Question 1 |
Two business owners Shveta and Ashok run their businesses in two different states. Each
of them, independent of the other, produces two products A and B, sells them at Rs.
2,000 per kg and Rs, 3,000 per kg, respectively, and uses Linear Programming to
determine the optimal quantity of A and B to maximize their respective daily revenue.
Their constraints are as follows: i) for each business owner, the production process is
such that the daily production of A has to be at least as much as B, and the upper
limit for production of B is 10 kg per day, and ii) the respective state regulations restrict
Shveta?s production of A to less than 20 kg per day and Ashok's production of A to
less than 15 kg per day. The demand of both A and B in both the states is very high
and everything produced is sold.
The absolute value of the difference in daily (optimal) revenue of Shveta and Ashok is ________ thousand Rupees (round off to 2 decimal places)
The absolute value of the difference in daily (optimal) revenue of Shveta and Ashok is ________ thousand Rupees (round off to 2 decimal places)
6 | |
7 | |
12 | |
10 |
Question 1 Explanation:
\begin{aligned} \text { Maximum } z &=2000 x_{1}+3000 x_{2} \\ A \rightarrow x_{1} \text { units } \quad x_{1} &\geq x_{2} \\ B \rightarrow x_{2} \text { units } \quad x_{2} &\geq 10 \\ x_{1} & \lt 20 \\ x_{1} & \lt 15 \\ \text { Shveta's Profit } &=\text { Rs. } 70000 \text { at }(20,10) \\ \text { Ashok's Profit } &=\text { Rs. } 60000 \text { at }(15,10) \end{aligned}
Difference Rs. 10000
Difference Rs. 10000
Question 2 |
Two models, P and Q, of a product earn profits of Rs. 100 and Rs. 80 per piece, respectively. Production times for P and Q are 5 hours and 3 hours, respectively, while the total production time available is 150 hours. For a total batch size of 40, to maximize profit, the number of units of P to be produced is ____________.
12 | |
15 | |
18 | |
20 |
Question 2 Explanation:
Max. Z=100 x P + 80 x Q
\begin{aligned} 5P+3Q&\leq 150\\ P+Q&\leq 40\\ \frac{P}{30}+\frac{Q}{30}&=1 \end{aligned}
Putting value of corner point in objective function
Z(A)=3200
Z(B)=3500
Z(C)=3000
Maximum at B,
P=15, Q=25
\begin{aligned} 5P+3Q&\leq 150\\ P+Q&\leq 40\\ \frac{P}{30}+\frac{Q}{30}&=1 \end{aligned}
Putting value of corner point in objective function
Z(A)=3200
Z(B)=3500
Z(C)=3000
Maximum at B,
P=15, Q=25
Question 3 |
Maximize Z=15X_{1}+20X_{2}
subject to
12X_{1}+4X_{2}\geq 36
12X_{1}-6X_{2}\leq24
X_{1},X_{2}\geq 0
The above linear programming problem has
subject to
12X_{1}+4X_{2}\geq 36
12X_{1}-6X_{2}\leq24
X_{1},X_{2}\geq 0
The above linear programming problem has
infeasible solution | |
unbounded solution | |
alternative optimum solutions | |
degenerate solution |
Question 3 Explanation:
\begin{aligned} 12X_1+4X_2&\geq 36 \\ 3X_1-X_2 & \geq 9 \\ \\ 12X_1-6X_2&\leq 24 \\ 2X_1-X_2&\leq 4 \\ \text{Let},&\\ 3X_1+X_2&=9 \\ X_1=0 & X_2=9\\ X_1=3 & X_2=0\\ \\ 2X_1-X_2&=4\\ X_1=0 & X_2=-4\\ X_1=2 & X_2=0 \end{aligned}
Cylindrical plug gauge is used to measure inside diameter of straight hole.
Cylindrical plug gauge is used to measure inside diameter of straight hole.
Question 4 |
For the linear programming problem:
Maximize Z=3X_{1}+2X_{2}
Subject to
-2X_{1}+3X_{2} \leq 9
X_{1}-5X_{2}\geq -20
X_{1},X_{2}\geq 0
The above problem has
Maximize Z=3X_{1}+2X_{2}
Subject to
-2X_{1}+3X_{2} \leq 9
X_{1}-5X_{2}\geq -20
X_{1},X_{2}\geq 0
The above problem has
unbounded solution | |
infeasible solution | |
alternative optimum solution | |
degenerate solution |
Question 4 Explanation:
\begin{aligned} z &=3 x_{1}+2 x_{2} \\-2 x_{1}+3 x_{2} & \leq 9 \\ x_{1}-5 x_{2} & \geq-20 \\ x_{1}, x_{2} & \leq 0 \end{aligned}
\therefore changing the inequalities to equalities
-2 x_{1}+3 x_{2}=9
x_{1}-5 x_{2}=-20
\therefore The soluction is unbounded.
\therefore changing the inequalities to equalities
-2 x_{1}+3 x_{2}=9
x_{1}-5 x_{2}=-20
\therefore The soluction is unbounded.
Question 5 |
Consider an objective function Z(x_{1},x_{2})=3x_{1}+9x_{2} and the constraints
x_{1}+x_{2}\leq 8,
x_{1}+2x_{2}\leq 4,
x_{1}\geq 0,x_{2}\geq 0.
The maximum value of the objective function is __________
x_{1}+x_{2}\leq 8,
x_{1}+2x_{2}\leq 4,
x_{1}\geq 0,x_{2}\geq 0.
The maximum value of the objective function is __________
98 | |
18 | |
48 | |
20 |
Question 5 Explanation:
\begin{aligned} (Z)_{(4,0)}&=3 \times 4+9 \times 0=12 \\ (Z)_{(0,2)}&=3 \times 0+9 \times 2=18 \\ (Z)_{(0,0)}&=0 \quad \text { (Trivial solution) } \\ Z_{\max }&=18 \end{aligned}
Question 6 |
A linear programming problem is shown below.
Maximize
3x+7y
Subject to
3x+7y\leq 10
4x+6y\leq 8
x,y\geq 0
It has
Maximize
3x+7y
Subject to
3x+7y\leq 10
4x+6y\leq 8
x,y\geq 0
It has
an unbounded objective function | |
exactly one optimal solution. | |
exactly two optimal solutions. | |
infinitely many optimal solutions. |
Question 6 Explanation:
\begin{aligned}\text { Max. } 3 x&+z y \\ \text { Sub to } 3 x&+7 y \leq 10\\ 4 x&+6 y \leq 8 \\ &x, y \geq 0 \end{aligned}
\Rightarrow It has exactly one optimal solution.
Question 7 |
One unit of product P_{1}
requires 3 kg of resource R_{1}
and 1kg of resource R_{2}. One unit of product P_{2}
requires 2kg of resource R_{1}
and 2kg of resource R_{2}
. The profits per unit by selling product P_{1}
and R_{2}
are Rs.2000 and Rs.3000 respectively. The manufacturer has 90kg of resource R_{1}
and 100kg of resource R_{2}
The manufacturer can make a maximum profit of Rs.
The manufacturer can make a maximum profit of Rs.
60000 | |
135000 | |
150000 | |
200000 |
Question 7 Explanation:
Since all Z_j-C_j\geq 0, an optimal basic feasible solution has been attained. Thus, the optimum solution to the given LPP is
Max \; Z=2000 \times 0+3000 \times 45=135000 \text{ with }P-1=0 \text{ and }P_2=45
Max \; Z=2000 \times 0+3000 \times 45=135000 \text{ with }P-1=0 \text{ and }P_2=45
Question 8 |
One unit of product P_{1}
requires 3 kg of resource R_{1}
and 1kg of resource R_{2}
. One unit of product P_{2}
requires 2kg of resource R_{1}
and 2kg of resource R_{2}
. The profits per unit by selling product P_{1}
and R_{2}
are Rs.2000 and Rs.3000 respectively. The manufacturer has 90kg of resource R_{1}
and 100kg of resource R_{2}
The unit worth of resource R_{2} i.e., dual price of resource R_{2} in Rs. Per kg is
The unit worth of resource R_{2} i.e., dual price of resource R_{2} in Rs. Per kg is
0 | |
1350 | |
1500 | |
2000 |
Question 8 Explanation:
Because the constraint on resource 2 has no effect on the feasible region.
Question 9 |
Simplex method of solving linear programming problem uses
all the points in the feasible region | |
only the corner points of the feasible region | |
intermediate points within the infeasible region | |
only the interior points in the feasible region. |
Question 9 Explanation:
Simplex method provides an algorithm which consists in moving from one point of the region of feasible solutions to another in such a manner that the value of the objective function at the succeeding point is less (or more, as the case may be) than at the preceding point. This procedure of jumping from one point to another is then repeated. Since the number of points is finite, the method leads to an optimal point in a finite number of steps.
Therefore simplex method only uses the interior points in the feasible region.
Therefore simplex method only uses the interior points in the feasible region.
Question 10 |
Consider the following Linear Programming Problem
(LPP): Maximize z=3x_{1}+2x_{2}
Subject to x_{1}\leq 4
x_{2}\leq 6
3x_{1}+2x_{2}\leq 18
x_{1}\geq 0 ,x_{2}\geq 0
(LPP): Maximize z=3x_{1}+2x_{2}
Subject to x_{1}\leq 4
x_{2}\leq 6
3x_{1}+2x_{2}\leq 18
x_{1}\geq 0 ,x_{2}\geq 0
The LPP has a unique optimal solution | |
The LPP is infeasible | |
The LPP is unbounded | |
The LPP has multiple optimal solutions |
Question 10 Explanation:
Linear Programming Problem (LPP)
Maximize, z=3 x_{1}+2 x_{2}
\begin{aligned} \text{Constraints }\quad x_{1} &\leq 4 &\quad \ldots(i)\\ x_{2} & \leq 6 &\quad \ldots(ii)\\ 3 x_{1}+2 x_{2} & \leq 18 &\quad \ldots(iii)\\ x_{1} & \leq 0, x_{2} \leq 0 &\quad \ldots(iv) \end{aligned}
Using graphical method
Because objective function have slope same as constraint (iii) i.e. objective function is parallel to constraint. Therefore the LPP has multiple optimal solution.
For example at Point B
\begin{aligned} \text { Maximum, } z &=3 x_{1}+2 x_{2} \\ &=3(4)+2(3)=18 \end{aligned}
and at point C,
Maximum, z=3(2)+2(6)=18
Maximize, z=3 x_{1}+2 x_{2}
\begin{aligned} \text{Constraints }\quad x_{1} &\leq 4 &\quad \ldots(i)\\ x_{2} & \leq 6 &\quad \ldots(ii)\\ 3 x_{1}+2 x_{2} & \leq 18 &\quad \ldots(iii)\\ x_{1} & \leq 0, x_{2} \leq 0 &\quad \ldots(iv) \end{aligned}
Using graphical method
Because objective function have slope same as constraint (iii) i.e. objective function is parallel to constraint. Therefore the LPP has multiple optimal solution.
For example at Point B
\begin{aligned} \text { Maximum, } z &=3 x_{1}+2 x_{2} \\ &=3(4)+2(3)=18 \end{aligned}
and at point C,
Maximum, z=3(2)+2(6)=18
There are 10 questions to complete.
6th question must give < 8. in question it has given 0.
Thank you for your suggestions. We have updated the correction suggested by You.
in question no 30…i think Z is max at x2 = 10
pardon….question no 20