Chapter 5: Service Processes. Generally classified according to who the customer is: äFinancial services äHealth care A contrast to manufacturing A service.

Chapter 5: Service Processes

Generally classified according to who the customer is:

Financial services

Health care

A contrast to manufacturing

A service business is the management of organizations whose primary business requires interaction with the customer to produce the service

Service Businesses

Service-System Design Matrix

Mail contact

Face-to-faceloose specs

Face-to-facetight specs

PhoneContact

Face-to-facetotal

customization

Buffered core (none)

Permeable system (some)

Reactivesystem (much)

High

LowHigh

Low

Degree of customer/server contact

Internet & on-site

technology

SalesOpportunity

ProductionEfficiency

Characteristics of Workers, Operations, and Innovations Relative to the Degree

of Customer/Service Contact

Queuing Theory

Waiting occurs in

Service facility Fast-food restaurants post office grocery store bank

Manufacturing

Equipment awaiting repair

Phone or computer network

Product orders

Why is there waiting?

Customer Service Population Sources

Population Source

Finite Infinite

Example: Number of machines needing repair when a company only has three machines.

Example: Number of machines needing repair when a company only has three machines.

Example: The number of people who could wait in a line for gasoline.

Example: The number of people who could wait in a line for gasoline.

Service Pattern

ServicePattern

Constant Variable

Example: Items coming down an automated assembly line.

Example: Items coming down an automated assembly line.

Example: People spending time shopping.

Example: People spending time shopping.

The Queuing System

Queue Discipline

Length

Number of Lines &Line Structures

Service Time Distribution

Queuing System

Examples of Line Structures

Single Channel

Multichannel

SinglePhase Multiphase

One-personbarber shop

Car wash

Hospitaladmissions

Bank tellers’windows

Measures of System Performance

Average number of customers waiting

In the queueIn the system

Average time customers waitIn the queueIn the system

System utilization

Number of ServersSingle Server . . .

Customers ServiceCenter

Multiple Servers

. . .

Customers

ServiceCenters

Multiple Single Servers

. . .

. . .

. . .

Customers ServiceCenters

Some Assumptions

Arrival Pattern: Poisson

Service pattern: exponential

Queue Discipline: FIFO

0 1 2 3 4 5 6 7 8 9 10 11 12 13

Cu stom ers p er time u nit

.02

.04

.06

.08

.10

.12

.14

.16

.18

Relative Frequency

Service Time

Relative Frequency (%)

. . .

CustomersServiceCenter

Some Models

1. Single server, exponential service time (M/M/1)

2. Multiple servers, exponential service time (M/M/s)

A TaxonomyM / M / s

Arrival Service Number ofDistribution Distribution Servers

whereM = exponential distribution (“Markovian”)

Given

= customer arrival rate = service rate (1/m = average service time)s = number of servers

Calculate

Lq = average number of customers in the queue

L = average number of customers in the system

Wq = average waiting time in the queue

W = average waiting time (including service)

Pn = probability of having n customers in the system

= system utilization

Note regarding Little’s Law: L = * W and Lq = * Wq

Model 1: M/M/1 Example

The reference desk at a library receives request for assistance at an average rate of 10 per hour (Poisson distribution). There is only one librarian at the reference desk, and he can serve customers in an average of 5 minutes (exponential distribution). What are the measures of performance for this system? How much would the waiting time decrease if another server were added?

M/M/s Queueing Model Template

Data 10 (mean arrival rate) 12 (mean service rate)s = 1 (# servers)

Prob(W > t) = 0.135335when t = 1

Prob(Wq > t) = 0.112779

0 when t = 1

ResultsL = 5 Number of customers in the system

Lq = 4.166666667 Number of customers in the queue

W = 0.5 Waiting time in the systemWq = 0.416666667 Waiting time in the queue

0.833333333 Utilization

P0 = 0.166666667 Prob zero customers in the system

Application of Queuing Theory

We can use the results from queuing theory to make the following types of decisions:

How many servers to employ

Whether to use one fast server or a number of slower servers

Whether to have general purpose or faster specific serversGoal: Minimize total cost = cost of servers + cost of waiting

Cost ofService Capacity

Cost of customerswaiting

Total Cost

OptimumService Capacity

Cost

Example #1: How Many Servers?In the service department of an auto repair shop, mechanics requiring parts for auto repair present their request forms at the parts department counter. A parts clerk fills a request while the mechanics wait. Mechanics arrive at an average rate of 40 per hour (Poisson). A clerk can fill requests in 3 minutes (exponential). If the parts clerks are paid $6 per hour and the mechanics are paid $18 per hour, what is the optimal number of clerks to staff the counter.

S = 4 IS THE SMALLEST

Data 40 (mean arrival rate) 20 (mean service rate) Resultss = 3 (# servers) L = 2.888888889

Lq = 0.888888889

Prob(W > t) = 2.0383E-08when t = 1 W = 0.072222222

Wq = 0.022222222Prob(Wq > t) = 1.8321E-09

0 when t = 1 0.666666667

P0 = 0.111111111

# Servers Service Cost Waiting Cost Total Cost

3 18.00$ 52.00$ 70.00$ 4 24.00$ 39.13$ 63.13$ 5 30.00$ 36.72$ 66.72$

Service Cost = s * CsWaiting Cost = * W * Cw

Example #2: How Many Servers?

Beefy Burgers is trying to decide how many

registers to have open during their busiest time,

the lunch hour. Customers arrive during the lunch

hour at a rate of 98 customers per hour (Poisson

distribution). Each service takes an average of 3

minutes (exponential distribution). Management

would not like the average customer to wait longer

than five minutes in the system. How many

registers should they open?

Need at least 5 (why?) Increment from there

Choose s = 6 since W = 0.0751 hour is less than 5 minutes.

For six servers

M/M/s Queueing Model Template

Data 98 (mean arrival rate) 20 (mean service rate)s = 6 (# servers)

Prob(W > t) = 1.19E-08when t = 1

Prob(Wq > t) = 2.77E-10

0 when t = 1

ResultsL = 7.359291808 Number of customers in the system

Lq = 2.459291808 Number of customers in the queue

W = 0.075094814 Waiting time in the systemWq = 0.025094814 Waiting time in the queue

0.816666667 Utilization

P0 = 0.00526507 Prob zero customers in the system

Example #3: One Fast Server or Many Slow Servers?

Beefy Burgers is considering changing the way that they serve

customers. For most of the day (all but their lunch hour), they

have three registers open. Customers arrive at an average

rate of 50 per hour. Each cashier takes the customer’s order,

collects the money, and then gets the burgers and pours the

drinks. This takes an average of 3 minutes per customer

(exponential distribution). They are considering having just

one cash register. While one person takes the order and

collects the money, another will pour the drinks and another

will get the burgers. The three together think they can serve a

customer in an average of 1 minute. Should they switch to one

register?

3 Slow Servers

1 Fast Server

W is less for one fast server, so choose this option.

Data 50 (mean arrival rate) 20 (mean service rate) Resultss = 3 (# servers) L = 6.011235955 Number of customers in the system

Lq = 3.511235955 Number of customers in the queue

Prob(W > t) = 6.38E-05when t = 1 W = 0.120224719 Waiting time in the system

Wq = 0.070224719 Waiting time in the queueProb(Wq > t) = 4.34E-05

0 when t = 1 0.833333333 Utilization

P0 = 0.04494382 Prob zero customers in the system

Data 50 (mean arrival rate) 60 (mean service rate) Resultss = 1 (# servers) L = 5 Number of customers in the system

Lq = 4.166666667 Number of customers in the queue

Prob(W > t) = 4.54E-05when t = 1 W = 0.1 Waiting time in the system

Wq = 0.083333333 Waiting time in the queueProb(Wq > t) = 3.78E-05

0 when t = 1 0.833333333 Utilization

P0 = 0.166666667 Prob zero customers in the system

Example 4: Southern RailroadThe Southern Railroad Company has been subcontracting for painting

of its railroad cars as needed. Management has decided the company might save money by doing the work itself. They are considering two alternatives. Alternative 1 is to provide two paint shops, where painting is to be done by hand (one car at a time in each shop) for a total hourly cost of $70. The painting time for a car would be 6 hours on average (assume an exponential painting distribution) to paint one car. Alternative 2 is to provide one spray shop at a cost of $175 per hour. Cars would be painted one at a time and it would take three hours on average (assume an exponential painting distribution) to paint one car. For each alternative, cars arrive randomly at a rate of one every 5 hours. The cost of idle time per car is $150 per hour.

Estimate the average waiting time in the system saved by alternative 2.

What is the expected total cost per hour for each alternative? Which is the least expensive?Answer: Alt 2 saves 1.87 hours. Cost of Alt 1 is: $421.25 / hour and cost of Alt 2 is $400.00 /hour.

Answer: Alt 2 saves 1.87 hours. Cost of Alt 1 is: $421.25 / hour and cost of Alt 2 is $400.00 /hour.

Example 5A large furniture company has a

warehouse that serves multiple stores. In the warehouse, a single crew with four members is used to load/unload trucks. Management currently is downsizing to cut costs and wants to make a decision about crew size.

Trucks arrive at the loading dock at a mean rate of one per hour. The time required by the crew to unload/and-or load a truck has an exponential distribution (regardless of crew size). The mean of the distribution for a four member crew is 15 minutes – i.e., 4 trucks per hour. If the crew size is changed, the service rate is proportional to its size – i.e., a three member crew could do 3 per hour, etc.

The cost of providing each member of the crew is $20 per hour and the cost for a truck waiting is $30 per hour. The company has a service goal such that the likelihood of a truck spending more than one hour being served is 5% or less.

a) For the current configuration, what is the average waiting time in the system? What is the average number of trucks waiting to be unloaded (not including the truck currently being unloaded? What is the probability that a truck waits more than one hour to be unloaded? What is the total cost of the four person crew?

b) Suppose the company is looking at alternatives. One is a three member crew. What is the cost of this crew? Compare the statistics mentioned in part a) with comparable statistics for the three member crew. Would you select the three member crew over the crew in part a)? Why or why not?

c) One person suggested that rather than have one four member crew, the firm should use two, two member crews, where each crew could load/unload two trucks per hour. What is the cost of this solution? What is the probability that a truck waits longer than one hour for loading/unloading? Would you recommend that they implement this solution? Why or why not?

Example 5 (Answer)part a)

W : 0.333 hours 20 minLq: 0.083 trucks (L = .33)

Pr(w>1 hour) = 0.05Total Cost = 90.00$

part b)W : 0.5 hours 30 minLq: 0.167 trucks (L = .5)

Pr(w>1 hour) = 0.14Total Cost = $75.00

The cost is less even though the service is worse. Based on costs, select the three person crew; o/w go with the 4 person crew

part c)Assume that there is one waiting line for the two, two member crews

Total Cost = $96.00 per unit so $100 totalPr(w>1 hour) = 0.220

No; the cost is greater as is the probability that a truck waits longeris over 20%

If assuming each crew has its own waiting line:

Cost for each: 50.00$ Total cost for 2: 100.00$ Pr(w>1 hour) = 0.220

1 2s = 2

Cs = 40.00$

Cw = 30.00$

1 4s = 1

Cs = 80.00$

Cw = 30.00$

1 3s = 1

Cs = 60.00$

Cw = 30.00$