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?
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
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