Waiting Line Models

Chapter 12

Waiting Line Models

1.a. = 5(0.4) = 2 customers per five minute period

b.

x / P(x)
0 / 0.1353
1 / 0.2707
2 / 0.2707
3 / 0.1804

c.P(Delay Problems) = P(x > 3) = 1 - P(x 3) = 1 - 0.8571 = 0.1429

2.µ = 0.6 customer per minute

a.P(T 1) = 1 - e-(0.6)1 = 0.4512

b.P(T 2) = 1 - e-(0.6)2 = 0.6988

c.P(T > 2) = 1 - 0.6988 = 0.3012

7.M/M/1 queuing situation

a.


b.

c.

d.

@QUE : bubba - problem 12.7

1

QUEUE 1

# SERVERS 1

SOURCE POP INF

ARR RATE 2.5

SERV DIST EXP

SERV TIME 0.2

SERV STD .

WAIT CAP .

# CUSTMERS .

WAIT COST .

COST/SERV .

LOSTCUST C .

bubba - problem 12.7

QUEUE 1 : M / M / c

QUEUE STATISTICS

Number of identical servers ...... 1

Mean arrival rate ...... 2.5000

Mean service rate per server ...... 5.0000

Mean server utilization (%) ...... 50.0000

Expected number of customers in queue . . . . 0.5000

Expected number of customers in system . . . 1.0000

Probability that a customer must wait . . . . 0.5000

Expected time in the queue ...... 0.2000

Expected time in the system ...... 0.4000

a. L = Expected number of customers in system = 1

b. Wq = Expected time in the queue = 0.2 hour

c. W = Expected time in the system = 0.4 hour

d. Pw = Probability that a customer must wait = 0.5

9.M/M/1 queuing situation

a.

b.

c.

d.

e.P(More than 2 waiting) = P(More than 3 are in system)

= 1 - (P0 + P1 + P2 + P3) = 1 - 0.9625 = 0.0375

f.

@QUE : bubba - problem 12.9

1

QUEUE 1

# SERVERS 1

SOURCE POP INF

ARR RATE 2.2

SERV DIST EXP

SERV TIME 0.2

SERV STD .

WAIT CAP .

# CUSTMERS .

WAIT COST .

COST/SERV .

LOSTCUST C .

bubba - problem 12.9

QUEUE 1 : M / M / c

QUEUE STATISTICS

Number of identical servers ...... 1

Mean arrival rate ...... 2.2000

Mean service rate per server ...... 5.0000

Mean server utilization (%) ...... 44.0000

Expected number of customers in queue . . . . 0.3457

Expected number of customers in system . . . 0.7857

Probability that a customer must wait . . . . 0.4400

Expected time in the queue ...... 0.1571

Expected time in the system ...... 0.3571

bubba - problem 12.9

QUEUE 1 : M / M / c

PROBABILITY DISTRIBUTION OF NUMBER IN SYSTEM

Number Prob 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

+----+----+----+----+----+----+----+----+----+----+

0 0.5600|**************************** |

1 0.2464|************+------|

2 0.1084|*****+------|

3 0.0477|**+------|

4 0.0210|*+------|

5 0.0092|+------|

OVER 0.0073|+------|

+----+----+----+----+----+----+----+----+----+----+

  1. P0 = 0.56
  1. P1 = 0.2464
  1. P2 = 0.1084
  1. P3 = 0.0477
  1. Pr(more than 2 customers waiting) = Pr(more than 3 customers in the system)

= 1 – Pr (0 or 1 or 2 or 3 customers) = 1 – (P0 + P1 + P2 + P3) = 0.0375

  1. Wq = Expected time in the queue = 0.1571 hour

17.M/M/2 model with

Average service time =

@QUE : bubba - problem 12.17

1

QUEUE 1

# SERVERS 2

SOURCE POP INF

ARR RATE 5

SERV DIST EXP

SERV TIME 0.1

SERV STD .

WAIT CAP .

# CUSTMERS .

WAIT COST .

COST/SERV .

LOSTCUST C .

bubba - problem 12.17

QUEUE 1 : M / M / c

QUEUE STATISTICS

Number of identical servers ...... 2

Mean arrival rate ...... 5.0000

Mean service rate per server ...... 10.0000

Mean server utilization (%) ...... 25.0000

Expected number of customers in queue . . . . 0.0333

Expected number of customers in system . . . 0.5333

Probability that a customer must wait . . . . 0.1000

Expected time in the queue ...... 6.6667E-03

Expected time in the system ...... 0.1067

bubba - problem 12.17

QUEUE 1 : M / M / c

PROBABILITY DISTRIBUTION OF NUMBER IN SYSTEM

Number Prob 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

+----+----+----+----+----+----+----+----+----+----+

0 0.6000|****************************** |

1 0.3000|***************+------|

2 0.0750|****------|

3 0.0188|*------|

OVER 0.0062|+------|

+----+----+----+----+----+----+----+----+----+----+

a.P0 = 0.60

b.Expected number of customers in queue = 0.0333 customer

c.Expected time in the queue = 0.0066667 hour  24 seconds

d.Expected time in the system = 0.1067 hour 6.4 minutes

e.This service is probably much better than is necessary with average waiting time being only 24 seconds. With P0 = 0.60, both channels will in fact be idle 60% of the time.

19.M/M/2 model with

Average service time =

@QUE : bubba - problem 12.19

1

QUEUE 1

# SERVERS 2

SOURCE POP INF

ARR RATE 14

SERV DIST EXP

SERV TIME 0.1

SERV STD .

WAIT CAP .

# CUSTMERS .

WAIT COST .

COST/SERV .

LOSTCUST C .

bubba - problem 12.19

QUEUE 1 : M / M / c

QUEUE STATISTICS

Number of identical servers ...... 2

Mean arrival rate ...... 14.0000

Mean service rate per server ...... 10.0000

Mean server utilization (%) ...... 70.0000

Expected number of customers in queue . . . . 1.3451

Expected number of customers in system . . . 2.7451

Probability that a customer must wait . . . . 0.5765

Expected time in the queue ...... 0.0961

Expected time in the system ...... 0.1961

bubba - problem 12.19

QUEUE 1 : M / M / c

PROBABILITY DISTRIBUTION OF NUMBER IN SYSTEM

Number Prob 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

+----+----+----+----+----+----+----+----+----+----+

0 0.1765|********* |

1 0.2471|************+------|

2 0.1729|*********------|

3 0.1211|******+------|

4 0.0847|****+------|

5 0.0593|***------|

6 0.0415|**+------|

7 0.0291|*+------|

8 0.0203|*+------|

9 0.0142|*------|

10 0.0100|+------|

11 0.0070|+------|

OVER 0.0163|*------|

+----+----+----+----+----+----+----+----+----+----+

a.P0 = 0.1765

b.L = Expected number of customers in system = 2.7451

c.Expected time in the queue = 0.0961 hour  5.77 minutes

d.W = Expected time in the system = 0.1961 hour  11.77 minutes

e.P0 = 0.1765

P1 = 0.2471

Pr(wait for service) = Pr(n 2)= 1 –Pr(n0 or 1)

= 1 –(0.1765 + 0.2471) = 0.5764

or

Pr(wait for service) = Probability that a customer must wait = 0.5765

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