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}

There are 5 questions to complete.