1 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3 Terminology: The characteristics of a queuing system is captured by five parameters:

1Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3

Terminology: The characteristics of a queuing system is captured by five parameters: Arrival pattern Service pattern Number of server Restriction on queue capacity The queue discipline

Terminology and Classification of Waiting Lines


M/M/1 Exponential interarrival times Exponential service times There is one server. No capacity limit

M/G/12/23 Exponential interarrival times General service times 12 servers Queue capacity is 23

Terminology and Classification of Waiting Lines


Coefficient of Variations

Interarrival Time or Processing Time distribution General Poisson Exponential Constant

Mean Interarrival Time or Processiong Time m m m m

Standard Deviation of interarrival or Processing Time s m m 0

Coeffi cient of Varriation of Interarrival or Processing Time s/m 1 1 0


Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has Exponential distribution. On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processor (with the

server)? On average how many customers are with the server? On average how many customers are in the system? On average how long a customer stay in the system?

Problem 1: M/M/1 Performance Evaluation


R = 6 customers per hourRp =1/5 customer per minute, or 60(1/5) = 12/hour r = R/Rp = 6/12 = 0.5On average how many customers are in the waiting

line?

M/M/1 Performance Evaluation

21

22)1(2pi

c

i

CCI

2

11

5.01

22)11(25.0

iI 5.05.0

5.0 2

iI


GAP Example

ii IRT 5.06 iT hour 6/5.0iT

minutes 5 )6/5.0(60 iT

How long a customer stays in the line?

How long a customer stays in the processor (with the server)?

minutes 5 pT

On average how many customers are with the server?

?pI 5.0 ely,Alternativ pI5.0)10/1)(5( TpRIp



min 1055 pi TTT

?I 15.05.0 pi III

On average how many customers are in the system?

On average how long a customer stay in the system?


Problem 2: M/M/1 Performance Evaluation

What if the arrival rate is 11 per hour?

08.10

1211

1

1211

1

2

2

iI

ii IRT 08.1011 iT

min 55or hours 91667.011/08.10 iT

12

11

Rp

R


As the utilization rate increases to 1 (100%) the number of customers in line (system) and the waiting time in line (in system) is increasing exponentially.


c R Rp Ii Ti (min) Tp T Ip I1 6 12 0.50 0.50 5 5 10 0.50 11 7 12 0.58 0.82 7 5 12 0.58 1.41 8 12 0.67 1.33 10 5 15 0.67 21 9 12 0.75 2.25 15 5 20 0.75 31 10 12 0.83 4.17 25 5 30 0.83 51 11 12 0.92 10.08 55 5 60 0.92 111 11.1 12 0.93 11.41 61.7 5 66.7 0.93 12.331 11.2 12 0.93 13.07 70 5 75 0.93 141 11.3 12 0.94 15.20 80.7 5 85.7 0.94 16.141 11.4 12 0.95 18.05 95 5 100 0.95 191 11.5 12 0.96 22.04 115 5 120 0.96 231 11.6 12 0.97 28.03 145 5 150 0.97 291 11.7 12 0.98 38.03 195 5 200 0.98 391 11.9 12 0.99 118.01 595 5 600 0.99 119


A local GAP store on average has 10 customers per hour for the checkout line. The inter-arrival time follows the exponential distribution. The store has two cashiers. The service time for checkout follows a normal distribution with mean equal to 5 minutes and a standard deviation of 1 minute.

On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processors (with the

servers)? On average how many customers are with the servers? On average how many customers are in the system ? On average how long a customer stay in the system ?

Problem 3: M/G/c


Arrival rate: R = 10 per hourAverage interarrival time: Ta = 1/R = 1/10 hr = 6 minStandard deviation of interarrival time: SaService rate per server: 12 per hourAverage service time: Tp = 1/12 hours = 5 minStandard deviation of service time: Sp = 1 min

Coefficient of variation for interarrivals : Ci= Sa /Ta = 1

Coefficient of variation for services: Cp = Sp /Tp = 1/5 =0.2

Number of servers: c =2Rp = c/Tp = 2/(1/12) = 24 per hourρ = R/Rp = 10/24 = 0.42

The Key Information


M/G/2

107.02

2.01

42.01

42.0

21

22)12(222)1(2

pic

i

CCI

ii IRT 107.010 iT

minutes 0.6hour 0107.0 iT

On average how many customers are in the waiting line?

How long a customer stays in the line?

How long a customer stays in the processors (with the servers)?

Average service time: Tp = 1/12 hours = 5 min


M/G/2

?T min 6.556.0 T

On average how many customers are with the servers?

?pI

84.0)42.0(2 ely,Alternativ cI p

84.0)10)(12/1( TpRIp

On average how many customers are in the system ?

?I 95.084.0107.0 pi III

On average how long a customer stay in the system ?


Approximation formula gives exact answers for M/M/1 system.

Approximation formula provide good approximations for M/M/2 system.

Comment on General Formula


A call center has 11 operators. The arrival rate of calls is 200 calls per hour. Each of the operators can serve 20 customers per hour. Assume interarrival time and processing time follow Poisson and Exponential, respectively. What is the average waiting time (time before a customer’s call is answered)?

Problem 4: M/M/c Example

91.0220

200 1iC 1pC


M/M/c Example

21

22)1(2pi

c

i

CCI

89.62

11

91.01

91.0 )12(2

iI

ii RTI iT20089.6

min 2.1or hour 0345.0iT


Suppose the service time is a constant What is the answer of the previous question?

In this case

Problem 5: M/D/c Example

0pC

21

22)1(2pi

c

i

CCI

45.3

2

01

91.01

91.0 12*2

iI

ii RTI iT20045.3

min 1.03or hour 017.0iT


Effect of Pooling

Ri

Server 1

Queue

Server 2

Ri

Server 2Queue 2

Ri/2

Server 1Queue 1

Ri/2

Ri =R= 10/minTp = 5 secs Interarrival time PoissonService time exponential


Effect of Pooling : 2M/M/1

Ri

Server 2Queue 2

Ri/2

Server 1Queue 1

Ri/2

Ri /2 = R= 5/minTp = 5 secs C = 1 Rp = 12 /min= 5/12= 0.417

3.0417.01

417.0 )11(2

iI

ii RTI iT53.0 sec 3.6or min 06.0iT

sec 6.856.3 pi TTT


Comparison of 2M/M/1 with M/M/2

Ri

Server 2Queue 2

Ri/2

Server 1Queue 1

Ri/2

3.0417.01

417.0 )11(2

iI

sec 6.856.3 pi TTT

6.0)3.0(22 iI

Server 1

Queue

Server 2

Ri


Effect of Pooling: M/M/2

Server 1

Queue

Server 2

Ri

Ri =R= 10/minTp = 5 secs C = 2Rp = 24 /min= 10/24= 0.417 AS BEFORE for each processor

2.0417.01

417.0 )12(2

iI

ii RTI iT102.0 sec 1.2or min 02.0iT

sec 2.652.1 pi TTT


Under Design A, We have Ri = 10/2 = 5 per minute, and TP= 5 seconds, c

=1, we arrive at a total flow time of 8.6 seconds Under Design B,

We have Ri =10 per minute, TP= 5 seconds, c=2, we arrive at a total flow time of 6.2 seconds

So why is Design B better than A? Design A the waiting time of customer is dependent on

the processing time of those ahead in the queue Design B, the waiting time of customer is only partially

dependent on each preceding customer’s processing time

Combining queues reduces variability and leads to reduce waiting times

Effect of Pooling


1. Decrease variability in customer inter-arrival and processing times.

2. Decrease capacity utilization.3. Synchronize available capacity with demand.

Performance Improvement Levers


Customers arrival are hard to control Scheduling, reservations, appointments, etc….

Variability in processing time Increased training and standardization processes Lower employee turnover rate more experienced work

force Limit product variety, increase commonality of parts

1. Variability Reduction Levers


If the capacity utilization can be decreased, there will also be a decrease in delays and queues.

Since ρ=R/Rp, to decrease capacity utilization there are two options Manage Arrivals: Decrease inflow rate Ri

Manage Capacity: Increase processing rate Rp

Managing Arrivals Better scheduling, price differentials, alternative

services Managing Capacity

Increase scale of the process (the number of servers) Increase speed of the process (lower processing time)

2. Capacity Utilization Levers


Capacity Adjustment Strategies Personnel shifts, cross training, flexible resources Workforce planning & season variability Synchronizing of inputs and outputs, Better

scheduling

3. Synchronizing Capacity with Demand

1 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3 Terminology: The characteristics of a queuing system is captured by five parameters:

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