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queueing_textbook.pptxTable of Contents Chapter 11 (Queueing Models)
Elements of a Queueing Model (Section 11.1) 11.2–11.12
Some Examples of Queueing Systems (Section 11.2) 11.13–11.15
Measures of Performance for Queueing Systems (Section 11.3) 11.16–11.19
A Case Study: The Dupit Corp. Problem (Section 11.4) 11.20–11.22
Some Single-Server Queueing Models (Section 11.5) 11.23–11.32
Some Multiple-Server Queueing Models (Section 11.6) 11.33–11.40
Priority Queueing Models (Section 11.7) 11.41–11.48
Some Insights about Designing Queueing Systems (Section 11.8) 11.49–11.51
Economic Analysis of the Number of Servers to Provide (Section 11.9) 11.52–11.55
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A Basic Queueing System
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Figure 11.1 A basic queueing system, where each customer is indicated by C and each server by S. Although this figure shows four servers, some queueing systems (including the example in this section) have only a single server.
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 | — | — | — |
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Table 11.1 The data for Herr Cutter’s first five customers.
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 = Expected 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
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Figure 11.2 The evolution of the number of customers in Herr Cutter’s barber shop over the first 100 minutes (from 8:00 to 9:40), given the data in Table 11.1.
The Exponential Distribution for Interarrival Times
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Figure 11.3 The shape of an exponential distribution, commonly used in queueing models as the distribution of interarrival times (and sometimes as the distribution of service times as well).
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 / m = Expected 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
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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: s = Mean
Constant Service Times
A better fit for systems that involve a fixed set of tasks.
Standard deviation: s = 0.
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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 are M = Exponential distribution (Markovian) D = Degenerate distribution (constant times) 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
Interarrival times are independent and identically distributed according to a specified probability distribution.
All arriving customers enter the queueing system and remain there until service has been completed.
The queueing system has a single infinite queue, so that the queue will hold an unlimited number of customers (for all practical purposes).
The queue discipline is first-come, first-served.
The queueing system has a specified number of servers, where each server is capable of serving any of the customers.
Each customer is served individually by any one of the servers.
Service times are independent and identically distributed according to a specified probability distribution.
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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 |
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Table 11.2 Examples of commercial service systems that are queueing systems.
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 |
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Table 11.3 Examples of internal service systems that are queueing systems.
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 |
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Table 11.4 Examples of transportation service systems that are queueing systems.
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
Since 1/m is the expected service time W = Wq + 1/m
Little’s formula states that L = lW and Lq = lWq
Combining the above relationships leads to L = Lq + l/m
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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
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
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The M/M/1 Model
Assumptions
Interarrival times have an exponential distribution with a mean of 1/l.
Service times have an exponential distribution with a mean of 1/m.
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)] = r2 / (1 – r)
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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) = re–m(1–r)t for t ≥ 0
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M/M/1 Queueing Model for the Dupit’s Current Policy
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Figure 11.4 This spreadsheet shows the results from applying the M/M/1 model with lambda = 3 and mu = 4 to the Dupit case study under the current policy.
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.
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M/M/1 Model for John Phixitt’s Suggested Approach (Reduce Machines/Rep)
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Figure 11.5 This application of the spreadsheet in Figure 11.4 shows that, when mu = 4, the M/M/1 model gives an expected waiting time to begin service of Wq = 0.25 days (the largest value that satisfies Dupit’s proposed new service standard) when lambda is changed from lambda = 3 to lambda = 2.
The M/G/1 Model
Assumptions
Interarrival times have an exponential distribution with a mean of 1/l.
Service times can have any probability distribution. You only need the mean (1/m) and standard deviation (s).
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 = [l2s2 + r2] / [2(1 – r)]
The expected number of customers in the system is
L = Lq + r
The expected waiting time in the queue is
Wq = Lq / l
The expected waiting time in the system is
W = Wq + 1/m
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The Values of s and Lq for the M/G/1 Model with Various Service-Time Distributions
| 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)] |
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Table 11.5 The values of s and Lq for the M/G/1 model with various service-time distributions.
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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Figure 11.6 This Excel template for the M/G/1 model shows the results from applying this model to the approach suggested by the Dupit’s vice president for engineering to use state-of-the-art equipment.
The M/M/s Model
Assumptions
Interarrival times have an exponential distribution with a mean of 1/l.
Service times have an exponential distribution with a mean of 1/m.
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.
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Values of L for the M/M/s Model for Various Values of s
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Figure 11.7 Values of L for the M/M/s model for various values of s, the number of servers.
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 = l = 6 (versus l = 3 before)
Mean service rate = m = 4 (as before)
Number of servers = s = 2 (versus s = 1 before)
Utilization factor = r = l/sm = 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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Figure 11.8 This Excel template for the M/M/s model shows the results from applying this model to the approach suggested by Dupit’s chief financial officer with two tech reps assigned to each territory.
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 = l = 9 (versus l = 3 before)
Mean service rate = m = 4 (as before)
Number of servers = s = 3 (versus s = 1 before)
Utilization factor = r = l/sm = 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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Figure 11.9 This Excel template modifies the results in Figure 11.8 by assigning three tech reps to each territory.
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) |
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Table 11.5 Comparison of Wq values with territories of different sizes for the Dupit problem.
Values of L for the M/D/s Model for Various Values of s
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Figure 11.10 Values of L for the M/D/s model for various values of s, the number of servers.
Priority Queueing Models
General Assumptions:
There are two or more categories of customers. Each category is assigned to a priority class. Customers in priority class 1 are given priority over customers in priority class 2. Priority class 2 has priority over priority class 3, etc.
After deferring to higher priority customers, the customers within each priority class are served on a first-come-fist-served basis.
Two types of priorities
Nonpreemptive priorities: Once a server has begun serving a customer, the service must be completed (even if a higher priority customer arrives). However, once service is completed, priorities are applied to select the next one to begin service.
Preemptive priorities: The lowest priority customer being served is preempted (ejected back into the queue) whenever a higher priority customer enters the queueing system.
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Preemptive Priorities Queueing Model
Additional Assumptions
Preemptive priorities are used as previously described.
For priority class i (i = 1, 2, … , n), the interarrival times of the customers in that class have an exponential distribution with a mean of 1/li.
All service times have an exponential distribution with a mean of 1/m, regardless of the priority class involved.
The queueing system has a single server.
The utilization factor for the server is
r = (l1 + l2 + … + ln) / m
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Nonpreemptive Priorities Queueing Model
Additional Assumptions
Nonpreemptive priorities are used as previously described.
For priority class i (i = 1, 2, … , n), the interarrival times of the customers in that class have an exponential distribution with a mean of 1/li.
All service times have an exponential distribution with a mean of 1/m, regardless of the priority class involved.
The queueing system can have any number of servers.
The utilization factor for the servers is
r = (l1 + l2 + … + ln) / sm
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VP of Marketing Approach (Priority for New Copiers)
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 of Marketing has proposed giving the printer-copiers priority over other machines for receiving service. The rationale for this proposal is that the printer-copier performs so many vital functions that its owners cannot tolerate being without it as long as other machines.
The mean arrival rates for the two classes of copiers are
l1 = 1 customer (printer-copier) per workday (now)
l2 = 2 customers (other machines) per workday (now)
The proportion of printer-copiers is expected to increase, so in a couple years
l1 = 1.5 customers (printer-copiers) per workday (later)
l2 = 1.5 customers (other machines) per workday (later)
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Nonpreemptive Priorities Model for VP of Marketing’s Approach (Current Arrival Rates)
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Figure 11.11 This Excel template applies the nonpreemptive priorities queueing model to the Dupit problem now under the approach suggested by the vice president for marketing to give priority to the printer-copiers.
Nonpreemptive Priorities Model for VP of Marketing’s Approach (Future Arrival Rates)
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Figure 11.12 The modification of Figure 11.11 that applies the same model to the later version of the Dupit problem.
Expected Waiting Times with Nonpreemptive Priorities
| s | When | l1 | l2 | m | r | Wq for Printer Copiers | Wq for Other Machines |
| 1 | Now | 1 | 2 | 4 | 0.75 | 0.25 workday (2 hrs.) | 1 workday (8 hrs.) |
| 1 | Later | 1.5 | 1.5 | 4 | 0.75 | 0.3 workday (2.4 hrs.) | 1.2 workday (9.6 hrs.) |
| 2 | Now | 2 | 4 | 4 | 0.75 | 0.107 workday (0.86 hr.) | 0.439 workday (3.43 hrs.) |
| 2 | Later | 3 | 3 | 4 | 0.75 | 0.129 workday (1.03 hrs.) | 0.514 workday (4.11 hrs.) |
| 3 | Now | 3 | 6 | 4 | 0.75 | 0.063 workday (0.50 hr.) | 0.252 workday (2.02 hrs.) |
| 3 | Later | 4.5 | 4.5 | 4 | 0.75 | 0.076 workday (0.61 hr.) | 0.303 workday (2.42 hrs.) |
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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 |
| VP for Marketing | Change to two-person territories with priority given to the printer-copiers for repairs | None, except disadvantages of larger territories |
Decision: Adopt fourth proposal (except for sparsely populated areas where second proposal should be adopted).
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Table 11.7 The four approaches being considered by Dupit management.
Some Insights About Designing Queueing Systems
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.
Decreasing the variability of service times (without any change in the mean) improves the performance of a queueing system substantially.
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.
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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Figure 11.13 This data table demonstrates Insight 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.
Effect of Decreasing s (Insight 2)
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Figure 11.14 This data table demonstrates Insight 2: Decreasing the variability of service times (without any change in the mean) improves the performance of a queueing system substantially.
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
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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.
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Excel Template for Choosing the Number of Servers
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Figure 11.16 This Excel template for using economic analysis to choose the number of servers with the M/M/s model is applied here to the Acme Machine Shop example with s = 3.
Comparing Expected Cost vs. Number of Clerks
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Figure 11.17 This data table compares the expected hourly costs with various alternative numbers of clerks assigned to the Acme Machine Shop tool crib.
Customers
Queue
Served Customers
Queueing System
Service
facility
S
S
S
S
C
C
C
C
C C C C C C C
Served Customers
20
40
60
80
1
2
3
4
Number of
Customers
in the
System
0
Time (in minutes)
100
M
e
a
n
T
i
m
e
0
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
G
H
Data
Results
l =
3
(mean arrival rate)
L =
3
m =
4
(mean service rate)
L
q
=
2.25
s =
1
(# servers)
W =
1
Pr(W > t) =
0.368
W
q
=
0.75
when t =
1
r =
0.75
Prob(W
q
> t) =
0.276
when t =
1
n
P
n
0
0.25
1
0.1875
2
0.1406
3
0.1055
4
0.0791
5
0.0593
6
0.0445
7
0.0334
8
0.0250
9
0.0188
10
0.0141
M|M|s
| Template for the M/M/s Queueing Model | ||||||||||
| Data | Results | Range Name | Cells | |||||||
| l = | 3 | (mean arrival rate) | 1 | L = | 3 | L | G4 | |||
| m = | 4 | (mean service rate) | 0 | Lq = | 2.25 | Lambda | C4 | |||
| s = | 1 | (# servers) | 0 | Lq | G5 | |||||
| 0 | W = | 1 | Mu | C5 | ||||||
| Pr(W > t) = | 0.368 | 0 | Wq = | 0.75 | n | F13:F38 | ||||
| when t = | 1 | 0 | 0 | P0 | G13 | |||||
| 0 | 0 | r = | 0.75 | Pn | G13:G38 | |||||
| Prob(Wq > t) = | 0.276 | 0 | Rho | G10 | ||||||
| when t = | 1 | 0 | n | Pn | s | C6 | ||||
| 0 | 0 | 0.25 | Time1 | C9 | ||||||
| 0 | 1 | 0.1875 | Time2 | C12 | ||||||
| 0 | 2 | 0.1406 | W | G7 | ||||||
| 0 | 3 | 0.1055 | Wq | G8 | ||||||
| 0 | 4 | 0.0791 | ||||||||
| 0 | 5 | 0.0593 | ||||||||
| 0 | 6 | 0.0445 | ||||||||
| 0 | 7 | 0.0334 | ||||||||
| 0 | 8 | 0.0250 | ||||||||
| 0 | 9 | 0.0188 | ||||||||
| 0 | 10 | 0.0141 | ||||||||
| 0 | 11 | 0.0106 | ||||||||
| 0 | 12 | 0.0079 | ||||||||
| 0 | 13 | 0.0059 | ||||||||
| 0 | 14 | 0.0045 | ||||||||
| 0 | 15 | 0.0033 | ||||||||
| 0 | 16 | 0.0025 | ||||||||
| 17 | 0.0019 | |||||||||
| 18 | 0.0014 | |||||||||
| 19 | 0.0011 | |||||||||
| 20 | 0.0008 | |||||||||
| 21 | 0.0006 | |||||||||
| 22 | 0.0004 | |||||||||
| 23 | 0.0003 | |||||||||
| 24 | 0.0003 | |||||||||
| 25 | 0.0002 |
Sheet1
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
G
H
Data
Results
l =
2
(mean arrival rate)
L =
1
m =
4
(mean service rate)
L
q
=
0.5
s =
1
(# servers)
W =
0.5
Pr(W > t) =
0.135
W
q
=
0.25
when t =
1
r =
0.5
Prob(W
q
> t) =
0.068
when t =
1
n
P
n
0
0.5
1
0.25
2
0.1250
3
0.0625
4
0.0313
5
0.0156
6
0.0078
7
0.0039
8
0.0020
9
0.0010
10
0.0005
M|M|s
| Template for the M/M/s Queueing Model | ||||||||||
| Data | Results | Range Name | Cells | |||||||
| l = | 2 | (mean arrival rate) | 1 | L = | 1 | L | G4 | |||
| m = | 4 | (mean service rate) | 0 | Lq = | 0.5 | Lambda | C4 | |||
| s = | 1 | (# servers) | 0 | Lq | G5 | |||||
| 0 | W = | 0.5 | Mu | C5 | ||||||
| Pr(W > t) = | 0.135 | 0 | Wq = | 0.25 | n | F13:F38 | ||||
| when t = | 1 | 0 | 0 | P0 | G13 | |||||
| 0 | 0 | r = | 0.5 | Pn | G13:G38 | |||||
| Prob(Wq > t) = | 0.068 | 0 | Rho | G10 | ||||||
| when t = | 1 | 0 | n | Pn | s | C6 | ||||
| 0 | 0 | 0.5 | Time1 | C9 | ||||||
| 0 | 1 | 0.25 | Time2 | C12 | ||||||
| 0 | 2 | 0.1250 | W | G7 | ||||||
| 0 | 3 | 0.0625 | Wq | G8 | ||||||
| 0 | 4 | 0.0313 | ||||||||
| 0 | 5 | 0.0156 | ||||||||
| 0 | 6 | 0.0078 | ||||||||
| 0 | 7 | 0.0039 | ||||||||
| 0 | 8 | 0.0020 | ||||||||
| 0 | 9 | 0.0010 | ||||||||
| 0 | 10 | 0.0005 | ||||||||
| 0 | 11 | 0.0002 | ||||||||
| 0 | 12 | 0.0001 | ||||||||
| 0 | 13 | 0.0001 | ||||||||
| 0 | 14 | 0.0000 | ||||||||
| 0 | 15 | 0.0000 | ||||||||
| 0 | 16 | 0.0000 | ||||||||
| 17 | 0.0000 | |||||||||
| 18 | 0.0000 | |||||||||
| 19 | 0.0000 | |||||||||
| 20 | 0.0000 | |||||||||
| 21 | 0.0000 | |||||||||
| 22 | 0.0000 | |||||||||
| 23 | 0.0000 | |||||||||
| 24 | 0.0000 | |||||||||
| 25 | 0.0000 |
Sheet1
3
4
5
6
7
8
9
10
11
12
B
C
D
E
F
G
Data
Results
l =
3
(mean arrival rate)
L =
1.163
1/m =
0.2
(expected service time)
L
q
=
0.563
s=
0.1
(standard deviation)
s =
1
(# servers)
W =
0.388
W
q
=
0.188
r =
0.6
P
0
=
0.4
M|G|1
| Template for the M/G/1 Queueing Model | |||||||||
| Data | Results | Range Name | Cell | ||||||
| l = | 3 | (mean arrival rate) | L = | 1.163 | L | G4 | |||
| 1/m = | 0.2 | (expected service time) | 0.563 | Lambda | C4 | ||||
| s= | 0.1 | (standard deviation) | Lq | G5 | |||||
| s = | 1 | (# servers) | W = | 0.388 | OneOverMu | C5 | |||
| 0.188 | Rho | G10 | |||||||
| s | C7 | ||||||||
| 0 | r = | 0.6 | Sigma | C6 | |||||
| 0 | W | G7 | |||||||
| 0.4 | Wq | G8 |
Sheet1
s = 1
s = 2
s = 3
s = 4
s = 5
s = 7
s = 10
s = 15
s = 20
s = 25
100
10
0.5
0.1
0.2
0
0.1
0.3
0.5
0.7
0.9
1.0
Utilization factor
r = l
s
m
Steady-state expected number of customers in the queueing system
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
G
H
Data
Results
l =
6
(mean arrival rate)
L =
3.4286
m =
4
(mean service rate)
L
q
=
1.9286
s =
2
(# servers)
W =
0.5714
Pr(W > t) =
0.169
W
q
=
0.3214
when t =
1
r =
0.75
Prob(W
q
> t) =
0.087
when t =
1
n
P
n
0
0.1429
1
0.2143
2
0.1607
3
0.1205
4
0.0904
5
0.0678
6
0.0509
7
0.0381
8
0.0286
9
0.0215
10
0.0161
M|M|s
| Template for the M/M/s Queueing Model | ||||||||||
| Data | Results | Range Name | Cells | |||||||
| l = | 6 | (mean arrival rate) | 1 | L = | 3.4286 | L | G4 | |||
| m = | 4 | (mean service rate) | 1.5 | Lq = | 1.9286 | Lambda | C4 | |||
| s = | 2 | (# servers) | 0 | Lq | G5 | |||||
| 0 | W = | 0.5714 | Mu | C5 | ||||||
| Pr(W > t) = | 0.169 | 0 | Wq = | 0.3214 | n | F13:F38 | ||||
| when t = | 1 | 0 | 0 | P0 | G13 | |||||
| 0 | 0 | r = | 0.75 | Pn | G13:G38 | |||||
| Prob(Wq > t) = | 0.087 | 0 | Rho | G10 | ||||||
| when t = | 1 | 0 | n | Pn | s | C6 | ||||
| 0 | 0 | 0.1429 | Time1 | C9 | ||||||
| 0 | 1 | 0.2143 | Time2 | C12 | ||||||
| 0 | 2 | 0.1607 | W | G7 | ||||||
| 0 | 3 | 0.1205 | Wq | G8 | ||||||
| 0 | 4 | 0.0904 | ||||||||
| 0 | 5 | 0.0678 | ||||||||
| 0 | 6 | 0.0509 | ||||||||
| 0 | 7 | 0.0381 | ||||||||
| 0 | 8 | 0.0286 | ||||||||
| 0 | 9 | 0.0215 | ||||||||
| 0 | 10 | 0.0161 | ||||||||
| 0 | 11 | 0.0120671817 | ||||||||
| 0 | 12 | 0.0090503863 | ||||||||
| 0 | 13 | 0.0067877897 | ||||||||
| 0 | 14 | 0.0050908423 | ||||||||
| 0 | 15 | 0.0038181317 | ||||||||
| 0 | 16 | 0.0028635988 | ||||||||
| 17 | 0.0021476991 | |||||||||
| 18 | 0.0016107743 | |||||||||
| 19 | 0.0012080807 | |||||||||
| 20 | 0.0009060606 | |||||||||
| 21 | 0.0006795454 | |||||||||
| 22 | 0.0005096591 | |||||||||
| 23 | 0.0003822443 | |||||||||
| 24 | 0.0002866832 | |||||||||
| 25 | 0.0002150124 |
Sheet1
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
G
H
Data
Results
l =
9
(mean arrival rate)
L =
3.9533
m =
4
(mean service rate)
L
q
=
1.7033
s =
3
(# servers)
W =
0.4393
Pr(W > t) =
0.090
W
q
=
0.1893
when t =
1
r =
0.75
Prob(W
q
> t) =
0.028
when t =
1
n
P
n
0
0.0748
1
0.1682
2
0.1893
3
0.1419
4
0.1065
5
0.0798
6
0.0599
7
0.0449
8
0.0337
9
0.0253
10
0.0189
M|M|s
| Template for the M/M/s Queueing Model | ||||||||||
| Data | Results | Range Name | Cells | |||||||
| l = | 9 | (mean arrival rate) | 1 | L = | 3.9533 | L | G4 | |||
| m = | 4 | (mean service rate) | 2.25 | Lq = | 1.7033 | Lambda | C4 | |||
| s = | 3 | (# servers) | 2.53125 | Lq | G5 | |||||
| 0 | W = | 0.4393 | Mu | C5 | ||||||
| Pr(W > t) = | 0.090 | 0 | Wq = | 0.1893 | n | F13:F38 | ||||
| when t = | 1 | 0 | 0 | P0 | G13 | |||||
| 0 | 0 | r = | 0.75 | Pn | G13:G38 | |||||
| Prob(Wq > t) = | 0.028 | 0 | Rho | G10 | ||||||
| when t = | 1 | 0 | n | Pn | s | C6 | ||||
| 0 | 0 | 0.0748 | Time1 | C9 | ||||||
| 0 | 1 | 0.1682 | Time2 | C12 | ||||||
| 0 | 2 | 0.1893 | W | G7 | ||||||
| 0 | 3 | 0.1419 | Wq | G8 | ||||||
| 0 | 4 | 0.1065 | ||||||||
| 0 | 5 | 0.0798 | ||||||||
| 0 | 6 | 0.0599 | ||||||||
| 0 | 7 | 0.0449 | ||||||||
| 0 | 8 | 0.0337 | ||||||||
| 0 | 9 | 0.0253 | ||||||||
| 0 | 10 | 0.0189 | ||||||||
| 0 | 11 | 0.0142099523 | ||||||||
| 0 | 12 | 0.0106574642 | ||||||||
| 0 | 13 | 0.0079930982 | ||||||||
| 0 | 14 | 0.0059948236 | ||||||||
| 0 | 15 | 0.0044961177 | ||||||||
| 0 | 16 | 0.0033720883 | ||||||||
| 17 | 0.0025290662 | |||||||||
| 18 | 0.0018967997 | |||||||||
| 19 | 0.0014225997 | |||||||||
| 20 | 0.0010669498 | |||||||||
| 21 | 0.0008002124 | |||||||||
| 22 | 0.0006001593 | |||||||||
| 23 | 0.0004501195 | |||||||||
| 24 | 0.0003375896 | |||||||||
| 25 | 0.0002531922 |
Sheet1
100
0.1
1.0
10
0
0.1
0.3
0.5
0.9
1.0
s = 2
s = 3
s = 4
s = 5
s = 7
s = 10
s = 15
s = 20
s = 25
s = 1
0.7
Utilization factor
Steady-state expected number of customers in the queueing system
r = l
s
m
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
B
C
D
E
F
G
Data
n =
2
(# of priority classes)
m =
4
(mean service rate)
s =
1
(# servers)
l
i
L
Lq
W
Wq
Priority Class 1
1
0.5
0.25
0.5
0.25
Priority Class 2
2
2.5
2
1.25
1
Priority Class 3
1
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
Priority Class 4
1
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
Priority Class 5
1
1.75
1.5
1.75
1.5
l =
3
r =
0.75
Results
Nonpreemptive Priorities
| Template for M/M/s Nonpreemptive Priorities Queueing Model | ||||||
| Data | ||||||
| 1 | n = | 2 | (# of priority classes) | |||
| 0 | m = | 4 | (mean service rate) | |||
| 0 | s = | 1 | (# servers) | |||
| 0 | ||||||
| 0 | Results | |||||
| 0 | L | Lq | W | Wq | ||
| 0 | Priority Class 1 | 1 | 0.5 | 0.25 | 0.5 | 0.25 |
| 0 | Priority Class 2 | 2 | 2.5 | 2 | 1.25 | 1 |
| 0 | Priority Class 3 | 1 | 0 | 0 | 0 | 0 |
| 0 | Priority Class 4 | 1 | 0 | 0 | 0 | 0 |
| 0 | Priority Class 5 | 1 | 1.75 | 1.5 | 1.75 | 1.5 |
| 0 | ||||||
| 0 | l = | 3 | 0 | |||
| 0 | r = | 0.75 | ||||
| 0 | 0 | |||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 |
Sheet1
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
B
C
D
E
F
G
Data
n =
2
(# of priority classes)
m =
4
(mean service rate)
s =
1
(# servers)
l
i
L
Lq
W
Wq
Priority Class 1
1.5
0.825
0.45
0.55
0.3
Priority Class 2
1.5
2.175
1.8
1.45
1.2
Priority Class 3
1
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
Priority Class 4
1
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
Priority Class 5
1
1.75
1.5
1.75
1.5
l =
3
r =
0.75
Results
Nonpreemptive Priorities
| Template for M/M/s Nonpreemptive Priorities Queueing Model | ||||||
| Data | ||||||
| 1 | n = | 2 | (# of priority classes) | |||
| 0 | m = | 4 | (mean service rate) | |||
| 0 | s = | 1 | (# servers) | |||
| 0 | ||||||
| 0 | Results | |||||
| 0 | L | Lq | W | Wq | ||
| 0 | Priority Class 1 | 1.5 | 0.825 | 0.45 | 0.55 | 0.3 |
| 0 | Priority Class 2 | 1.5 | 2.175 | 1.8 | 1.45 | 1.2 |
| 0 | Priority Class 3 | 1 | 0 | 0 | 0 | 0 |
| 0 | Priority Class 4 | 1 | 0 | 0 | 0 | 0 |
| 0 | Priority Class 5 | 1 | 1.75 | 1.5 | 1.75 | 1.5 |
| 0 | ||||||
| 0 | l = | 3 | 0 | |||
| 0 | r = | 0.75 | ||||
| 0 | 0 | |||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 | ||||||
| 0 |
Sheet1
3
4
5
B
C
D
E
G
H
Data
Results
l =
0.5
(mean arrival rate)
L =
1
m =
1
(mean service rate)
L
q
=
0.5
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
A
B
C
D
E
Data Table Demonstrating the Effect of
Increasing
r
on Lq and L for M/M/1
l = r
L
q
L
1
0.5
1
0
0.01
0.0001
0.0101
0
0.25
0.0833
0.3333
0
0.5
0.5
1
0
0.6
0.9
1.5
0
0.7
1.6333
2.3333
0
0.75
2.25
3
0
0.8
3.2
4
0
0.85
4.8167
5.6667
0
0.9
8.1
9
0
0.95
18.05
19
0
0.99
98.01
99
0
0.999
998.001
999
0
20
40
60
80
100
00.20.40.60.81
System Utilization (r)
Average Line Length (L)
Chart1
| 0.01 |
| 0.25 |
| 0.5 |
| 0.6 |
| 0.7 |
| 0.75 |
| 0.8 |
| 0.85 |
| 0.9 |
| 0.95 |
| 0.99 |
| 0.999 |
M|M|s
| Template for M/M/s Queueing Model | ||||||||||||
| Data | Results | Range Name | Cells | |||||||||
| l = | 0.5 | (mean arrival rate) | L = | 1 | L | G4 | ||||||
| m = | 1 | (mean service rate) | 0.5 | Lambda | C4 | |||||||
| s = | 1 | (# servers) | Lq | G5 | ||||||||
| 0 | W = | 2 | ||||||||||
| 0 | 1 | |||||||||||
| Data Table Demonstrating the Effect of | ||||||||||||
| r = | 0.5 | |||||||||||
| l = r | L | n | ||||||||||
| 1 | 0.5 | 1 | 0 | 0.5 | ||||||||
| 0 | 0.01 | 0.0001 | 0.0101 | |||||||||
| 0 | 0.25 | 0.0833 | 0.3333 | |||||||||
| 0 | 0.5 | 0.5 | 1 | |||||||||
| 0 | 0.6 | 0.9 | 1.5 | |||||||||
| 0 | 0.7 | 1.6333 | 2.3333 | |||||||||
| 0 | 0.75 | 2.25 | 3 | |||||||||
| 0 | 0.8 | 3.2 | 4 | |||||||||
| 0 | 0.85 | 4.8167 | 5.6667 | |||||||||
| 0 | 0.9 | 8.1 | 9 | |||||||||
| 0 | 0.95 | 18.05 | 19 | |||||||||
| 0 | 0.99 | 98.01 | 99 | |||||||||
| 0 | 0.999 | 998.001 | 999 | |||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 |
M|M|s
M|M|s
| Template for the M/M/s Queueing Model | ||||||||||
| Data | Results | Range Name | Cells | |||||||
| l = | 0.5 | (mean arrival rate) | 1 | L = | 1 | L | G4 | |||
| m = | 1 | (mean service rate) | 0 | 0.5 | Lambda | C4 | ||||
| s = | 1 | (# servers) | 0 | Lq | G5 | |||||
| 0 | W = | 2 | Mu | C5 | ||||||
| Pr(W > t) = | 0.6065306597 | 0 | 1 | n | F13:F38 | |||||
| when t = | 1 | 0 | 0 | P0 | G13 | |||||
| 0 | 0 | r = | 0.5 | Pn | G13:G38 | |||||
| 0.3032653299 | 0 | Rho | G10 | |||||||
| when t = | 1 | 0 | n | s | C6 | |||||
| 0 | 0 | 0.5 | Time1 | C9 | ||||||
| 0 | 1 | 0.25 | Time2 | C12 | ||||||
| 0 | 2 | 0.125 | W | G7 | ||||||
| 0 | 3 | 0.0625 | Wq | G8 | ||||||
| 0 | 4 | 0.03125 | ||||||||
| 0 | 5 | 0.015625 | ||||||||
| 0 | 6 | 0.0078125 | ||||||||
| 0 | 7 | 0.00390625 | ||||||||
| 0 | 8 | 0.001953125 | ||||||||
| 0 | 9 | 0.0009765625 | ||||||||
| 0 | 10 | 0.0004882813 | ||||||||
| 0 | 11 | 0.0002441406 | ||||||||
| 0 | 12 | 0.0001220703 | ||||||||
| 0 | 13 | 0.0000610352 | ||||||||
| 0 | 14 | 0.0000305176 | ||||||||
| 0 | 15 | 0.0000152588 | ||||||||
| 0 | 16 | 0.0000076294 | ||||||||
| 17 | 0.0000038147 | |||||||||
| 18 | 0.0000019073 | |||||||||
| 19 | 0.0000009537 | |||||||||
| 20 | 0.0000004768 | |||||||||
| 21 | 0.0000002384 | |||||||||
| 22 | 0.0000001192 | |||||||||
| 23 | 0.0000000596 | |||||||||
| 24 | 0.0000000298 | |||||||||
| 25 | 0.0000000149 |
M|M|s
| Template for M/M/s Queueing Model | ||||||||||||
| Data | Results | Range Name | Cells | |||||||||
| l = | 0.5 | (mean arrival rate) | L = | 1 | L | G4 | ||||||
| m = | 1 | (mean service rate) | 0.5 | Lambda | C4 | |||||||
| s = | 1 | (# servers) | Lq | G5 | ||||||||
| 0 | W = | 2 | ||||||||||
| 0 | 1 | |||||||||||
| Data Table Demonstrating the Effect of | ||||||||||||
| r = | 0.5 | |||||||||||
| l = r | L | n | ||||||||||
| 1 | 0.5 | 1 | 0 | 0.5 | ||||||||
| 0 | 0.01 | 0.0001 | 0.0101 | |||||||||
| 0 | 0.25 | 0.0833 | 0.3333 | |||||||||
| 0 | 0.5 | 0.5 | 1 | |||||||||
| 0 | 0.6 | 0.9 | 1.5 | |||||||||
| 0 | 0.7 | 1.6333 | 2.3333 | |||||||||
| 0 | 0.75 | 2.25 | 3 | |||||||||
| 0 | 0.8 | 3.2 | 4 | |||||||||
| 0 | 0.85 | 4.8167 | 5.6667 | |||||||||
| 0 | 0.9 | 8.1 | 9 | |||||||||
| 0 | 0.95 | 18.05 | 19 | |||||||||
| 0 | 0.99 | 98.01 | 99 | |||||||||
| 0 | 0.999 | 998.001 | 999 | |||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 | ||||||||||||
| 0 |
M|M|s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
A
B
C
D
E
F
G
H
Template for the M/G/1 Queueing Model
Data
Results
l =
0.5
(mean arrival rate)
L =
0.8125
1/m =
1
(expected service time)
L
q
=
0.3125
s=
0.5
(standard deviation)
s =
1
(# servers)
W =
1.625
W
q
=
0.625
r =
0.5
P
0
=
0.5
Data Table Demonstrating the Effect of Decreasing s on Lq for M/G/1
Body of Table Shows L
q
Values
s
0.3125
1
0.5
0
0.5
0.500
0.313
0.250
r (=l)
0.75
2.250
1.406
1.125
0.9
8.100
5.063
4.050
0.99
98.010
61.256
49.005
M|G|1
| Template for the M/G/1 Queueing Model | |||||||||||
| Data | Results | Range Name | Cell | ||||||||
| l = | 0.5 | (mean arrival rate) | L = | 0.8125 | Lambda | C4 | |||||
| 1/m = | 1 | (expected service time) | 0.3125 | Lq | G5 | ||||||
| s= | 0.5 | (standard deviation) | Sigma | C6 | |||||||
| s = | 1 | (# servers) | W = | 1.625 | |||||||
| 0.625 | |||||||||||
| 0 | r = | 0.5 | |||||||||
| 0 | |||||||||||
| 0.5 | |||||||||||
| Data Table Demonstrating the Effect of Decreasing s on Lq for M/G/1 | |||||||||||
| s | |||||||||||
| 0.3125 | 1 | 0.5 | 0 | ||||||||
| 0.5 | 0.500 | 0.313 | 0.250 | ||||||||
| r (=l) | 0.75 | 2.250 | 1.406 | 1.125 | |||||||
| 0.9 | 8.100 | 5.063 | 4.050 | ||||||||
| 0.99 | 98.010 | 61.256 | 49.005 |
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
B
C
D
E
F
G
Data
Results
l =
120
(mean arrival rate)
L =
1.736842105
m =
80
(mean service rate)
L
q
=
0.236842105
s =
3
(# servers)
W =
0.014473684
Pr(W > t) =
0.02581732
W
q
=
0.001973684
when t =
0.05
r =
0.5
Prob(W
q
> t) =
0.00058707
when t =
0.05
n
P
n
0
0.210526316
1
0.315789474
Cs =
$20.00
(cost / server / unit time)
2
0.236842105
Cw =
$48.00
(waiting cost / unit time)
3
0.118421053
4
0.059210526
Cost of Service
$60.00
5
0.029605263
Cost of Waiting
$83.37
6
0.014802632
Total Cost
$143.37
7
0.007401316
Economic Analysis:
Economic Analysis
| Template for Economic Analysis of M/M/s Queueing Model | |||||||||
| Data | Results | Range Name | Cells | ||||||
| l = | 120 | (mean arrival rate) | L = | 1.7368421053 | CostOfService | C18 | |||
| m = | 80 | (mean service rate) | 0.2368421053 | CostOfWaiting | C19 | ||||
| s = | 3 | (# servers) | Cs | C15 | |||||
| W = | 0.0144736842 | Cw | C16 | ||||||
| Pr(W > t) = | 0.0258173221 | 0.0019736842 | L | G4 | |||||
| when t = | 0.05 | Lambda | C4 | ||||||
| r = | 0.5 | Lq | G5 | ||||||
| 0.0005870729 | Mu | C5 | |||||||
| when t = | 0.05 | n | n | F13:F38 | |||||
| 1 | 0 | 0.2105263158 | P0 | G13 | |||||
| 1.5 | Economic Analysis: | 1 | 0.3157894737 | Pn | G13:G38 | ||||
| 1.125 | Cs = | $20.00 | (cost / server / unit time) | 2 | 0.2368421053 | Rho | G10 | ||
| 0 | Cw = | $48.00 | (waiting cost / unit time) | 3 | 0.1184210526 | s | C6 | ||
| 0 | 4 | 0.0592105263 | Time1 | C9 | |||||
| 0 | Cost of Service | $60.00 | 5 | 0.0296052632 | Time2 | C12 | |||
| 0 | Cost of Waiting | $83.37 | 6 | 0.0148026316 | TotalCost | C20 | |||
| 0 | Total Cost | $143.37 | 7 | 0.0074013158 | W | G7 | |||
| 0 | 8 | 0.0037006579 | Wq | G8 | |||||
| 0 | 9 | 0.0018503289 | |||||||
| 0 | 10 | 0.0009251645 | |||||||
| 0 | 11 | 0.0004625822 | |||||||
| 0 | 12 | 0.0002312911 | |||||||
| 0 | 13 | 0.0001156456 | |||||||
| 0 | 14 | 0.0000578228 | |||||||
| 0 | 15 | 0.0000289114 | |||||||
| 0 | 16 | 0.0000144557 | |||||||
| 0 | 17 | 0.0000072278 | |||||||
| 0 | 18 | 0.0000036139 | |||||||
| 0 | 19 | 0.000001807 | |||||||
| 0 | 20 | 0.0000009035 | |||||||
| 0 | 21 | 0.0000004517 | |||||||
| 0 | 22 | 0.0000002259 | |||||||
| 0 | 23 | 0.0000001129 | |||||||
| 0 | 24 | 0.0000000565 | |||||||
| 0 | 25 | 0.0000000282 |
1
2
3
4
5
6
7
8
9
10
H
I
J
K
L
M
N
Data Table for Expected Total Cost of Alternatives
Cost of
Cost of
Total
s
r
L
Service
Waiting
Cost
0.50
1.74
$60.00
$83.37
$143.37
1
1.50
#N/A
$20.00
#N/A
#N/A
2
0.75
3.43
$40.00
$164.57
$204.57
3
0.50
1.74
$60.00
$83.37
$143.37
4
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
$250
012345
Number of Servers (s)
Cost ($/hour)
Cost of
Service
Cost of
Waiting
Total Cost
Economic Analysis
| Economic Analysis of Acme Machine Shop Example | Data Table for Expected Total Cost of Alternatives | ||||||||||||
| Data | Results | Cost of | Cost of | Total | |||||||||
| l = | 120 | (mean arrival rate) | L = | 1.7368421053 | s | r | L | Service | Waiting | Cost | |||
| m = | 80 | (mean service rate) | 0.2368421053 | 0.50 | 1.74 | $60.00 | $83.37 | $143.37 | |||||
| s = | 3 | (# servers) | 1 | 1.50 | 0.00 | $20.00 | $0.00 | $0.00 | |||||
| W = | 0.0144736842 | 2 | 0.75 | 3.43 | $40.00 | $164.57 | $204.57 | ||||||
| Pr(W > t) = | 0.0258173221 | 0.0019736842 | 3 | 0.50 | 1.74 | $60.00 | $83.37 | $143.37 | |||||
| when t = | 0.05 | 0 | 4 | 0.38 | 1.54 | $80.00 | $74.15 | $154.15 | |||||
| 0 | r = | 0.5 | 5 | 0.30 | 1.51 | $100.00 | $72.41 | $172.41 | |||||
| 0.0005870729 | |||||||||||||
| when t = | 0.05 | n | |||||||||||
| 1 | 0 | 0.2105263158 | |||||||||||
| 1.5 | Economic Analysis: | 1 | 0.3157894737 | ||||||||||
| 1.125 | Cs = | $20.00 | (cost / server / unit time) | 2 | 0.2368421053 | ||||||||
| 0 | Cw = | $48.00 | (waiting cost / unit time) | 3 | 0.1184210526 | ||||||||
| 0 | 4 | 0.0592105263 | |||||||||||
| 0 | Cost of Service | $60.00 | 5 | 0.0296052632 | |||||||||
| 0 | Cost of Waiting | $83.37 | 6 | 0.0148026316 | |||||||||
| 0 | Total Cost | $143.37 | 7 | 0.0074013158 | |||||||||
| 0 | 8 | 0.0037006579 | |||||||||||
| 0 | 9 | 0.0018503289 | |||||||||||
| 0 | 10 | 0.0009251645 | |||||||||||
| 0 | 11 | 0.0004625822 | |||||||||||
| 0 | 12 | 0.0002312911 | |||||||||||
| 0 | 13 | 0.0001156456 | |||||||||||
| 0 | 14 | 0.0000578228 | |||||||||||
| 0 | 15 | 0.0000289114 | |||||||||||
| 0 | 16 | 0.0000144557 | |||||||||||
| 0 | 17 | 0.0000072278 | |||||||||||
| 0 | 18 | 0.0000036139 | |||||||||||
| 0 | 19 | 0.000001807 | |||||||||||
| 0 | 20 | 0.0000009035 | |||||||||||
| 0 | 21 | 0.0000004517 | |||||||||||
| 0 | 22 | 0.0000002259 | |||||||||||
| 0 | 23 | 0.0000001129 | |||||||||||
| 0 | 24 | 0.0000000565 | |||||||||||
| 0 | 25 | 0.0000000282 | |||||||||||
| Range Name | Cells | ||||||||||||
| CostOfService | C18 | ||||||||||||
| CostOfWaiting | C19 | ||||||||||||
| Cs | C15 | ||||||||||||
| Cw | C16 | ||||||||||||
| L | G4 | ||||||||||||
| Lambda | C4 | ||||||||||||
| Lq | G5 | ||||||||||||
| Mu | C5 | ||||||||||||
| n | F13:F38 | ||||||||||||
| P0 | G13 | ||||||||||||
| Pn | G13:G38 | ||||||||||||
| Rho | G10 | ||||||||||||
| s | C6 | ||||||||||||
| Time1 | C9 | ||||||||||||
| Time2 | C12 | ||||||||||||
| TotalCost | C20 | ||||||||||||
| W | G7 | ||||||||||||
| Wq | G8 |
Chart1
| 1 | 1 | 1 |
| 2 | 2 | 2 |
| 3 | 3 | 3 |
| 4 | 4 | 4 |
| 5 | 5 | 5 |
Economic Analysis
| Economic Analysis of Acme Machine Shop Example | Data Table for Expected Total Cost of Alternatives | ||||||||||||
| Data | Results | Cost of | Cost of | Total | |||||||||
| l = | 120 | (mean arrival rate) | L = | 1.7368421053 | s | r | L | Service | Waiting | Cost | |||
| m = | 80 | (mean service rate) | 0.2368421053 | 0.50 | 1.74 | $60.00 | $83.37 | $143.37 | |||||
| s = | 3 | (# servers) | 1 | 1.50 | 0.00 | $20.00 | $0.00 | $0.00 | |||||
| W = | 0.0144736842 | 2 | 0.75 | 3.43 | $40.00 | $164.57 | $204.57 | ||||||
| Pr(W > t) = | 0.0258173221 | 0.0019736842 | 3 | 0.50 | 1.74 | $60.00 | $83.37 | $143.37 | |||||
| when t = | 0.05 | 0 | 4 | 0.38 | 1.54 | $80.00 | $74.15 | $154.15 | |||||
| 0 | r = | 0.5 | 5 | 0.30 | 1.51 | $100.00 | $72.41 | $172.41 | |||||
| 0.0005870729 | |||||||||||||
| when t = | 0.05 | n | |||||||||||
| 1 | 0 | 0.2105263158 | |||||||||||
| 1.5 | Economic Analysis: | 1 | 0.3157894737 | ||||||||||
| 1.125 | Cs = | $20.00 | (cost / server / unit time) | 2 | 0.2368421053 | ||||||||
| 0 | Cw = | $48.00 | (waiting cost / unit time) | 3 | 0.1184210526 | ||||||||
| 0 | 4 | 0.0592105263 | |||||||||||
| 0 | Cost of Service | $60.00 | 5 | 0.0296052632 | |||||||||
| 0 | Cost of Waiting | $83.37 | 6 | 0.0148026316 | |||||||||
| 0 | Total Cost | $143.37 | 7 | 0.0074013158 | |||||||||
| 0 | 8 | 0.0037006579 | |||||||||||
| 0 | 9 | 0.0018503289 | |||||||||||
| 0 | 10 | 0.0009251645 | |||||||||||
| 0 | 11 | 0.0004625822 | |||||||||||
| 0 | 12 | 0.0002312911 | |||||||||||
| 0 | 13 | 0.0001156456 | |||||||||||
| 0 | 14 | 0.0000578228 | |||||||||||
| 0 | 15 | 0.0000289114 | |||||||||||
| 0 | 16 | 0.0000144557 | |||||||||||
| 0 | 17 | 0.0000072278 | |||||||||||
| 0 | 18 | 0.0000036139 | |||||||||||
| 0 | 19 | 0.000001807 | |||||||||||
| 0 | 20 | 0.0000009035 | |||||||||||
| 0 | 21 | 0.0000004517 | |||||||||||
| 0 | 22 | 0.0000002259 | |||||||||||
| 0 | 23 | 0.0000001129 | |||||||||||
| 0 | 24 | 0.0000000565 | |||||||||||
| 0 | 25 | 0.0000000282 | |||||||||||
| Range Name | Cells | ||||||||||||
| CostOfService | C18 | ||||||||||||
| CostOfWaiting | C19 | ||||||||||||
| Cs | C15 | ||||||||||||
| Cw | C16 | ||||||||||||
| L | G4 | ||||||||||||
| Lambda | C4 | ||||||||||||
| Lq | G5 | ||||||||||||
| Mu | C5 | ||||||||||||
| n | F13:F38 | ||||||||||||
| P0 | G13 | ||||||||||||
| Pn | G13:G38 | ||||||||||||
| Rho | G10 | ||||||||||||
| s | C6 | ||||||||||||
| Time1 | C9 | ||||||||||||
| Time2 | C12 | ||||||||||||
| TotalCost | C20 | ||||||||||||
| W | G7 | ||||||||||||
| Wq | G8 |
