

15
A
4
−
16
−
8
72 72 0
B
8
56
24 16
102 46 8 = 16 − 8
C
8
−
16 24
41 8 = 16 − 8
Demand 56 0 82 77 5 215
Penalty 4 = 8 − 4 8 = 24 − 16 8 = 24 − 16
The maximum penalty is 8, occurs in row B.
It satisfy demand of W and adjust the supply of B from 102 to 46 (102 − 56 = 46).

15
A
4
−
16
−
8
72 72 0
B
8
56
24 16
102 46 8 = 24 − 16
C
8
−
16 24
41 8 = 24 − 16
Demand 56 82 77 5 215
Penalty 8 = 24 − 16 8 = 24 − 16
The maximum penalty is 8, occurs in row C.
It satisfy supply of C and adjust the demand of X from 82 to 41 (82 − 41 = 41).

15
A
4
−
16
−
8
72 72 0
B
8
56
24 16
102 46 8 = 24 − 16
C
8
−
16
41
24
− 41 0 8 = 24 − 16
Demand 56 82 41 77 5 215
Penalty 8 = 24 − 16 8 = 24 − 16
The maximum penalty is 8, occurs in row C.
It satisfy supply of C and adjust the demand of X from 82 to 41 (82 − 41 = 41).

15
A
4
−
16
−
8
72 72 0
B
8
56
24 16
102 46 8 = 24 − 16
C
8
−
16
41
24
− 41 0
Demand 56 82 41 77 5 215
Penalty 24 16
The maximum penalty is 24, occurs in column X.
The minimum cij in this column is c22 = 24.
It satisfy demand of X and adjust the supply of B from 46 to 5 (46 − 41 = 5).

15
A
4
−
16
−
8
72 72 0
B
8
56
24
41
16
102 46 8 = 24 − 16
C
8
−
16
41
24
− 41 0
Demand 56 82 41 77 5 0 215
Penalty 24 16
The maximum penalty is 24, occurs in column X.
The minimum cij in this column is c22 = 24.
It satisfy demand of X and adjust the supply of B from 46 to 5 (46 − 41 = 5).

15
Step… Once all the allocations are over, i.e., both rim requirement (column and row i.e., availability
A
4
−
16
−
8
72 72 0
B
8
56
24
41
16
5 102 46
C
8
−
16
41
24
− 41 0
Demand 56 82 41 0 77 5 0 215
Penalty

16
Step… Once all the allocations are over, i.e., both rim requirement (column and row i.e., availability
Rim Requirements
table we have
Transportation cost
A Y 72 72 × 8 = 576
B W 56 56 × 8 = 448
B X 41 41 × 24 = 984
B Y 5 5 × 16 = 80
C X 41 41 × 16 = 656
Total 2744

17
# Stepping stone method of optimality test #
Once, we get the basic feasible solution for a transportation problem, the next duty is to test
whether the solution got is an optimal one or not? This can be done by using Stepping Stone
methods.

17
Step‚ Starting from the empty cell draw a loop moving horizontally and vertically from loaded cell
to loaded cell. We have to take turn only at loaded cells and move to vertically downward or
upward or horizontally to reach another loaded cell. In between, if we have a loaded cell, where we
cannot take a turn, ignore that and proceed to next loaded cell in that row or column. After
completing the loop, mark minus (−) and plus (+) signs alternatively, but begin with the + mark in
the empty cell, and calculate the cost change.

17
Step‚ Starting from the empty cell draw a loop moving horizontally and vertically from loaded cell
to loaded cell. We have to take turn only at loaded cells and move to vertically downward or
upward or horizontally to reach another loaded cell. In between, if we have a loaded cell, where we
cannot take a turn, ignore that and proceed to next loaded cell in that row or column. After
completing the loop, mark minus (−) and plus (+) signs alternatively, but begin with the + mark in
the empty cell, and calculate the cost change.
A −→ Y : +AY − BY + BX − AX = +8 − 16 + 24 − 16 = 0

17
Stepƒ Repeat these steps again until all the empty cells get evaluated.

17
B −→ W : +BW − BX + AX − AW = +8 − 24 + 16 − 4 = −4

17
C → W : +CW−CY+BY−BX+AX−AW = +8−24+16−24+16−4 = −12

17
C → X : +CX − CY + BY − BX = +16 − 24 + 16 − 24 = −16

18
Step„ Evaluate the cost change. If the cost change is positive, it means that if we include the
evaluated cell in the program, the cost will increase. If the cost change is negative, it means that
the total cost will decrease, by including the evaluated cell in the program.
Now, if all the cost changes are positive or are equal to or greater than zero, then the optimal
solution has been reached. But in case, if any, value comes to be negative, then there is a scope to
reduce the transportation cost further. Then, select that loop which has the most negative of cost
change.
S.No. Empty Cell Evalution Loop formation C. change
B → W +BW − BX + AX − AW +8 − 24 + 16 − 4 −4
C → W +CW − CY + BY − BX + AX − AW +8 − 24 + 16 − 24 + 16 − 4 −12
C → X +CX − CY + BY − BX +16 − 24 + 16 − 24 -16
A → Y +AY − BY + BX − AX +8 − 16 + 24 − 16 0

18
Step„ Evaluate the cost change. If the cost change is positive, it means that if we include the
evaluated cell in the program, the cost will increase. If the cost change is negative, it means that
the total cost will decrease, by including the evaluated cell in the program.
Now, if all the cost changes are positive or are equal to or greater than zero, then the optimal
solution has been reached. But in case, if any, value comes to be negative, then there is a scope to
reduce the transportation cost further. Then, select that loop which has the most negative of cost
change.
C −→ X

18
Step… Identify the lowest load in the cells marked with negative sign. This number is to be added to
the cells where plus sign is marked and subtract from the load of the cell where negative sign is
marked. Do not alter the loaded cells, which are not in the loop. The process of adding and
subtracting at each turn or corner is necessary to balance the demand and supply requirements.
-
66
+
36
+
−
-
41

18
Step† Construct a table of empty cells and work out the cost change for a shift of load from loaded
cell to loaded cell.
-
66
+
36
+
−
-
41
=⇒
41 − 66
25
41 + 36
77
41 + 0
41
41 − 41
−

19
New table
W X Y Demand
A
4
56
16
16
8
− 72
B
8
−
24
25
16
77 102
C
8
−
16
41
24
− 41
Supply 56 82 77 215
Rim Requirements
Transportation cost
A W 56 56 × 4 = 224
A X 16 16 × 16 = 256
B X 25 25 × 24 = 600
B Y 77 77 × 16 = 1232
C X 41 41 × 16 = 656
Total 2968

20
Second iteration
A −→ Y : +AY − BY + BX − AX = +8 − 16 + 24 − 16 = 0

20
Second iteration
B −→ W : +BW − AW + AX − BX = +8 − 4 + 16 − 24 = −4

20
Second iteration
C −→ W : +CW − CX + AX − AW = +8 − 16 + 16− = 4

20
Second iteration
C −→ Y : +CY − BY + BX − CX = +24 − 16 + 24 − 16 = 16

21
Empty boxes
S.No. Empty Cell Evalution Loop formation Cost change in $
A → Y +AY − BY + BX − AX +8 − 16 + 24 − 16 0
B → W +BW − AW + AX − BX +8 − 4 + 16 − 24 -4
C → W +CW − CX + AX − AW +8 − 16 + 16 − 4 4
C → Y +CY − BY + BX − CX +24 − 16 + 24 − 16 16
-
56
+
16
+
−
-
25
=⇒
56 − 25
31
16 + 25
41
25 + 0
25
25 − 25
−

22
New table
W X Y Supply
A
4
31
16
41
8
− 72
B
8
25
24
−
16
77 102
C
8
−
16
41
24
− 41
Demand 56 82 77 215
Rim Requirements
Transportation cost
A W 31 31 × 4 = 124
A X 41 41 × 16 = 656
B X 25 25 × 8 = 200
B Y 77 77 × 16 = 1232
C X 41 41 × 16 = 656
Total 2868

23
Third iteration
A → Y +AY − BY + BW − AW +8 − 16 + 8 − 4 -4
B → X +BX − AX + AW − BW +24 − 16 + 4 − 8 4
C → W +CW − AW + AX − CX +8 − 4 + 16 − 16 4
C → Y +CY − BY + BW − AW + AX − CX 24 − 16 + 8 − 4 + 16 − 16 12
-
31
+
−
+
25
-
77
=⇒
− 31
56 46

24
New table
W X Y Supply
A
4
−
16
41
8
31 72
B
8
56
24
−
16
46 102
C
8
−
16
41
24
− 41
Demand 56 82 77 215
Rim Requirements
Transportation cost
A Y 31 31 × 8 = 248
A X 41 41 × 16 = 656
B W 56 56 × 8 = 448
B Y 46 46 × 16 = 736
C X 41 41 × 16 = 656
Total 2744

25
Final iteration
Empty boxes
B → X +BX − AX + AY − BY +24 − 16 + 8 − 16 0
C → W +CW − BW + BY − AY + AX − CX +8 − 8 + 16 − 8 + 16 − 16 8
C → Y +CY − AY + AX − CX +24 − 8 + 16 − 16 16
A → W +AW − BW + BY − AY +4 − 8 + 16 − 8 4
Then the optimal solution of the transportation cost is 2744$.

Transportation problem

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