Transportation problem in operation reasearch

1. A special case in LPP

2. A firm owns facilities at seven places. It has manufacturing plants at places
A, B and C with daily output of 500, 300 and 200 units of an item,
respectively. It has warehouses at places P, Q, R and S with daily
requirements of 180, 150, 350 and 320 units, respectively. Per unit shipping
charges on different routes are given below.
TRANSPORTATION PROBLEM
To: P Q R S
From A : 12 10 12 13
From B : 7 11 8 14
From C : 6 16 11 7
The firm wants to send the output from various plants to warehouses involving
minimum transportation cost. How should it route the product so as to achieve
its objective?

3. 3

4. FROM TO
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
Minimise Z = 12x11 + l0x12 + l2x13 + 13x14 + 7x21 + 11x22
+ 8x23 + 14x24 + 6x31 + 16x32 + l lx31 + 7x34

5. FROM
`
TO
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
Method 1: NORTH-WEST CORNER RULE
180 150
200
120
180
170
Total Cost = 12 * 180 + 10 * 150 + 12 * 170 + 8 * 180 + 14 * 120 + 7 * 200 = Rs. 10,220
0 0 180
0 200
0
320
170
0
120
0
0

6. FROM TO
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
Method 2: LEAST COST METHOD (LCM)
180
150
20
300
300
50
Total Cost = 10 * 150 + 12 * 50 + 13 * 300 + 8 * 300 + 6 * 180 + 7 * 20 = Rs. 9,620
0 300
20
0
50
0
0
0
0 0

7. FROM TO
P Q R S Supply
Iteration
I II III
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
I
II
III
Method 3 : VOGEL'S APPROXIMATION METHOD (VAM)
12
(2nd Min)
- 10
(Min)
= 2
1
1
6
3
1
1

8. FROM TO
P Q R S Supply
Iteration
I II III
A 12 10 12 13 500 2
B 7 11 8 14 300 1
C 6 16 11 7 200 1
Demand 180 150 350 320 1000
I 1 1 3 6
II
III
Method 3 : VOGEL'S APPROXIMATION METHOD (VAM)
180
120
230
150
200
Largest Cost
Difference
0
120
0
120
230
0
0
12
(2nd Min)
- 7
(Min)
= 5
120
0
0
0
1
4
1
-
1
2
3
-
2
- 1 1
4

9. FROM TO
P Q R S Supply
Iteration
I II III
A 12 10 12 13 500 2 2 2
B 7 11 8 14 300 1 1 3
C 6 16 11 7 200 1 - -
Demand 180 150 350 320 1000
I 1 1 3 6
II 5 1 4 1
III - 1 4 1
Method 3 : VOGEL'S APPROXIMATION METHOD (VAM)
180
120
230
150
200
120
Total Cost = 10 * 150 + 12 * 230 + 13 * 120 + 7 * 180 + 8 * 120 + 7 * 200 = Rs. 9,440

10. FROM TO P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
METHOD 1: STEPPING – STONE METHOD
180 150
200
120
180
170
Total Cost = 12 * 180 + 10 * 150 + 12 * 170 + 8 * 180 + 14 * 120 + 7 * 200 = Rs. 10,220
Note:
This is the initial basic feasible
solution obtained by North-West
Corner Rule
-
+ -
+
For Unoccupied Cell AS:
Closed Loop: AS-AR-BR-BS

11. FROM TO P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
METHOD 1: STEPPING – STONE METHOD
180 150
200
120
180
170
Total Cost = 12 * 180 + 10 * 150 + 12 * 170 + 8 * 180 + 14 * 120 + 7 * 200 = Rs. 10,220
Note:
This is the initial basic feasible
solution obtained by North-West
Corner Rule
+
-
For Unoccupied Cell BP:
Closed Loop: BP-BR-AR-AP
-
+

12. FROM TO P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
METHOD 1: STEPPING – STONE METHOD
180 150
200
120
180
170
Total Cost = 12 * 180 + 10 * 150 + 12 * 170 + 8 * 180 + 14 * 120 + 7 * 200 = Rs. 10,220
Note:
This is the initial basic feasible
solution obtained by North-West
Corner Rule
+
-
For Unoccupied Cell CP :
Closed Loop: CP-CS-BS-BR-AR-AP
-
+
+
-

13. FROM TO P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
METHOD 1: STEPPING – STONE METHOD
180 150
200
120
180
170
Total Cost = 12 * 180 + 10 * 150 + 12 * 170 + 8 * 180 + 14 * 120 + 7 * 200 = Rs. 10,220
Note:
This is the initial basic feasible
solution obtained by North-West
Corner Rule
+
-
For Unoccupied Cell CQ :
Closed Loop: CQ-CS-BS-BR-AR-AQ
-
+
+
-

14. FROM TO P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
METHOD 1: STEPPING – STONE METHOD
180 150
200
120
180
170
Total Cost = 12 * 180 + 10 * 150 + 12 * 170 + 8 * 180 + 14 * 120 + 7 * 200 = Rs. 10,220
Note:
This is the initial basic feasible
solution obtained by North-West
Corner Rule
- +
-
+
For Unoccupied Cell CR:
Closed Loop: CR-CS-BS-BR

15. METHOD 1: STEPPING – STONE METHOD
Cell Closed Loop Net Cost Change
Opportunity
Cost
AS AS-AR-BR-BS 13-12+8-14 = -5 5
BP BP-BR-AR-AP 7- 8 + 12-12 = -1 1
BQ BQ-BR-AR-AQ ll-8+12-10 = 5 -5
CP CP-CS-BS-BR-AR-AP 6 - 7 + 14- 8 + 12 - 12 = 5 -5
CQ CQ-CS-BS-BR-AR-AQ 16 - 7 + 14 - 8 + 12 - 10 = 17 -17
CR CR-CS-BS-BR 11-7+14-8 = 10 -10

16. FROM TO
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
METHOD 1: STEPPING – STONE METHOD
180 150
200
120
180
170
Total Cost = 12 * 180 + 10 * 150 + 12 * 50 + 8 * 300 + 13 * 120 + 7 * 200 = Rs. 9620
-
+ -
+
50
300
120

17. METHOD 1: STEPPING – STONE METHOD
Cell Closed Loop Net Cost Change
Opportunity
Cost
BS BS-AS-AR-BR 14-13+12-8 = 5 -5
BP BP-BR-AR-AP 7- 8 + 12-12 = -1 1
BQ BQ-BR-AR-AQ ll-8+12-10 = 5 -5
CP CP-CS-AS-AP 6-7+13-12 = 0 0
CQ CQ-CS-AS-AQ 16-7+13-10 = 12 -12
CR CR-CS-AS-AR 11-7+13-12 = 5 -5

18. FROM TO
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
METHOD 1: STEPPING – STONE METHOD
150
200
180
300
50
Total Cost = 10 * 150 + 12 * 230 + 13 * 120 + 7 * 180 + 8 * 120 + 7 * 200 = Rs. 9,440
-
+
-
+
230
120
180
120

19. METHOD 1: STEPPING – STONE METHOD
Cell Closed Loop Net Cost Change
Opportunity
Cost
BS BS-AS-AR-BR 14-13+12-8 = 5 -5
AP AP-AR-BR-BP 12-12+8-7 = 1 -1
BQ BQ-BR-AR-AQ ll-8+12-10 = 5 -5
CP CP-CS-AS-AR-BR-BP 6-7+13-12+8-7 = 1 -1
CQ CQ-CS-AS-AQ 16-7+13-10 = 12 -12
CR CR-CS-AS-AR 11-7+13-12 = 5 -5
Since the opportunity costs of all the empty cells
are negative, the solution obtained is optimal.
Total Cost = 10 * 150 + 12 * 230 + 13 * 120 + 7 * 180 + 8 * 120 + 7 * 200 = Rs. 9,440

20. ui
vj
FROM TO
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320
1000
METHOD 2: MODIFIED DISTRIBUTION METHOD (MODI)
180 150
200
120
180
170
-
+
u1 = 0
u3 = -11
u2 = -4
v1 = 12 v3 = 12
v2 = 10 v4 = 18

21. FROM TO
P Q R S Supply ui
A 12 10 12 13 500 0
B 7 11 8 14 300 -4
C 6 16 11 7 200 -11
Demand 180 150 350 320
1000
vj 12 10 12 18
METHOD 2: MODIFIED DISTRIBUTION METHOD (MODI)
180 150
200
120
180
170
-
+
∆14 = 0 + 18 - 13 = 5, ∆21 =-4 + 12 – 7 = 1,
∆22 = -4 + 10 - 11 = -5, ∆ 31 =-11 + 12 – 6 = -5,
∆ 32 = -11 + 10 – 16 = -17 and ∆ 33 =-11 + 12 -11 = -10.
5
-10
-15
-5
-5
1
Here, the cell AS has the largest opportunity cost (∆ 14 = 5) and, therefore, we select ∆ 14 for inclusion
as a basic variable.

22. FROM TO
P Q R S Supply ui
A 12 10 12 13 500 0
B 7 11 8 14 300 -4
C 6 16 11 7 200 -11
Demand 180 150 350 320
1000
vj 12 10 12 18
METHOD 2: MODIFIED DISTRIBUTION METHOD (MODI)
180 150
200
120
180
170
-
-
+ -
+
300
50 120

23. FROM TO
P Q R S Supply ui
A 12 10 12 13 500 0
B 7 11 8 14 300 -4
C 6 16 11 7 200 -11
Demand 180 150 350 320
1000
vj 12 10 12 18
METHOD 2: MODIFIED DISTRIBUTION METHOD (MODI)
180 150
200
120
300
50
+
-5
-5
-12
0
-5
1
Recalculating ∆’s Values..
-
+
-
+
120
230
180

24. FROM TO
P Q R S Supply ui
A 12 10 12 13 500 0
B 7 11 8 14 300 -4
C 6 16 11 7 200 -11
Demand 180 150 350 320
1000
vj 12 10 12 18
METHOD 2: MODIFIED DISTRIBUTION METHOD (MODI)
180
150
200
120
120
230
-5
-5
-12
-1
-5
-1
Recalculating ∆’s Values..
This solution is found to be optimal. It involves a total cost of Rs. 9,440.
Further, since all ∆ij values are negative, the solution is unique.
Total Cost = 10 * 150 + 12 * 230 + 13 * 120 + 7 * 180 + 8 * 120 + 7 * 200 = Rs. 9,440

25. 1. Maximization Prob
2. Prohibited Rout
3. Unbalanced Prob
4. Degeneracy in Transportation Prob

26. P Q R SSS SUPPL
Y
A 5 7 11 6 1
B 8 5 9 6 1
C 4 7 10 7 1
D 10 4 8 3 1
DEMA
ND
1 1 1 1

27. THANK YOU