3. (20 Points) The Seaboard Shipping company owns a warehouse terminal that is c

Question

3. (20 Points) The Seaboard Shipping company owns a warehouse terminal that is configured with three loading docks that operates 24/7. As trucks enter the terminal grounds, drivers are given a number and when one of the three docks becomes available, the truck with the lowest number enters the vacant dock for loading/unloading. The arrival times for the trucks and the service times for loading and unloading are both exponentially distributed. Trucks arrive at the rate of 5 trucks/hour and the service rate at each of the three docks is 2 trucks/hour. a. On average, how long must a truck wait to be called to an available dock? b. Management has decided that it must reduce the amount of times trucks are waiting at the warehouse. Management is considering two options. The company can choose either to open a fourth dock or to hire additional employees. Hiring additional employees will reduce average service time from 30 minutes/truck to 23 minutes/truck at each of the three loading docks. Because the cost of each alternative is roughly the same, management will choose the one that results in the smallest total time a truck spends at the warehouse terminal. Which alternative should management select? the existing

Explanation / Answer

This is an application of M/M/C Queue system, where C = 3.

Back-up Theory

An M/M/C queue system is characterized by arrivals following Poisson pattern with average rate , service time following Exponential Distribution with average service rate of µ and multiple (> 1) 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 no customers in the system is given by

P0 = 1/(S1 + S2), where S1 = [0,c - 1](n/n!) and S2 = (c)/[c!{1 - (/c)}]……………(1)

The steady-state probability of n customers in the system is given by

Pn = P0 x n/n! for n < c …………………………………………………………………(2a)

Pn = P0 x n/{c!(cn-c) for n c .…………………………………………………………(2b)

Average queue length = E(m) = P0{(µ)(/µ)c}/{(c - 1)!(cµ - )2}……………………..(3)

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

Average waiting time = E(w) = E(m)/()…… ……………………………………………..(5)

Average time spent in the system = E(v) = E(w) + (1/µ)..…………………………………..(6)

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

Now, to work out the solution,

Given C = 3, = 5 and µ = 2, = 5/2 = 2.5.

Vide (1) above,

S1 = [0,c - 1](n/n!)

= 1 + 2.5 + (2.52/2)

= 6.625

S2 = (c)/[c!{1 - (/c)}]

= (2.53)/[6{(1 – (2.5/3)}]

= 15.625

P0 = 1/(6.625 + 15.625)

= 1/22.25

= 0.0449

Part (a)

Average waiting time = E(w) = E(m)/() [vide (5) above]

E(m) = P0{(µ)(/µ)c}/{(c - 1)!(cµ - )2} [vide (3) above]

= 0.0449{10(2.53)}/{2(1)2}

= 14.03125

E(w) = 14.03125/5

= 2.8 hours

= 2 hours 48 minutes. ANSWER

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