Chapter 16 Capacity Planning and Queuing Models

McGraw-Hill/Irwin Service Management: Operations, Strategy, and Information Technology, 6e Copyright © 2008 by The McGraw-Hill Companies, Inc. All Rights Reserved.

Learning Objectives

     Discuss the strategic role of capacity planning.

Describe a queuing model using A/B/K notation.

Use queuing models to calculate system performance measures.

Describe the relationships between queuing system characteristics.

Use queuing models and various decision criteria for capacity planning.

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Capacity Planning Challenges

    Inability to create a steady flow of demand to fully utilize capacity Idle capacity always a reality for services.

Customer arrivals fluctuate and service demands also vary.

Customers are participants in the service and the level of congestion impacts on perceived quality.

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Strategic Role of Capacity Decisions

    Using long range capacity as a preemptive strike where market is too small for two competitors (e.g. building a luxury hotel in a mid-sized city) Lack of short-term capacity planning can generate customers for competition (e.g. restaurant staffing) Capacity decisions balance costs of lost sales if capacity is inadequate against operating losses if demand does not reach expectations.

Strategy of building ahead of demand is often taken to avoid losing customers. 16-4

Analytical Queuing Models

 A/B/K System A = Arrival Process B = Service Process K = number of servers Where    M = Exponential inter-arrival or service times (or Poisson arrival or service rates) D= Deterministic or constant service times G= General service times

Analytical Queuing Models (Continued)

 A/B/K System assumes the following - FCFS - no reneging, no balking - customers form one line - the arrival rate is less than the service rate - the arrival and service rates do not change - the system is in a steady state

Single Server : M/M/1 Poisson Arrival and Service Rates

1. Mean arrival rate: 2. Mean service rate:   3. Mean number in service:     4. Probability of exactly “n” customers in the system: 5. Probability of “k” or more customers in the system:

P n

6. Mean number of customers in the system: 7. Mean number of customers in queue: 8. Mean time in system:

Ws

 1

Lq Ls

         

n

( 1  

k

)   ) 

k

9. Mean time in queue:

Wq

    16-7

Example 1: White & Sons

White & Sons wholesale fruit distributions employ a single crew whose job is to unload fruit from farmer’s trucks. Trucks arrive at the unloading dock at an average rate of 5 per hour Poisson distributed. The crew is able to unload a truck in approximately 10 minutes with exponential distribution.

1. On the average, how many trucks are waiting in the queue to be unloaded ?

2.

What is the average number of trucks in the system? 3.

What is the average waiting time in queue and average time spent in the system?

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Example 1: White & Sons (Cont.)

5.

What is the probability that there is no truck in the system?

6. The management has received complaints that waiting trucks have blocked the alley to the business next door. If there is room for 2 trucks at the loading dock before the alley is blocked, how often will this problem arise?

7. What is the probability that an arriving truck will find space available at the unloading dock and not block the alley?

M/G/1, M/M/1 and M/D/1 Single Server Queues with Poisson Arrival Rate

Single Server General Service Distribution Model: M/G/1

Lq

  2   s  2  ) 2 where s  standard deviation of the service times Single Server Exponential Service Distribution Model: M/M/1

Lq

  ( 1  2  ) where s = 1/ Single Server Constant Service Distribution Model: M/D/1

Lq

  2 ( 1 2   ) 16-10

M/M/1: Congestion as

  10 100 10 8 6 4 2 0 0 1.0 With: Then:

L s

 1    

L s

0 0 0.2 0.25

0.5 1 0.8 4 0.9 9 0.99 99 16-11

Relationships among system characteristics M/G/1 and M/M/c:

Mean number of customers in queue for two servers: M/M/2

Lq

 4   3  2 Relationships among system characteristics M/G/1 and M/M/c:

Ls



Lq

 

Ws



Wq

 1 

Ws Wq

  1  1 

Ls Lq

Queuing System Cost Tradeoff

Let: C w = Cost of one customer waiting in queue for an hour C s = Hourly cost per server C = Number of servers Total Cost/hour = Hourly Service Cost + Hourly Customer Waiting Cost Total Cost/hour = C s C + C w L s 16-13

Example 2: Foto-Mat

On average 2 customers arrive per hour at a Foto-Mat to process film, Poisson distributed. There is one clerk in attendance that on average spends 15 minutes per customer, exponentially distributed.

1. What is the average length in the system?

2. If the clerk is paid $4 per hour and a customer’s waiting cost in queue is considered $6 per hour. What is the total system cost per hour?

3. What would be the total system cost per hour, if a second clerk who works separately, were added? Assume a single queue were used.

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Capacity Analysis of Robot Maintenance and Repair Service

1.

A large service facility’s robots periodically break down and require service. The average time between breakdowns is three days and is exponentially distributed. The average time to repair a robot is two days and also exponentially distributed. Currently, one mechanic repairs the robots in the order in which they fail. The cost of having a robot out of service is $500 per day and mechanic is paid $100 per day. What is the average number of robots out of service?

2. What is the total cost of the current system?

3.

Management is considering a preventive maintenance (PM) program at a daily cost of $100 which will extend the average breakdowns to six days. Would you recommend this program? Explain.

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Capacity Analysis of Robot Maintenance and Repair Service (Continued)

Now assume that all the robots were replaced and the average breakdowns is 6 days, exponentially distributed. The breakdown and mechanic costs remain the same. Answer the following questions.

4. Suppose another mechanic can be hired at the same cost? If s/he is hired, then the service rate is doubled. Both mechanics work as a team and the service time for the team is exponentially distributed. Would you hire the second mechanic? Explain.

5. What is the optimal number of mechanics to hire, if you can hire additional workers to work as a team? Assume exponential service times.

6.

Reconsider Q4 but now the two mechanics work separately, where each mechanic works alone and there are two queues. Which system is better? Explain.

7. Consider Q6. What is the optimal decision if there is only one queue?

Determining Number of Mechanics to Serve Robot Line

Q 4, 5. Assume mechanics work together as a team  

M

2

Ls

    in Crew (M) Mechanic cost Robot idleness Cost per day   1 1/2 2 1 1  3 3/2    16-17

Determining Number of Mechanics to Serve Robot Line

Q6. Assume Robots divided equally among mechanics working alone Identical $100 n $500 Ls (n) Total System    /

n

cost

Ls

   /

n

1 1/ 6 $100 $250 $350 2 1/ 12 $200 500 (1/5) 2=$200 $400 / / 16-18

Determining Number of Mechanics to Serve Robot Line

Q7. Assume two mechanics work alone from a single queue.

Note:     1 3

Ls

 4    3 2   = 0.01 + 0.33

= 0.34

Total Cost/day = 100(2) + 500(.34) = 200 + 170 =$370 16-19

Comparisons of System Performance for Two Mechanics

System

Ls Ws



Ls



Wq



Ws

 1  Single Queue with Team Service 6/ 5 =1.2 days 0.2 days Single Queue Servers 4   3  2 Multiple Queue    and Multiple 12 (1/5) =2.4 days 0.4 days Servers 16-20

General Queuing Observations

1. Variability in arrivals and service times contribute equally to congestion as measured by L q

.

2. Service capacity must exceed demand.

3. Servers must be idle some of the time.

4. Single queue preferred to multiple queue unless jockeying is permitted.

5. Large single server (team) preferred to multiple-servers if minimizing mean time in system, W S.

6. Multiple-servers preferred to single large server (team) if minimizing mean time in queue, W Q.

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L

q

for Various Values of C and

  0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1 C=1 C=2 C=3 C=4 0.026

0.001

0.05

0.083

0.129

0.188

0.002

0.004

0.007

0.011

0.267

0.368

0.5

0.672

0.017

0.024

0.033

0.045

0.002

0.003

0.004

0.006

0.9

1.207

1.633

2.25

0.059

0.077

0.098

0.123

0.008

0.011

0.015

0.019

3.2

4.817

8.1

18.05

0.152

0.187

0.229

0.277

0.333

0.024

0.03

0.037

0.045

0.003

0.004

0.005

0.007

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