California State Polytechnic University, Pomona
Department of Industrial and Manufacturing Engineering
IE 417 (Operations Research II)
Dr. Sima Parisay
Winter 2011
February 10, 2011
The M/M/s/GD/inf/inf Queuing System
(Multiple Servers)
Chapter 20.6 Extra Credit
Team #9 Members:
Kristianto Andersen
Kevin Meono
Kaveh (Kevin) Shamuilian
Contents
General Notation& Equations 3
Ch. 20.6 - Table 6 5
Ch. 20.6 (Page 1089) Example 9: Bank Tellers 6
Ch. 20.6 Problem 1 (Page 1094) – Grocery Store 8
Ch. 20.6 Problem 4 (Page 1094) – MacBurger 13
WinQSB calculations 17
Sensitivity analysis 25
Suggested sensitivity analysis from Prof Parisay 26
Summary Tables 28
Report to Manager 29
General Notation& Equations
M/M/s/GD/inf/inf system
Inter-arrival times = exponential
Service times = exponential
j = # of customers in system
s = # of servers
If j = s, all customers are being served. The next customer to arrive will wait in the queue
If j > s, all servers are busy, and a line has been formed. If this is the case, the arriving customers will wait in the queue.
If j < s, there are some idle servers in the system, and any arriving customer may be served immediately.
If ρ < 1, steady state exists
If ρ ≥1, steady state does not exist and the system will “Blow up”
Important Equations:
ρ= λsμ Note: this is not utilization
Steady State Probabilities
Probability that all servers are busy:
Average # of customers in the queue:
Average waiting time in the queue:
Average total number in system:
Another way of writing average total number in system:
Average service time:
Average number in service:
Average total time a customer spends in the system:
Ch. 20.6 - Table 6
Ch. 20.6 (Page 1089) Example 9: Bank Tellers
Problem Statement:
Consider a bank with two tellers. An average of 80 customers per hour arrive at the bank
and wait in a single line for an idle teller. The average time it takes to serve a customer is
1.2 minutes. Assume that inter-arrival times and service times are exponential. Determine
1 .The expected number of customers present in the bank
2. The expected length of time a customer spends in the bank
3. The fraction of time that a particular teller is idle
Example 9 (Page 1089) Solution:
This is an M/M/2/GD/inf/inf system
λ=80 customer/hour
μ=50 customer/hour
.80 < 1, so steady state DOES exist
From Table 6,
PJ ≥2= .71
Ch. 20.6 Problem 1 (Page 1094) – Grocery Store
Problem Statement
A supermarket is trying to decide how many cash registers to keep open. Suppose an average of 18 customers arrive each hour, and the average checkout time for a customer is 4 minutes. Interarrival times and service times are exponential, and the system may be modeled as an M/M/s/GD/inf/inf∞ queuing system. It costs $20 per hour to operate a cash register, and a cost of 25¢ is assessed for each minute the customer spends in the cash register area. How many registers should the store open?
Solution – Problem 1
System: M/M/S/GD/∞/∞
Given Information (Input):
λ (inter arrival rate)=18 customerhour
Service cost: 20 $/hour for each cash register
μ (service rate) = 604 = 15 customershour
Waiting cost: .25 $/minute customer = 15 $/hour customer
For steady state, we want: ρ ( traffic intensity) = λ (cust/hr)sμ < 1 ,
How many registers should the store open?
Find information (Output):
Expected Delay cost/customer = 15($/ hours customer) Wq
Expected Delay cost/hour= (18 customerhour)(15$customer hour )(Wq)= 270($/hr^2)Wq
Service cost= $20/server
What is the least amount of cash registers needed?
Steady state = λsμ= 18 cust/hrs(15cust/hr(server))≤ 1 , so let s > 1.2 Server ßHere we solve for s
So the steady state at 1.2 server means that we need at least 2 servers (always round up) because anything less than 1.2 servers would result in an unsteady state and lead to a blow up.
Cost of having 2 servers
ρ= λ2µ= 18 cust/hr2 server15cust/hr(server)= .6
Lq= Pj≥sρ1-ρ= .45*.61-.6= .675 customer
Wq=Lqλ= .675 cust18cust/hr= .0375 hours
Expected Delay CostHour= 270 Wq= (270$ hour^2) (.0375 hours) = $10.125/ hour
Total expected costHour= $20/hr server (2 servers) + $10.125/ hour = $50.125/ hour
WinQSB Input:
WinQSB Output :
Cost of having 3 servers
ρ= λsµ= 18cust/hr3serv(15cust/hr(serv))= .4
Lq= Pj≥sρ1-ρ= .14*.41-.4= .093 customer
Wq=Lqλ= .093 customer18 customer/hr= .0052 hr
Expected Delay CostHour= 270Wq= (270$ hour^2) (.0052hours) = 1.4 $/hr
Total expected costHour= 20s + 1.4= $20 hr/server (3server) + $1.4/ hour= $61.4/ hour
WinQSB Input:
WinQSB Output:
Since opening 3 servers would cost more than having 2 registers and will result in lower utilization, one would only open 2 registers and the supermarket should only open 2 cashiers.
Ch. 20.6 Problem 4 (Page 1094) – MacBurger
M/M/S/GD//
µ = 50 customers per hour
l= 100 customers per hour
Service cost = $ 5
Cost waiting in line = $ 20
1 using 1 server
In this scenario we can’t have one server
If there’s only 1 server
= l / sm = 100 customer per hours /( 1 server * 50 customer per (hours server) ) = 2
This process is not in steady state condition which means that the delay will increase over time
Steady state condition does not hold.
2 servers
In this scenario we can’t have 2 servers
If there’s 2 servers
= l / m = 100 customer per hours /(2 server *50 customer per (hours server)) = 1
Steady state condition does not hold.
3 servers
In this scenario we can have 3 servers
= l / sm = 100 customer per hours/ (3 (server)*50 (customer per (hours server)) = .66667
Total cost per hour = Expected service cost per hour+ Expected delay cost per hour
Expected service cost per hour:
Number of server * cost for a server per hour
3*$ 5/ hour = $15/ hour
Expected delay cost per hours = expected customer in line* delay cost
= 11+.667+ .889+4 = .152 (probability there will be 0 customer is 15.2%)
= 3* .6673 ..12533 (1-.667) = .405
(Probability there will be more customer than server is 40.5%)
Wq= .405% / ((3server*50 (customer per (hours server)) -100 customer per hours)
= .0081hr per customer
Since there are 100 people coming each hours
Expected delay cost = (Wq) (l) cost of delays per hours
= .0081hr per customer (100 customer/ hour) ($20/ hour) = $ 16. 2/ hour
Total cost = $15 / hour + $16.2/ hour = $31.2/ hour
Since the expected delay cost is more than total fixe cost it means (be specific. How much for what) that it would be beneficial to add more servers in the system
4 servers
= l / sm = 100 customer per hours /(4 (server)*50 (customer per (hours server)) = .5
Number of server * cost for a server per hour
4*$ 5/ hour = $20/ hour
Expected delay cost per hours = expected customer in line* delay cost
Using table 6 we can find out right away that
= .17
(Probability there will be more customer than server is 17%)
Wq = .17/ ((4*50 customer per hours) -100 customer per hours) = .0017 hr per customer
Since there are 100 people coming each hours
Expected delay cost = (Wq) (l) cost of delays per hours
= .0017 hours per customer (100 customers per hour) ($20 per hour) = $ 3.4 per hour
Total cost = $23.4 per hour
Having 4 servers is the best solution (had to talk in terms of total cost) to reduce cost since its expected service cost already less than the fix cost. If we add another server; the total fix cost itself is already (5*$5 per hour = $25 per hour). The fix cost to have 5 servers is more expensive than the total cost of having 4 servers (rewrite)
WinQSB calculations
Using 1 server (no need to try unsteady scenarios)
WInQSB input
WinQSB output
Explain what you are trying to do here.
This analysis shows that if traffic intensity is more than 1; the delay will increase over time which makes the equation shown on the supplementary notes and textbook not applicable. WinQSB solves this problems using simulation to get the number of customer in line (Lq) so that this problems can be solved. In addition to that; the cost of having traffic intensity more than 1 is so enormous in which case should be avoided.
Using 2 servers (no need for this)
WInQSB input
WinQSB output
Traffic intensity is still more than 1 in which cause huge cost in customer waiting in line
3 servers
WinQSB input
WinQSB output
Probability summary
When traffic intensity is less than 1; now the total cost significantly a lot (not clear) . In this case; by having 3 servers; the company has a total cost of $32.78 per hour. The utilization becomes 66.6% and the delay cost becomes 17.78. The calculation delay cost is $16.2 per hour; but its difference is less than 10%. The only reason we can come up with what? is we round up the number which make the calculation slightly off from its actual value.
In the probability table it is shown that the probability of having more than 3 customers in the system is 29.6 percent. This data is also important for the manager to think about when it comes time for decision making. Explain how it is important.
4 servers
WinQSB input
WinQSB output
Probability summary
When traffic intensity is less than 1; now the total cost will drop a lot. In this case; by having 4 servers; the company has a total cost of $23.48 per hour. The utilization becomes 50% and the delay cost becomes $3.48 per hour.
Having 4 servers is the best solution to reduce cost since its expected service cost already less than the fix cost. If we add another server; the total fix cost itself is already (5*$5 per hour = $25 per hour). The fix cost to have server is more expensive than the total cost of having 4 servers.
In the probability table it is shown that the probability of having more than 4 customers in the system is 8.7 percent. This data is also important for the manager to think about when decision making.
Sensitivity analysis using WinQSB
Changes in waiting cost (could include input window)
In this sensitivity analysis; we want to change the waiting cost per hours between 18 to 25 dollars when we have 4 servers. Based on the results; it shows that the total waiting cost can never exceed $5 per hour. This means having 4 servers is the best choice if the actual waiting cost per hours is near $20 per hour
Suggested sensitivity analysis from Prof Parisay
We want to show what will happen if the cost of server per hours change. Will having 4 servers is still the best option? We actually want to present this on the presentation; but we face difficulties doing it. The problem with doing this is that WinQSB assign fixed cost into 2 types. (good explanation)
· Idle server cost
· Busy server cost
When we want to do the sensitivity analysis; WinQSB only allowed us to change 1 input.
The way to make this sensitivity analysis is by adding the waiting cost with the new fix cost per ours using Microsoft excel. We can find the total fixed cost by doing sensitivity analysis to the number of server. As we can see below; having 1 or 2 servers will make the system blow up. In this case the sensitivity analysis said that the system is unstable
After finding the total fixed cost for specific server; now is just adding the total fixed cost with the desirable total fixed cost.
Example calculation (server cost per hours = $4per hour)
After all the data is calculated; the summary table for change in fixed cost will be shown as a table below
The sensitivity analysis will be similar with what Prof Parisay drew
Summary Tables
By having 3 servers; the analysis leads to the following results:
Performance measure / Observe value / Acceptable valueL / 2.89 / 5
Lq / .89 / 4
W / .0289 / .25 (15 min)
Π(j>s) / 29.6% / 25%
Wq / .0089 / .167 ( 10min)
Util / 66% / 70- 80%
Total cost per hours / $32.78 / $45
By having 4 servers; the company has these results.
Performance measure / Observe value / Acceptable valueL / 2.179 / 5
Lq / .179 / 4
W / .02017 / .25 (15 min)
Π(j>s) / 8.7% / 25%
Wq / .00017 / .167 ( 10min)
Util / 50% / 70- 80%
Total cost per hours / $23.48 / $45
Report to Manager
Despite the fact that having 4 servers makes the server idle more; having 4 servers is the best solution. It results in the company having the least amount of cost per hour ($23.4 per hour). By having 4 servers the company will be able to fulfill other company policies. If the actual waiting cost per hour deviates from $20 per hour; having 4 servers is still the best choice. If for some reason the cost of one server deviates from $5 per hour. The sensitivity analysis graph still shows that having 4 servers is still the best option In addition to that; having 4 servers fulfills the company policies (acceptable values) that require the probability of having more customers than servers should be less than 25% of the time.