Additional Problems - Chapter 10 Operations Research 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

AdditionalProblems
I. One Mark Questions
Choose the correct Answer
Question 1.
Which of the following methods is used to verify the optimality of the current solution of the transportation problem?
(a) Least cost method
(b) Vogel's method
(c) North-west comer rule
(d) None of these
Answer:
(a) Least cost method
Question 2.
The degeneracy'in the transportation problem indicates that
(a) Dummy allocations need to be added
(b) The problem has no feasible solution
(c) Multiple optimal solutions exist
(d) All of the above
Answer:
(c) Multiple optimal solutions exist
Question 3.
The Hungarian method can also be used to solve
(a) Transportation problem
(b) Travelling salesman problem
(c) A linear programming problem
(d) All the above
Answer:
(b) Travelling salesman problem
Question 4.
An optimal solution of an assignment problem can be obtained only if,
(a) each row and column has only one zero element
(b) each row and column has at least one zero element
(c) The data is arranged in a square matrix
(d) None of the above

Answer:

(d) None of the above
Question 5 .
Say True or False.
1. In a transportation problem, a single source may supply something to all destinations.
2. A transportation model must have the same number of rows and columns.
3. It is usually possible to find an optimal solution to a transportation problem that is degenerate.
4. In a transportation problem, a dummy source is given a zero cost, while in an assignment problem, a dummy source is given a very high cost.
5. The Hungarian method operates on the principle of matrix reduction, whereby the cost table is reduced to a set of opportunity costs.
Answer:
1. True
2. False
3. True
4. False
5. True
Question 6.
Fill in the blanks.
1. In a transportation problem, we must make the number of _________ and ________ equal.
2. ________or ___________ are used to balance an assignment problem.
3. The method of finding an initial solution based on opportunity costs is called ___________
4. ___________ occurs when the number of occupied squares is less than the number of rows plus the number of columns minus one.
5. Both transportation and assignment problems are members of a category of LP problems called _____________
6. In the case of an unbalanced problem, shipping cost coefficients of _____________are assigned to each dummy factory or warehouse.
Answer:
1. units supplied, units demanded
2. Dummy rows, dummy columns
3. Vogel's approximation method
4. Degeneracy
5. Network flow problems
6. zero

Question 7.
Match the following.

Answer:
(a) - (iii)
(b) $-$ (i)
(c) $-$ (iv)
(d) $-(\mathrm{v})$
(e) - (ii)
Question 8.
The objective function of transportation problem is to
(a) Maximise total cost
(b) Minimise the total cost
(c) Total cost should be zero
(d) All the above
Answer:
(b) Minimise the total cost
Question 9.
In transportation problem, optimal solution can be verified by using
(a) NWC
(b) $\mathrm{LCM}$
(c) MODI method
(d) Matrix method
Answer:
(c) MODI method
Question 10.
The cells in the transportation problem can be classified as
(a) assigned cells and empty cells
(b) allocated cells and unallocated cells
(c) occupied and unoccupied cells
(d) assigned and unoccupied cells
Answer:

(c) occupied and unoccupied cells

Question 11.
In transportation problem if total supply $>$ total demand we add
(a) dummy row with cost 0
(b) dummy column with cost 0
(c) dummy row with cost 1
(d) dummy column with cost 1
Answer:
(b) dummy column with cost 0
Question 12.
In an LPP the objective function is to be
(a) Minimised
(b) Maximised
(c) (a) or (b)
(d) only (b)
Answer:
(c) (a) or (b)
Question 13.
The method used for solving an assignment problem is called
(a) Reduced matrix method
(b) MODI method
(c) Hungarian method
(d) Graphical method
Answer:
(c) Hungarian method
II. 2 Mark Questions
Question 1.
Consider 3 jobs to be assigned to 3 machines. The cost for each combination is shown in the table below. Find the minimal job machine combinations.

Solution:

Step 1:

Step 2:

Step 3: (Assignment)

Optimal assignment:

Question 2.
Find an initial basic feasible solution by LCM.

Solution:

Total cost $=(1 \times 2)+(6 \times 1)+(4 \times 4)+(4 \times 6)$ $=2+6+16+24$ $=48$

Question 3.
Find an initial basic feasible solution by the North West Corner Rule (NWC).

Solution:
$
\text { Total demand }=\text { Total supply }=60
$

$
\begin{aligned}
& \text { Total cost }=(10 \times 9)+(11 \times 6)+(12 \times 8)+(2 \times 3)+(25 \times 11) \\
& =90+66+96+6+275 \\
& =533
\end{aligned}
$
Question 4.
Find an initial basic feasible solution using Least cost method.

Solution:
Total Demand $=5+8+7+14=34$
Total Supply $=7+9+18=34$
Since they are equal, problem is balanced.

The minimum total transportation cost is $=(7 \times 10)+(2 \times 70)+(7 \times 40)+(3 \times 40)+(8 \times 8)$
$
\begin{aligned}
& +(7 \times 20) \\
& =70+140+280+120+64+140 \\
& =\text { Rs. } 814
\end{aligned}
$
Question 5.
Find the investment option using Maximin rule for the following:

Solution:

$\operatorname{Max}(5,-13,-5)=5$. Since the maximum payoff is 5 , by maximin criteria, the decision is to invest in bonds.
III. 3 and 5 Marks Questions
Question 1.
Find an optimal solution to the following transportation problem by North West Corner Method.

Solution:
Total supply $=65=$ Total demand. So the given problem is balanced. First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Total Transportation cost

Question 2.
Find an initial basic solution for the following transportation problem by Vogel's Approximation method.

Solution:
Total demand $=72+102+41=215$ and
Total supply $=76+82+77=235$.
Total supply $>$ Total demand. So we add a dummy constraint with 0 unit cost and with allocation $20(235-215)$. The modified table is

First allocation:

The maximum penalty is 16 . Allot 20 units to cell $\left(\mathrm{S}_2, \mathrm{D}_{\text {dummy }}\right)$
Second allocation:

Third allocation: 

Fourth allocation:

The final allocation table is given below.

The minimum total cost $=(76 \times 8)+(21 \times 24)+(41 \times 16)+(20 \times 0)+(72 \times 8)+(5 \times 16)=$ 2424
Question 3.
A company has 4 men available for 4 separate jobs. Only one man can work on anyone job. The cost of assigning each man to each job is given below. Find the optimal solution by the Hungarian method.

Solution:
The number of rows and columns are equal. So the given problem is a balanced assignment problem and we can get an optimal solution.
Step 1:

Step 2:

Step 3: (Assignment) 

We are not able to assign job for person B. Proceed as follows. Draw a minimum number of vertical and horizontal lines to cover all the zeros.

Subtract the smallest element 1 from all the uncovered elements and add it to the elements which lie at the intersection of two lines. Thus we obtain another reduced matrix for fresh assignment.

Tital cost is

Question 4.
There are five machines and five jobs are to be assigned and the cost matrix is given below.
Find the proper assignment.

Solution:
Step 1: (Row-reduction)

Step 2: (Column - reduction)

Step 3: (Assignment)

We are not able to assign a machine to job D. We proceed as follows.

The smallest uncovered element is 2 . Subtract 2 from all those elements which are not covered. Add 2 all elements which are at the intersection of two lines. Then proceed with the new matrix.

The assignment is as follows

Question 5.
The cost of transportation from 3 sources to four destinations are given in the follow-ing table. Obtain an initial basic feasible solution using
(i) North West Corner Rule (NWC)
(ii) Least Cost Method (LCM) and
(iii) Vogel's Approximation Method (VAM)

Solution:
(i) North West Corner Rule
We start by allotting the units to the North -West Comer cell. We show all the allocations in a single table.

Total transportation cost is $(200 \times 4)+(50 \times 2)+(350 \times 7)+(100 \times 5)+(200 \times 3)+(1 \times$
$
\begin{aligned}
& \text { 300) } \\
& =800+100+2450+500+600+300 \\
& =\text { Rs. } 4750
\end{aligned}
$
(ii) Least cost method (LCM)

Transportation cost is $=(250 \times 2)+(200 \times 3)+(150 \times 7)+(100 \times 5)+(200 \times 3)+(300 \times 1)$
$
\begin{aligned}
& =500+600+1050+500+600+300 \\
& =\text { Rs. } 3550
\end{aligned}
$

$
\begin{aligned}
& \text { Transportation cost is }=(250 \times 2)+(200 \times 3)+(150 \times 7)+(100 \times 5)+(200 \times 3)+(300 \times 1) \\
& =500+600+1050+500+600+300 \\
& =\text { Rs. } 3550
\end{aligned}
$
(iii) Vogel Approximation Method (VAM)

There are five penalties which have the maximum value 2 . The cell with the least cost is row 3 and hence select cell (3, D) for allocation.

There are four penalties which have maximum value 2 . Select cell $(1, B)$ which has the least cost for allocation.

$\text { The largest penalty is } 6 \text {. Allot units to cell }(2, \mathrm{~A})$

The largest penalty is 3 . Allot units to cell $(3, \mathrm{~B})$

We first allot 50 units to cell $(3$, C) which has less cost. Then the balance units we allot to cell $(2, \mathrm{C})$. We get the final allocation table as follows.

Transportation cost is $=(250 \times 2)+(200 \times 3)+(250 \times 5)+(150 \times 4)+(50 \times 3)+(300 \times 1)$ $=500+600+1250+600+150+300$
$=$ Rs. 3400