California State Polytechnic University, Pomona Department of Industrial and Manufacturing Engineering

IE 417 (Operations Research II)Dr. Sima Parisay

Winter 2011February 10, 2011

The M/M/s/GD/inf/inf Queuing System(Multiple Servers)

Chapter 20.6 Extra Credit

Team #9 Members:Kristianto Andersen

Kevin MeonoKaveh (Kevin) Shamuilian

ContentsGeneral 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 systemInter-arrival times = exponentialService times = exponential

j = # of customers in systems = # of serversIf j = s, all customers are being served. The next customer to arrive will wait in the queueIf 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 existsIf 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 TellersProblem Statement:Consider a bank with two tellers. An average of 80 customers per hour arrive at the bankand wait in a single line for an idle teller. The average time it takes to serve a customer is1.2 minutes. Assume that inter-arrival times and service times are exponential. Determine

1 .The expected number of customers present in the bank2. The expected length of time a customer spends in the bank3. The fraction of time that a particular teller is idle

Example 9 (Page 1089) Solution:This is an M/M/2/GD/inf/inf systemλ=80customer /hourμ=50customer /hour

.80 < 1, so steady state DOES exist

From Table 6,P (J ≥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 = 15customershour

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= 18cust /hr

s (15cust /hr (server))≤ 1 , so let s > 1.2 Server Here 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= 18cust /hr

2 server (15 ) cust /hr (server)= .6

Lq= P ( j≥ s ) ρ1−ρ = .45∗.6

1−.6 = .675 customer

Wq=Lqλ = .675 cust

18cust /hr= .0375 hours

ExpectedDelay CostHour = 270 Wq= (270

$hour2 ¿ (.0375 hours) = $10.125/ hour

Totalexpected costHour = $20/hr server (2 servers) + $10.125/ hour = $50.125/ hour

WinQSB Input:

WinQSB Output :

Cost of having 3 servers

ρ= λsµ= 18cust /hr

3 serv(15cust /hr (serv ))= .4

Lq= P ( j≥ s ) ρ1−ρ = .14∗.4

1−.4 = .093 customer

Wq=Lqλ = .093 customer

18 customer /hr = .0052 hr

ExpectedDelay CostHour = 270Wq= (270

$hour2 ¿ (.0052hours) = 1.4 $/hr

Totalexpected 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

= 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

ρ = / s = 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

ρ = / = 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

ρ = / s = 100 customer per hours/ (3 (server)*50 (customer per (hours server)) = .66667

ρ= λsμ

<1

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

P( j≥s )=( sρ )s π0

s !(1−ρ)

π0 = 11+ .667+.889+4 = .152 (probability there will be 0 customer is 15.2%)

P( j≥s ) = (3∗.667 )3 ..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) () 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

ρ = / s = 100 customer per hours /(4 (server)*50 (customer per (hours server)) = .5

ρ= λsμ

<1

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

P( j≥s ) = .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) () 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 calculationsUsing 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 WinQSBChanges 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

3 4 5 6 70

10

20

30

40

50

60

fixed cost $3fixed cost $4fixed cost $ 5fixed cost $6fixed cost $7fixed cost $8

Summary TablesBy having 3 servers; the analysis leads to the following results:

Performance measure Observe value Acceptable valueL 2.89 5Lq .89 4W .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 5Lq .179 4W .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 ManagerDespite 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.