(i) North west corner rule
The given transportation problem is

Total supply = Total Demand = 80 
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 :

Sixth allocation:

Seventh Allocation :

Transportation schedule: A-I, A-II, B-II, B-IV, CIII, D-III, D-IV
= \(21 \times 5+13 \times 1+12 \times 3+3 \times 5+12 \times 4+2 \times 4+17 \times 5\)
The transportation cost = Rs. 310
(ii) Least cost method
Total Demand = Total supply = 80
The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
Given transportation problem is

The least cost is 1 corresponds to the cells (A,I) and (D,I) Take the cell (D, II)
Allocate min (25, 34) = 25 units to this cell

The reduced table is

The least cost is 3 corresponds to the cells (A, III), (A, IV), (B, I), (C, IV) 
Allocate min (9, 17) = 9 units to this cell

The reduced table is

The least cost is 3 corresponds to the cells (B, I), (C, IV).
Allcoate min (12,17) = 12 units to this cell

The reduced table is

The least cost is 3 corresponds to the cells (B, I) Allocate min (21, 15) = 15 units to this cell

The reduced table is

The least cost is 4 corresponds to the cells (D, III) Allocate min (8,13) = 8 units to this cell

The reduced table is

Here allocate 5 units in the cell (D, IV)

Thus we have the following allocations:

Transportation schedule = A-I, B-I, C-IV, D-I, D-II, D-III, D-IV
\( =9 \times 3+15 \times 3+12 \times 3+6 \times 4+25 \times 1+8 \times 4+5 \times 5 \)
 = 27 + 45 + 36 + 24 + 25 + 32 + 25
Transportation cost = Rs. 214
 

Total demand = Total supply Penalty
It is a balanced transportation problem
Hence it has a feasible solution
(i) North west corner :

Total cost 
\( =(35 \times 8)+(10 \times 9)+(20 \times 12)+ (20 \times 13)+(10 \times 16)+(30 \times 5) \)
\(= 280+90+240+260+160+150 \)
= 1180
(ii) Least cost method:

Total optimum cost
\( = (30 \times 9)+(20 \times 6)+(15 \times 8)+ (20 \times 13)+(10 \times 16)+(30 \times 5) \)
=  270+120+120+260+160+150 
=  1080
(iii) Vogel's approximation method:

Penalty 1   3   3    2
Penalty 1   3   3    -
Penalty 1   6   3    -

Penalty 1      3
Penalty  -      3
Penalty  -      3
The final allotment is

Total transportation cost
\( = (10 \times 6)+(25 \times 10)+(45 \times 9)+ (5 \times 13)+(10 \times 9)+(30 \times 5) \)
60 + 250 + 405 + 65 + 90 + 150 = 1020

Total supply = Total demand
It is a balanced Transportation problem


Penalty 1 4
The final allotment is

Total cost = (11 x 13) + (3 x 14) +(6 x 17)+ (4 x 23) +(10 x 17) + (9 x 18)
Total cost = 143 + 42 + 102 + 92 + 170 + 162
= 711

Total requirements = Total supply
It is a balanced transportation problem
(i) North west corner Rule:

Total optimal cost
\( = (20 \times 1)+(10 \times 2)+(30 \times 3)+ (20 \times 2)+(10 \times 5)+(10 \times 9)+ (50 \times 6)+(5 \times 2)+(20 \times 6) \)
= 20 + 20 + 90 + 40 + 50 + 90 + 300 + 10 + 120
= 740
(ii) Least cost method:

\( = (20 \times 1)+(10 \times 1)+(20 \times 2)+(10 \times 1)+ (20 \times 4)+(20 \times 2)+(30 \times 6)+ (25 \times 2)+(20 \times 1) \)
= 20 + 10 + 40+ 10 + 80 + 40 + 180 + 50 + 20
= 450

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

Here only 3 tasks can be assigned to 3 men.
Step 1 :
Select the smallest element in each row and subtract it will all the elements in its row.

Here each row and column has atleast one zero.
Step 2:
Examine the row with only one zero, mark that zero by \(\Box\) and draw a vertical line.
After examining all the rows, examine the column with single zero, mark that zero by \(\Box\) and draw a horizontal line.
Step 3:
Only two assignment have been made.
The elements not lying on the line are
\(\begin{matrix} 6 & 10 & 14 \\ 5 & 9 & 11 \\ 5 & 5 & 12 \end{matrix}\)
and minimum is 5.
Subtract 5 from all these numbers. Other numbers remains the same.
A new cost matrix will be found .and repeat step 2.
∴ The new cost matrix is

Thus, 3 assignments have been made.
The optical assignment schedule and total cost is

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

Here column IV has 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:
Examine the row with exactly one zero. Mark the zero by \(\Box \) and draw a vertical line. After examining all the rows examine the column with one zero. mark the zero by \(\Box\) and draw a horizontal line.

Here only 4 assignments have been made.
The numbers not lying on the line are

and min. of these numbers is 1.
Now subtract 1 from all these numbers and add 1 to the numbers on the intersecting line (ie. 6, 7, 2). Other numbers remains the same.
∴ The new cost matrix is

Now, repeat Step 3.
Thus, all the 5 assignments have been made.
The optimal assignment schedule and total cost is

Here Σa_i = 22 + 15 + 8 = 45
Σbj = 7 + 12 + 17 + 9 = 45
Σa_{i =} Σbj
∴ The given problem is balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I-allocation:

[∵ the max penalty is 4. In II, least cost is 2 & min (12,22) = 12]

II-allocation:

[∵ the max penalty is 3. In B, least cost is 1 & min (17,15) = 15]
III-allocation:


[∵ the max penalty is 3. In III, least cost is 4 & min (2, 10) = 2]
IV-allocation:

[∵ the max penalty is 2. In A, least cost is 3 & min (9, 8) = 8]
V-allocation:

[∵ the max penalty is 1. In C, least cost is 4 & min (7, 8) = 7]
VI-allocation:

[∵ min (1,1) = 1]
Thus, the allocations are

∴ The transportation schedule is
A → II, A → III, A → IV, B → III, C → I and C → IV
Hence, the total transportation cost is
= 12(2) + 2(4) + 8(3) + 15(1) + 7(4) + 1(5)
= 24 + 8 + 24 + 15 + 28 + 5 = Rs.104