waiting line models (1) pertemuan 11 matakuliah: k0442-metode kuantitatif tahun: 2009

Waiting Line Models (1) Pertemuan 11

Matakuliah : K0442-Metode Kuantitatif

Tahun : 2009

Material Outline

• Structure of a Waiting Line System• Single-Channel Waiting Line Model

with Poisson Arrivals and Exponential Service Times

• Multiple-Channel Waiting Line Model with Poisson Arrivals and Exponential Service Times

• Queuing theory is the study of waiting lines. • Four characteristics of a queuing system are:

– the manner in which customers arrive– the time required for service– the priority determining the order of service– the number and configuration of servers in the

system.

Structure of a Waiting Line System


• Queue Discipline– Most common queue discipline is first come, first served

(FCFS). – An elevator is an example of last come, first served (LCFS)

queue discipline.– Other disciplines assign priorities to the waiting units and

then serve the unit with the highest priority first.

• Distribution of Arrivals– Generally, the arrival of customers into the system is a

random event. – Frequently the arrival pattern is modeled as a Poisson

process.

• Distribution of Service Times– Service time is also usually a random variable. – A distribution commonly used to describe service time is

the exponential distribution.


• Single Service Channel

• Multiple Service Channels

SS11SS11

SS11SS11

SS22SS22

SS33SS33

Customerleaves

Customerleaves

Customerarrives

Customerarrives

Waiting line

Waiting line

System

System

M/M/1 Queuing System• Single channel• Poisson arrival-rate distribution• Exponential service-time distribution• Unlimited maximum queue length• Infinite calling population• Examples:

– Single-window theatre ticket sales booth– Single-scanner airport security station

Example: SJJT, Inc. (A)

• M/M/1 Queuing SystemJoe Ferris is a stock trader on the floor of the

New York Stock Exchange for the firm of Smith, Jones, Johnson, and Thomas, Inc. Stock transactions arrive at a mean rate of 20 per hour. Each order received by Joe requires an average of two minutes to process.

MM//MM/1 Queuing System/1 Queuing SystemOrders arrive at a mean rate of 20 per hour or one

order every 3 minutes. Therefore, in a 15 minute interval the average number of orders arriving will be = 15/3 = 5.

Example: SJJT, Inc. (A)• Arrival Rate Distribution

Question

What is the probability that no orders are received within a 15-minute period?Answer

P (x = 0) = (50e -5)/0! = e -5 = .0067

Example: SJJT, Inc. (A)• Arrival Rate Distribution

Question

What is the probability that exactly 3 orders are received within a 15-minute period?Answer P (x = 3) = (53e -5)/3! = 125(.0067)/6 = .1396

• Arrival Rate DistributionQuestion

What is the probability that more than 6 orders arrive within a 15-minute period? Answer P (x > 6) = 1 - P (x = 0) - P (x = 1) - P (x = 2)

- P (x = 3) - P (x = 4) - P (x = 5)

- P (x = 6) = 1 - .762 = .238


• Service Rate DistributionQuestion

What is the mean service rate per hour?

AnswerSince Joe Ferris can process an order in an

average time of 2 minutes (= 2/60 hr.), then the mean service rate, µ, is µ = 1/(mean service time), or 60/2.

= 30/hr.

Example: SJJT, Inc. (A)• Service Time Distribution

QuestionWhat percentage of the orders will take less

than one minute to process?

AnswerSince the units are expressed in hours, P (T < 1 minute) = P (T < 1/60 hour).

Using the exponential distribution, P (T < t ) = 1 - e-

µt. Hence, P (T < 1/60) = 1 - e-30(1/60)

= 1 - .6065 = .3935 = 39.35%

Example: SJJT, Inc. (A)• Service Time Distribution

QuestionWhat percentage of the orders will be

processed in exactly 3 minutes?

AnswerSince the exponential distribution is a

continuous distribution, the probability a service time exactly equals any specific value is 0.


• Service Time DistributionQuestion

What percentage of the orders will require more than 3 minutes to process?

AnswerThe percentage of orders requiring more

than 3 minutes to process is: P (T > 3/60) = e-30(3/60) = e -1.5 = .2231 = 2.31%

Example: SJJT, Inc. (A)• Average Time in the System

QuestionWhat is the average time an order must wait

from the time Joe receives the order until it is finished being processed (i.e. its turnaround time)?

AnswerThis is an M/M/1 queue with = 20 per hour

and = 30 per hour. The average time an order waits in the system is: W = 1/(µ - )

= 1/(30 - 20) = 1/10 hour or 6 minutes


• Average Length of QueueQuestion

What is the average number of orders Joe has waiting to be processed?

AnswerAverage number of orders waiting in the queue

is: Lq = 2/[µ(µ - )] = (20)2/[(30)(30-20)]

= 400/300

= 4/3

Example: SJJT, Inc. (A)• Utilization Factor

QuestionWhat percentage of the time is Joe

processing orders?

AnswerThe percentage of time Joe is processing

orders is equivalent to the utilization factor, /. Thus, the percentage of time he is processing orders is:

/ = 20/30 = 2/3 or 66.67%

M/M/k Queuing System• Multiple channels (with one central waiting line)• Poisson arrival-rate distribution• Exponential service-time distribution• Unlimited maximum queue length• Infinite calling population• Examples:

– Four-teller transaction counter in bank– Two-clerk returns counter in retail store

Example: SJJT, Inc. (B)• M/M/2 Queuing System

Smith, Jones, Johnson, and Thomas, Inc. has begun a major advertising campaign which it believes will increase its business 50%. To handle the increased volume, the company has hired an additional floor trader, Fred Hanson, who works at the same speed as Joe Ferris.

Note that the new arrival rate of orders, , is 50% higher than that of problem (A). Thus, = 1.5(20) = 30 per hour.

Example: SJJT, Inc. (B)

• Sufficient Service RateQuestion

Why will Joe Ferris alone not be able to handle the increase in orders?

AnswerSince Joe Ferris processes orders at a

mean rate of µ = 30 per hour, then = µ = 30 and the utilization factor is 1.

This implies the queue of orders will grow infinitely large. Hence, Joe alone cannot handle this increase in demand.

Example: SJJT, Inc. (B)• Probability of n Units in System

QuestionWhat is the probability that neither Joe nor Fred will

be working on an order at any point in time?

AnswerGiven that = 30, µ = 30, k = 2 and ( /µ) = 1, the

probability that neither Joe nor Fred will be working is:

= 1/[(1 + (1/1!)(30/30)1] + [(1/2!)(1)2][2(30)/(2(30)-30)]

= 1/(1 + 1 + 1) = 1/3 = .333

1

0

0

)(!

)/(!

)/(

1k

n

kn

kk

kn

P

1

0

0

)(!

)/(!

)/(

1k

n

kn

kk

kn

P

Example: SJJT, Inc. (B)• Probability of n Units in System (continued)

Answer

Given that = 30, µ = 30, k = 2 and ( /µ) = 1, the probability that neither Joe nor Fred will be working is:

= 1/[(1 + (1/1!)(30/30)1] + [(1/2!)(1)2][2(30)/(2(30)-30)]

= 1/(1 + 1 + 1) = 1/3 = .333

1

0

0

)(!

)/(

!

)/(

1k

n

kn

k

k

kn

P

1

0

0

)(!

)/(

!

)/(

1k

n

kn

k

k

kn

P


• Average Time in System

QuestionWhat is the average turnaround time for

an order with both Joe and Fred working?

• Average Time in System (continued)

Answer

The average turnaround time is the

average waiting time in the system, W.

L = Lq + ( /µ) = 1/3 + (30/30) = 4/3

W = L/(4/3)/30 = 4/90 hr. = 2.67 min.


2

02 2

( ) (30)(30)(30 30) 1( ) (1/ 3)

( 1)!( ) (1!)(2(30) 30) 3

k

qL Pk k

2

02 2

( ) (30)(30)(30 30) 1( ) (1/ 3)

( 1)!( ) (1!)(2(30) 30) 3

k

qL Pk k

Example: SJJT, Inc. (B)• Average Length of Queue

QuestionWhat is the average number of orders

waiting to be filled with both Joe and Fred working?

AnswerThe average number of orders waiting to be

filled is Lq. This was calculated earlier as 1/3.


• Creating Special Excel Function to Compute P0

Select the Tools pull-down menuSelect the Macro optionChoose the Visual Basic EditorWhen the Visual Basic Editor appears Select the Insert pull-down menu Choose the Module option

When the Module sheet appears Enter Function Po (k,lamda,mu) Enter Visual Basic program (on next slide)

Select the File pull-down menuChoose the Close and Return to MS Excel option


• Visual Basic Module for P0 Function

Function Po(k, lamda, mu)Sum = 0For n = 0 to k - 1 Sum = Sum + (lamda/mu) ^ n / Application.Fact(n)NextPo = 1/(Sum+(lamda/mu)^k/Application.Fact(k))*

(k*mu/(k*mu-lamda)))End Function

Example: SJJT, Inc. (C)• Economic Analysis of Queuing Systems

The advertising campaign of Smith, Jones, Johnson and Thomas, Inc. (see problems (A) and (B)) was so successful that business actually doubled. The mean rate of stock orders arriving at the exchange is now 40 per hour and the company must decide how many floor traders to employ. Each floor trader hired can process an order in an average time of 2 minutes.

Example: SJJT, Inc. (C)• Economic Analysis of Queuing Systems

Based on a number of factors the brokerage firm has determined the average waiting cost per minute for an order to be $.50. Floor traders hired will earn $20 per hour in wages and benefits. Using this information compare the total hourly cost of hiring 2 traders with that of hiring 3 traders.

Example: SJJT, Inc. (C)• Economic Analysis of Waiting Lines

Total Hourly Cost = (Total salary cost per hour) + (Total hourly cost for orders in the system) = ($20 per trader per hour) x (Number of traders)

+ ($30 waiting cost per hour) x (Average number of orders in the system) = 20k + 30L.

Thus, L must be determined for k = 2 traders and for k = 3 traders with = 40/hr. and = 30/hr. (since the average service time is 2 minutes (1/30 hr.).

Example: SJJT, Inc. (C)• Cost of Two Servers

P0 = 1 / [1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))]

= 1 / [1 + (4/3) + (8/3)]

= 1/5

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

Example: SJJT, Inc. (C)• Cost of Two Servers (continued)

Thus,

L = Lq + ( /µ) = 16/15 + 4/3 = 2.40

Total Cost = (20)(2) + 30(2.40) = $112.00 per hour

2

02 2

( ) (40)(30)(40 30) 16( ) (1/ 5)

( 1)!( ) (1!)(2(30) 40) 15

k

qL Pk k

2

02 2

( ) (40)(30)(40 30) 16( ) (1/ 5)

( 1)!( ) (1!)(2(30) 40) 15

k

qL Pk k

Example: SJJT, Inc. (C)

• Cost of Three Servers

P0 = 1/[[1+(1/1!)(40/30)+(1/2!)(40/30)2]+

[(1/3!)(40/30)3(90/(90-40))] ]

= 1 / [1 + 4/3 + 8/9 + 32/45]

= 15/59

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

P

n kk

k

n k

n

k0

0

1

1

( / )!

( / )!

( )

Example: SJJT, Inc. (C)

• Cost of Three Servers (continued)

Thus, L = .1446 + 40/30 = 1.4780

Total Cost = (20)(3) + 30(1.4780) = $104.35 per hour

3

02 2

( ) (30)(40)(40 30)( ) (15/ 59) .1446

( 1)!( ) (2!)(3(30) 40)

k

qL Pk k

3

02 2

( ) (30)(40)(40 30)( ) (15/ 59) .1446

( 1)!( ) (2!)(3(30) 40)

k

qL Pk k

Example: SJJT, Inc. (C)• System Cost Comparison

Wage Waiting Total

Cost/Hr Cost/Hr Cost/Hr 2 Traders $40.00 $82.00

$112.00 3 Traders 60.00 44.35 104.35

Thus, the cost of having 3 traders is less than that of 2 traders.

M/D/1 Queuing System

• Single channel• Poisson arrival-rate distribution• Constant service time• Unlimited maximum queue length• Infinite calling population• Examples:

– Single-booth automatic car wash– Coffee vending machine

Example: Ride ‘Em Cowboy!• M/D/1 Queuing System

The mechanical pony ride machine at the entrance to a very popular J-Mart store provides 2 minutes of riding for $.50. Children wanting to ride the pony arrive (accompanied of course) according to a Poisson distribution with a mean rate of 15 per hour.

What fraction of the time is the pony idle?

= 15 per hour = 60/2 = 30 per hourUtilization = / = 15/30 = .5Idle fraction = 1 – Utilization = 1 - .5 =

.5

• What is the average number of children waiting to ride the pony?

• What is the average time a child waits for a ride?

Example: Ride ‘Em Cowboy!

2 2(15) = = .25 children

2 ( - ) 2(30)(30 - 15)

qL2 2(15)

= = .25 children2 ( - ) 2(30)(30 - 15)

qL

15 = = .01667 hours

2 ( - ) 2(30)(30 - 15)

qW15

= = .01667 hours2 ( - ) 2(30)(30 - 15)

qW

(or 1 minute)(or 1 minute)

M/G/k Queuing System

• Multiple channels• Poisson arrival-rate distribution• Arbitrary service times• No waiting line• Infinite calling population• Example:

– Telephone system with k lines. (When all k lines are being used, additional callers get a busy signal.)

Example: Allen-Booth

• M/G/k Queuing SystemAllen-Booth (A-B) is an OTC market maker. A broker

wishing to trade a particular stock for a client will call on a firm like Allen-Booth to execute the order. If the market maker's phone line is busy, a broker will immediately try calling another market maker totransact the order.

A-B estimates that on the average, a broker will try to call to execute a stock transaction every two minutes.The time required to complete the transaction averages 75 seconds. A-B has four traders staffing its phones. Assume calls arrive according to a Poisson distribution.

Example: Allen-BoothThis problem can be modeled as an M/G/k system with block customers cleared with:

1/ = 2 minutes = 2/60 hour = 60/2 = 30 per hour 1/µ = 75 sec. = 75/60 min. = 75/3600

hr. µ = 3600/75 = 48 per hour


• % of A-B’s Customers Lost Due to Busy Line

First, we must solve for P0

where k = 4

1

P0 = 1 + (30/48) + (30/48)2/2! + (30/48)3/3! +

(30/48)4/4!

P0 = .536

continued

0

0

1

( ) !k

i

i

Pi

0

0

1

( ) !k

i

i

Pi

• % of A-B’s Customers Lost Due to Busy Line

Now,

Thus, with four traders 0.3% of the potential customers are lost.


4

4 0

( ) (30 48)(.536) .003

! 4!

k

P Pk

4

4 0

( ) (30 48)(.536) .003

! 4!

k

P Pk

waiting line models (1) pertemuan 11 matakuliah: k0442-metode kuantitatif tahun: 2009

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