Question 1 |
Which one of the options given represents the feasible region of the linear
programming model:
\begin{aligned} Maximize\;\; 45X_1&+60X_2 \\ X_1&\leq 45 \\ X_2&\leq 50 \\ 10X_1+10X_2& \geq 600 \\ 25X_1+5X_2&\leq 750 \end{aligned}
\begin{aligned} Maximize\;\; 45X_1&+60X_2 \\ X_1&\leq 45 \\ X_2&\leq 50 \\ 10X_1+10X_2& \geq 600 \\ 25X_1+5X_2&\leq 750 \end{aligned}
Region P | |
Region Q | |
Region R | |
Region S |
Question 1 Explanation:
\begin{aligned}
x_1&=45 &...(i)\\
x_2&= 50&...(ii)\\
10x_1+10x_2&=600 \\
or\; x_1+x_2&=60&...(iii) \\
25x_1+5x_2&=750 \\
or\; 5x_1+x_2&=150&...(iv) \\
\end{aligned}
By drawing the curve we get 3 values of x_1 and x_2 as (10, 50), (20, 50), (22.5, 37.5)
So, Z_{max}=45x_1+60x_2 for (10,50)
Z_{max}=450+3000=4450
for (20,50)
Z_{max}=45\times 20+50 \times 60=3900
for (22.5, 37.5)
Z_{max}=45\times 22.5+60 \times 37.5=3262.5
So, Z_{max}=3900\; for \; (x_1,x_2)=(20,50)
By drawing the curve we get 3 values of x_1 and x_2 as (10, 50), (20, 50), (22.5, 37.5)
So, Z_{max}=45x_1+60x_2 for (10,50)
Z_{max}=450+3000=4450
for (20,50)
Z_{max}=45\times 20+50 \times 60=3900
for (22.5, 37.5)
Z_{max}=45\times 22.5+60 \times 37.5=3262.5
So, Z_{max}=3900\; for \; (x_1,x_2)=(20,50)
Question 2 |
With reference to the Economic Order Quantity (EOQ) model, which one of the
options given is correct?
Curve P1: Total cost, Curve P2: Holding cost, Curve P3: Setup cost, and Curve P4: Production cost. | |
Curve P1: Holding cost, Curve P2: Setup cost,
Curve P3: Production cost, and Curve P4: Total cost. | |
Curve P1: Production cost, Curve P2: Holding cost, Curve P3: Total cost, and Curve P4: Setup cost. | |
Curve P1: Total cost, Curve P2: Production cost, Curve P3: Holding cost, and Curve P4: Setup cost. |
Question 2 Explanation:
Question 3 |
A machine produces a defective component with a probability of 0.015. The
number of defective components in a packed box containing 200 components
produced by the machine follows a Poisson distribution. The mean and the
variance of the distribution are
3 and 3, respectively | |
\sqrt{3} and \sqrt{3} , respectively | |
0.015 and 0.015, respectively | |
3 and 9, respectively |
Question 3 Explanation:
P = 0.015
n = 200
mean =\lambda =np= 200 \times 0.015 = 3
variance =\sigma ^2=\lambda =3
n = 200
mean =\lambda =np= 200 \times 0.015 = 3
variance =\sigma ^2=\lambda =3
Question 4 |
A project consists of five activities (A, B, C, D
and E). The duration of each activity follows beta
distribution. The three time estimates (in weeks)
of each activity and immediate predecessor(s) are
listed in the table. The expected time of the project
completion is __________ weeks (in integer).
\begin{array}{|c|c|c|c|c|} \hline \text{Activity} & \text{Optimistic time (week)} & \text{Most likely time (week)} &\text{Pessimistic time (week)} &\text{Immediatepredecessor(s)} \\ \hline A& 4 & 5 &6 & None\\ \hline B& 1& 3 & 5 & A\\ \hline C& 1& 2 & 3 & A\\ \hline D& 2& 4 & 6 & C\\ \hline E& 3& 4 & 5 &B,D \\ \hline \end{array}
\begin{array}{|c|c|c|c|c|} \hline \text{Activity} & \text{Optimistic time (week)} & \text{Most likely time (week)} &\text{Pessimistic time (week)} &\text{Immediatepredecessor(s)} \\ \hline A& 4 & 5 &6 & None\\ \hline B& 1& 3 & 5 & A\\ \hline C& 1& 2 & 3 & A\\ \hline D& 2& 4 & 6 & C\\ \hline E& 3& 4 & 5 &B,D \\ \hline \end{array}
12 | |
15 | |
16 | |
18 |
Question 4 Explanation:
\begin{array}{|c|c|}
\hline
\text{Activity}& t_e=\frac{t_0+4t_m+t_p}{6}\\ \hline
A&5\\ \hline
B&3\\ \hline
C&2\\ \hline
D&4\\ \hline
E&4 \\ \hline
\end{array}
\begin{array}{|c|c|} \hline \text{Path}& \text{Duration}\\ \hline 1-2-4-5(A-B-E)&12\\ \hline 1-2-3-4-5 (A-C-D-E)&15\\ \hline \end{array}
Expected completion time of the project = critical path duration = 15 weeks
\begin{array}{|c|c|} \hline \text{Path}& \text{Duration}\\ \hline 1-2-4-5(A-B-E)&12\\ \hline 1-2-3-4-5 (A-C-D-E)&15\\ \hline \end{array}
Expected completion time of the project = critical path duration = 15 weeks
Question 5 |
A manufacturing unit produces two products P1
and P2. For each piece of P1 an P2, the table below
provides quantities of materials M1, M2, and M3
required, and also the profit earned. The maximum
quantity available per day for M1, M2 and M3 is
also provided. The maximum possible profit per
day is Rs __________.
\begin{array}{|c|c|c|c|c|} \hline &\text{M1 } &\text{M2}&\text{M3}&\text{Profit per piece (Rs.)} \\ \hline P1 & 2 &2 &0&150\\ \hline P2 & 3& 1&2&100\\ \hline \text{Maximum quantity available per day}& 70& 50&40\\ \hline \end{array}
\begin{array}{|c|c|c|c|c|} \hline &\text{M1 } &\text{M2}&\text{M3}&\text{Profit per piece (Rs.)} \\ \hline P1 & 2 &2 &0&150\\ \hline P2 & 3& 1&2&100\\ \hline \text{Maximum quantity available per day}& 70& 50&40\\ \hline \end{array}
5000 | |
4000 | |
3000 | |
6000 |
Question 5 Explanation:
Let,
x_1= No. of units of product P_1
x_2= No. of units of product P_2
Max, Z=150x_1+100x_2
\begin{aligned} s.t.\;\; 2x_1+3x_2 &\leq 70 \\ 2x_1+3x_2&\leq 50 \\ 2x_2 &\leq 40 \\ x_1\geq 0, x_2&\geq 0 \end{aligned}
Point B: Intersection of
\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_2 &\leq 40 \\ x_2=20, x_1&=5 \end{aligned}
Point C: Intersection of
\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_1+x_2 &\leq 50 \\ \text{By sloving }2x_2 &=20 \\ \Rightarrow x_2=10, x_1&=20 \end{aligned}
\begin{aligned} Z_{max}&=150x_1+100x_2\\ Z|_{A(0,20)}&=150(0)+100(20)=2000\\ Z|_{B(5,20)}&=150(5)+100(20)=2750\\ Z|_{C(20,10)}&=150(20)+100(10)=4000\\ Z|_{D(25,0)}&=150(25)+100(0)=3750\\ \Rightarrow x_1&=20,x_2=10\\ Z_{max}&=4000 \end{aligned}
x_1= No. of units of product P_1
x_2= No. of units of product P_2
Max, Z=150x_1+100x_2
\begin{aligned} s.t.\;\; 2x_1+3x_2 &\leq 70 \\ 2x_1+3x_2&\leq 50 \\ 2x_2 &\leq 40 \\ x_1\geq 0, x_2&\geq 0 \end{aligned}
Point B: Intersection of
\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_2 &\leq 40 \\ x_2=20, x_1&=5 \end{aligned}
Point C: Intersection of
\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_1+x_2 &\leq 50 \\ \text{By sloving }2x_2 &=20 \\ \Rightarrow x_2=10, x_1&=20 \end{aligned}
\begin{aligned} Z_{max}&=150x_1+100x_2\\ Z|_{A(0,20)}&=150(0)+100(20)=2000\\ Z|_{B(5,20)}&=150(5)+100(20)=2750\\ Z|_{C(20,10)}&=150(20)+100(10)=4000\\ Z|_{D(25,0)}&=150(25)+100(0)=3750\\ \Rightarrow x_1&=20,x_2=10\\ Z_{max}&=4000 \end{aligned}
There are 5 questions to complete.
Hello,
I believe the answer to question 8 is option D (CNC Milling Machines). Kindly make the changes accordingly.
Love your effort.
Dear Suraj,
We have updated the answer as D.
149 answers 400 hoga