QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Business Maths Subject - Operations Research, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

Download Tamil Nadu 12th Standard Business Maths and Statistics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Generic 3

20 x 3 = 60

A person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay-off matrix based on three potential economic conditions is given in the following table:

Alternative	Economic conditions
Alternative	High growth(Rs.)	Normal growth(Rs.)	Slow growth (Rs.)s
Stocks	10000	7000	3000
Bonds	8000	6000	1000
Debentures	6000	6000	6000

Determine the best investment plan using each of following criteria i) Maxmin ii) Minimax.

The following table summarizes the supply, demand and cost information for four factors S₁, S₂, S₃, S₄. shipping goods to three warehouses D₁, D₂, D₃.

Find an initial solution by using north west corner rule. What is the total cost for this solution?

Following pay-off matrix, which is the optimal decision under each of the following rule
(i) maxmin
(ii) minimax

Act	States of nature
Act	S₁	S₂	S₃	S₄
A₁	14	9	10	5
A₂	11	10	8	7
A₃	9	10	10	11
A₄	8	10	11	13

The research department of Hindustan Ltd. has recommended to pay marketing department to launch a shampoo of three different types. The marketing types of shampoo to be launched under the following estimated pay-offs for various level of sales.
Types of shampoo Estimated Sales (in Units) 15000 10000 5000 Egg shampoo 30 10 10 Clinic Shampoo 40 15 5 Deluxe Shampoo 55 20 3 What will be the marketing manager’s decision if
(i) Maximin and
(ii) Minimax principle applied?

A farmer wants to decide which of the three crops he should plant on his 100-acre farm. The profit from each is dependent on the rainfall during the growing season. The farmer has categorized the amount of rainfall as high medium and low. His estimated profit for each is shown in the table.

Rainfall	Estimated Conditional Profit(Rs.)
Rainfall	crop A	crop B	crop C
High	8000	3500	5000
Medium	4500	4500	5000
Low	2000	5000	4000

If the farmer wishes to plant only crop, decide which should be his best crop using
(i) Maximin
(ii) Minimax

Given the following pay-off matrix(in rupees) for three strategies and two states of nature.

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

Select a strategy using each of the following rule
(i) Maximin
(ii) Minimax

Three jobs A, B and C one to be assigned to three machines U, V and W. The processing cost for each job machine combination is shown in the matrix given below.
Determine the allocation that minimizes the overall processing cost.

(cost is in Rs. per unit)

Find the initial basic feasible solution of the following transportation problem :

Using (i) North West Corner rule
(ii) Least Cost method
(iii) Vogel’s approximation method

Determine basic feasible solution to the following transportation problem using North west Corner rule.

10.

Obtain an initial basic feasible solution to the following transportation problem by using least- cost method.

11.

Determine an initial basic feasible solution of the following transportation problem by north west corner method

12.

Find an initial basic feasible solution of the following problem using north west corner rule.

13.

Consider the following pay-off matrix

Alternative	Pay – offs (Conditional events)
Alternative	A₁	A₂	A₃	A₄
E₁	7	12	20	27
E₂	10	9	10	25
E₃	23	20	14	23
E₄	32	24	21	17

Using minmax principle, determine the best alternative.

14.

A business man has three alternatives open to him each of which can be followed by any of the four possible events. The conditional pay offs for each action - event combination are given below:

Alternative	Pay – offs (Conditional events)
Alternative	A	B	C	D
X	8	0	-10	6
Y	-4	12	18	-2
A₃	14	6	0	8

Determine which alternative should the businessman choose, if he adopts the maximin principle.

15.

Consider the following pay-off (profit) matrix Action States

Action	States
Action	(s₁)	(s₂)	(s₃)	(s₄)
A₁	5	10	18	25
A₂	8	7	8	23
A₃	21	18	12	21
A₄	30	22	19	15

Determine best action using maximin principle.

16.

Solve the following assignment problem.

17.

Determine how much quantity should be stepped from factory to various destinations for the following transportation problem using the least cost method

Cost are expressed in terms of rupees per unit shipped.

18.

Obtain an initial basic feasible solution to the following transportation problem using least cost method.

Here O_iand D_j denote i^th origin and j^th destination respectively.

19.

Determine an initial basic feasible solution to the following transportation problem using North West corner rule.

Here O_i and D_j represent i^th origin and j^th destination.

20.

Obtain the initial solution for the following problem

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 3

20 x 3 = 60

	Economic Conditions
Alternative	High growth	Normal growth	Slow growth	Minimum payoff	Maximum payoff
Stocks	10000	7000	3000	3000	10000
Bonds	8000	6000	1000	1000	8000
Debentures	6000	6000	6000	6000	6000

(i) Max (3000, 1000, 6000) = 6000
∴ Debentures is the best using maximin principle
(ii) Max (10000, 8000, 6000) = 6000
∴ Debentures is the best using minimax principle.

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

Act	States of Nature				Minimum payoff	Maximum payoff
Rainfall	S₁	S₂	S₃	S₄
_A1	14	9	10	5	5	14
_A2	11	10	8	7	7	11
_A3	9	10	10	11	9	11
_A4	8	10	11	13	8	13

(i) Max (5, 7, 9, 8) = 9
∴ A₃ is the optimal decision according to maximin principle
(ii) Min (14, 11, 11, 13) = 11
∴ A₂and A₃ are optimal decisions according to minimax principle

Estimated Sales (in Units) Types of Shampoo 15000 10000 5000 Minimum payoff Maximum payoff Egg Shampoo 30 10 10 10 30 Clinic Shampoo 40 15 5 5 40 Deluxe Shampoo 55 20 3 3 55 (i) Max (10, 5, 3) = 10.
Since the maximum pay-off is 10 units, the marketing manager has to choose Egg shampoo by Maximin rule.
(ii) Min (30, 40, 55) = 30.
Since the minimum pay-off is 30 units, the marketing manager has to choose Egg shampoo by minimax rule.

	Estimated Conditional Profit 0
Rainfall	High	Medium	Low	Minimum payoff	Maximum payoff
Crop A	8000	4500	2000	2000	8000
Crop B	3500	4500	5000	3500	5000
Crop C	5000	5000	4000	4000	5000

(i) Max (2000,3500,4000) = 4000
∴ Crop C is the best according to maximin criteria
(ii) Min (8000,5000,5000) = 5000
∴ Crop B and C are best according to minimax criteria

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.

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

A	0	8	14	17
B	0	15	6	10
C	1	3	0	11

Here column V has no zero. Go to Step 2.
Step 2 : Select the minimum element in each column and subtract this from all the elements in its column.

	U	V	W
A	0	5	14
B	0	12	6
C	1	0	0
Column minimum	3

New, each row and column contains atleast one zero. Hence, assignments can be made.
Step 3 : Examine the rows with exactly one zero. Mark it by Ԡ and draw a vertical line.
After examining the row, examine the column with one zero mark it by Ԡ and draw a horizontal line.

New the elements not lying on the line are

5	14
12	6

and min is 5 subtract 5 from all these number and add 5 to 1which lies in the intersecting lines. Other numbers remains the same.
A new cost matrix will be formed and repeat step 3.

Thus, all the 3 assignments have been made.
∴The optimal assignment schedule and total cost is

Job	Machine	Cost
A	V	25
B	U	10
C	W	11

Total cost = Rs. 46

i) The total transportation cost is
x₁₁= 7, x₂₁= 3, x₂₂= 9, x₃₂= 1, x₃₃= 10
= (7 \(\times\)1) + (3\(\times\)0) + (9\(\times\)4) + (1\(\times\)1) + (10\(\times\)5)
Total Cost = 94
ii) Total Transportation cost
x₁₁= 7, x₂₁= 10, x₂₃= 2, x₃₂= 10, x₃₃= 1
\( =(7 \times 6)+(10 \times 0)+(2 \times 2)+(10 \times 1)+(1 \times 5) \)
= 42 + 0 + 4 + 10 + 5
Total Cost = 61
iii) Total transportation cost
x₁₁= 7, x₂₁= 2, x₂₃= 10, x₃₁= 1, x₃₂= 10
\( =7 \times 1+2 \times 0+10 \times 2+1 \times 3+10 \times 1 \)
= 7 + 20 + 3 + 10
Total Cost = 40

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

10.

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

11.

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

12.

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Final allocation:

The trasportation cost:
\( =(16 \times 5)+(3 \times 3)+(15 \times 7)+(22 \times 9) +(9 \times 7)+(25 \times 5) \)
\(=80+9+105+198+63+125=Rs. 580\)

13.

Alternative	Pay – offs (Conditional events)				Minimum pay off
Alternative	A₁	A₂	A₃	A₄
E₁	7	12	20	27	27
E₂	10	9	10	25	25
E₃	23	20	14	23	23
E₄	32	24	21	17	32

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

14.

Alternative	Pay – offs (Conditional events)				Minimum Pay off
Alternative	A	B	C	D
X	8	0	-10	6	-10
Y	-4	12	18	-2	-4
A₃	14	6	0	8	0

Max (–10,–4, 0) = 0. Since the maximum payoff is 0, the alternative Z is selected by the businessman

15.

Action	States				Minimum
Action	(s₁)	(s₂)	(s₃)	(s₄)	Minimum
A₁	5	10	18	25	5
A₂	8	7	8	23	7
A₃	21	18	12	21	12
A₄	30	22	19	15	15

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

16.

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

Task	Men	cost
P	1	9
Q	3	6
R	2	20
s	d	0
Total cost		35

The optimal assignment (minimum) cost = Rs. 35

17.

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

18.

Total Supply = Total Demand = 24
\(\therefore\)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 (O₁, D₁) and (O3, D₄)
Take the Cell (O₁, D₁) arbitrarily.
Allocatemin (6,4) = 4 units to this cell.

The reduced table is

The least cost corresponds to the cell (O₃, D₄). Allocate min (10,6) = 6 units to this cell.

The reduced table is

The least costis 2 corresponds to the cells (O₁, D₂), (O₂, D₃), (O₃, D₂), (O₃, D₃)
Allocate min (2,6) = 2 units to this cell.

The reduced table is

The least cost is 2 corresponds to the cells (O₂, D₃), (O₃, D₂), (O₃, D₃)
Allocate min ( 8,8) = 8 units to this cell.

The reduced table is

Here allocate 4 units in the cell (O₃, D₂)

Thus we have the following allocations:

Transportation schedule :
O₁⟶D₁, O₁⟶D₂,O₂⟶D₃,O₃⟶D₂,O₃⟶D₄
Total transportation cost
= (4×1)+ (2×2)+(8×2)+(4×2)+(6×1)
= 4+4+16+8+6
= Rs. 38.

19.

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

20.

Here total supply = 5 + 8 + 7 + 14 = 34, Total demand = 7 + 9 + 18 = 34
(i.e) Total supply =Total demand
Therefore The given problem is balanced transportation problem.
\(\therefore\) we can findan initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North West Corner. Here the cell is (1, A)
Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.
i.e. x₁₁ = min (5, 7) = 5

Reduced transportation table is

Now the cell in the North west corner is (2, A)
Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted.
i.e. x₁₂ = min (2, 8) = 2

Reduced transportation table is

Here north west corner cell is (2, B) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the second column is exhausted.
i.e. x₂₂ = min (6, 9) = 6

Reduced transportation table is

Here north west corner cell is (3,B).
Allocate as much as possible in the first cell so that either the capacity of third row is exhausted or the destination requirement of the second column is exhausted.
i.e. x₃₂ = min (7, 3) = 3

Reduced transportation table is

Here north west corner cell is (3,C) Allocate as much as possible in the first cell so that either the capacity of third row is exhausted or the destination requirement of the third column is exhausted.
i.e. x₃₃ = min (4, 18) = 4

Reduced transportation table and final allocation is x₄₄ = 14

Thus we have the following allocations

Transportation schedule : 1⟶A, 2⟶B, 3⟶B, 3⟶C, 4⟶C
The total transportation cost.
= (5 \(\times\) 2) + (2 x\(\times\) 3) + (6 \(\times\) 3) + (3 \(\times\) 4) + (4 \(\times\) 7) + (14 \(\times\) 2)
= Rs. 102