# 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:
$\mu>\lambda$
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.