Queuing Theory and Transportation

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
A
A-E-D-F-H-C-G-B
B
A-D-E-F-H-C-G-B
C
G-E-D-F-H-C-A-B
D
B-G-C-H-F-D-E-A
GATE ME 2021 SET-1   Industrial Engineering
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
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____.
A
30
B
35
C
40
D
50
GATE ME 2017 SET-2   Industrial Engineering
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
A
0
B
15
C
35
D
105
GATE ME 2017 SET-2   Industrial Engineering
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}
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?
A
6 per hour
B
10 per hour
C
12 per hour
D
24 per hour
GATE ME 2017 SET-2   Industrial Engineering
Question 4 Explanation: 
For steady state,
\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.
A
6.5s
B
2.5s
C
1.8s
D
2.3s
GATE ME 2016 SET-2   Industrial Engineering
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
A
service time distribution and queue discipline
B
number of servers and size of calling source
C
number of servers and queue discipline
D
service time distribution and maximum number allowed in system
GATE ME 2015 SET-3   Industrial Engineering
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
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 _______
A
1.67
B
2.65
C
3.21
D
8.65
GATE ME 2014 SET-4   Industrial Engineering
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
A
n^{2}
B
2n
C
2n-1
D
n
GATE ME 2014 SET-4   Industrial Engineering
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
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
A
m + n
B
m+n+1
C
m+n-1
D
m
GATE ME 2014 SET-2   Industrial Engineering
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
A
3
B
4
C
5
D
6
GATE ME 2013   Industrial Engineering
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}
There are 10 questions to complete.

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