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Assignment 1: M/M/1 Performance Evaluation. Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has exponential distribution. On average how many customers are in the waiting line?

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assignment 1 m m 1 performance evaluation
Assignment 1: M/M/1 Performance Evaluation

Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has exponential distribution.

  • On average how many customers are in the waiting line?
  • How long does a customer stay in the line?
  • How long does a customer stay in the processor (with the server)?
  • On average how many customers are there with the server?
  • On average how many customers are in the system?
  • On average how long does a customer stay in the system?
m m 1 performance evaluation
M/M/1 Performance Evaluation

R = 6 customers per hour

Rp =1/5 customer per minute, or 60(1/5) = 12/hour

U= R/Rp = 6/12 = 0.5

a) On average how many customers are there in the waiting line?

gap example
GAP Example

b) How longdoes a customer stays in the line?

d) On average how many customers are there with the server?

c) How long does a customer stay in the processor (with the server)?

m m 1 performance evaluation4
M/M/1 Performance Evaluation

e) On average how many customers are in the system?

f) On average how long does a customer stay in the system?

assignment 2 m m 1 performance evaluation
Assignment 2: M/M/1 Performance Evaluation
  • What if the arrival rate is 11 per hour? Processing rate is still Rp=12
m m 1 performance evaluation6
M/M/1 Performance Evaluation

As the utilization rate increases to 1 (100%) the number of customers in line (system) and the waiting time in line (in system) is increasing exponentially.

assignment 3 m g c
Assignment 3: M/G/c

A local GAP store on average has 10 customers per hour for the checkout line. The inter-arrival time follows the exponential distribution. The store has two cashiers. The service time for checkout follows a normal distribution with mean equal to 5 minutes and a standard deviation of1 minute.

  • On average how many customers are in the waiting line?
  • How long does a customer stay in the line?
  • How long does a customer stay in the processors (with the servers)?
  • On average how many customers are there with the servers?
  • On average how many customers are in the system ?
  • On average how long does a customer stay in the system ?
key information
Key Information

Arrival rate: R = 10 per hour

Average inter-arrival time: Ta = 1/R = 1/10 hr = 6 min

Standard deviation of inter-arrival time: Sa = Ta

Service rate per server: 12 per hour

Average service time: Tp= 1/12 hours = 5 min

Standard deviation of service time: Sa= 1 min

Coefficient of variation for inter-arrival time: Ca = Sa/Ta = 1

Coefficient of variation for service time: Cp= Sp/Tp = 1/5 =0.2

Number of servers: c =2

Rp = c/Tp = 2/(1/12) = 24 per hour

U= R/Rp = 10/24 = 0.42

m g 2 example
M/G/2 Example

a) On average how many customers are there in the waiting line?

b) How long does a customer stay in the line?

c) How long does a customer stay in the processors (with the servers)?

  • Average service time: Tp= 1/12 hours = 5 min
m g 2 example10
M/G/2 Example

d) On average how many customers are therewith the servers?

e) On average how many customers are there in the system?

f) On average how long does a customer stay in the system?

comment on general formula
Comment on General Formula
  • Approximation formula gives exact answers for M/M/1 system.
  • Approximation formula provide good approximations for M/M/2 system.
assignment 4 m m c example
Assignment 4: M/M/c Example

A call center has 11 operators. The arrival rate of calls is 200 calls per hour. Each of the operators can serve 20 customers per hour. Assume inter-arrival time and processing time follow Poisson and Exponential, respectively. What is the average waiting time (time before a customer’s call is answered)?

assignment 5 m d c example
Assignment 5: M/D/c Example
  • Suppose the service time is a constant
  • What is the answer to the previous question?
    • In this case
problem 6
Problem 6

Students arrive at the Administrative Services Office on the average of one every 15 minutes, and their request take on average 10 minutes to be processed. The service counter clerk works 8 hours per day andis staffed by only 1 clerk, Judy Gumshoe. Assume Poisson arrivals and exponential service times.

M/M/1 Queuing System

R = 4 customers/hour, Poisson (Ca =1)

Rp = 6 customers/hour, Exponential (Cp =1)

a) What percentage of time is Judy idle?

b) How much time, on average, does a student spend waiting in line?

problem 6 m m 1
Problem 6; M/M/1

a) What percentage of time is Judy idle?

U = R/Rp = 4/6 = 66.67% of time she is busy

1- U = 33.33% of time idle

b) How much time, on average, does a student spend waiting in line?

Ti R = Ii  Ti = Ii/R  1.33/4 = 0.33 hours

= 0.33 hours or 20 minutes

problem 7
Problem 7

You are working at a bank and doing resource requirements planning. You think that there should be six tellers working in the bank. Tellers take fifteen minutes per customer with a standard deviation of five minutes. On average one customer arrives in every three minutes according to an exponential distribution.

a) On average how many customers would be waiting in line?

b) On average how long would a customer spend in the bank?

slide19

c = 6, R = 20, Rp= c/Tp = 6/15 /min, 60(6/15) = 24 /hr

U = R/Rp = 20/24 = 0.83

Ci = 1, Cp = 5/15 = 0.33

a) On average how many customers are in line?

Problem 7; M/G/c

  • b) On average how long would a customer spend in the bank?

Ti = Ii/R  1.62/20 = 0.081 hours, or 4.86 minutes

Tp = 15 minutes

T = Ti+Tp = 4.86+15 = 19.86 minutes

slide20

Consider a call center that employs 8 agents. Past data collected has shown that the mean time between customer arrivals is 1 minute, and has a standard deviation of 1/2 minute. The amount of time in minutes the past 10 callers have spent talking to an agent is as follows: 4.1, 6.2, 5.5, 3.5, 3.2, 7.3, 8.4, 6.3, 2.6, 4.9.

a) What is the coefficient of variation for the inter-arrival times?

b) What is the mean time a caller spends talking to an agent?

c) What is the standard deviation of the time a caller spends talking to an agent?

d) What is the coefficient of variation for the times a caller spends talking to an agent?

e) What is the expected number of callers on hold, waiting to talk to an agent?

Problem 8

slide21

a) What is the coefficient of variation for the inter-arrival times?

Ca = Sa/Ta = 0.5/1 = 0.5 

b) What is the mean time a caller spends talking to an agent?

= average (4.1, 6.2, 5.5, 3.5, 3.2, 7.3, 8.4, 6.3, 2.6, 4.9) = 5.2 minutes.

c) What is the standard deviation of the time a caller spends talking to an agent?

= stdev(4.1, 6.2, 5.5, 3.5, 3.2, 7.3, 8.4, 6.3, 2.6, 4.9) =1.88 minutes

Problem 8; G/G/1

slide22

d) What is the coefficient of variation for the times a caller spends talking to an agent?

(standard deviation)/mean = 1.88/5.2 = 0.36

e) What is the expected number of callers on hold, waiting to talk to an agent?

R= 1 per minute, c = 8, processing rate for one agent is = 1/5.2. For c=8 agents, Rp = 8/5.2 = 1.54 /min

U = R/Rp= 1/(1.54) = 0.65

Problem 8

slide23

f) What is the expected number of callers either on hold or talking to an agent?

I = Ii + Ip= 0.09 + 0.65(8) =5.29

g) What is the expected amount of time a caller must wait to talk to an agent?

RTi = Ii  Ti = 0.09/1 = 0.09 minutes

h) What is the expected amount of time between when a caller first arrives to the system, and when that caller finishes talking to an agent?

T = I/R = 5.29/1 = 5.29 minutes

Alternatively; T= Ti+Tp = 0.09+5.2 = 5.29

Problem 8

slide24

Wells Fargo operates one ATM machine in a certain Trader Joe’s. There is on average 8 customers that use the ATM every hour, and each customer spends on average 6 minutes at the ATM. Assume customer arrivals follow a Poisson process, and the amount of time each customer spends at the ATM follows an exponential distribution.

Problem 9

a) What is the percentage of time the ATM is in use?

R = 8 per hour, Rp = 10 per hour

U = 8/10 = 0.8 =80%

slide25

b) On average, how many customers are there in line waiting to use the ATM?

Problem 9; M/M/1

c) Suppose that the number of customers in line waiting to use the ATM is 3. (This may or may not be the answer you found in part b). All information remains as originally stated. What is the average time a customer must wait to use the ATM? State your answer in minutes.

  • TiR = Ii Ti = 3/8 hrTi = 3/8 (60) = 22.5 minutes
slide26

The Matador housing office has one customer representative for walk-in students. The arrival rate is 10 customers per hour and the average service time is 5 minutes. Both inter-arrival time and service time follow exponential distributions.

a) What is the average waiting time in line?

R = 10 /hr, Rp = 20 /hr, U = 10/12= 0.83

Problem 10

Ti = Ii/R = 4.17/10 = 0.42 hr

slide27

The Monterey post station has 7 tellers from Monday to Saturday. Customers arrive to the station following a Poisson process with a rate of 36 customers per hour. The service time is exponentially distributed with mean 10 minutes.

a) What is the utilization rate of the tellers?

R = 36, Rp = 7/10=0.7 /min or 42 /hr

U = R/Rp= 36/42 = 6/7 = 85.7%

b) What is the average number of customers waiting in line?

Problem 11

slide28

On Sunday, instead of tellers, the post station only opens 3 auto-mail machines to provide automatic service. Each machine can weight the different size of packages, print self-adhesive labels and accept payments. Arrival is Poisson with rate 20 customers per hour. The service time is 4 minutes with probability 0.75 and 16 minutes with probability 0.25.

c) What is the mean service time?

Tp = 0.75(4) + 0.25(16) = 7 min

d) What is the coefficient of variation of service time?

Sp = sqrt[0.75(4-7)2+0.25(16-7)2] = sqrt[0.75(3)2+0.25(9)2] =6.93

Cp = 6.93 / 7 = 0.866

Problem 11

slide29

e) What is the utilization rate?

R = 20 customers/hr, Rp = 3/8=0.375 /min or 22.5 /hr

U =R/Rp= 20/(22.5) = 88.9%

f) What is the average number of customers waiting in line?

Problem 11

slide30

Bank of San Pedro has only 1 teller. On average, 1customer comes every 6 minutes, and it takes the teller an average of 3 minutes to serve a customer. To improve customer satisfaction, the bank is going to implement a unique policy called, “We Pay While You Wait.” Once implemented, the bank will pay each customer $3 per minute while a customer waits in line. (So the clock starts when a customer joins the line, and stops when the customer begins to talk to the teller.) Bank of San Pedro hired you as a consultant and you are responsible for estimating how much the “We Pay While You Wait” program will cost. Your preliminary study indicates there are, on average, 0.5 customers waiting in line. Assume linear cost. If a customer waits for ten seconds in line, Bank of San Pedro will pay $0.5. Assume that arrival follows Poisson and service time follows exponential distribution.

Problem 12

slide31

Problem 12

a) Compute the capacity of the teller

  • 10 customers/hour
  • 3.33 customers/hour
  • 20 customers/hour
  • 30 customers/hour
  • Cannot be determined

It takes the teller an average of 3 minutes to serve a customer

60/3 = 20 customers per hour

slide32

Problem 12

b) Calculate the proportion of the time the teller is busy.

  • 100%
  • 80%
  • 62%
  • 50%
  • 40%

R=10, Rp= 20

R/Rp= 10/20 = 0.5

= 50%

slide33

Problem 12

c) How long, on average, does a customer wait in line?

  • 6 minutes
  • 4.8 minutes
  • 3 minutes
  • 2.6 minutes
  • 2 minutes

Indeed Ii was even given in the problem.

Ti= Ii/R

Ti= 0.5/10 = .05

0.05(60) = 3 minutes

slide34

Problem 12

d) Calculate the expected “hourly” cost of the “We Pay While You Wait” program.

  • $9
  • $36
  • $60
  • $90
  • $140

Ii =0.5. Therefore, a half of a customer is always there. For each hour one customer gets 60(3) = $180. Thus 0.5 customer gets $90.

Perhaps you do not believe me.

Each customer waits, on average, 3 minutes.

So he or she receives, on average, 3(3) =$9.

There are 10 customers arriving per hour.

So the overall cost of this program is 9*10=$90.

slide35

Problem 12

e) Suppose each additional clerk costs X dollars per hour (including all other clerk related costs such as benefits, space and equipment hourly costs). Compute the maximum value of X if it is at our benefit to hire one additional clerk?

If we have two clerks, Rp increases from 20 to 40, and utilization drops from 0.5 to 10/40 = 0.25

Ii reduced from 0.5 customers to 0.045 customers = 0.455

slide36

Problem 12

The number of customers waiting in the line reduced by 0.455. This means each hour, there are 0.455 less customers waiting in line.

The cost of each hour waiting per 1 customer is $180

The waiting cost of 0.455 customers is 180(0.455) = 81.96

If the additional clerk costs less than $81.96 per hour it is at our benefit to hire her.

  • e) Suppose each additional clerk costs $30 per hour. How many new clerk should we hire, one or two?
  • Obviously, it is at hour benefit to hire one clerk.
  • If we hire two clerks (to have 3 clerks), Rp increases to 60, and utilization drops 10/60 = 0.167
slide37

Problem 12

  • Ii is reduced from 0.045 customers to 0.008, a 0.042 customer reduction.
  • By adding the third clerk, there are 0.042 less customers waiting in line (each hour and always)
  • 180(0.042) = about $7-$8
  • It is not at our benefit to hire the second clerk, pay $30 per hour capacity cost to reduce waiting cost by $8 per hour.
slide38

Problem 12

We did not need to do this much computations for the third clerk. With two clerks the total number of customers waiting in line was:

Even if we reduce the number of customers waitingto 0,we have reduced the line by 0.045 customers.

0.045(180) = 8.1

It is not worth the cost $30 to benefit $8

slide39

Problem 13

A local grocery store has 2 cashier stations, and 1 experienced cashier and 1 novice cashier. During a typical working day (8 hours), 120 customers will show up. The novice cashier will serve 48 customers and the experienced cashier will serve 72 customers. On average it takes 6 minutes for the novice cashier to serve one customer and 3 minutes for the experienced cashier to serve one customer.

slide40

Problem 12

a) During the rush hours, approximate 25 customers will show up in an hour. Does the store have enough capacity for the rush hour?

  • Yes, the capacity of the store is 25 customers per hour.
  • Yes, the capacity of the store is 30 customers per hour.
  • No, the capacity of the store is 10 customers per hour.
  • No, the capacity of the store is 20 customers per hour.
  • None of the above.

On average it takes 6 minutes for the novice cashier to serve a customer  60/6 = 10 customers/hr

  • On average it takes 3 minutes for the experienced cashier to serve a customer  60/3 = 20 customers/hr
  • Capacity = 10+20 = 30
slide41

Problem 12

b) During the day, both cashier stations on average have 2 customers waiting. On average, how long does a customer stay in the novice cashier’s waiting line?

  • 8 minutes
  • 11 minutes
  • 16 minutes
  • 20 minutes
  • None of the above
  • The novice cashier will serve 48/8 = 6 customers/hr
  • R = 6 /hr
  • TR = I
  • 6T = 2
  • T = 1/3 hr or 20 min
slide42

Problem 13

c) On average, how long does a customer stay in the experienced cashier’s line?

  • 7 minutes
  • 9.4 minutes
  • 13.3 minutes
  • 15 minutes
  • None of the above
  • On the same token
  • the novice cashier will serve 72/8 = 9 customers/hr
  • R = 9 /hr
  • TR = I  9T = 2  T=2/9 hr or 13.33 min
slide43

Problem 13

d) On average, how long does a customer spend in the store?

The novice and experienced cashier serve 6, and 9 customersper hour, respectively.

Customers served by Novice: 6 min service time, 20 min waiting time. T = 26 min for 6/15 customers.

Customers served by Experienced: 3 min service time, 13.33 min waiting time. T = 16.33 for 9/15 customers.

(6/15)(26) + (9/15)(16.33) = 20.2

slide44

Problem 13

To shorten the waiting time, the manager does a detailed study and finds that half of the customers purchase 5 items or less, and half of the customers purchase more than 5 items. The manager decides to let the novice cashier only serve the customers that purchase 5 items or less. After the change, it turns out that both cashier stations have 1.75 customers waiting on average. Assume that the novice cashier serves all of the customers purchasing 5 items or less and the experienced cashier serves all of the customers purchasing more than 5 items.

slide45

Problem 13

e) What is the average waiting time in the novice cashier’s line?

  • 13 minutes
  • 14 minutes
  • 17 minutes
  • 19 minutes
  • None of the above
  • Each of the two cashiers will serve 120/2 = 60 customers per day or 60/8 =7.5 customers/hr
  • TR = I
  • 7.5T = 1.75
  • T = 1.75/7.5 = 0.2333 hr or 14 min
slide46

Problem 13

f) What is the average waiting time in the experienced cashier’s line?

  • 9 minutes
  • 11 minutes
  • 12 minutes
  • 14 minutes
  • None of the above
  • R and I are the same for Naïve and Experience
  • Therefore, T is the same; 14 min
slide47

Problem 14

American Vending Inc. (AVI) supplies vending food to a large university. Because students often kick the machines out of anger and frustration, management has a constant repair problem. The machines break down on an average of 3/hr, and the breakdowns are distributed in a Poisson manner. Downtime costs the company $25/hr/machine, and each maintenance worker gets $4 per hr. One worker can service machines at an average rate of 5/hr, distributed exponentially; 2 workers working together can service 7/hr, distributed exponentially; and a team of 3 workers can do 8/hr, distributed exponentially. What is the optimal maintenance crew size for servicing the machines?

problem 14
Problem 14
  • c=????
  • c = 1
  • U=??

Down time cost = 25Ii

Capacity Cost= 4(# of team members)

Total Cost

slide49

Problem 14

Have I made any mistakes?

Downtime costs the company $25 /hr/machine.

When the machine is down?

Until it is up.

In the waiting line it is down. In the processor until the end of the process it is down.

There fore, besides Ii, I also need Ip

slide50

Problem 14

Lets check by using Ti and Tp instead of Ii and Ip

Ti = Ii/R

Wait Cost /Machine =

25Ti

Wait Cost /hr =

R(Wait Cost /Machine) =

3(Wait Cost /Machine)

In Process Time = Tp = 1/Rp

In Process Cost /Machine =

25Tp

In Process Cost /hr =

3(25Tp)