West Corner Method :
Here total supply = 5 + 8 + 7 + 14 = 34
total demand = 7 + 9 + 18 = 34
total supply = total demand
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I - allocation:

[∵ min (5, 7) = 5]
II - allocation:

[∵ min (2, 8) = 2]
III - allocation:

 

[∵ min (9, 6) = 6]
IV - allocation:

[∵ min (3, 7) = 3]
V - allocation:

[∵ min (18, 4) = 4]
VI - allocation:

[∵ min (14, 14) = 14]
Thus, the allocation are

∴ The transportation schedule is
S₁ → D₁, S₂ → D₁, S₂ → D₂, S₃ → D₂, S₃ → D₃, S₄ → D₄
Hence total transportation cost
= 5 (2) + 2 (3) + 6 (3) + 3 (4) + 4 (7) + 14 (2)
= 10 + 6 + 18 + 12 + 28 + 28 = Rs. 102

Strategy	States-of-nature		Minimum payoff	Maximum payoff
	E1	E2
S1	40	60	40	60
S2	10	-20	-20	10
S3	-40	150	-40	150

(i) Max (40,-20,-40) = 40
∴ Strategy S₁ is the best accordirtg to maximum criteria
(ii) Min (60,10,150) = 10
∴ Strategy S₂ is the best accordirtg to minimax principle.

 First allocation:

Second allocation:

Third allocation:

 Fourth allocation:

 Final allocation:

The total transportation cost is
\( =(15 \times 9)+(10 \times 5)+(35 \times 4) +(15 \times 7)+(25 \times 6) \)
= 135 + 50 + 140 + 105 + 150 = Rs. 580

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Final allocation:

The transportation cost is
\( = (30 \times 6)+(5 \times 5)+(28 \times 11)+ (7 \times 9)+(25 \times 7)+(25 \times 13) \)
= 180 + 25 + 308 + 63 + 175 + 325  
= Rs. 1076

Here total supply = 25 + 35 + 40 = 100
total requirement = 30 + 25 + 45 100
total supply = total requirement
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution for the given transportation problem.
North West Corner Rule:
I - allocation:

[∵ min (25, 30) = 25]
II - allocation:

[∵ min (5, 35) = 5]
III - allocation:

[∵ min (25, 30) = 25]
IV - allocation:

[∵ min (5, 45) = 5]
V - allocation:

[∵ min (40, 40) = 40]
Thus, the allocations are

∴ The transportation schedule is
S₁ → D₁, S₂ → D₁, S₂ → D₂, S₂ → D₃, S₃ → D₃
Hence, the total transportation cost
= 25(9) + 5(6) + 25(8) + 5(4) + 40(9)
= 225 + 30 + 200 + 20 + 360
= Rs. 835

Here, the number of rows and columns are equal
∴ The given assignment problem is balanced.
Step I : Select a smallest element in each row and subtract this from all the elements in its row.
∴ The given assignment problem is

Here column 2 has no zero. Go to Step 2.
Step 2 : Select the smallest element (10) and subtract it from all the elements in its column.

Step 3 : Examine the rows with only one zero mark that zero by Ԡ. Mark other zeros in its column by X.
Row 1 and Row 3 contains only one zero. Mark the other zeros by X
Column 2 contains exactly one zero. Mark it by Ԡ
Thus, all the 3 assignments have been made.
Hence, the optimal assignment schedule and total cost is

Programmers	Programmes	Cost
1	R	80
2	Q	90
3	P	110

Total cost = Rs. 280
Thus, the optimal assignment (minimum) cost  =  Rs. 280

First allocation :

Second allocation :

Third allocation :

Fourth allocation :

Fifth allocation :

The transportation cost is 
\( =(200 \times 11)+(50 \times 13)+(175 \times 18)+ (125 \times 14)+(150 \times 13)+(250 \times 10) \)
= 2200 + 650 + 3150 + 1750 + 650 + 2500 
= 10900

The assignment problem is a special case of transportation problem where the number of sources and destinations are equal. Here, jobs represent sources and machines represent destinations.

The mathematical form of assignment problem is minimize Z = \(\overset { n }{ \underset { i=1 }{ \Sigma } } \overset { n }{ \underset { j=1 }{ \Sigma } } { { C }_{ ij } }{ x }_{ ij }\) subject to the
Constraints \(\overset { n }{ \underset { i=1 }{ \Sigma } }{ x }_{ ij } = 1, j=1,2,.....n\)
\(\overset { n }{ \underset { j=1 }{ \Sigma } }{ x }_{ ij } = 1, i=1,2,.....n\) and x_ij = 0 or 1 for all i, j