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.