By QB365 on 03 Sep, 2022
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.
12th Standard
Business Maths
Obtain the initial solution for the following problem
Determine an initial basic feasible solution to the following transportation problem using North West corner rule.
Here Oi and Dj represent ith origin and jth destination.
Obtain an initial basic feasible solution to the following transportation problem using least cost method.
Here Oi and Dj denote ith origin and jth destination respectively.
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.
Solve the following assignment problem.
Consider the following pay-off (profit) matrix Action States
Action | States | |||
(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.
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) | |||
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.
Consider the following pay-off matrix
Alternative | Pay – offs (Conditional events) | |||
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.
Find an initial basic feasible solution of the following problem using north west corner rule.
Determine an initial basic feasible solution of the following transportation problem by north west corner method
Obtain an initial basic feasible solution to the following transportation problem by using least- cost method.
Determine basic feasible solution to the following transportation problem using North west Corner rule.
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
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)
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
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.) | ||
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
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?
Following pay-off matrix, which is the optimal decision under each of the following rule
(i) maxmin
(ii) minimax
Act | States of nature | |||
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 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?
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 | ||
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.
Answers
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
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
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.
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
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
Action | States | Minimum | |||
(s1) | (s2) | (s3) | (s4) | ||
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
Alternative | Pay – offs (Conditional events) | Minimum Pay off | |||
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
Alternative | Pay – offs (Conditional events) | Minimum pay off | |||
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.
First allocation:
Second allocation:
Third allocation:
Fourth allocation:
Fifth allocation:
Final allocation:
The trasportation cost:
\( =(16 \times 5)+(3 \times 3)+(15 \times 7)+(22 \times 9) +(9 \times 7)+(25 \times 5) \)
\(=80+9+105+198+63+125=Rs. 580\)
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
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 :
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
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
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
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.
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
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.
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
∴ A2 and A3 are optimal decisions according to 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
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.