1. Waiting Lines (Queuing Theory) Service Analysis Tutorial 1
Compiled by:
Alex J. Ruiz-Torres, Ph.D.
From information developed by many.

2. Objectives/ Structure
This tutorial will illustrate all the steps required to analyze a service system based on the queuing theory modeling concepts.
Setting up the problem and variables
Calculating system performance
Performing sensitivity analysis

3. Problem Description Part 1. Measures
The service is a medical clinic that provides non-emergency care to walk-in patients. A patient walks in about every 9 minutes and the initial screening takes one average 400 seconds. A single nurse performs the screening.
Part 1. Measures
What is the utilization of the screening nurse?
What is the average size of the line waiting for screening?
The clinic has a policy that the average waiting time cannot exceed 30 minutes, are they meeting that requirement?
What is the probability there are 4 or more patients waiting?
What is the probability there are 1 or 0 patients waiting?
Part 2. Changes (Sensitivity Analysis)
Due to changes in the economy the clinic expects an increase of 20% in the number of patients. Review the answers a-e given this increase in demand.

4. Problem’s parameters
Start the solution process by clearly determining what are the entities, what are the servers, the arrival rate, and the service rate.
Entity = patients;
Server = nurse.
arrival rate (how many patients arrive): l;
service rate (how many patients can be processed per nurse): m;
By clipartkid (clipartkid.com) [CC BY-SA 4.0 ( via Wikimedia Commons

5. The inverse of the inter-arrival time is the arrival rate: l
Problem’s parameters
The problem states that a patient walks in every 9 minutes.
This is the inter-arrival time: 9 minutes/patient.
Remember that m and l must be in entities/time thus it needs to be transformed.
The inverse of the inter-arrival time is the arrival rate: l

6. The inverse of the inter-arrival time is the arrival rate: l
Problem’s parameters
The inverse of the inter-arrival time is the arrival rate: l
l = 1 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 9 𝑚𝑖𝑛𝑢𝑡𝑒 = 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑚𝑖𝑛𝑢𝑡𝑒
To make it more understandable, we move up one level of time unit. (from minutes to hours)
l = 1 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 9 𝑚𝑖𝑛𝑠 × 60 𝑚𝑖𝑛𝑠 1 ℎ𝑜𝑢𝑟 = 60 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 9 ℎ𝑜𝑢𝑟𝑠 = 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 ℎ𝑜𝑢𝑟

7. The inverse of the service time is the service rate: m
Problem’s parameters
The problem states that the screening process takes 400 seconds.
This is the service time: 400 second/patient.
Remember that m and l must be in entities/time thus it needs to be transformed.
The inverse of the service time is the service rate: m

8. The inverse of the service time is the service rate: m
Problem’s parameters
The inverse of the service time is the service rate: m
m = 1 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 400 𝑠𝑒𝑐𝑜𝑛𝑑 = patient / second
However, is not in the same units as l (in patient/ hour).
Need to modify units in m.
m = 1 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 400 𝑠𝑒𝑐𝑜𝑛𝑑 × 𝑠𝑒𝑐𝑜𝑛𝑑 1 ℎ𝑜𝑢𝑟 =9 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 ℎ𝑜𝑢𝑟

9. Calculating System Performance
Will use the equations for a single server Utilization: λ/μ Probability of 0 entities in the line: P0 = 1 – λ/μ Probability of n entities in the line: Pn = (λ/μ)n P0 (n > 0)

10. System Performance Yes, they are meeting the requirement
Part 1. Answer the following
a) What is the utilization of the screening nurse?
Utilization: λ/μ = 𝑝𝑎𝑡𝑖𝑒𝑛𝑡/ℎ𝑜𝑢𝑟 9 𝑝𝑎𝑡𝑖𝑒𝑛𝑡/ℎ𝑜𝑢𝑟 = 74.1% busy
b) What is the average size of the line waiting for screening?
Avg. size of the line: Lq .
Use ratio = λ/μ = 0.74 round up to 0.8. Lq = 3.2 patients
c) The clinic has a policy that the average waiting time cannot exceed 30 minutes, are they meeting that requirement?
Avg time to wait per entity: Wq =Lq/λ = 3.2 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑝𝑎𝑡𝑖𝑒𝑛𝑡/ℎ𝑜𝑢𝑟 = 0.48 hour
0.48 hour 𝑥 60 𝑚𝑖𝑛𝑢𝑡𝑒 1 ℎ𝑜𝑢𝑟 = 28.8 minute.
Yes, they are meeting
the requirement

11. System Performance P(patients waiting ≥ 4) = P4 + P5 + …. + P
Part 1. Answer the following
d) What is the probability there are 4 or more patients waiting?
P(patients waiting ≥ 4) = P4 + P5 + …. + P
= 1 – (P0 + P1 + P2 + P3 )
Using these two equations we generate this table
P0 = 1 – λ/μ ;
Pn = (λ/μ)n P0
= 1 – (0.698) = = 30.2%
Parameter
Value P0
0.259 P1
0.192 P2
0.142 P3
0.105

12. System Performance P(patients waiting ≤ 1) = P0 + P1
Part 1. Answer the following
e) What is the probability there are 1 or 0 patients waiting?
P(patients waiting ≤ 1) = P0 + P1
= = 0.451= 45.1%
Parameter
Value P0
0.259 P1
0.192 P2
0.142 P3
0.105

13. Sensitivity Analysis
Part 2. Changes (Sensitivity Analysis) Due to changes in the economy the expect more patients to use their service, an increase of 20%. Review the answers a-e given this increase in demand. We need to determine the new arrival time by: l =6.667 𝑝𝑎𝑡𝑖𝑒𝑛𝑡𝑠 ℎ𝑜𝑢𝑟 × 120% = 8 𝑝𝑎𝑡𝑖𝑒𝑛𝑡𝑠 ℎ𝑜𝑢𝑟 Thus more than one additional patient per hour.

14. Sensitivity Analysis
Re- Answer the following
a) What is the utilization of the screening nurse?
Utilization: λ/μ = 8 𝑝𝑎𝑡𝑖𝑒𝑛𝑡/ℎ𝑜𝑢𝑟 9 𝑝𝑎𝑡𝑖𝑒𝑛𝑡/ℎ𝑜𝑢𝑟 = 88.8% busy
b) What is the average size of the line waiting for screening?
Avg. size of the line: Lq .
Use ratio = λ/μ = 0.88 round up to 0.9. Lq = 8.1 patients
c) The clinic has a policy that the average waiting time cannot exceed 30 minutes, are they meeting that requirement?
Avg time to wait per entity: Wq =Lq/λ = 8.1 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 8 𝑝𝑎𝑡𝑖𝑒𝑛𝑡/ℎ𝑜𝑢𝑟 = hour
1.014 hour 𝑥 60 𝑚𝑖𝑛𝑢𝑡𝑒 1 ℎ𝑜𝑢𝑟 = 60.8 minute.
If demand increases by
20%, they will no meet
this requirement

15. Sensitivity Analysis New table
Re-Answer the following
d) What is the probability there are 4 or more patients waiting?
P(patients waiting ≥ 4) = P4 + P5 + …. + P
= 1 – (P0 + P1 + P2 + P3 )
New table
P0 = 1 – λ/μ ;
Pn = (λ/μ)n P0
= 1 – (0.376) = 0.624= 62.4%
Parameter
Value P0
0.111 P1
0.099 P2
0.088 P3
0.078

16. Sensitivity Analysis P(patients waiting ≤ 1) = P0 + P1
Re-Answer the following
e) What is the probability there are 1 or 0 patients waiting?
P(patients waiting ≤ 1) = P0 + P1
= = = 21.0%
Parameter
Value P0
0.111 P1
0.099 P2
0.088 P3
0.078

17. Increased in demand by 20%
Summary of Results
Parameter
Current
demand
Increased in demand by 20%
Nurse utilization
74.1%
88.9%
Average Line
3.2 patients
8.1 patients
Average Wait Time
28.8 minutes
60.8 minutes
Meet 30 min. requirement
Yes No
Prob (4 or more waiting)
30.2%
62.4%
Prob ( 1 or 2 waiting)
45.1%
21.0%
A 20% increase in demand results in more than
double the average waiting time!!!!!