A Basic Queueing System
Served Customers
Queueing System
Queue
Customers
CCCCCCC
C
C
C
C
Served Customers
1
S
S
S
S
Service
facility
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.
Customer
Time of
Arrival
Haicut
Begins
Duration
of Haircut
Haircut
Ends
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
—
—
—
2
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
l = Mean arrival rate for customers coming to the queueing system
where l is the Greek letter lambda.
•
The mean of the probability distribution of interarrival times is
1 / l = Mean interarrival time
•
Most queueing models assume that the form of the probability distribution of
interarrival times is an exponential distribution.
3
Evolution of the Number of Customers
4
Number of
Customers
in the
System
3
2
1
0
20
40
60
Time (in minutes)
4
80
100
The Exponential Distribution for Interarrival Times
0
Mean
Time
5
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.
6
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 lastcome, first-served.
7
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 / m = Mean service time
where m is the Greek letter mu.
•
The interpretation of m itself is the mean service rate.
m = Expected service completions per unit time for a single busy server
8
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: s = Mean
•
Constant Service Times
– A better fit for systems that involve a fixed set of tasks.
– Standard deviation: s = 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, s = mean / (k)
9
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 / 22) (Mean)
Erlang, k = 16
(1 / 4) (Mean)
10
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 are
M = 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)
11
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.
12
Examples of Commercial Service Systems
That 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
13
Examples of Internal Service Systems
That 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
14
Examples of Transportation Service Systems
That 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
15
Choosing a Measure of Performance
•
Managers who oversee queueing systems are mainly concerned with two
measures of performance:
–
–
•
When customers are internal to the organization, the first measure tends to be
more important.
–
•
How many customers typically are waiting in the queueing system?
How long do these customers typically have to wait?
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.
16
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.
17
Relationship between L, W, Lq, and Wq
•
Little’s formula states that
L = lW
and
Lq = lWq
•
Since 1/m is the expected service time
W = Wq + 1/m
•
Combining the above relationships leads to
L = Lq + l/m
18
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
19
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.
20
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.
21
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.)
22
Notation for Single-Server Queueing Models
•
l
= Mean arrival rate for customers
= Expected number of arrivals per unit time
1/l = expected interarrival time
•
m
= Mean service rate (for a continuously busy server)
= Expected number of service completions per unit time
1/m = expected service time
•
r
= the utilization factor
= the average fraction of time that a server is busy serving customers
=l/m
23
The M/M/1 Model
•
Assumptions
1. Interarrival times have an exponential distribution with a mean of 1/l.
2. Service times have an exponential distribution with a mean of 1/m.
3. The queueing system has one server.
•
The expected number of customers in the system is
L = r / (1 – r) = l / (m – l)
•
The expected waiting time in the system is
W = (1 / l)L = 1 / (m – l)
•
The expected waiting time in the queue is
Wq = W – 1/m = l / [m(m – l)]
•
The expected number of customers in the queue is
Lq = lWq = l2 / [m (m – l)]
24
The M/M/1 Model
•
The probability of having exactly n customers in the system is
Pn = (1 – r)rn
Thus,
•
P0 = 1 – r
P1 = (1 – r)r
P2 = (1 – r)r2
:
:
The probability that the waiting time in the system exceeds t is
P(W > t) = e–m (1–r)t for t ≥ 0
•
The probability that the waiting time in the queue exceeds t is
P(Wq > t) = r e–m (1–r)t for t ≥ 0
25
M/M/1 Queueing Model for the Dupit’s Current Policy
l=
m=
s =
Pr(W > t) =
when t =
Prob(W q > t) =
when t =
Data
3
4
1
(mean arrival rate)
(mean service rate)
(# servers)
0.368
L=
Lq =
Results
3
2.25
W =
Wq =
1
0.75
r =
0.75
1
0.276
1
n
0
1
2
3
4
5
6
7
8
9
10
26
Pn
0.25
0.1875
0.1406
0.1055
0.0791
0.0593
0.0445
0.0334
0.0250
0.0188
0.0141
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 r = l / m, until Wq ≤ 1/4 day,
where
l = (Number of machines assigned to tech rep) / 50.
27
M/M/1 Model for John Phixitt’s Suggested Approach
(Reduce Machines/Rep)
B
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
l
m
s=
Pr(W > t) =
when t =
Prob(W q > t) =
when t =
C
Data
2
4
1
D
E
(mean arrival rate)
(mean service rate)
(# servers)
0.135
1
0.068
1
28
G
L=
Lq =
H
Results
1
0.5
W=
Wq =
0.5
0.25
r
0.5
n
0
1
2
3
4
5
6
7
8
9
10
Pn
0.5
0.25
0.1250
0.0625
0.0313
0.0156
0.0078
0.0039
0.0020
0.0010
0.0005
The M/G/1 Model
•
Assumptions
1. Interarrival times have an exponential distribution with a mean of 1/l.
2. Service times (T) can have any probability distribution.
E(T) = 1/m , Var(T) = s2.
3. The queueing system has one server.
•
The probability of zero customers in the system is
P0 = 1 – r
•
The expected number of customers in the queue is
Lq = l2[Var(T)+ E(T)2] / [2(1 – lE(T))]
•
The expected number of customers in the system is
L = Lq + l/m
The expected waiting time in the queue is
Wq = Lq / l
•
The expected waiting time in the system is
W = Wq + 1/m
29
The Values of s and Lq for the M/G/1 Model
with Various Service-Time Distributions
•
The expected number of customers in the queue is
Lq = l2[Var(T)+ E(T)2] / [2(1 – lE(T))]= [l2s2 + r2] / [2(1 – r)]
Distribution
Mean
s
Model
Lq
Exponential
1/m
1/m
M/M/1
r2 / (1 – r)
Degenerate (constant)
1/m
0
M/D/1
(1/2) [r2 / (1 – r)]
Erlang, with shape parameter k
1/m
(1/k) (1/m)
M/Ek/1
(k+1)/(2k) [r2 / (1 – r)]
30
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.
31
M/G/1 Model for the VP of Engineering Approach
(New Equipment)
B
3
4
5
6
7
8
9
10
11
12
l
1/m 
s
s=
C
Data
3
0.2
0.1
1
D
E
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
32
F
L=
Lq =
G
Results
1.163
0.563
W=
Wq =
0.388
0.188
r
0.6
P0 =
0.4
The M/M/s Model
•
Assumptions
1. Interarrival times have an exponential distribution with a mean of 1/l.
2. Service times have an exponential distribution with a mean of 1/m.
3. Any number of servers (denoted by s).
•
With multiple servers, the formula for the utilization factor becomes
r = l / sm
but still represents that average fraction of time that individual servers are
busy.
33
Values of L for the M/M/s Model for Various Values of s
Steady-state expected number of customers in the queueing system
100
10
s = 25
s = 20
s = 15
s = 10
s =7
s =5
s =4
s =3
0.5
s =2
s =1
0.2
0.1
0
0.1
0.3
34
0.5
0.7
Utilization factor
0.9
1.0
rl
sm
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
Mean arrival rate = l = 6
Mean service rate = m = 4
Number of servers = s = 2
Utilization factor = r = l/sm = 0.75
(versus 150 before)
(versus l = 3 before)
(as before)
(versus s = 1 before)
(as before)
35
M/M/s Model for the CFO’s Suggested Approach
(Combine Into Teams of Two)
B
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
l
m
s=
C
Data
6
4
2
Pr(W > t) =
when t =
0.169
1
Prob(W q > t) =
when t =
D
E
(mean arrival rate)
(mean service rate)
(# servers)
0.087
1
36
G
L=
Lq =
H
Results
3.4286
1.9286
W=
Wq =
0.5714
0.3214
r
0.75
n
0
1
2
3
4
5
6
7
8
9
10
Pn
0.1429
0.2143
0.1607
0.1205
0.0904
0.0678
0.0509
0.0381
0.0286
0.0215
0.0161
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
Mean arrival rate = l = 9
Mean service rate = m = 4
Number of servers = s = 3
Utilization factor = r = l/sm = 0.75
(versus 150 before)
(versus l = 3 before)
(as before)
(versus s = 1 before)
(as before)
37
M/M/s Model for the CFO’s Suggested Approach
(Combine Into Teams of Three)
B
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
l
m
s=
C
Data
9
4
3
Pr(W > t) =
when t =
0.090
1
Prob(W q > t) =
when t =
D
E
(mean arrival rate)
(mean service rate)
(# servers)
0.028
1
38
G
L=
Lq =
H
Results
3.9533
1.7033
W=
Wq =
0.4393
0.1893
r
0.75
n
0
1
2
3
4
5
6
7
8
9
10
Pn
0.0748
0.1682
0.1893
0.1419
0.1065
0.0798
0.0599
0.0449
0.0337
0.0253
0.0189
Comparison of Wq with Territories of Different Sizes
Number of
Tech Reps
Number of
Machines
l
m
s
r
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)
39
Values of L for the M/D/s Model for Various Values of s
Steady-state expected number of customers in the queueing system
100
s = 25
10
s = 20
s = 15
s = 10
s =7
s =5
1.0
s =4
s =3
s =2
s =1
0.1
0
0.1
0.3
0.5
0.7
Utilization factor
40
0.9
1.0
rl
sm
Values of L for the M/Ek/2 Model for Various Values of k
Steady-state expected number of customers in the queueing system
100
k=1
10
k=2
k=8
1.0
0.1
0
0.2
0.4
41
0.6
0.8
Utilization factor
1.0
rl
sm
The Four Approaches Under Considerations
Proposer
Proposal
Additional Cost
John Phixitt
Maintain one-person territories, but $300 million per year
reduce number of machines assigned
to each from 150 to 100
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
42
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.
43
Effect of High-Utilization Factors (Insight 1)
B
3
4
5
l
m
A
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
C
D
E
G
(mean arrival rate)
(mean service rate)
D
H
Results
L=
Lq =
1
0.5
E
Data Table Demonstrating the Effect of
Increasing r on Lq and L for M/M/1
l r
1
0 0.01
0 0.25
0
0.5
0
0.6
0
0.7
0 0.75
0
0.8
0 0.85
0
0.9
0 0.95
0 0.99
0 0.999
Lq
0.5
0.0001
0.0833
0.5
0.9
1.6333
2.25
3.2
4.8167
8.1
18.05
98.01
998.001
Average Line Length (L)
9
10
B
C
Data
0.5
1
L
1
0.0101
0.3333
1
1.5
2.3333
3
4
5.6667
9
19
99
999
100
80
60
40
20
0
0
0.2
0.4
0.6
0.8
System Utilization (r)
44
1
Effect of Decreasing s (Insight 2)
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
F
G
H
Template for the M/G/1 Queueing Model
l
1/m 
s
s=
Data
0.5
1
0.5
1
(mean arrival rate)
(expected service time)
(standard deviation)
(# servers)
Results
L=
0.8125
Lq =
0.3125
W=
Wq =
1.625
0.625
r
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
r (l)
0.3125
0.5
0.75
0.9
0.99
1
0.500
2.250
8.100
98.010
s
0.5
0.313
1.406
5.063
61.256
45
0
0.250
1.125
4.050
49.005
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 time
SC = Expected service cost per unit time
WC = Expected waiting cost per unit time
The objective is then to choose the number of servers so as to
Minimize 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
46
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 l and the mean service rate (per server)
m are
l = 120 customers per hour
m = 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.
47
Excel Template for Choosing the Number of Servers
B
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
l
m
s=
C
Data
120
80
3
D
E
(mean arrival rate)
(mean service rate)
(# servers)
Pr(W > t) = 0.02581732
when t =
0.05
Prob(W q > t) = 0.00058707
when t =
0.05
Economic Analysis:
Cs =
Cw =
Cost of Service
Cost of Waiting
Total Cost
$20.00
$48.00
(cost / server / unit time)
(waiting cost / unit time)
$60.00
$83.37
$143.37
48
F
L=
Lq =
G
Results
1.736842105
0.236842105
W=
Wq =
0.014473684
0.001973684
r
0.5
n
0
1
2
3
4
5
6
7
Pn
0.210526316
0.315789474
0.236842105
0.118421053
0.059210526
0.029605263
0.014802632
0.007401316
Comparing Expected Cost vs. Number of Clerks
H
J
K
L
M
N
Data Table for Expected Total Cost of Alternatives
s
1
2
3
4
5
r
0.50
1.50
0.75
0.50
0.38
0.30
L
1.74
#N/A
3.43
1.74
1.54
1.51
Cost of
Service
$60.00
$20.00
$40.00
$60.00
$80.00
$100.00
Cost of
Waiting
$83.37
#N/A
$164.57
$83.37
$74.15
$72.41
Total
Cost
$143.37
#N/A
$204.57
$143.37
$154.15
$172.41
$250
Cost ($/hour)
1
2
3
4
5
6
7
8
9
10
I
Cost of
Service
$200
$150
Cost of
Waiting
$100
Total Cost
$50
$0
0
1
2
3
4
Number of Servers (s)
49
5