NUMERICAL PROBLEMS ON NORTH WASTE CORNER

1. A transportation problem is a special type of linear programming problem which establish a
least cost for transportation goods source to destination.
There are two parts
Part 1: Find feasibility test
1. North –West Corner Method (NWCM)
2. Least Cost Method (LCM)
3. Vogel’s Approximation Method (VCM)
Part 2: To find optimum solution
MODI Method (Modified Distribution Method)
1. NORTH WEST CORNER METHOD
Q.1 Find initial basic feasible solution by using NWCM for the following transportation
minimization problem.
D1 D2 D3 Supply
S1 3 2 1 20
S2 2 4 1 50
S3 3 5 2 30
S4 4 8 7 25
Demand 40 30 55
Step 1: Check total demand and total supply
Total demand = 40+30+55=125
Total supply=20+50+30+25=125
Here Total Demand = Total Supply so we can calculate initial basic feasible solution
Step 2: Choose the north west corner that is (S1,D1), compare the supply and demand that
cell and allocated 20 units to it. After allocating 20 units to (S1,D1) reduce the corresponding
supply and demand. Cover row S1 as the supply has been fully used.
D1 D2 D3 Supply
S1 3 20 2 1 0
S2 2 4 1 50
S3 3 5 2 30
S4 4 8 7 25
Demand 40 20 30 55
Step 3: Choose the north west corner that is (S2,D1), compare the supply and demand that
cell and allocated 20 units to it. After allocating 20 units to (S2,D1) reduce the corresponding
supply and demand. Cover column D1 as the Demand has been fully used.
D1 D2 D3 Supply
S1 3 20 2 1 00

2. S2 2 20 4 1 30
S3 3 5 2 30
S4 4 8 7 25
Demand 0 30 55
Step 4: Choose the north west corner that is (S2,D2), compare the supply and demand that
cell and allocated 30 units to it. After allocating 30 units to (S2,D2) reduce the corresponding
supply and demand. Cover column D2 as the Demand has been fully used and S2 also has
been fully used.
D1 D2 D3 Supply
S1 3 20 2 1 00
S2 2 20 4 30 1 00
S3 3 5 2 30
S4 4 8 7 25
Demand 0 00 55
Step 5 : Allocate reaming demand and supply which 30, 25 and 55 on S3,S4and D3
respectively. Then S3 ,S4 and D3 fully used their demand and supply.
D1 D2 D3 Supply
S1 3 20 2 1 00
S2 2 20 4 30 1 00
S3 3 5 2 30 00
S4 4 8 7 25 00
Demand 0 00 00
Total cost= 3*20+2*20+4*30+2*30+7*25=60+40+120+60+105=385 unit
2.
D E F Supply
A 5 8 4 50
B 6 6 3 40
C 3 9 6 60
Demand 20 95 35
Step 1: Check total demand and total supply
Total demand =20+95+35= 150
Total supply=50+40+60=150
Here Total Demand = Total Supply so we can calculate initial basic feasible solution
Step 2: Choose the north west corner that is (A,D), compare the supply and demand that cell
and allocated 20 units to it. After allocating 20 units to (A,D) reduce the corresponding
supply and demand. Cover column D as the demand has been fully used.
D E F Supply

3. A 5 20 8 4 30
B 6 6 3 40
C 3 9 6 60
Demand 00 95 35
Step 3: Choose the north west corner that is (A,E), compare the supply and demand that cell
and allocated 30 units to it. After allocating 30 units to (A, E) reduce the corresponding
supply and demand. Cover row A as the supply has been fully used.
D E F Supply
A 5 20 8 30 4 00
B 6 6 3 40
C 3 9 6 60
Demand 00 65 35
Step 4: Choose the north west corner that is (B,E), compare the supply and demand that cell
and allocated 40 units to it. After allocating 40 units to (B, E) reduce the corresponding
supply and demand. Cover row B as the supply has been fully used.
D E F Supply
A 5 20 8 30 4 00
B 6 6 40 3 00
C 3 9 6 60
Demand 00 25 35
Step 5: Allocating remaining elements or cell (C,E) and (C,F) by supply (60)and demand
(25,35) so matrix will be
D E F Supply
A 5 20 8 30 4 00
B 6 6 40 3 00
C 3 9 25 6 35 00
Demand 00 00 00
Total Cost= 5*20+8*30+6*40+9*25+6*35=100+240+240+225+210=1015 unit.