Operations Management Help

PREVIOUS EXAM QUESTIONS & SOLUTIONS ON Queueing Models:

Fall 2012, Final, Problem 2

Suppose that you operate your own deli store in Rolla, MO that serves its customers delicious

sandwiches. The customers give their sandwich orders by calling your store. You observed that

the time between two consecutive order calls is exponentially distributed with mean inter-arrival

time equal to 15 minutes. Once a customer gives a sandwich order via phone, he/she comes to

your store to pick his/her sandwich. There is one employee who is preparing any sandwich orders

that you received on phone. You observed that preparation time for any sandwich is

exponentially distributed with average preparation time equal to 10 minutes. Answer the

following questions based on the above information.

a) (5 points) Describe the above system as a queueing system by defining the customers

and the arrival rate, servers and the service rate, and express the queueing model

using Kendall’s notation.

b) (5 points) Calculate L using the formula for L for M/M/1 model (that is, do not use

excel template). Then using little’s law and the relationship between measure of

performance, calculate LQ, W, WQ (that is, do not use excel template and do not use

the specific formulas given for each measure of performance under M/M/1). Only use

the little’s law and the relationships between performance measures to calculate LQ,

W, WQ after you calculate L using the formula under M/M/1.

Customers The customer calls for sandwiches

60/15=4 4 customers per hour is the arrival rate

Inter-arrivals are noted to be exponential

Server(s) The employee preparing the sandwiches (single server as there is one employee preparing the sandwiches)

60/10=6 6 customers are served per hour (5 sandwiches are prepared per hour)

The service time is noted to be exponential

Using Kendall's notation, the above queueing systems can be denoted as

M/M/1 because we have exponential interarrivals, exponential service time, 1 server, and no queue capacity.

l = 4

m = 6

c) (5 points) A sandwich is in preparation from the time you received its order via phone

until the time your employee finishes its preparation. As the owner of the store, you

think that there should be less than 3 sandwiches in preparation at least 70% of the

time. Determine whether this criterion is being satisfied currently using the formulas

for M/M/1. That is, do not use the excel template for your calculation.

d) (5 points) Since sandwiches are standard, each sandwich costs you $3. Suppose you

sell sandwiches for $10 each. Hence, your hourly revenue is the number of orders that

In this part, we are asked to determine various measures of performance

using the explicity formulas derived for M/M/1 systems.

Particularly, we are first asked to determine L, the expected number of people in the system

l = 4 from part a

m = 6 from part a

Thus, L=4/(6-4) = 2 customers in the system on average

From Little's Law, we know that

L = lW

Thus, W=2/4 = 0.5 hour is the expected time a customer waits until he/she get the order (or 30 minutes)

this might be interpreted as the expected time after a order call received until the sandwich is ready

From the relationship between L and Lq, we have

Thus, L q = 2-4/6=1.33 customer orders are in the line waiting on average

From Little's Law, we know that

L q = l W q Thus, Wq = 1.33/4=0.333 hours is the expected time a customer order waits until the sandwich ordered starts to be prepared

(or 18 minutes)

lm

l





 =

 =

1 L

L = Lq + l/m

As the owner of the store, you think that there should be less than 3 sandwiches in preparation at least 70% of the time.

So you want to have P0+P1+P2>=0.7

Using the formulas, we have the following probability calculations

P0= 1l /m = 1 - 4/6 = 0.333

P1= rho(1-rho)=4/6(1-4/6)= 0.222222

P2= rho^2(1-rho)=16/25(1-4/5)= 0.148148

P0+P1+P2= 0.703 >0.7

thus, this criterion is currently satisfied

you get in one hour multiplied by 10$. You pay $4 per hour to your employee. What

is the current hourly profit of your store?

e) (10 points) Suppose that you are considering the following promotion to increase the

number of your customers. If the sandwich is not ready to pick up in 30 minutes after

the order call, you sell the sandwich for its half price, i.e., for $5 instead of $10. You

expect that, with this promotion, more customers will call your store. In particular,

you expect that the time between two consecutive calls for sandwiches will be

exponentially distributed with mean inter-arrival time equal to 12 minutes. Calculate

the hourly profit of your store with this promotion. Recall that each sandwich costs

you $3 and you pay $4 per hour to your employee. Is it more profitable to offer this

promotion compared to the case without the promotion? The case without the

promotions has the hourly profit that you found in part e). You can use excel

templates provided.

So, we sell 4 sandwiches per hour on average, each costs $3 each brings $10

revenues 40 per hour

sandwich cost 12 per hour

labor 4 per hour

profit 24 per hour

We still have M/M/1

Data Results

1 l = 5 (mean arrival rate) L = 5

0 m = 6 (mean service rate) Lq = 4.166667

0 s = 1 (# servers)

0 W = 1

0 Pr(W > t) = 0.606531 Wq = 0.833333

0 when t = 0.5

0  = 0.833333

0 Prob(Wq > t) = 0.306566

0 when t = 1 n Pn

0 0 0.166667

0 1 0.138889

0 2 0.115741

0 3 0.096451

Fall 2013, Final, Problem 3

Dr. Konur decided to open a Turkish restaurant in Rolla, MO. Since he is the only Turkish cook in

the area, he will be the chef of the restaurant. Dr. Konur will offer a standard dish in his

restaurant: Turkish kabob. Preparing and serving a Turkish kabob after a customer walks into the

restaurant takes 5 minutes on average and the preparation and serving time is exponentially

distributed. He estimates that customers will arrive with a rate of 6 per hour and the interarrival

time is exponentially distributed. Each Turkish kabob costs $5 and is sold for $15. Dr. Konur

observed that the tips that customers leave depends on the time they wait until they get their

kabobs. Specifically, if it takes the customer have his/her kabob prepared and served within less

than 5 minutes, the customer leaves a tip for $6; if it takes the customer have his/her kabob

prepared and served within 5 to 10 minutes, the customer leaves a tip for $4; and if it takes the

customer have his/her kabob prepared and served within more than 10 minutes, the customer

leaves a tip for $2.

a) (3 points) Explain the queueing system for Ala-Turca by defining customers, arrival rate,

servers, service rate and use Kendall’s notation to define it. See below.

Now, let's calculate the cost of this system

labor cost $4 per hour it did not change

What is changing is our revenues

Now we have 5 sales in an hour

However while some of these sales are $10 per sandwich

some of these sales are $5 because the sandwich was not ready in 30 minutes

in particular, probability of having a sandwich ready in more than 30 minutes is

P(W>30minutes)

And it is equal to 0.606531

So, around 61% of the customers are getting $5 sandwiches

thus 5*0.39*10+5*0.62*0 is the hourly revenues

34.83673 is hourly revenues

you still need to prepare the sandwiches so you pay

5*3=15 dolars and 4 to the employe

your hourly revenue with the promotion is

15.83673 dollars per hour

not good compared to part d

so do not do the promotion

b) (5 points) Calculate the hourly expected profit of Ala-turca. Profit is equal to profit gained

from kabob sales plus the tips customers leave.

Expected Profit per hour = Expected profit from sales per hour + expected tips per hour

 Expected profit from sales/ hour = Expected number of customer/ hour*profit from one

customer= mean arrival rate * 10 = 6/hour * $10 = $60/hour

 Expected tips/hour = expected tip from one customer * expected number of

customers/hour

o Expected tip from one customer = $6 * (probability of a customer leaving $6 tip) +

$4* (probability of a customer leaving a $4 tip) + $2* (probability of a customer

leaving $2 tip)

 We are given that the customer leaves a tip when his/her kabob is ready

and served within 5 minutes, i.e., if his/her total waiting time in the system

is < 5 minutes. Therefore, a customer’s probability of leaving a $6 tip is

equal to P(W<5)=1-P(W>5)

 Similarly, we can discuss that probability of a customer leaving a $4 tip is

equal to P(5<W<10) = 1 - [P(W<5) + P(W>10)] = 1 - [1 - P(W>5) + P(W>10)]

= P(W>5) - P(W>10).

 Similarly, we can discuss that probability of customer leaving a $2 tip is

equal to P(W>10).

o So, expected tip from one customer is = 6*[1-P(W>5)]+4*[P(W>5)-

P(W>10)]+2*P(W>10)

 We know from template how to calculate P(W>t). Note that, the above

numbers in the probabilities are minutes, if you have put arrival and

service rates in terms of per hour, you need to convert the minutes to

hours as well.

o Once we know the expected tip from one customer, expected tips/hour = arrival

rate * expected tip from one customer

Now, suppose that Dr. Konur can hire two more additional cooks.

c) (10 points) If an additional cook costs $3 per hour, how many additional cooks Dr. Konur

should hire to maximize the hourly profit of Ala-turca. That is, should he hire 0 or 1 or 2

additional cooks?

MM1 customers are the customers, server is Dr. Konur

Data

1 l = 6 (mean arrival rate)

0 m = 12 (mean service rate)

0 s = 1 (# servers)

0

0 Pr(W > t) = 0.606531 0.39346934 Pr(W > t) = 0.367879441

0 when t = 0.083333 5 minutes when t = 0.166666667 10 minutes

0

0 Prob(Wq > t) = 0.18394

0 when t = 0.166667 Tip

0 P(W<5)= 0.39346934 6

0 P(5<W<10)= 0.238651219 4

0 P(W>10)= 0.367879441 2

0 Profit Calculation:

0

0 Profit from Sales

0 60

0

0 Tips 24.30708

0

0

0 Hourly Profit 84.30708

1 additional worker MM2

Data

l = 6 (mean arrival rate)

m = 12 (mean service rate)

s = 2 (# servers)

Pr(W > t) = 0.396829 0.603170703 Pr(W > t) = 0.152444926

when t = 0.083333 5 minutes when t = 0.166666667 10 minutes

Prob(Wq > t) = 0.004979

when t = 0.166667 Tip

P(W<5)= 0.603170703 6

P(5<W<10)= 0.244384371 4

P(W>10)= 0.152444926 2

Profit Calculation:

Profit from Sales

60

Tips 29.40871

Addition Cook Cost 3

Hourly Profit 86.40871 Better than no additional cook

2 additional workers MM3

Data

l = 6 (mean arrival rate)

m = 12 (mean service rate)

s = 3 (# servers)

Pr(W > t) = 0.370766 0.629233746 Pr(W > t) = 0.136634246

when t = 0.083333 5 minutes when t = 0.166666667 10 minutes

Prob(Wq > t) = 0.000102

when t = 0.166667 Tip

P(W<5)= 0.629233746 6

P(5<W<10)= 0.234132007 4

P(W>10)= 0.136634246 2

Profit Calculation:

Profit from Sales

60

Tips 29.91119

Addition Cook Cost 6

Hourly Profit 83.91119 Worse than no additional cook

Worse than 1 addition cook

Now, suppose that Dr. Konur does not hire any additional cooks.

d) (3 points) If he prepares and serves the kabobs faster, will he be collecting more tips?

Explain your answer briefly.

e) (4 points) Suppose that with additional $2 per hour, Dr. Konur can prepare and serve

kabobs with a mean service time equal to 4 minutes. Should Dr. Konur pay this additional

$2 per hour if he wants to maximize profits?

No additional Cook 1 additional cook 2 additional cooks

Hourly Profit 84.30707879 86.40870932 83.911194

So, hire 1 additional cook

d)

Yes, since he is faster, P(W<5) will increase

and P(W>10) will decrease.

Note that summation of P(W<5)+P(5<W<10)+P(W>10)=1

Since the most tip is given for P(W<5), the expected tips will increase

MM1 e) service rate is now 15

Data

l = 6 (mean arrival rate)

m = 15 (mean service rate)

s = 1 (# servers)

Pr(W > t) = 0.472367 0.527633447 Pr(W > t) = 0.22313016

when t = 0.083333 5 minutes when t = 0.166666667 10 minutes

Prob(Wq > t) = 0.089252

when t = 0.166667 Tip

P(W<5)= 0.527633447 6

P(5<W<10)= 0.249236393 4

P(W>10)= 0.22313016 2

Profit Calculation:

Profit from Sales

60

Tips 27.65404

Additional cost oer hour 2

Hourly Profit 85.65404 > 84.30707879

So, pay the additional cost

Fall 2013, Final, Problem 4

Suppose that you have an M/M/1 queueing sytem. You are considering a set of changes in your

system. Please answer the following questions independent of each other and explain your

answers briefly.

Spring 2013, Final, Problem 1

Dr. Konur is an assistant professor in the Engineering Management and Systems Engineering department at the

Missouri University of Science and Technology. Aside from his work related to the university, he runs a car-

mechanic shop in Rolla, MO called Dino-Mec of Rolla. He works as a mechanic there repairing cars and he is the

only mechanic of the shop. Using his knowledge in statistics, Dr. Konur observed that cars are coming to Dino-Mec

for repair at a rate of 10 per month and the interarrival time is exponentially distributed. Dr. Konur can repair cars

at a rate of 12 per month. The cars are repaired in a first-come first-served manner.

a) Express Dr. Konur’s Dino-Mec shop as a queueing system by defining customers, arrival rate and expected

interarrival time, server or servers, service rate and expected service time. Use Kendall’s notation to define

the Dino-Mec queueing system.

a) (3 points) If you increase the service rate, the server will be busier. True or False? Explain why?

False, the server will be less busier, rho will decrease

b) (3 points) If the mean interarrival time increases, expected number of customers in the system will increase. True or False? Explain why?

False, it means that arrival rate will decrease, hence, L will decrease

c) (3 points) If the arrival rate increases, the probability of having 0 people in the system will decrease. True or False? Explain why?

True, P0=1-rho=1-lambda/mu so if lambda increases, P0 decreases

d) (3 points) If you increase the mean service time, the average waiting time in line for customers will increase. True or False? Explain why?

True, it means that service rate will decrease, hence, utilization factor will increase, hence Wq will increase

e) (3 points) If you want customers to spend less time in the system, you should increase the mean service time. True or False? Explain why?

False, if mean service time increases, service rate decreases; hence, utilization factor increases, so W increases

Customers: Cars coming to be repaired

Arrrival Rate 10 per month

Mean Interaarival Time 3 days (assuming 30 days in a month)

Servers Mechanic (Dr. Konur)

Service Rate 12 per month

Mean Service Time 2.5 days (assuming 30 days in a month)

M/M/1 (Kendall's Notation)

Dr. Konur parks the cars waiting to be repaired inside the shop; but, if there is no space left inside the shop, he

parks them in the outside parking area. The shop can have 2 cars parked inside the shop plus the car that is

currently being repaired.

b) Since parking in the outside parking area is problematic, Dr. Konur wishes to have the cars parked inside the

shop at least 75% of the time on average. Does Dino-Mec satisfy this criterion? Show your calculations on how

you reached your answer; otherwise you will not get any credit. You can use excel template if needed.

From each repair, Dr. Konur earns $1,000 if the repair is completed within a month. The repair time is the time

between the car’s arrival for repair and the car is completely repaired. On the other hand, if the repair time is

more than a month, Dr. Konur charges the half price of the regular repair; hence, he earns $500.

c) If Dr. Konur earns less than $10,000 per month on average, he will consider closing Dino-Mec. Will Dr. Konur

consider closing Dino-Mec? Show your calculations on how you reached your answer; otherwise you will not

get any credit. That is, calculate the expected monthly earnings and compare to $10,000. You can use excel

template if needed.

Apparently, one of his students in his EMGT 365 class analyzed Dr. Konur’s Dino-Mec shop and he found out that

Dr. Konur will consider closing the shop. Dr. Konur does not know if this student is right or wrong, but he heard

the rumors so decided to take action. He is considering the following two actions:

 Action 1: Increase repair charge to $1,500 and continue on the promotion that if the repair is not

completed within a month, charge the half price, i.e., $750. With this new price, the number of

customers coming to the shop will reduce and the estimated arrival rate will be 7 per month

(interarrival time will still be exponential).

Template for the M/M/s Queueing Model

Data Results

1 l = 10 (mean arrival rate) L = 5

0 m = 12 (mean service rate) Lq = 4.166666667

0 s = 1 (# servers)

0 W = 0.5

0 Pr(W > t) = 0.135335283 Wq = 0.416666667

0 when t = 1

0  = 0.833333333

0 Prob(Wq > t) = 0.112779403

0 when t = 1 n Pn So, if there are less than or equal to 3 cars in the

0 0 0.166666667 queueing system, then there are no cars parked

0 1 0.138888889 outside. Thus, the percentage of time cars being

0 2 0.115740741 parked out side is:

0 Expected earning from 1 customer Note that probability of earning 3 0.096450617 P0+P1+P2+P3 = 0.51774691

0 regular price = $1000 is equal to Pr(W<1 month) 4 0.080375514

0 1000 Probability of earning $1000 x $1000 and probability of earning $500 is 5 0.066979595 So, 51.78% of the time, the cars

0 discounted price + equal to Pr(W>1 month) 6 0.055816329 are parked inside on average.

0 500 Probability of earning $500 x $500 We know Pr(W<1 month) from 7 0.046513608 This does not satisfy the criteria

0 = the template and 8 0.03876134

0 932.33$ Pr(W>1 month)=1-Pr(W<1 month) 9 0.032301117

0 10 0.026917597

0 11 0.022431331

0 Expected earning in a month 12 0.018692776

0 = 13 0.015577313

0 Expected earning from a customer 14 0.012981094

0 x 15 0.010817579

0 Expected number of customers arriving in a month 16 0.009014649

= 17 0.007512207

9,323.32$ < $10,000 18 0.006260173

Therefore, Dr. Konur will consider 19 0.005216811

closing Dino-Mec 20 0.004347342

Problem 1-c

Problem 1-b

 Action 2: Remove the promotion, that is charge $1,000 for any repair. Without the discount

promotion, the number of customers coming to the shop will reduce and the estimated arrival

rate will be 9 per month (interarrival time will still be exponential).

Dr. Konur will take one of these actions.

d) If he wants to have the shop open, he should earn more than $10,000 per month on average. Would Action 1

guarantee that the shop will remain open? Show your calculations on how you reached your answer;

otherwise you will not get any credit. That is, calculate the expected monthly earnings with Action 1 and

compare to $10,000. You can use excel template if needed.

e) If he wants to have the shop open, he should earn more than $10,000 per month on average. Would Action 2

guarantee that the shop will remain open? Show your calculations on how you reached your answer;

otherwise you will not get any credit. That is, calculate the expected monthly earnings with Action 2 and

compare to $10,000. You can use excel template if needed.

Template for the M/M/s Queueing Model

Data Results

1 l = 7 (mean arrival rate) L = 1.4

0 m = 12 (mean service rate) Lq = 0.816666667

0 s = 1 (# servers)

0 W = 0.2

0 Pr(W > t) = 0.006737947 Wq = 0.116666667

0 when t = 1

0  = 0.583333333

0 Prob(Wq > t) = 0.003930469

0 when t = 1 n Pn

0 0 0.416666667

0 1 0.243055556

0 2 0.141782407

0 Expected earning from 1 customer Note that probability of earning 3 0.082706404 P0+P1+P2+P3 = 0.88421103

0 regular price = $1500 is equal to Pr(W<1 month) 4 0.048245403

0 1500 Probability of earning $1500 x $1500 and probability of earning $750 is 5 0.028143151 So, 88.42% of the time, the cars

0 discounted price + equal to Pr(W>1 month) 6 0.016416838 are parked inside on average.

0 750 Probability of earning $750 x $750 We know Pr(W>1 month) from 7 0.009576489 Action 1 satisfies the criteria

0 = the template and 8 0.005586285

0 1,494.95$ Pr(W<1 month)=1-Pr(W>1 month) 9 0.003258666

0 10 0.001900889

0 11 0.001108852

0 Expected earning in a month 12 0.00064683

0 = 13 0.000377318

0 Expected earning from a customer 14 0.000220102

0 x 15 0.000128393

0 Expected number of customers arriving in a month 16 7.48958E-05

= 17 4.36892E-05

10,464.63$ > $10,000 18 2.54854E-05

Therefore, Dr. Konur will not 19 1.48665E-05

close Dino-Mec 20 8.67211E-06

Problem 1-c

Will be needed in part f)

f) (5 points) Which action should Dr. Konur take if he wants to keep Dino-Mec open? Will this action satisfy the

criterion that cars are being parked inside the shop at least 75% of the time? You can use excel template if

needed.

Spring 2013, Final, Problem 2

Dr. Konur is an assistant professor in the Engineering Management and Systems Engineering department at the

Missouri University of Science and Technology. Aside from his work related to the university, he manages a local

bank in Rolla, MO called Dino-Bank of Rolla. He must now decide on how many tellers are needed in the bank.

Specifically, Dr. Konur hires his students to work as tellers. The interarrival time between customers is

exponentially distributed with a mean interarrival time of 30 seconds. Customers arriving at the bank wait in a

single line and go to the tellers in a first-come-first served manner. Dr. Konur estimates that his students can

complete a customer’s transaction in 2 minutes on average and the service time is exponentially distributed.

Template for the M/M/s Queueing Model

Data Results

1 l = 9 (mean arrival rate) L = 3

0 m = 12 (mean service rate) Lq = 2.25

0 s = 1 (# servers)

0 W = 0.333333333

0 Pr(W > t) = 0.049787068 Wq = 0.25

0 when t = 1

0  = 0.75

0 Prob(Wq > t) = 0.037340301

0 when t = 1 n Pn

0 0 0.25

0 1 0.1875

0 2 0.140625

0 Earning from 1 customer 3 0.10546875 P0+P1+P2+P3 = 0.68359375

0 regular price = 4 0.079101563

0 1000 $1,000 5 0.059326172 So, 68.36% of the time, the cars

0 discounted price 6 0.044494629 are parked inside on average.

0 no discount Expected earning in a month 7 0.033370972 Action 2 does not satisfy the criteria

0 = 8 0.025028229

0 Earning from a customer 9 0.018771172

0 x 10 0.014078379

0 Expected number of customers arriving in a month 11 0.010558784

0 = 12 0.007919088

0 9,000.00$ < $10,000 13 0.005939316

0 Therefore, Dr. Konur will 14 0.004454487

0 close Dino-Mec 15 0.003340865

Problem 1-c

Will be needed in part f)

As is clear from parts d and e, Dr. Konur should

take Action 1 to keep Dino-Mec open.

Furthermore, this action satisfies the criteria.

Cost Inside Parking Cost Inside Parking

10,464.63$ 88.42% 9,000.00$ 68.36%

So, pick Action 1

Action 1 Action 2

a) What is the minimum number of students (tellers) Dr. Konur has to hire so that Dino-Bank will reach a steady-

state, i.e., the Dino-Bank will not ‘blow up’? Briefly explain why.

Now, Dr. Konur decided to hire the minimum number of tellers needed. Each teller is paid at a rate of $12 per

hour. And, Dr. Konur believes that each customer standing in the line for a minute has a delay cost of 10¢.

b) Calculate the expected hourly cost of Dino-Bank with the minimum number of tellers needed. Expected hourly

cost is equal to teller costs plus expected delay costs. You can use excel template if needed.

The interarrival time between customers is exponentially distributed with a mean interarrival time of 30 seconds.

This means that the arrival rate is 2 per minute

Dr. Konur estimates that his students can complete a customer’s transaction in 2 minutes on average and the service time is exponentially distributed.

This means that the service rate is 0.5 per minute

Recall that to reach a steady-state the utilization factor should be less than 1. In multiple server queues, utilization factor is

l where s is the number of servers. Thus, to have

s m  <1 we should have s>4

Therefore, the minimum number of tellers needed is 5

 =

c) Would adding one teller decrease or increase the expected hourly cost of Dino-Bank? You can use excel

template if needed.

Template for the M/M/s Queueing Model

Data Results

l = 2 (mean arrival rate) L = 6.216450216

m = 0.5 (mean service rate) Lq = 2.216450216

s = 5 (# servers)

W = 3.108225108

Pr(W > t) = 0.774573786 Wq = 1.108225108

when t = 1

 = 0.8

Prob(Wq > t) = 0.336086253

when t = 1 n Pn

0 0.012987013

1 0.051948052

2 0.103896104

Expected delay cost for 1 customer Note that probability of earning 3 0.138528139

= expected delay for 1 customer is 4 0.138528139

Delay Cost Expected delay time for 1 customer (minutes) the expected time a customer waits 5 0.110822511

Per Minute x in the line. This is given by Wq 6 0.088658009

0.10$ Delay cost per minute 7 0.070926407

= 8 0.056741126

0.1108225$ per customer 9 0.0453929

10 0.03631432

11 0.029051456

Expected delay cost per minute 12 0.023241165

= 13 0.018592932

Expected delay cost for 1 customer 14 0.014874346

x 15 0.011899476

Expected number of customers arriving in minute 16 0.009519581

= 17 0.007615665

0.2216450$ per minute 18 0.006092532

19 0.004874026

Expected delay cost per hour 20 0.00389922

= 21 0.003119376

13.2987$ per hour 22 0.002495501

Teller Cost 23 0.001996401

Per hour Teller Cost per hour 24 0.001597121

$12 = 25 0.001277697

Per teller cost per hour

x

Number of tellers

=

$60

Expected Hourly Cost

=

Expected Hourly Delay Cost

+

Hourly Tellers Cost

=

73.30$

Dr. Konur thinks that adding another teller will not help. After thinking about Dino-Bank for days and days, he

came up with the following idea: instead of having a single line with minimum number of tellers needed, he will

have 5 different lines such that each line is for each individual teller. In this case, he will manage the customer

arrivals such that customers arrive at each teller’s line with a mean interarrival time of 2.5 minutes, which is

Template for the M/M/s Queueing Model

Data Results

l = 2 (mean arrival rate) L = 4.569521691

m = 0.5 (mean service rate) Lq = 0.569521691

s = 6 (# servers)

W = 2.284760845

Pr(W > t) = 0.674489182 Wq = 0.284760845

when t = 1

 = 0.666666667

Prob(Wq > t) = 0.104757661

when t = 1 n Pn

0 0.016685206

Problem 1-c 1 0.066740823

2 0.133481646

Expected delay cost for 1 customer Note that probability of earning 3 0.177975528

= expected delay for 1 customer is 4 0.177975528

Delay Cost Expected delay time for 1 customer (minutes) the expected time a customer waits 5 0.142380423

Per Minute x in the line. This is given by Wq 6 0.094920282

0.10$ Delay cost per minute 7 0.063280188

= 8 0.042186792

0.0284761$ per customer 9 0.028124528

10 0.018749685

11 0.01249979

Expected delay cost per minute 12 0.008333193

= 13 0.005555462

Expected delay cost for 1 customer 14 0.003703642

x 15 0.002469094

Expected number of customers arriving in minute 16 0.001646063

= 17 0.001097375

0.0569522$ per minute 18 0.000731584

19 0.000487722

Expected delay cost per hour 20 0.000325148

= 21 0.000216765

3.4171$ per hour 22 0.00014451

Teller Cost 23 9.63402E-05

Per hour Teller Cost per hour 24 6.42268E-05

$12 = 25 4.28179E-05

Per teller cost per hour

x

Number of tellers

=

$72

Expected Hourly Cost

= Note that expected delay costs decreased

Expected Hourly Delay Cost with the additional teller, however, the cost increase

+ due to adding one more teller is greater than the

Hourly Tellers Cost cost decrease in delays due to adding one more teller.

= Hence, the costs will increase with the additional

75.42$ teller.

exponentially distributed. Each teller will still have exponential service time distribution with mean service time

of 2 minutes. Still, each customer standing in the line for a minute has a delay cost of 10¢.

d) Will the idea of having separate lines reduce expected costs compared to the expected costs with the single

line and the minimum number of tellers needed (i.e., the expected cost calculated in part b)? You can use

excel template if needed. (hint: each teller is a separate but identical queueing system under the new idea).

Suppose that Dr. Konur decided that the new idea is not good and he will continue with a single line and the

minimum number of servers. Now, answer the following questions about Dino-Bank independent of each other

and briefly explain your answers:

Template for the M/M/s Queueing Model

Data Results

1 l = 0.4 (mean arrival rate) L = 4

0 m = 0.5 (mean service rate) Lq = 3.2

0 s = 1 (# servers)

0 W = 10

0 Pr(W > t) = 0.904837418 Wq = 8

0 when t = 1

0  = 0.8

0 Prob(Wq > t) = 0.723869934

0 when t = 1 n Pn

0 0 0.2

0 Problem 1-c 1 0.16

0 2 0.128

0 Expected delay cost for 1 customer Note that probability of earning 3 0.1024

0 = expected delay for 1 customer is 4 0.08192

0 Delay Cost Expected delay time for 1 customer (minutes) the expected time a customer waits 5 0.065536

0 Per Minute x in the line. This is given by Wq 6 0.0524288

0 0.10$ Delay cost per minute 7 0.04194304

0 = 8 0.033554432

0 0.8000000$ per customer 9 0.026843546

0 10 0.021474836

0 11 0.017179869

0 Expected delay cost per minute 12 0.013743895

0 = 13 0.010995116

0 Expected delay cost for 1 customer 14 0.008796093

0 x 15 0.007036874

0 Expected number of customers arriving in minute 16 0.0056295

= 17 0.0045036

0.3200000$ per minute 18 0.00360288

19 0.002882304

Expected delay cost per hour 20 0.002305843

= 21 0.001844674

19.2000$ per hour 22 0.00147574

Teller Cost 23 0.001180592

Per hour Teller Cost per hour 24 0.000944473

$12 = 25 0.000755579

Per teller cost per hour

x

Number of tellers

=

$12

Expected Hourly Cost Now, this cost is the expected hourly

= cost of a single line-single teller

Expected Hourly Delay Cost We will have 5 of these, which are all

+ identical.

Hourly Tellers Cost

= Then the total cost will be

31.20$ 156.00$

Compared to single-line-multiple tellers of part b

separating lines increases costs

e) If service time of each teller is reduced by the same amount and still exponentially distributed, will the average

number of people waiting in the line increase or decrease? Explain why?

Since service time is reduced, the service rate will increase, hence, the utilization factor will

decrease. This means that the average number of people in the line will decrease

f) If one more additional server, with the same characteristics with the other tellers, is hired, will Dino-Bank be

subject to higher or lower delay costs? Explain why?

g) Suppose that Fridays are busier than the other days; hence, the arrival rate of the customers increases. Will

this increase or decrease the expected delay costs? Explain why?

This will increase the utilization factor

This will increase average waiting time in line; hence, this will increase

expected delay cost for 1 customer. Thus, Dino-Bank will have higher delay costs.

h) Instead of working with multiple student tellers, Dr. Konur decided to be the single teller of Dino-Bank as he

can complete a customer transaction way faster than his students and he works for lower salaries. He can

complete 3 customer transactions per minute. The customer interarrival time is still exponentially distributed

with mean of 30 seconds. Since Dr. Konur does not like staying idle, whenever there is no customer in Dino-

Bank, he plays Angry-birds. What percentage of time Dr. Konur is playing Angry birds? Explain why?

i) Still suppose that Dr. Konur is the single teller and Fridays are busier than the other days. Will this increase

the difference between the expected number of people in Dino-Bank and the expected number of people in

the line? Explain why?

We already know the answer to this by comparing parts b and c. Additional server will decrease the

utilization factor. This will reduce average waiting time in line; hence, this will reduce

expected delay cost for 1 customer. Thus, Dino-Bank will have lower delay costs.

Now, we have a single server system and the idle time percentage of the server is 1-utilization factor.

arrival rate = 2 per minute

service rate = 3 per minute

utilization factor = 2/3 = 0.667

Thus, Dr. Konur will be idle for 0.333, or 33.33% of the time playing Angry-birds.

We know that

L = L q + l /m

That is, the utilization factor is the difference between the expected number of people in Dino-Bank and the

expected number of the people in line.

Since Fridays are busier, it means that the arrival rate increases, hence, the utilization factor increases. Therefore,

This will increase the difference between the expected number of people in Dino-Bank and the expected number of

people in the line.

Spring 2014, Final, Problem 2

Suppose that you are the manager of a workstation that pre-processes a part of the product your

company sells. Specifically, you are in control of three machines that can be used in the pre-

processing. These machines can be programmed such that a machine can complete the whole or

some of the pre-processing operation of a part. Currently, you are considering re-designing your

workstation and you have the following alternative setups:

 Setup 1: Program each machine so that each machine can complete a whole pre-process

for a part. In this case, each machine can complete a whole pre-process of a part in 12

minutes on average and the pre-process time is exponentially distributed. Furthermore,

you will set three sub-workstations; one for each machine such that each machine will

have its own sub-workstation; and, each sub-workstation will have its own inventory

space for parts waiting to be pre-processed. In this setting, a part will arrive in a sub-

workstation at every 15 minutes on average and the interarrival time is exponentially

distributed.

o In this case, we have 3 identical M/M/1 systems, each of which has service

rate=5/hour and arrival rate=4/hour.

 Setup 2: Similar to Setup 1, program each machine so that each machine can complete a

whole pre-process for a part. In this case, each machine can complete a whole pre-process

of a part in 12 minutes on average and the pre-process time is exponentially distributed.

Different than Setup 1, you will not have sub-workstations. Instead, you will have a single

work-station and it will have its own inventory space for parts waiting to be pre-

processed. In this setting, a part will arrive in the workstation at every 5 minutes on

average and the interarrival time is exponentially distributed. A part waiting will be pre-

processed by the next available machine.

o In this case, we have one M/M/3 system with service rate=5/hour for a server and

the arrival rate=12/hour.

 Setup 3: Program each machine such that they complete a specific part of the pre-process

of a part. In this case, all three machines will simultaneously work on completing a pre-

process operation. With three machines simultaneously working on the same pre-process

operation of a part, it takes 4 minutes to complete a pre-process operation on average

and the pre-process time is exponentially distributed. Similar to Setup 2, you will have a

single work-station and it will have its own inventory space for parts waiting to be pre-

processed. In this setting, a part will arrive in the workstation at every 5 minutes on

average and the interarrival time is exponentially distributed.

o In this case, we have one M/M/1 system with service rate=15/hour and the arrival

rate=12/hour.

You want to select one of these alternative setups based on their hourly average cost. The hourly

average cost consists of the machine costs and the inventory holding costs of the parts waiting

to be pre-processed. In particular, a machine costs $10 per hour for the times it is actively working

on pre-processing and only $2 per hour for the times it is idle. A part waiting in the inventory to

be pre-processed has an inventory holding cost of $4 per hour.

 Hourly Cost = Machine Cost/hour + Inventory Cost/hour

Machine Cost/hour = ($10*p+2*(1-p))*(number of machines)

Inventory Cost/hour = $4*Wq*lambda

a) (6 points) Calculate the average hourly cost of Setup 1.

We have total cost/hour is equal to $21.2 for one M/M/1. We have 3 identical ones.

Therefore, total cost/hour is $63.6.

b) (6 points) Calculate the average hourly cost of Setup 2.

We have total cost/hour is equal to $35.55.

Data Results Total

l = 4 (mean arrival rate) Machine Working Cost L = 4 12 <=10?? No!!!

m = 5 (mean service rate) 10 Lq = 3.2

s = 1 (# servers) Machine Idle Cost

2 W = 1

Pr(W > t) = 0.367879 Inventory Cost per hour per part Wq = 0.8

when t = 1 4

 = 0.8

Prob(Wq > t) = 0.677185 <0.5?? No!!

when t = 0.166667 n Pn

0 0.2

1 0.16

Machine Cost/ hour 8.4 2 0.128

Inventory Cost/hour 12.8 3 0.1024

4 0.08192

Total Cost/hour 21.2 5 0.065536

Problem 2 - a

Data Results

l = 12 (mean arrival rate) Machine Working Cost L = 4.988764045 <=10?? Yes!!

m = 5 (mean service rate) 10 Lq = 2.588764045

s = 3 (# servers) Machine Idle Cost

2 W = 0.415730337

Pr(W > t) = 0.07639 Inventory Cost per hour per part Wq = 0.215730337

when t = 1 4

 = 0.8

Prob(Wq > t) = 0.392541 <0.5?? Yes!!

when t = 0.166667 n Pn

0 0.056179775

1 0.134831461

Machine Cost/ hour 25.2 2 0.161797753

Inventory Cost/hour 10.35505618 3 0.129438202

4 0.103550562

Total Cost/hour 35.55505618 5 0.082840449

Problem 2 - b

c) (6 points) Calculate the average hourly cost of Setup 3.

We have total cost/hour is equal to $38.

d) (2 points) Based on the average hourly cost of the setups, which setup alternative you

would select?

We would select Setup 2.

e) (6 points) Now, you realize that average hourly cost is not the only thing you are

concerned with. Specifically, you want that at most 50 percent of the parts wait in the

inventory more than 10 minutes. Which setup alternatives satisfy this criterion?

This criterion implies that P(Wq>10minutes)<0.5. Setups 2 and 3 satisfy it.

f) (6 points) Furthermore, you do not want to have more than 10 parts on average, either

waiting in inventory or being pre-processed, in the whole work-station. Which setup

alternatives satisfy this criterion?

This criterion implies that L<10. Note that in Setup 1, total will be 3*L. Hence, only Setups

2 and 3 satisfy it, setup 1 does not satisfy it.

g) (3 points) If the busy and idle time costs of each machine increase, would this change

your selection in part d, i.e., the selection you made based on average hourly costs?

Explain without solving the problem.

No, because Setups 1, 2, and 3 have the same utilization factor, hence, the same machine

costs per hour.

Data Results

l = 12 (mean arrival rate) Machine Working Cost L = 4 <=10?? Yes!!

m = 15 (mean service rate) 10 Lq = 3.2

s = 1 (# servers) Machine Idle Cost

2 W = 0.333333333

Pr(W > t) = 0.049787 Inventory Cost per hour per part Wq = 0.266666667

when t = 1 4

 = 0.8

Prob(Wq > t) = 0.485225 <0.5?? Yes!!

when t = 0.166667 n Pn

0 0.2

1 0.16

Machine Cost/ hour 25.2 2 0.128

Inventory Cost/hour 12.8 3 0.1024

4 0.08192

Total Cost/hour 38 5 0.065536

Problem 2 - c