(i) Max (5, 7, 9, 8) = 9
∴ A₃ is the optimal decision according to maximin principle
(ii) Min (14, 11, 11, 13) = 11
∴ A₂ and A₃ are optimal decisions according to minimax principle

First allocation :

Second allocation :

Third allocation :

Fourth allocation :

Fifth allocation :

Sixth allocation :

\( =(3 \times 2)+(1 \times 11)+(2 \times 4)+(4 \times 7)+ (2 \times 2)+(3 \times 8)+(6 \times 12) \)
= 6 + 11 + 8 + 28 + 4 + 24 + 72 = Rs. 153 
Total Cost = 153

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

min( 27, 25, 23, 32) = 23. Since the minimum cost is 23, the best alternative is E₃ according to minimax principle.

Max (5,7,12,15) = 15 ஃ Action A4 is the best

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.
Step 1: Its not necessary, since each row contains zero entry. Go to Step 2.
Step 2:

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 35

Total Capacity = Total Demand
\(\therefore\) The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
Given Transportation Problem is

First Allocation:

Second Allocation:

Third Allocation:

Fourth Allocation:

Fifth Allocation:

Sixth Allocation:

Transportation schedule :
T⟶H,T⟶P,B⟶C,B⟶H,M⟶H,M⟶K
The total Transportation cost = ( 5×8) + (25×5)+ (35×5) + (5×11)+ (18×9) + (32×7)
= 40+125+175+55+162+224
= Rs.781

Given transportation table is

Total Availability = Total Requirement 
Therefore the given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First allocation :

Second allocation :

Third Allocation :

Fourth Allocation :

Fifth allocation :

Final allocation :

Transportation schedule : O₁⟶D₁, O₁⟶D₂, O₂⟶D₂, O₂⟶D₃, O₃⟶D₃,O₃⟶D₃.
The transportation cost
= (6\(\times\)6)+(8\(\times\)4)+(2\(\times\)9)+(14\(\times\)2)+(1\(\times\)6)+(4\(\times\)2)
= Rs.128

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.

To assign the different jobs to the different machines (one job per machine) to minimize the overall cost is known as assignment problem.

If the total supply = total demand, then the given problem is a balanced transportation problem.

A feasible solution to a transportation problem is a set of non negative values x_ij (i = 1, 2, m, j = 1, 2, ....... n) that satisfies the constraints.
If a basic feasible solution to a transportation problem contains exactly m + n - l allocations. in independent positions, it is called a non degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

A transportation problem is to determine the amount to be transported from each origin to each destinations such that the total transportation cost is minimized.

Here, the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 : Select a smallest element in each row and subtract this from all the elements in its row.
∴ The cost matrix: of the given assignment problem is

Column 1 contains no zero. Go to step 2.
Step 2 : Select the smallest element (1) in column 1 and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3 : Examine the rows with only one zero.
Row P, Q and S contains exactly one zero, mark them by 0 and mark the other zeros in the column byX.
Thus, all the four assignments have been made.
∴ The optimal assignment schedule and total cost is

Here the number of rows and columns are equal.
\(\therefore\) The given assignment problem is balanced.
Now let us find the solution.
Step 1: Select a smallest element in each row and subtract this from all the elements in its row.
The cost matrix of the given assignment problem is

Column 3 contains no zero. Go to Step 2.
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3: (Assignment):
Examine the rows with exactly one zero. Row B contains exactly one zero. Mark that zero by \(\square\) (i.e) Person B is assigned to Job 1. Mark other zeros in its column by ×.

Now, Row C contains exactly one zero. Mark that zero by \(\square\). Mark other zeros in its column by × .

Now, Row D contains exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its column by × .

Row E contains more than one zero, now proceed column wise. In column 1, there is an assignment. Go to column 2. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

There is an assignment in Column 3 and column 4. Go to Column 5. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 9