1 A Basic Queueing System. 2 Herr Cutter’s Barber Shop Herr Cutter is a German barber who runs a one-man barber shop. Herr Cutter opens his shop at 8:00.
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1
A Basic Queueing System
CustomersQueue
Served Customers
Queueing System
Service facility
SSSS
CCCC
C C C C C C C
Served Customers
2
Herr Cutter’s Barber Shop
• Herr Cutter is a German barber who runs a one-man barber shop.
• Herr Cutter opens his shop at 8:00 A.M.
• The table shows his queueing system in action over a typical morning.
CustomerTime ofArrival
HaicutBegins
Durationof Haircut
HaircutEnds
1 8:03 8:03 17 minutes 8:20
2 8:15 8:20 21 minutes 8:41
3 8:25 8:41 19 minutes 9:00
4 8:30 9:00 15 minutes 9:15
5 9:05 9:15 20 minutes 9:35
6 9:43 — — —
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Arrivals
• The time between consecutive arrivals to a queueing system are called the interarrival times.
• The expected number of arrivals per unit time is referred to as the mean arrival rate.
• The symbol used for the mean arrival rate is
= Mean arrival rate for customers coming to the queueing system
where is the Greek letter lambda.
• The mean of the probability distribution of interarrival times is
1 / = Mean interarrival time
• Most queueing models assume that the form of the probability distribution of interarrival times is an exponential distribution.
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Evolution of the Number of Customers
20 40 60 80
1
2
3
4
Number of Customers in the System
0
Time (in minutes)100
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The Exponential Distribution for Interarrival Times
Mean Time0
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Properties of the Exponential Distribution
• There is a high likelihood of small interarrival times, but a small chance of a very large interarrival time. This is characteristic of interarrival times in practice.
• For most queueing systems, the servers have no control over when customers will arrive. Customers generally arrive randomly.
• Having random arrivals means that interarrival times are completely unpredictable, in the sense that the chance of an arrival in the next minute is always just the same.
• The only probability distribution with this property of random arrivals is the exponential distribution.
• The fact that the probability of an arrival in the next minute is completely uninfluenced by when the last arrival occurred is called the lack-of-memory property.
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The Queue
• The number of customers in the queue (or queue size) is the number of customers waiting for service to begin.
• The number of customers in the system is the number in the queue plus the number currently being served.
• The queue capacity is the maximum number of customers that can be held in the queue.
• An infinite queue is one in which, for all practical purposes, an unlimited number of customers can be held there.
• When the capacity is small enough that it needs to be taken into account, then the queue is called a finite queue.
• The queue discipline refers to the order in which members of the queue are selected to begin service.
– The most common is first-come, first-served (FCFS).– Other possibilities include random selection, some priority procedure, or even last-
come, first-served.
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Service
• When a customer enters service, the elapsed time from the beginning to the end of the service is referred to as the service time.
• Basic queueing models assume that the service time has a particular probability distribution.
• The symbol used for the mean of the service time distribution is
1 / = Mean service time
where is the Greek letter mu.
• The interpretation of itself is the mean service rate.
= Expected service completions per unit time for a single busy server
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Some Service-Time Distributions
• Exponential Distribution– The most popular choice.
– Much easier to analyze than any other.
– Although it provides a good fit for interarrival times, this is much less true for service times.
– Provides a better fit when the service provided is random than if it involves a fixed set of tasks.
– Standard deviation: = Mean
• Constant Service Times– A better fit for systems that involve a fixed set of tasks.
– Standard deviation: = 0.
• Erlang Distribution– Fills the middle ground between the exponential distribution and constant.
– Has a shape parameter, k that determines the standard deviation.
– In particular, = mean / (k)
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Standard Deviation and Mean for Distributions
Distribution Standard Deviation
Exponential mean
Degenerate (constant) 0
Erlang, any k (1 / k) (Mean)
Erlang, k = 2 (1 / 2) (Mean)
Erlang, k = 4 (1 / 2) (Mean)
Erlang, k = 8 (1 / 22) (Mean)
Erlang, k = 16 (1 / 4) (Mean)
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Labels for Queueing Models
To identify which probability distribution is being assumed for service times (and for interarrival times), a queueing model conventionally is labeled as follows:
Distribution of service times
— / — / — Number of Servers
Distribution of interarrival times
The symbols used for the possible distributions areM = Exponential distribution (Markovian)D = Degenerate distribution (constant times)Ek = Erlang distribution (shape parameter = k)GI = General independent interarrival-time distribution (any distribution)G = General service-time distribution (any arbitrary distribution)
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Summary of Usual Model Assumptions
1. Interarrival times are independent and identically distributed according to a specified probability distribution.
2. All arriving customers enter the queueing system and remain there until service has been completed.
3. The queueing system has a single infinite queue, so that the queue will hold an unlimited number of customers (for all practical purposes).
4. The queue discipline is first-come, first-served.
5. The queueing system has a specified number of servers, where each server is capable of serving any of the customers.
6. Each customer is served individually by any one of the servers.
7. Service times are independent and identically distributed according to a specified probability distribution.
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Examples of Commercial Service SystemsThat Are Queueing Systems
Type of System Customers Server(s)
Barber shop People Barber
Bank teller services People Teller
ATM machine service People ATM machine
Checkout at a store People Checkout clerk
Plumbing services Clogged pipes Plumber
Ticket window at a movie theater People Cashier
Check-in counter at an airport People Airline agent
Brokerage service People Stock broker
Gas station Cars Pump
Call center for ordering goods People Telephone agent
Call center for technical assistance People Technical representative
Travel agency People Travel agent
Automobile repair shop Car owners Mechanic
Vending services People Vending machine
Dental services People Dentist
Roofing Services Roofs Roofer
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Examples of Internal Service SystemsThat Are Queueing Systems
Type of System Customers Server(s)
Secretarial services Employees Secretary
Copying services Employees Copy machine
Computer programming services Employees Programmer
Mainframe computer Employees Computer
First-aid center Employees Nurse
Faxing services Employees Fax machine
Materials-handling system Loads Materials-handling unit
Maintenance system Machines Repair crew
Inspection station Items Inspector
Production system Jobs Machine
Semiautomatic machines Machines Operator
Tool crib Machine operators Clerk
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Examples of Transportation Service SystemsThat Are Queueing Systems
Type of System Customers Server(s)
Highway tollbooth Cars Cashier
Truck loading dock Trucks Loading crew
Port unloading area Ships Unloading crew
Airplanes waiting to take off Airplanes Runway
Airplanes waiting to land Airplanes Runway
Airline service People Airplane
Taxicab service People Taxicab
Elevator service People Elevator
Fire department Fires Fire truck
Parking lot Cars Parking space
Ambulance service People Ambulance
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Choosing a Measure of Performance
• Managers who oversee queueing systems are mainly concerned with two measures of performance:
– How many customers typically are waiting in the queueing system?
– How long do these customers typically have to wait?
• When customers are internal to the organization, the first measure tends to be more important.
– Having such customers wait causes lost productivity.
• Commercial service systems tend to place greater importance on the second measure.
– Outside customers are typically more concerned with how long they have to wait than with how many customers are there.
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Defining the Measures of Performance
L = Expected number of customers in the system, including those being served (the symbol L comes from Line Length).
Lq = Expected number of customers in the queue, which excludes customers being served.
W = Expected waiting time in the system (including service time) for an individual customer (the symbol W comes from Waiting time).
Wq = Expected waiting time in the queue (excludes service time) for an individual customer.
These definitions assume that the queueing system is in a steady-state condition.
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Relationship between L, W, Lq, and Wq
• Little’s formula states that
L = W
and
Lq = Wq
• Since 1/ is the expected service time
W = Wq + 1/
• Combining the above relationships leads to
L = Lq +
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Using Probabilities as Measures of Performance
• In addition to knowing what happens on the average, we may also be interested in worst-case scenarios.
– What will be the maximum number of customers in the system? (Exceeded no more than, say, 5% of the time.)
– What will be the maximum waiting time of customers in the system? (Exceeded no more than, say, 5% of the time.)
• Statistics that are helpful to answer these types of questions are available for some queueing systems:
– Pn = Steady-state probability of having exactly n customers in the system.
– P(W ≤ t) = Probability the time spent in the system will be no more than t.
– P(Wq ≤ t) = Probability the wait time will be no more than t.
• Examples of common goals:– No more than three customers 95% of the time: P0 + P1 + P2 + P3 ≥ 0.95
– No more than 5% of customers wait more than 2 hours: P(W ≤ 2 hours) ≥ 0.95
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The Dupit Corp. Problem
• The Dupit Corporation is a longtime leader in the office photocopier marketplace.
• Dupit’s service division is responsible for providing support to the customers by promptly repairing the machines when needed. This is done by the company’s service technical representatives, or tech reps.
• Current policy: Each tech rep’s territory is assigned enough machines so that the tech rep will be active repairing machines (or traveling to the site) 75% of the time.
– A repair call averages 2 hours, so this corresponds to 3 repair calls per day.
– Machines average 50 workdays between repairs, so assign 150 machines per rep.
• Proposed New Service Standard: The average waiting time before a tech rep begins the trip to the customer site should not exceed two hours.
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Alternative Approaches to the Problem
• Approach Suggested by John Phixitt: Modify the current policy by decreasing the percentage of time that tech reps are expected to be repairing machines.
• Approach Suggested by the Vice President for Engineering: Provide new equipment to tech reps that would reduce the time required for repairs.
• Approach Suggested by the Chief Financial Officer: Replace the current one-person tech rep territories by larger territories served by multiple tech reps.
• Approach Suggested by the Vice President for Marketing: Give owners of the new printer-copier priority for receiving repairs over the company’s other customers.
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The Queueing System for Each Tech Rep
• The customers: The machines needing repair.
• Customer arrivals: The calls to the tech rep requesting repairs.
• The queue: The machines waiting for repair to begin at their sites.
• The server: The tech rep.
• Service time: The total time the tech rep is tied up with a machine, either traveling to the machine site or repairing the machine. (Thus, a machine is viewed as leaving the queue and entering service when the tech rep begins the trip to the machine site.)
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Notation for Single-Server Queueing Models
• = Mean arrival rate for customers= Expected number of arrivals per unit time
1/ = expected interarrival time
• = Mean service rate (for a continuously busy server)= Expected number of service completions per unit time
= expected service time
• = the utilization factor= the average fraction of time that a server is busy serving customers=
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The M/M/1 Model
• Assumptions1. Interarrival times have an exponential distribution with a mean of 1/.
2. Service times have an exponential distribution with a mean of 1/.
3. The queueing system has one server.
• The expected number of customers in the system is
L = 1 – = –
• The expected waiting time in the system is
W = (1 / )L = 1 / ( – )
• The expected waiting time in the queue is
Wq = W – 1/ = / [( – )]
• The expected number of customers in the queue is
Lq = Wq = 2 / [( – )]
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The M/M/1 Model
• The probability of having exactly n customers in the system is
Pn = (1 – )n
Thus,P0 = 1 – P1 = (1 – )P2 = (1 – )2
::
• The probability that the waiting time in the system exceeds t is
P(W > t) = e–(1–)t for t ≥ 0• The probability that the waiting time in the queue exceeds t is
P(Wq > t) = e–(1–)t for t ≥ 0
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M/M/1 Queueing Model for the Dupit’s Current Policy
Data Results= 3 (mean arrival rate) L = 3= 4 (mean service rate) Lq = 2.25
s = 1 (# servers)W = 1
Pr(W > t) = 0.368 Wq = 0.75
when t = 1= 0.75
Prob(Wq > t) = 0.276
when t = 1 n Pn
0 0.251 0.18752 0.14063 0.10554 0.07915 0.05936 0.04457 0.03348 0.02509 0.0188
10 0.0141
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John Phixitt’s Approach (Reduce Machines/Rep)
• The proposed new service standard is that the average waiting time before service begins be two hours (i.e., Wq ≤ 1/4 day).
• John Phixitt’s suggested approach is to lower the tech rep’s utilization factor sufficiently to meet the new service requirement.
Lower = / , until Wq ≤ 1/4 day,where
= (Number of machines assigned to tech rep) / 50.
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M/M/1 Model for John Phixitt’s Suggested Approach(Reduce Machines/Rep)
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B C D E G HData Results
= 2 (mean arrival rate) L = 1= 4 (mean service rate) Lq = 0.5s = 1 (# servers)
W = 0.5Pr(W > t) = 0.135 Wq = 0.25
when t = 1= 0.5
Prob(Wq > t) = 0.068when t = 1 n Pn
0 0.51 0.252 0.12503 0.06254 0.03135 0.01566 0.00787 0.00398 0.00209 0.0010
10 0.0005
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The M/G/1 Model
• Assumptions1. Interarrival times have an exponential distribution with a mean of 1/.
2. Service times (T) can have any probability distribution.
E(T) = 1/ , Var(T) = 2.
3. The queueing system has one server.
• The probability of zero customers in the system isP0 = 1 –
• The expected number of customers in the queue isLq = 2[Var(T+ E(T)2] / [2(1 – E(T))]
• The expected number of customers in the system isL = Lq + /
The expected waiting time in the queue isWq = Lq /
• The expected waiting time in the system isW = Wq + 1/
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The Values of and Lq for the M/G/1 Modelwith Various Service-Time Distributions
Distribution Mean Model Lq
Exponential 1/ 1/ M/M/1 2 / (1 – )
Degenerate (constant) 1/ 0 M/D/1 (1/2) [2 / (1 – )]
Erlang, with shape parameter k 1/ (1/k) (1/) M/Ek/1 (k+1)/(2k) [2 / (1 – )]
• The expected number of customers in the queue isLq = 2[Var(T+ E(T)2] / [2(1 – E(T))]= 22 + 2] / [2(1 – )]
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VP for Engineering Approach (New Equipment)
• The proposed new service standard is that the average waiting time before service begins be two hours (i.e., Wq ≤ 1/4 day).
• The Vice President for Engineering has suggested providing tech reps with new state-of-the-art equipment that would reduce the time required for the longer repairs.
• After gathering more information, they estimate the new equipment would have the following effect on the service-time distribution:
– Decrease the mean from 1/4 day to 1/5 day.
– Decrease the standard deviation from 1/4 day to 1/10 day.
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M/G/1 Model for the VP of Engineering Approach(New Equipment)
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B C D E F GData Results
= 3 (mean arrival rate) L = 1.163= 0.2 (expected service time) Lq = 0.563
= 0.1 (standard deviation)s = 1 (# servers) W = 0.388
Wq = 0.188
= 0.6
P0 = 0.4
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The M/M/s Model
• Assumptions1. Interarrival times have an exponential distribution with a mean of 1/.
2. Service times have an exponential distribution with a mean of 1/3. Any number of servers (denoted by s).
• With multiple servers, the formula for the utilization factor becomes
= / s
but still represents that average fraction of time that individual servers are busy.
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Values of L for the M/M/s Model for Various Values of s
s = 1
s = 2
s = 3s = 4s = 5s = 7s = 10
s = 15s = 20s = 25
100
10
0.5
0.1
0.2
0 0.1 0.3 0.5 0.7 0.9 1.0Utilization factor =
s
Steady-state expected number of customers in the queueing system
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CFO Suggested Approach (Combine Into Teams)
• The proposed new service standard is that the average waiting time before service begins be two hours (i.e., Wq ≤ 1/4 day).
• The Chief Financial Officer has suggested combining the current one-person tech rep territories into larger territories that would be served jointly by multiple tech reps.
• A territory with two tech reps:– Number of machines = 300 (versus 150 before)
– Mean arrival rate = = 6 (versus = 3 before)
– Mean service rate = = 4 (as before)
– Number of servers = s = 2 (versus s = 1 before)
– Utilization factor = = /s = 0.75 (as before)
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M/M/s Model for the CFO’s Suggested Approach(Combine Into Teams of Two)
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B C D E G HData Results
= 6 (mean arrival rate) L = 3.4286= 4 (mean service rate) Lq = 1.9286s = 2 (# servers)
W = 0.5714Pr(W > t) = 0.169 Wq = 0.3214
when t = 1= 0.75
Prob(Wq > t) = 0.087when t = 1 n Pn
0 0.14291 0.21432 0.16073 0.12054 0.09045 0.06786 0.05097 0.03818 0.02869 0.0215
10 0.0161
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CFO Suggested Approach (Teams of Three)
• The Chief Financial Officer has suggested combining the current one-person tech rep territories into larger territories that would be served jointly by multiple tech reps.
• A territory with three tech reps:– Number of machines = 450 (versus 150 before)
– Mean arrival rate = = 9 (versus = 3 before)
– Mean service rate = = 4 (as before)
– Number of servers = s = 3 (versus s = 1 before)
– Utilization factor = = /s = 0.75 (as before)
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M/M/s Model for the CFO’s Suggested Approach(Combine Into Teams of Three)
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B C D E G HData Results
= 9 (mean arrival rate) L = 3.9533= 4 (mean service rate) Lq = 1.7033s = 3 (# servers)
W = 0.4393Pr(W > t) = 0.090 Wq = 0.1893
when t = 1= 0.75
Prob(Wq > t) = 0.028when t = 1 n Pn
0 0.07481 0.16822 0.18933 0.14194 0.10655 0.07986 0.05997 0.04498 0.03379 0.0253
10 0.0189
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Comparison of Wq with Territories of Different Sizes
Number ofTech Reps
Number ofMachines s Wq
1 150 3 4 1 0.75 0.75 workday (6 hours)
2 300 6 4 2 0.75 0.321 workday (2.57 hours)
3 450 9 4 3 0.75 0.189 workday (1.51 hours)
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Values of L for the M/D/s Model for Various Values of s
100
0.1
1.0
10
0 0.1 0.3 0.5 0.9 1.0
s = 2
s = 3s = 4
s = 5s = 7
s = 10s = 15
s = 20
s = 25
s = 1
0.7Utilization factor
Steady-state expected number of customers in the queueing system
=s
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Values of L for the M/Ek/2 Model for Various Values of k
100
10
1.0
0.10 0.2 0.4 0.6 0.8 1.0
k = 1
k = 2
k = 8
Utilization factor =s
Steady-state expected number of customers in the queueing system
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The Four Approaches Under Considerations
Proposer Proposal Additional Cost
John Phixitt Maintain one-person territories, but reduce number of machines assigned to each from 150 to 100
$300 million per year
VP for Engineering Keep current one-person territories, but provide new state-of-the-art equipment to the tech-reps
One-time cost of $500 million
Chief Financial Officer Change to three-person territories None, except disadvantages of larger territories
Decision: Adopt the third proposal
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Some Insights About Designing Queueing Systems
1. When designing a single-server queueing system, beware that giving a relatively high utilization factor (workload) to the server provides surprisingly poor performance for the system.
2. Decreasing the variability of service times (without any change in the mean) improves the performance of a queueing system substantially.
3. Multiple-server queueing systems can perform satisfactorily with somewhat higher utilization factors than can single-server queueing systems. For example, pooling servers by combining separate single-server queueing systems into one multiple-server queueing system greatly improves the measures of performance.
4. Applying priorities when selecting customers to begin service can greatly improve the measures of performance for high-priority customers.
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Effect of High-Utilization Factors (Insight 1)
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B C D E G HData Results
= 0.5 (mean arrival rate) L = 1= 1 (mean service rate) Lq = 0.5
9
10
1112
13
141516171819202122232425
A B C D EData Table Demonstrating the Effect ofIncreasing on Lq and L for M/M/1
= Lq L
1 0.5 1
0 0.01 0.0001 0.01010 0.25 0.0833 0.33330 0.5 0.5 10 0.6 0.9 1.50 0.7 1.6333 2.33330 0.75 2.25 30 0.8 3.2 40 0.85 4.8167 5.66670 0.9 8.1 90 0.95 18.05 190 0.99 98.01 990 0.999 998.001 999
0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
System Utilization (r)A
vera
ge
Lin
e L
en
gth
(L
)
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Effect of Decreasing (Insight 2)
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A B C D E F G H
Template for the M/G/1 Queueing Model
Data Results= 0.5 (mean arrival rate) L = 0.8125
= 1 (expected service time) Lq = 0.3125= 0.5 (standard deviation)s = 1 (# servers) W = 1.625
Wq = 0.625
= 0.5
P0 = 0.5
Data Table Demonstrating the Effect of Decreasing s on Lq for M/G/1
Body of Table Shows L q Values
0.3125 1 0.5 0
0.5 0.500 0.313 0.250= 0.75 2.250 1.406 1.125
0.9 8.100 5.063 4.0500.99 98.010 61.256 49.005
46
Economic Analysis of the Number of Servers to Provide
• In many cases, the consequences of making customers wait can be expressed as a waiting cost.
• The manager is interested in minimizing the total cost.TC = Expected total cost per unit timeSC = Expected service cost per unit timeWC = Expected waiting cost per unit time
The objective is then to choose the number of servers so as toMinimize TC = SC + WC
• When each server costs the same (Cs = cost of server per unit time),SC = Cs s
• When the waiting cost is proportional to the amount of waiting (Cw = waiting cost per unit time for each customer),
WC = Cw L
47
Acme Machine Shop
• The Acme Machine Shop has a tool crib for storing tool required by shop mechanics.
• Two clerks run the tool crib.
• The estimates of the mean arrival rate and the mean service rate (per server) are
= 120 customers per hour = 80 customers per hour
• The total cost to the company of each tool crib clerk is $20/hour, so Cs = $20.
• While mechanics are busy, their value to Acme is $48/hour, so Cw = $48.
• Choose s so as to Minimize TC = $20s + $48L.
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Excel Template for Choosing the Number of Servers
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B C D E F GData Results
= 120 (mean arrival rate) L = 1.736842105= 80 (mean service rate) Lq = 0.236842105s = 3 (# servers)
W = 0.014473684Pr(W > t) = 0.02581732 Wq = 0.001973684
when t = 0.05= 0.5
Prob(Wq > t) = 0.00058707when t = 0.05 n Pn
0 0.2105263161 0.315789474
Cs = $20.00 (cost / server / unit time) 2 0.236842105Cw = $48.00 (waiting cost / unit time) 3 0.118421053
4 0.059210526Cost of Service $60.00 5 0.029605263Cost of Waiting $83.37 6 0.014802632
Total Cost $143.37 7 0.007401316
Economic Analysis:
49
Comparing Expected Cost vs. Number of Clerks
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10
H I J K L M N
Data Table for Expected Total Cost of Alternatives
Cost of Cost of Totals r L Service Waiting Cost
0.50 1.74 $60.00 $83.37 $143.371 1.50 #N/A $20.00 #N/A #N/A2 0.75 3.43 $40.00 $164.57 $204.573 0.50 1.74 $60.00 $83.37 $143.374 0.38 1.54 $80.00 $74.15 $154.15
5 0.30 1.51 $100.00 $72.41 $172.41
$0
$50
$100
$150
$200
$250
0 1 2 3 4 5
Number of Servers (s)
Cos
t ($/
hour
)
Cost ofService
Cost ofWaiting
Total Cost
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