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Q5.In a waiting line situation, arrivals occur at a rate of 2 per minute, and the...

Q5.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 [Hint: convert service time 18 seconds to service rate].JosiahA

a.   What is l?

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.

Solutions

Expert Solution

Solution

Q5

The solution is based on the theory of M/M/1 queue system

Back-up Theory

An M/M/1 queue system is characterized by

arrivals following Poisson pattern with average rate λ,

[this is also the same as exponential arrival with average inter-arrival time = 1/λ]

service time following Exponential Distribution with average service time of (1/µ)

[this is also the same as Poisson service with average service rate = µ]

and single service channel.

NOTE:

Time unit of both λ and µ must be the same.

Let n = number of customers in the system

m = number of customers in the queue.

[Trivially, n = m + number of customers under service.]

w = waiting time i.e., the time spent in waiting to get the service (does not include the time spent in service )

v = total time spent in the system, i.e., time in queue + time in service.

Traffic Intensity = ρ = (λ/µ) ……………………...............................................................………………………......……………..(A)

The steady-state probability of n customers in the system is given by Pn = ρn(1 - ρ)ρ = (λ/µ)..................................…………(1)

steady state probability of no customers in the system is given by P0 = (1 - ρ) …......................................................….……(2)

Average queue length = E(m) = (λ2)/{µ(µ - λ)} …………………….......................................................……………………..…..(3)

Average number of customers in the system = E(n) = (λ)/(µ - λ) = ρ/(1 – ρ)………....................................................………..(4)

Average waiting time = E(w) = (λ)/{µ(µ - λ)} = (1/µ)ρ/(1 – ρ) = E(n)/µ ……….....................................................……………..(5)

Average time spent in the system = E(v) = {1/(µ - λ)}…………………………......................................................……………..(6)

Proportion idle time of service channel = P0 = (1 - ρ) …………………...................................................…..…………………..(7)

Probability the system is empty = P0 = (1 - ρ) …………………...................................................…………....…………..........(7a)

Now to work out the solution,

Arrivals occur at a rate of 2 per minute, and the service times average 18 seconds. Assume the Poisson and exponential distributions. => λ = 2, µ = 60/18 = 3.33 and hence ρ = (2/3.33) = 0.6 ................................................................................ (8)

Part (a)

Vide (8), λ = 2 Answer 1

Part (b)

Vide (8), µ = 3.33 Answer 2

Part (c)

Probability of no units in the system

= P0

= (1 - ρ) [vide (7a)]

= 0.4 [vide (8)] Answer 3

Part (d)

Average number of units in the system

= E(n)

= ρ/(1 – ρ) [vide (4)]

= 0.6 x 0.4 [vide (8)]

= 0.24 Answer 4

Part (e)

Average time in the waiting line.

= E(w)

= (2)/(3.33 x 1.33) [vide (5)]

= 0.45 minute Answer 5

Part (f)

Average time in the system.

= E(v)

= 1/(3.33 - 2) [vide (6)]

= 0.75 minute Answer 6

Part (g)

Probability that there is one person waiting.

= P2 [one waiting => 2 in the system]

= 0.62 x 0.4

= 0.144 Answer 7

Part (h)

Probability an arrival will have to wait

= P(n ≥ 1)

= 1 – P0

= 0.6 [vide Answer 1] Answer 8

DONE


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