Solution Set-A&C

7/26/2019 Solution Set-A&C

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Question 1 (A)

firm of RKS, Inc. Stock transactions arrive at a mean rate of 20 per hour. Each order received by Rajesh

requires an average of two minutes to process.

Orders arrive at a mean rate of 20 per hour or one order every 3 minutes. Therefore, in a 15 minute

interval the average number of orders arriving will be

= 15/3 = 5.

(a)

What is the probability that no orders are received within a 15-minute period?

Answer

P (x = 0) = (50e -5)/0! = e -5 = .0067

What is the probability that exactly 3 orders are received within a 15-minute period?

Answer

P (x = 3) = (53e -5)/3! = 125(.0067)/6 = .1396

(b) What is the probability that more than 6 orders arrive within a 15-minute period?

Answer

P (x > 6) = 1 - P (x = 0) - P (x = 1) - P (x = 2)

- P (x = 3) - P (x = 4) - P (x = 5)

- P (x = 6)

= 1 - .762 = .238


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Question

What is the mean service rate per hour?

Answer

Since Rajesh can process an order in an average time of 2 minutes (= 2/60 hr.), then the mean service

rate, , is = 1/(mean service time), or 60/2. = 30/hr.

Question

What percentage of the orders will take less than one minute to process?

Answer

Since the units are expressed in hours,

P (T < 1 minute) = P (T < 1/60 hour).

Using the exponential distribution, P (T < t ) = 1 - e-t.

Hence, P (T < 1/60) = 1 - e-30(1/60)

= 1 - .6065 = .3935 = 39.35%

Question

(c) What percentage of the orders will be processed in exactly 3 minutes?

Answer

Since the exponential distribution is a continuous distribution, the probability a service time exactly

equals any specific value is 0.

Question

(d) What percentage of the orders will require more than 3 minutes to process?

Answer

The percentage of orders requiring more than 3 minutes to process is:

P (T > 3/60) = e-30(3/60) = e -1.5 = .2231 = 22.31%


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Question

(e) What is the average time an order must wait from the time Rajesh receives the order until it is

finished being processed (i.e. its turnaround time)?

Answer

This is an M/M/1 queue with = 20 per hour and = 30 per hour. The average time an order waits in

the system is: W = 1/( - )

= 1/(30 - 20)

= 1/10 hour or 6 minutes

Question

What is the average number of orders Rajesh has waiting to be processed?

Answer

Average number of orders waiting in the queue is:

Lq = 2/[( - )]

= (20)2/[(30)(30-20)]

= 400/300

= 4/3

Utilization Factor

Question

What percentage of the time is Rajesh processing orders?

Answer

The percentage of time Rajesh is processing orders is equivalent to the utilization factor, /.

Thus, the percentage of time he is processing orders is:

/ = 20/30

= 2/3 or 66.67%


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Question-2

RKS, Inc. has begun a major advertising campaign which it believes will increase its business 50%. To

handle the increased volume, the company has hired an additional floor trader, Shyam, who works at

the same speed as Rajesh. Note that the new arrival rate of orders, , is 50% higher than that ofproblem (A).

Thus, = 1.5(20) = 30 per hour.

Question

(a) Why will Rajesh alone not be able to handle the increase in orders?

Answer

Since Rajesh processes orders at a mean rate of = 30 per hour, then = = 30 and the utilization

factor is

1. This implies the queue of orders will grow infinitely large. Hence, Rajesh alone cannot handle

this increase in demand.

em

Question

(b)

What is the probability that neither Rajesh nor Shyam will be working on an order at any pointin time?

Answer

Given that = 30, = 30, s =2 and ( /) = 1, the probability that neither Rajesh nor Shyam will be

working is:

= 1/[(1 + (1/1!)(30/30)1] + [(1/2!)(1)2][2(30)/(2(30)-30)]

= 1/(1 + 1 + 1) = 1/3 = .333


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Question

(c.) What is the average turnaround time for an order with both Rajesh and Shyam working?

Average Time in System (continued)

Answer

The average turnaround time is the average waiting time in the system, W.

L = Lq + ( /) = 1/3 + (30/30) = 4/3

W = L/ = (4/3)/30 = 4/90 hr. = 2.67 min.

Question

(d)

What is the average number of orders waiting to be filled with both Rajesh and Shyam working?

Answer

The average number of orders waiting to be filled is Lq. This was calculated earlier as 1/3.

uing Systems

The advertising campaign of RKS, Inc. was so successful that business actually doubled. The mean rate of

stock orders arriving at the exchange is now 40 per hour and the company must decide how many floor

traders to employ. Each floor trader hired can process an order in an average time of 2 minutes.

mic Analysis of Queuing Systems Based on a number of factors the brokerage firm has

determined the average waiting cost per minute for an order to be $.50. Floor traders hired will earn

$20 per hour in wages and benefits. Using this information compare the total hourly cost of hiring 2

traders with that of hiring 3 traders.

Let 1 USD = INR 60.

Total Hourly Cost

= (Total salary cost per hour)

+ (Total hourly cost for orders in the system)

= ($20 per trader per hour) x (Number of traders) + ($30 waiting cost per hour) x (Average number of

orders in the system)

= 20k + 30L.


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Thus, L must be determined for k = 2 traders and for k = 3 traders with = 40/hr. and = 30/hr. (since

the average service time is 2 minutes (1/30 hr.).

P0 = 1 / [1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))]

= 1 / [1 + (4/3) + (8/3)]

= 1/5

Thus,

L = Lq + ( /) = 16/15 + 4/3 = 2.40

Total Cost = (20)(2) + 30(2.40) = $112.00 per hour

Cost of Three Servers

P0 = 1/[[1+(1/1!)(40/30)+(1/2!)(40/30)2]+

[(1/3!)(40/30)3(90/(90-40))] ]

= 1 / [1 + 4/3 + 8/9 + 32/45]

= 15/59

st of Three Servers (continued) Thus, L = .1446 + 40/30 = 1.4780

Total Cost = (20)(3) + 30(1.4780) = $104.35 per hour

System Cost ComparisonWage Waiting Total

Cost/Hr Cost/Hr Cost/Hr

2 Traders $40.00 $82.00 $112.00

3 Traders 60.00 44.35 104.35

Thus, the cost of having 3 traders is less than that of 2 traders.

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Solution Set-A&C