A single overhead crane attends to re-load ten machines. When a machine finishes

Question

A single overhead crane attends to re-load ten machines. When a machine finishes its load, the overhead crane is called to unload the machine and to provide it with a new load from an adjacent storage area. The machine time per load is assumed exponential with mean 30 minutes. The time from the moment the crane moves to service a machine until a new load is installed is also exponential with mean 10 minutes. 2. a. What percentage of time is the crane idle? b. What is the expected number of machines waiting for crane service?

Explanation / Answer

This is a typical application of M/M/1Queue system, where the machines form the ‘customers’ and the crane serves as the service channel.

Back-up Theory

An M/M/1 queue system is characterized by arrivals following Poisson pattern with average rate , [this is also the same as exponential arrival with average inter-arrival time = 1/ ] 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.

Let n = number of customers in the system and m = number of customers in the queue.

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

Let (/µ) =

The steady-state probability of n customers in the system is given by Pn = n(1 - ) …………(1)

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

Average queue length = E(m) = (2)/{µ(µ - )} ……………………………………………..(3)

Preparatory work

Given ‘machine time per load is assumed exponential with mean 30 minutes’ => arrival is Poisson with = 2 per hour and ‘the time from the moment the crane moves to service a machine till a new load is installed is exponential with mean 10 minutes’ => service time is exponential with mean µ = 6 per hour.

So, = /µ = 2/6 = 1/3. ……………………………………………………………..(4)

Part (a)

The crane is idle so far as no machine is there waiting for its service. Hence, percentage of time the crane is idle

= 100 x P(No machine in the system)

= 100 x P0

= 100 x (1 - ) [vide (2)]

= 100 x {1 – (1/3)} [vide (4)]

= 66.67% ANSWER.

Part (b)

Expected number of machines waiting for crane service

= Average queue length

= E(m)

= (2)/{µ(µ - )} [vide (3)]

= (22)/{6(6 - 2)}

= 1/6 ANSWER

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