Answer the below question from Operation Research I QUESTION ONE Arrival rate of telephone calls at a telephone both are according to a Poisson distribution with an average time of 9 minutes between two consecutive arrivals The length of the telephone call is assumed to be exponentially distributed with mean 3 minutes Determine the probability that the person arriving at the booth will have to wait Find the average queue length that is formed from time to time The telephone company will install a second booth when convinced that an arrival would expect to have to wait at least four minutes for the phone Find the increase in flow rate of arrivals which will justify a second booth What is the probability that an arrival will have to wait for more than 10 minutes before the phone is free What is the probability that he will have to wait for more than 10 minutes before the phone is available and the call is also complete Find the fraction of a day that the phone will be in use Given Arrival rate, 1 9 0 11per minute service rate, 1 3 0 33 per minute Answer 1 p (1 9) (1 3) 0 3333 the probability that the person arriving at the booth will have to wait 0 33 Answer 2 Average queue length that forms from time to time ( ) Average queue length that forms from time to time 0 33 (0 33 0 11) Average queue length that forms from time to time 1 5 Answer 3 Wq ( ( )) waiting time to install a second booth Wq 4minutes so we can put the value of Wq in the above formula 4 (0 33 (0 33 ) 4 0 33 (0 33 ) 0 4356 1 32 0 4356 2 32 0 187 or 0 19 arrivals minute increase in flow of arrivals 0 19 0 11 0 08 per minute Answer 4 the probability that arrival will have to wait for more than 10 minutes before the phone is free P(Wait time 10) (0 11 0 33) e 10(0 330 11) P(Wait time 10) 1 30 P(Wait time 10) 0 033 What is the probability that an arrival will have to wait for more than 10 minutes before the phone is free What is the probability that he will have to wait for more than 10 minutes before the phone is available and the call is also complete Find the fraction of a day that the phone will be in use

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Question: Answer the below question from Operation Research I QUESTION ONE Arrival rate of telephone calls at a telephone both are according to a Poisson distribution

Answer the below question from Operation Research I

QUESTION ONE

1. Arrival rate of telephone calls at a telephone both are according to a Poisson distribution with an average time of 9 minutes between two consecutive arrivals. The length of the telephone call is assumed to be exponentially distributed with mean 3 minutes.
Determine the probability that the person arriving at the booth will have to wait.
Find the average queue length that is formed from time to time
The telephone company will install a second booth when convinced that an arrival would expect to have to wait at least four minutes for the phone. Find the increase in flow rate of arrivals which will justify a second booth.
What is the probability that an arrival will have to wait for more than 10 minutes before the phone is free?
What is the probability that he will have to wait for more than 10 minutes before the phone is available and the call is also complete?
Find the fraction of a day that the phone will be in use

Given

Arrival rate, =1/9 =0.11per minute

service rate, =1/3=0.33 per minute

Answer 1=

p=/=(1/9)/(1/3) =0.3333

the probability that the person arriving at the booth will have to wait =0.33

Answer 2= Average queue length that forms from time to time =/(-)

Average queue length that forms from time to time = 0.33/(0.33-0.11)

Average queue length that forms from time to time =1.5

Answer 3=

Wq=/(*(-))

waiting time to install a second booth Wq= 4minutes

so we can put the value of Wq in the above formula

4=/(0.33*(0.33-)

=4*0.33*(0.33-) =0.4356-1.32

=0.4356/2.32

=0.187 or 0.19 arrivals/minute

increase in flow of arrivals=0.19-0.11=0.08 per minute

Answer 4=

the probability that arrival will have to wait for more than 10 minutes before the phone is free

P(Wait time >10) = (0.11/0.33)*e^10(0.330.11)

P(Wait time >10) =1/30

P(Wait time >10) =0.033

What is the probability that an arrival will have to wait for more than 10 minutes before the phone is free?
What is the probability that he will have to wait for more than 10 minutes before the phone is available and the call is also complete?
Find the fraction of a day that the phone will be in use

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