Question

# In a waiting line situation, arrivals occur at a rate of 2 per minute, and the...

In a waiting line situation, arrivals occur at a rate of 2 per minute, and the service times average 18 seconds. Assume the Poisson and exponential distributions. a. What is ?? b. What is µ? c. Find probability of no units in the system. d. Find average number of units in the system. e. Find average time in the waiting line. f. Find average time in the system. g. Find probability that there is one person waiting. h. Find probability an arrival will have to wait. SOLVE THE PROBLEM, SHOW YOUR CALCULATIONS if you show me numerical results only I will assume you got it from somewhere else.

Thanks

1. Arrival rate ( lambda ) =a = 2 / minute
2. Service rate ( miu) @ 18 seconds =s = 60 /18 = 3.33 / minute
3. Probability of no units in the system = Po = ( 1 – a/s ) = 1 – 2/3.33 = 0.4
4. Average number of units in the system

= a^2/ S x ( s – a ) + a/s

= 2x2/ 3.33 x ( 3.33 – 2 )+ 2/3.33

= 4/( 3.33 x 1.33)+ 2/3.33

= 3/3.33 + 2/3.33

= 5/3.33

= 1.50

1. Average time in waiting line

= 2 / 3.33 x ( 3.33 – 2 )

= 2/ ( 3.33 x 1.33 )

=0.451 minute

= 27.06 seconds

1. Average time in the system

= Average time in the waiting line + a/s minutes

= 0.451 minutes + 0.600 minutes

= 1.051 minutes

= 63.06 seconds

1. Probability of no person waiting = Po = ( 1 – 2/3.33) = 0.4

Probability that one person waiting = P1 = a/s) x Po = ( 2/3.33) x 0.4 = 0.24

1. Probability that an arrival has to wait

= 1 – Probability that no person waiting

= 1 – Po

= 1 – 0.4

= 0.60

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