Embed Size (px)
Citation preview
ADVANCED OPERATIONS
RESEARCH
By: - Hakeem–Ur–Rehman
IQTM–PU 1
R A O QUEUEING THEORY
QUEUEING THEORY: AN INTRODUCTION Queuing theory is one of the most widely used quantitative analysis
techniques. The three basic components are: Arrivals, Service facilities & Actual waiting line
7:00 7:10 7:20 7:30 7:40 7:50 8:00
Patient
Arrival Time
Service Time
1
2
3
4
5
6
7
8
9
10
11
12
0
5
10
15
20
25
30
35
40
45
50
55
4
4
4
4
4
4
4
4
4
4
4
4
A SOMEWHAT ODD SERVICE PROCESS
Patient
Arrival Time
Service Time
1
2
3
4
5
6
7
8
9
10
11
12
0
7
9
12
18
22
25
30
36
45
51
55
5
6
7
6
5
2
4
3
4
2
2
3
Time
7:10 7:20 7:30 7:40 7:50 8:00 7:00
Patient 1 Patient 3 Patient 5 Patient 7 Patient 9 Patient 11
Patient 2 Patient 4 Patient 6 Patient 8 Patient 10 Patient 12
0
1
2
3
2 min. 3 min. 4 min. 5 min. 6 min. 7 min.
Service times
Num
ber
of ca
ses
A MORE REALISTIC SERVICE PROCESS
7:00 7:10 7:20 7:30 7:40 7:50
Inventory (Patients at
lab)
5
4
3
2
1
0
8:00
7:00 7:10 7:20 7:30 7:40 7:50 8:00
Wait time
Service time
Patient
Arrival Time
Service Time
1 2 3 4 5 6 7 8 9 10 11 12
0 7 9 12 18 22 25 30 36 45 51 55
5 6 7 6 5 2 4 3 4 2 2 3
VARIABILITY LEADS TO WAITING TIME
Input: • Unpredicted Volume swings • Random arrivals (randomness is the rule, not the exception) • Incoming quality • Product Mix
Resources: • Breakdowns / Maintenance • Operator absence • Set-up times
Tasks: • Inherent variation • Lack of SOPs • Quality (scrap / rework)
Routes: • Variable routing • Dedicated machines
Buffer Processing
Especially relevant in service operations (what is different in service industries?): • emergency room • air-line check in • call center • check-outs at cashier
VARIABILITY: WHERE DOES IT COME FROM?
TEST FOR STATIONARY ARRIVAL Call
Arrival
Time, ATi
1
2
3
4
5
6
7
6:00:29
6:00:52
6:02:16
6:02:50
6:05:14
6:05:50
6:06:28
Time 6:01 6:02 6:03 6:04 6:05 6:06 6:00
Call 1
The concept of inter-arrival times
Call 2 Call 3 Call 4 Call 5 Call 6 Call 7
IA1 IA2 IA3 IA4 IA5 IA6
Inter-Arrival
Time, IAi=ATi+1 -ATi
00:23
01:24
00:34
02:24
00:36
00:38
0
20
40
60
80
100
120
140
160
0:1
5
2:0
0
3:4
5
5:3
0
7:1
5
9:0
0
10:4
5
12:3
0
14:1
5
16:0
0
17:4
5
19:3
0
21:1
5
23:0
0
Seasonality over the course of a day
Time
Number of customers Per 15 minutes
0
20
40
60
80
100
120
140
160
0:1
5
2:0
0
3:4
5
5:3
0
7:1
5
9:0
0
10:4
5
12:3
0
14:1
5
16:0
0
17:4
5
19:3
0
21:1
5
23:0
0 Time
Number of customers Per 15 minutes
TEST FOR STATIONARY ARRIVAL (Cont…)
Test for stationary arrivals
0
6:00:00 7:00:00 8:00:00 9:00:00 10:00:00
0
100
200
300
400
500
600
700
Cumulative
Customers
6:00:00 7:00:00 8:00:00 9:00:00 10:00:00 6:00:00 7:00:00 8:00:00 9:00:00 10:00:00
Time
Expected arrivals
if stationary
Actual, cumulative
arrivals
Test for stationary arrivals
0
7:15:00 7:18:00 7:21:00 7:24:00 7:27:00 7:30:00
0
10
20
30
40
50
60
70
7:15:00 7:18:00 7:21:00 7:24:00 7:27:00 7:30:00 Time
Cumulative
Customers
Stationary
Arrivals?
How to analyze a demand / arrival process?
YES NO
Break arrival process up into
smaller time intervals
YES
Exponentially distributed inter-
arrival times?
NO
Compute a: average interarrival time
CVa=1
Compute a: average interarrival time
CVa= St.dev. of interarrival times / a
Blocked calls (busy signal)
Abandoned calls (tired of waiting)
Calls on Hold
Sales reps processing
calls
Answered Calls
Incoming calls
Call center
Financial consequences
Lost throughput Holding cost Lost goodwill Lost throughput (abandoned)
$$$ Revenue $$$ Cost of capacity Cost per customer
At peak, 80% of calls dialed received a busy signal.
Customers getting through had to wait on average 10 minutes
Extra telephone expense per day for waiting was $25,000.
AN EXAMPLE OF A SIMPLE QUEUING SYSTEM
ARRIVAL CHARACTERISTICS: POISSON DISTRIBUTION
! P(X)
X
e x
.00
.05
.10
.15
.20
.25
.30
.35
0 1 2 3 4 5 6 7 8 9 10 11 X
P(X
) P(X), = 2
.00
.05
.10
.15
.20
.25
.30
0 1 2 3 4 5 6 7 8 9 10 11 X
P(X
)
P(X), = 4
SERVICE TIME CHARACTERISTICS: EXPONENTIAL DISTRIBUTION
Pro
babili
ty
(for
Inte
rvals
of
1 M
inute
)
30 60 90 120 150 180
Average Service Time of 1 Hour
Average Service Time of 20 Minutes
X
MinutePer ServedNumber Average μ
0 μ 0, for x μef(x) μx
KENDALL NOTATION FOR QUEUING MODELS
Kendall notation consists of a basic three-symbol form.
Arrival Service Time Number of Service Distribution Distribution Channels Open
Where, M = Poisson distribution for the number of occurrences (or exponential times) D = Constant (deterministic rate) G = General distribution with mean and variance known
M/M/2 Single channel with Poisson arrivals and
exponential service times and two channels
Average flow time T
Theoretical Flow Time
Utilization 100%
Increasing Variability
Waiting Time Formula
Service time factor
Utilization factor
Variability factor
21
22
pa CVCV
nutilizatio
nutilizatioTimeActivity queue in Time
Inflow Outflow
Inventory waiting Iq
Entry to system Departure Begin Service
Time in queue Tq Service Time p
Flow Time T=Tq+p
Flow rate
THE WAITING TIME FORMULA
THE CONCEPT OF POOLING
Independent Resources
2x(m=1)
Pooled Resources
(m=2)
The concept of pooling
How pooling can reduce waiting time
0.00
10.00
20.00
30.00
40.00
50.00
60.00
70.00
60% 65%
m=1
m=2
m=5
m=10
70% 75% 80% 85% 90% 95%
Waiting
Time Tq
[sec]
Utilization u
21
221)1(2pa
m CVCV
nutilizatio
nutilizatio
m
time Activityqueue in Time
Waiting Time Formula for Multiple (m) Servers
Inflow Outflow
Inventory waiting Iq
Entry to system Departure Begin Service
Time in queue Tq Service Time p
Flow Time T=Tq+p
Inventory in service Ip
Flow rate
WAITING TIME FORMULA FOR MULTIPLE, PARALLEL RESOURCES
1. Collect the following data:
- number of servers, m
- activity time, p
- interarrival time, a
- coefficient of variation for interarrival (CVa ) and processing time (CVp )
2. Compute utilization: u= ma
p
3. Compute expected waiting time
Tq
21
221)1(2pa
m CVCV
nutilizatio
nutilizatio
m
timeActivity
4. Based on Tq , we can compute the remaining performance measures as
Flow time T=Tq+p
Inventory in service Ip=m*u
Inventory in the queue=Iq=Tq/a
Inventory in the system I=Ip+Iq
QUEUEING THEORY FORMULAS
qp
p
q
pam
q
III
muI
Ta
I
pTT
CVCV
u
u
m
pT
am
p
pm
au
*
*1
21
1*
1
221)1(2
Server
Inflow Outflow
Inventory waiting Iq
Entry to system
Departure Begin Service
Waiting Time Tq Service Time p
Flow Time T=Tq+p
Flow unit
Inventory in service Ip
Utilization (Note: make sure <1)
Time related measures
Inventory related measures (Flow rate=1/a)
SUMMARY OF QUEUING ANALYSIS
Customers send e-mails to a help desk of an online retailer every 2 minutes, on average, and the standard deviation of the inter-arrival time is also 2 minutes. The online retailer has three employees answering e-mails. It takes on average 4 minutes to write a response e-mail. The standard deviation of the service times is 2 minutes. a. Estimate the average customer wait before being
served. b. How many e-mails would there be, on average,
that have been submitted to the online retailer but not yet answered?
QUESTION: ONLINE RETAILER
My-law.com is a recent start-up trying to cater to customers in search of legal services who are intimidated by the idea of talking to a lawyer or simply too lazy to enter a law office. Unlike traditional law firms, My-law.com allows for extensive interaction between lawyers and their customers via telephone and the Internet. This process is used in the upfront part of the customer interaction, largely consisting of answering some basic customer questions prior to entering a formal relationship. In order to allow customers to interact with the firm's lawyers, customers are encouraged to send e-mails to [email protected]. From there, the incoming e-mails are distributed to the lawyer who is currently “on call.” Given the broad skills of the lawyers, each lawyer can respond to each incoming request. E-mails arrive from 8 a.m. to 6 p.m. at a rate of 10 e-mails per hour (coefficient of variation for the arrivals is 1). At each moment in time, there is exactly one lawyer “on call,” that is, sitting at his or her desk waiting for incoming e-mails. It takes the lawyer, on average, 5 minutes to write the response e-mail. The standard deviation of this is 4 minutes. λ= 10 emails per hour Inter-arrival time (a): = 10 emails / 60 min a = 6 min,
CVa = 1 ; processing time: p = 5 min, CVp = 4 min / 5 min = 0.8
a. What is the average time a customer has to wait for the response to his/her e-mail, ignoring any transmission times? Note: This includes the time it takes the lawyer to start writing the e-mail and the actual writing time.
TS = ? (Tq) waiting time = 5 min * [5/6 / (1-5/6)] * [(12 + 0.82)/2] = 20.5 min Total Response Time (Ts) = waiting time + processing time = 25.5 min
b. How many e-mails will a lawyer have received at the end of a 10-hour day? emails per day = 10 emails per hour * 10 hours = 100 emails per day
QUESTION: (My-Law.Com)
c. When not responding to e-mails, the lawyer on call is encouraged to actively pursue cases that potentially could lead to large settlements. How much time on a 10-hour day can a My-law.com lawyer dedicate to this activity (assume the lawyer can instantly switch between e-mails and work on a settlement)?
Idle Time = Probability that there in no customer in the system=? idle time = (1- utilization) * 10 hours = 1.66 hours
To increase the responsiveness of the firm, the board of My-law.com proposes a new operating policy. Under the new policy, the response would be highly standardized, reducing the standard deviation for writing the response e-mail to 0.5 minute. The average writing time would remain unchanged. d. How would the amount of time a lawyer can dedicate to the search for large
settlement cases change with this new operating policy? The average amount of time would not change, since utilization is not dependent on the variance or standard deviation, but on the average processing and inter-arrival time.
e. How would the average time a customer has to wait for the response to his/her e-mail change? Note: This includes the time until the lawyer starts writing the e-mail and the actual writing time. processing time: p = 5 min, CVp = 0.5 min / 5 min = 0.1 waiting time = 5 min * [5/6 / (1-5/6)] * [(12 + 0.12)/2] = 12.63 min total response time = waiting time + processing time = 17.63 min
QUESTION: (My-Law.Com) Cont…
The airport branch of a car rental company maintains a fleet of 50 SUVs. The interarrival time between requests for an SUV is 2.4 hours, on average, with a standard deviation of 2.4 hours. There is no indication of a systematic arrival pattern over the course of a day. Assume that, if all SUVs are rented, customers are willing to wait until there is an SUV available. An SUV is rented, on average, for 3 days, with a standard deviation of 1 day.
We approach this problem as though the rental car is the “server”. a = 2.4 hours, p = 72 hours, CVa = (2.4/2.4) = 1, CVp = (24/72) = 0.33, and m = 50 cars.
a. What is the average number of SUVs parked in the company's lot?
To determine the number of cars on the lot, we can look at the utilization rate of our “servers” = (1/a) / (m/p) = (1/2.4) / (50/72) = 60%. Therefore, on average 60% of the cars are in use or 30 cars, so on average 20 cars are in the lot.
QUESTION: (Car Rental Company)
b. Through a marketing survey, the company has discovered that if it reduces its daily rental price of $80 by $25, the average demand would increase to 12 rental requests per day and the average rental duration will become 4 days. Is this price decrease warranted? Provide an analysis! We assume that the standard deviations DO NOT change. If the average demand is increased to 12 rentals per day, then a = 2 hours. If the average rental duration increases to 4 days, then p = 96 hours. These values raise the utilization rate to (1/2) / (50/96) = 96%. This means that 48 cars are rented on average. With the initial rate average revenue per day = $80*30 cars = $2400. With the proposed rate average revenue per day = $55*48 = $2640. Therefore, the company should make the proposed changes.
c. What is the average time a customer has to wait to rent an SUV? Please
use the initial parameters rather than the information in (b). the average wait time = 0.019 hours or 1.15 minutes.
d. How would the waiting time change if the company decides to limit all
SUV rentals to exactly 4 days? Assume that if such a restriction is imposed, the average inter-arrival time will increase to 3 hours, with the standard deviation changing to 3 hours.
the average wait time is computed as 0.046 hours.
QUESTION: (Car Rental Company)
The following situation refers to Tom Opim, a first-year MBA student. In order to pay the rent, Tom decides to take a job in the computer department of a local department store. His only responsibility is to answer telephone calls to the department, most of which are inquiries about store hours and product availability. As Tom is the only person answering calls, the manager of the store is concerned about queuing problems. Currently, the computer department receives an average of one call every 3 minutes, with a standard deviation in this inter-arrival time of 3 minutes. Tom requires an average of 2 minutes to handle a call. The standard deviation in this activity time is 1 minute. The telephone company charges $5.00 per hour for the telephone lines whenever they are in use (either while a customer is in conversation with Tom or while waiting to be helped). Assume that there are no limits on the number of customers that can be on hold and that customers do not hang up even if forced to wait a long time.
inter-arrival time (λ): 1 call / 3 min a = 3 min, CVa = 3 min / 3 min = 1
processing time: p = 2 min, CVp = 1 min / 2 min = 0.5
a. For one of his courses, Tom has to read a book (The Pole, by E. Silvermouse). He can
read 1 page per minute. Tom's boss has agreed that Tom could use his idle time for studying, as long as he drops the book as soon as a call comes in. How many pages can Tom read during an 8-hour shift?
idle time = (1 – U) = (1 – 2/3) =(1 / 3) * 8 hours = 160 min pages read = 160 min
* 1 page/min = 160 pages
QUESTION: (Tom Opim)
b. How long does a customer have to wait, on average, before talking to Tom? waiting time = 2 min * [2/3 / (1-2/3)] * [(12 + 0.52)/2] = 2.5 min
c. What is the average total cost of telephone lines over an 8-hour shift? Note that
the department store is billed whenever a line is in use, including when a line is used to put customers on hold.
The average total time a line is used per customer = average wait time + average processing time.
In this case, the average total time per customer = 2.5 + 2 = 4.5 minutes per customer. There are an average of 20 customers per hour, so the average number of minutes per
hour = 20*4.5 = 90 minutes. Thus, the total per hour charge = (90/60) * $5 = $7.50 per hour or $60 for 8 hours.
Another way to approach the same problem: Average number of callers at any given time (Is) = average number of callers on hold at any
given time (Iq) + average number of callers talking to Tom at any given time (IP) . We can calculate Iq = R * Tq where the flow rate = 1 call / 3 minutes and Tq = 2.5
minutes. Thus, Iq = 0.83 calls. The average number of callers talking to Tom must be a number
between 0 and 1, and is equal to Tom’s utilization = 0.67. So, the average number of callers at any given time = 0.83 + 0.67 = 1.5 callers. The
line charge for 8 hours = $5*8 = $40 per line. Therefore the total cost over an 8-hour shift = 1.5*40 = $60.
QUESTION: (Tom Opim)
Atlantic Video, a small video rental store in Philadelphia, is open 24 hours a day, and—due to its proximity to a major business school—experiences customers arriving around the clock. A recent analysis done by the store manager indicates that there are 30 customers arriving every hour, with a standard deviation of interarrival times of 2 minutes. This arrival pattern is consistent and is independent of the time of day. The checkout is currently operated by one employee, who needs on average 1.7 minutes to check out a customer. The standard deviation of this check-out time is 3 minutes, primarily as a result of customers taking home different numbers of videos.
inter-arrival time: 30 customers per hour = 1 customer / 2 min a = 2 min, CVa = 1
process time: p = 1.7 min, CVp = 3 min / 1.7 min = 1.765 a. If you assume that every customer rents at least one video (i.e., has to go
to the checkout), what is the average time a customer has to wait in line before getting served by the checkout employee, not including the actual checkout time (within 1 minute)?
waiting time = 1.7 min * [(1.7/2)/(1-1.7/2)] * [(12 + 1.7652)/2] = 19.82 min
QUESTION: (Atlantic Video)
b. If there are no customers requiring checkout, the employee is sorting returned videos, of which there are always plenty waiting to be sorted. How many videos can the employee sort over an 8-hour shift (assume no breaks) if it takes exactly 1.5 minutes to sort a single video?
utilization = 1.7 min / 2 min = 0.85 idle time = 0.15 * 8 hrs = 72 min / 1.5 min per sort = 48 sorts
c. What is the average number of customers who are at the checkout desk, either waiting or
currently being served (within 1 customer)? Using R = minimum of 1/a and 1/p, we calculate R = 0.5. Thus, the average number of customers in line waiting I = R*Tq = 0.5*19.8 = 9.9
customers. In addition, an average of 0.85 customers (equal to the utilization u) are being served
at any given time. So the average number of customers at the check-out desk = 9.9 + .85 = 10.75.
d. Now assume for this question only that 10 percent of the customers do not rent a video at all
and therefore do not have to go through checkout. What is the average time a customer has to wait in line before getting served by the checkout employee, not including the actual checkout time (within 1 minute)? Assume that the coefficient of variation for the arrival process remains the same as before. Because only 90% of customers go through checkout, the inter-arrival time of paying
customers changes: 27 customers per hour = 1 customer / 2.22 min waiting time = 1.7 min * [(1.7/2.22)/(1-1.7/2.22)] * [(12 + 1.7652)/2] = 11.38 min
QUESTION: (Atlantic Video)
e. As a special service, the store offers free popcorn and sodas for customers waiting in line at the checkout desk. (Note: The person who is currently being served is too busy with paying to eat or drink.) The store owner estimates that every minute of customer waiting time costs the store 75 cents because of the consumed food. What is the optimal number of employees at checkout? Assume an hourly wage rate of $10 per hour. The average person waits 19.81 minutes, and 30 customers arrive in
one hour, so there are approximately 19.82*30=594.6 minutes of wait time per hour.
This costs the store .75*594.6 = $445.95 for wait time. If 2 servers are used, we can apply the formula, and using the same
methodology, calculate a cost of .75*(0.88*30) = $19.79. Finally, if 3 servers are used, the cost is .75*(0.162*30) = $3.65. Adding any more employees would not be cost effective. The most cost
effective number of employees is 3.
QUESTION: (Atlantic Video)
RentAPhone is a new service company that provides European mobile phones to American visitors to Europe. The company currently has 80 phones available at Charles de Gaulle Airport in Paris. There are, on average, 25 customers per day requesting a phone. These requests arrive uniformly throughout the 24 hours the store is open. (Note: This means customers arrive at a faster rate than 1 customer per hour.) The corresponding coefficient of variation is 1. Customers keep their phones on average 72 hours. The standard deviation of this time is 100 hours. Given that RentAPhone currently does not have a competitor in France providing equally good service, customers are willing to wait for the telephones. Yet, during the waiting period, customers are provided a free calling card. Based on prior experience, RentA-Phone found that the company incurred a cost of $1 per hour per waiting customer, independent of day or night. To answer this question, we must first set up the problem where we consider the phones as “servers”. a. What is the average number of telephones the company has in its store? m = 80, a = 24 hours/25 customers = 0.96 hours average inter-arrival time p = 72 hours. So, utilization = p / (m*a) = 72/(80*.96) = 94%. This means that .94*80 phones = 75 phones are in use, and 5 phones are available
on average.
QUESTION: (RentAPhone)
b. How long does a customer, on average, have to wait for the phone? We are given that CVa = 1 and CVp = 100/72 = 1.39. Using the wait time formula, we
calculate an average wait time of 9.89 hours.
c. What are the total monthly (30 days) expenses for telephone cards? We first need to calculate the average number of people in the queue. We know that the flow rate R = demand rate = 1/.96 = 1.04 customers/hour. Time in the queue Tq = 9.89 hours, so Iq = Tq * R = 9.89 * 1.04 = 10.31 people in
the queue on average. So we can multiply $1 * 24 hours/day * 30 days * 10.31 people in the queue =
$7419.87.
d. Assume RentAPhone could buy additional phones at $1,000 per unit. Is it worth it to buy one additional phone? Why? We can repeat the same calculations for m=81 and obtain Iq=7.379. The new
expenses would be 5312, thus it would pay to buy at least one additional phone.
e. How would waiting time change if the company decides to limit all rentals to exactly 72 hours? Assume that if such a restriction is imposed, the number of customers requesting a phone would be reduced to 20 customers per day.
QUESTION: (RentAPhone) (Cont…)
QUESTIONS
29
