REQUIREMENT #3 (25 points): Electronex experiences random breakdowns in their ma

Question

REQUIREMENT #3 (25 points): Electronex experiences random breakdowns in their machines that machine is inoperable when awaiting repair. Machines break down at a rate of 5 per hour produce electronic circuits. A and unavailable to produce circuits whenever it is being repaired and also . The company currently ne certitied technician. On average, it takes the technician 8 minutes to repair a machine breakdowns and the repair times are assumed to follow the machine. The time exponential distribution. (SHOW YOUR WORK) a. (5 points) For this repair system, what is the average number of inoperable machines in the repair process? b. (5 points) For repair after it breaks down? sy repir areriro the current system, on average how long does it take for a machine to begin (5 points) For the current system, what fraction of time will the technician be busy? c. d. (10 points) Electronex could decide to hire a second technician. If hired, the two technicians will work together to repair a machine until it is returned to service. Electronex pays certified technicians $60 per hour. It costs the company $250 per hour for each inoperable machine due to lost production. Because they work together to repair the machines, on average the technicians can repair 10 machines each hour, exponentially distributed. Sam asks you, "Is it cost effective to hire a second technician? What do you recommend and why?"

Explanation / Answer

This problem is a direct application of M/M/1 system.

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 = 1/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)

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

Average waiting time = E(w) = ()/{µ(µ - )} ……………………………………………..(5)

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

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

Percentage time service channel is busy = 1 - P0 = ..…………………………………….(8)

Preparatory Work

Given = 5 per hour, µ = 7.5 per hour and hence = 5/7.5 = 2/3. ……………………..(9)

[Given on an average it takes 8 minutes for a technician to repair a machine => 7.5 machines are repaired per hour]

Part (a)

A machine is inoperable if it is waiting to get repaired or getting repaired. So, the required answer is: E(n) = ()/(µ - ) [vide (4) above]

= 5/2.5

= 2 machines on an average are inoperable. ANSWER

Part (b)

Time a machine takes to begin repair = the time it is waiting to get repaired =

E(w) = ()/{µ(µ - )} [vide (5) above]

= 2/7.5 hour = 16 minutes. ANSWER

Part (c)

Fraction of time technician is busy = = 2/3 [vide (8) and (9) above] ANSWER

Part (d)

When a second technician is employed, µ is given to be 10 per hour and hence the total idle time of broken-down machines = waiting time + service time

= E(v) = {1/(µ - )} [vide (6) above]

= 1/(10 - 5)

= 1/5 hour = 12 minutes …………………………………………………………………..(10)

Average number of machines in the system = E(n) = ()/(µ - ) [vide (4) above]

= 5/5 = 1 ………………………………………………………………………………..(11)

(10) and (11) => on an average, total machine time lost = 1 x 12 = 12 minutes. ………(12)

With one technician, E(v) = 1/(7.5 - 5)

= 1/2.5 hour = 24 minutes. ………………………………………………………………..(13)

Part(a) answer and (12) =>

on an average, total machine time lost = 2 x 24 = 48 minutes………………………. .(14)

(12) and (14) => savings in machine down-time = 36 minutes and hence given cost of machine down-time is $250, financial saving due to second technician = $150.

Since cost of second technician is only $60 [< 150], it is worthwhile employing the second technician. ANSWER

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