Question 1 |
A set of jobs A, B, C, D, E, F, G, H arrive at time t = 0 for processing on turning and grinding machines. Each job needs to be processed in sequence
- first on the turning machine and second on the grinding machine, and the grinding must occur immediately after turning. The processing times of the jobs are given below.
\begin{array}{|l|l|l|l|l|l|l|l|l|}\hline \text{Job} & \text{A} & \text{B} & \text{C} & \text{D} & \text{E} & \text{F} & \text{G} & \text{H} \\ \hline \text{Turning (minutes)} & \text{2} & \text{4} &\text{8}&\text{9} &\text{7} &\text{6} &\text{5} &\text{10} \\ \hline \text{Grinding (minutes)} & \text{6} & \text{1} &\text{3} &\text{7} &\text{9}&\text{5} &\text{2} &\text{4} \\ \hline \end{array}
If the makespan is to be minimized, then the optimal sequence in which these jobs must be processed on the turning and grinding machines is
\begin{array}{|l|l|l|l|l|l|l|l|l|}\hline \text{Job} & \text{A} & \text{B} & \text{C} & \text{D} & \text{E} & \text{F} & \text{G} & \text{H} \\ \hline \text{Turning (minutes)} & \text{2} & \text{4} &\text{8}&\text{9} &\text{7} &\text{6} &\text{5} &\text{10} \\ \hline \text{Grinding (minutes)} & \text{6} & \text{1} &\text{3} &\text{7} &\text{9}&\text{5} &\text{2} &\text{4} \\ \hline \end{array}
If the makespan is to be minimized, then the optimal sequence in which these jobs must be processed on the turning and grinding machines is
A-E-D-F-H-C-G-B | |
A-D-E-F-H-C-G-B | |
G-E-D-F-H-C-A-B | |
B-G-C-H-F-D-E-A |
Question 1 Explanation:
Sequencing,
\begin{array}{|c|c|c|} \hline A & (2) & 6 \\ \hline B & 4 & (1) \\ \hline C & 8 & (3) \\ \hline D & 9 & (7) \\ \hline E & (7) & 9 \\ \hline F & 6 & (5) \\ \hline G & 5 & (2) \\ \hline H & 10 & (4) \\ \hline \end{array}
AEDFHCGB
\begin{array}{|c|c|c|} \hline A & (2) & 6 \\ \hline B & 4 & (1) \\ \hline C & 8 & (3) \\ \hline D & 9 & (7) \\ \hline E & (7) & 9 \\ \hline F & 6 & (5) \\ \hline G & 5 & (2) \\ \hline H & 10 & (4) \\ \hline \end{array}
AEDFHCGB
Question 2 |
Maximize Z=5x_{1}+3x_{2},
subject to
x_{1}+2x_{2}\leq 10 ,
x_{1}-x_{2}\leq 8 ,
x_{1},x_{2}\geq 0 ,
In the starting Simplex tableau, x_{1} and x_{2} are non-basic variables and the value of Z is zero. The value of Z in the next Simplex tableau is____.
subject to
x_{1}+2x_{2}\leq 10 ,
x_{1}-x_{2}\leq 8 ,
x_{1},x_{2}\geq 0 ,
In the starting Simplex tableau, x_{1} and x_{2} are non-basic variables and the value of Z is zero. The value of Z in the next Simplex tableau is____.
30 | |
35 | |
40 | |
50 |
Question 2 Explanation:
\begin{aligned} \operatorname{Max} . Z &=5 x_{1}+3 x_{2}+0 \times S_{1}+0 \times S_{2} \\ x_{1}+2 x_{1}+S_{1} &=10 \\ x_{1}-x_{2}+S_{2} &=8 \end{aligned}
Question 3 |
A product made in two factories P and Q, is transport to two destinations, R and S. The per unit costs of transportation (in Rupees) from factories to destinations are as per the following matrix:
Factory P produces 7 units and factory Q produces 9 units of the product. Each destination required 8 units. If the north-west corner method provides the total transportation cost as X (in Rupees) and the optimized (the minimum) total transportation cost Y (in Rupees), then (X-Y), in Rupees, is
Factory P produces 7 units and factory Q produces 9 units of the product. Each destination required 8 units. If the north-west corner method provides the total transportation cost as X (in Rupees) and the optimized (the minimum) total transportation cost Y (in Rupees), then (X-Y), in Rupees, is
0 | |
15 | |
35 | |
105 |
Question 3 Explanation:
As per GATE official answer key MTA (Marks to All)
X=105
Using penalty corner method and following modi method we get
-4 water value indicates that its optimum table, so
\begin{aligned} Y &=7 \times 7+3 \times 8+4 \\ &=49+24+4=78 \\ X-Y &=105-77=28 \end{aligned}
X=105
Using penalty corner method and following modi method we get
-4 water value indicates that its optimum table, so
\begin{aligned} Y &=7 \times 7+3 \times 8+4 \\ &=49+24+4=78 \\ X-Y &=105-77=28 \end{aligned}
Question 4 |
For a single with Poisson arrival and exponential service time, the arrival rate is 12 per hour. Which one of the following service rates will provide a steady state finite queue length?
6 per hour | |
10 per hour | |
12 per hour | |
24 per hour |
Question 4 Explanation:
For steady state,
\mu>\lambda
as \lambda=12\text{cust/hr}, we should go with option (D) 24/hr
\mu>\lambda
as \lambda=12\text{cust/hr}, we should go with option (D) 24/hr
Question 5 |
In a single-channel queuing model, the customer arrival rate is 12 per hour and the serving rate is
24 per hour. The expected time that a customer is in queue is _______ minutes.
6.5s | |
2.5s | |
1.8s | |
2.3s |
Question 5 Explanation:
\begin{aligned} \lambda&=12 \mathrm{hr}^{-1} \\ u&=24 \mathrm{hr}^{-1} \\ w_{s}&=\frac{\lambda}{\mu(u-\lambda)}=\frac{12}{24 \times 12} \times 60 \\ &=2.5\text{seconds} \end{aligned}
Question 6 |
In the notation (a/b/c) : (d/e/f) for summarizing the characteristics of queueing situation, the letters 'b' and 'd' stand respectively for
service time distribution and queue discipline | |
number of servers and size of calling source | |
number of servers and queue discipline | |
service time distribution and maximum number allowed in system |
Question 6 Explanation:
The general form of notation is
(a/b/c) : (d/e/f)
where,
a : Probability distribution for arrival pattern
b : departure (or service time) distribution
c : No. of server within system
d : service discipline
e : Size and capacity of system
f : Size or capacity of calling population
(a/b/c) : (d/e/f)
where,
a : Probability distribution for arrival pattern
b : departure (or service time) distribution
c : No. of server within system
d : service discipline
e : Size and capacity of system
f : Size or capacity of calling population
Question 7 |
At a work station, 5 jobs arrive every minute. The mean time spent on each job in the work station is 1/8 minute. The mean steady state number of jobs in the system is _______
1.67 | |
2.65 | |
3.21 | |
8.65 |
Question 7 Explanation:
\begin{aligned} \lambda(\text { arrival rate })&=5 \text { jobs/minute } \\ \mu \text { (service rate) }&=8 \text { jobs/minute } \\ \qquad L_{S}&=\frac{\lambda}{\mu-\lambda}=\frac{5}{8-5}=\frac{5}{3}=1.67 \end{aligned}
Question 8 |
The total number of decision variables in the objective function of an assignment problem of size n x n (n jobs and n machines) is
n^{2} | |
2n | |
2n-1 | |
n |
Question 8 Explanation:
A decision variable is an unknown in an optimization problem. It has a domain, which is a compact representation of the set of all possible values for the variable. So for assignment problem of size n \times n total no. of decision variable are total possibilities to assign value in assignment matrix.
Total no. of decision variables =n \times n=n^2
Total no. of decision variables =n \times n=n^2
Question 9 |
If there are m sources and n destinations in a transportation matrix, the total number of basic variables in a basic feasible solution is
m + n | |
m+n+1 | |
m+n-1 | |
m |
Question 9 Explanation:
no of basic variables = n+m-1
Question 10 |
Customers arrive at a ticket counter at a rate of 50 per hr and tickets are issued in the order of their arrival. The average time taken for issuing a ticket is 1 min. Assuming that customer arrivals form a Poisson process and service times are exponentially distributed, the average waiting time in queue in min is
3 | |
4 | |
5 | |
6 |
Question 10 Explanation:
\lambda=50 \mathrm{cust} / \mathrm{hr}=\frac{50}{60} \mathrm{cust} / \mathrm{min} \\ =0.833 \mathrm{cust} / \mathrm{min}
\mu=1 \mathrm{min} \quad \therefore \rho=\frac{\lambda}{\mu}=0.833
As we know,
L_{q}=\lambda \cdot \omega_{q}
\begin{aligned} \frac{\rho^{2}}{1-\rho}&=\lambda \omega_{q} \\ \Rightarrow \frac{(0.833)^{2}}{1-0.833}&=0.833 \omega_{q} \\ \Rightarrow \omega_{q}&=5 \mathrm{min} \end{aligned}
\mu=1 \mathrm{min} \quad \therefore \rho=\frac{\lambda}{\mu}=0.833
As we know,
L_{q}=\lambda \cdot \omega_{q}
\begin{aligned} \frac{\rho^{2}}{1-\rho}&=\lambda \omega_{q} \\ \Rightarrow \frac{(0.833)^{2}}{1-0.833}&=0.833 \omega_{q} \\ \Rightarrow \omega_{q}&=5 \mathrm{min} \end{aligned}
There are 10 questions to complete.