WAITING LINE MANAGEMENT

MNGT 25

Berenguel, Carmilyn ᴥ Ponciano, Riza

LearningLearning ObjectivesObjectives• Emphasize the importance of providing fast service as a

competitive advantage to companies.

• Demonstrate how service managers can design their operations and train their employees to provide faster service without incurring any additional cost.

NotionNotionCustomer’s encounters with waiting lines, which are often Customer’s encounters with waiting lines, which are often their initial encounter with the firm, can significantly affect their initial encounter with the firm, can significantly affect

their overall level of satisfaction with the organization.their overall level of satisfaction with the organization.

Waiting Lines or QueuesWaiting Lines or Queues• One of the most

important areas in operations management

• Basic• Pervasive

Economics of the Waiting Line ProblemEconomics of the Waiting Line Problem• The central problem is

a trade-off decisiontrade-off decision.

• Weighing the added cost of providing more rapid service against the inherent cost of waiting.

Practical View of Waiting LinesPractical View of Waiting Lines

Practical View of Waiting LinesPractical View of Waiting Lines

Practical View of Waiting Lines

The Queuing The Queuing PhenomenonPhenomenonThree major componentsThree major components:a. The source population and the way customers

arrive at the systemb. The servicing systemc. The condition of the customer exiting the system

(back to source population or not?)

Components of the Queuing PhenomenonComponents of the Queuing Phenomenon

Customer Customer ArrivalsArrivals• Finite PopulationFinite Population – refers to limited-size

customer pool that will use the service and, at times, form a line.

• Infinite PopulationInfinite Population – one large enough in relation to the service system so that the changes in the population size caused by subtractions or additions to the population does not significantly affect the system probabilities.

Distribution of ArrivalsDistribution of ArrivalsWaiting line formulas generally

require an arrival rate.arrival rate.Arrival RateArrival Rate – number of units per

period (10 units per hour)Constant Arrival Rate - periodicVariable Arrival Rate - random

Exponential DistributionExponential Distribution• When arrivals occur random fashion, a plot of

interarrival times yields an exponential distribution.exponential distribution.

*where λ is the mean number of arrivals per time period

• Allows us to compute the probabilities of arrivals within a specified time

Exponential DistributionExponential DistributionEx. Single arrivals to waiting line (λ=1)

(minutes)(minutes)

Probability that the Next Probability that the Next Arrival Will Occur in Arrival Will Occur in tt Minutes or More (eMinutes or More (e-t-t))

Probability that the Next Probability that the Next Arrival Will Occur in Arrival Will Occur in tt Minutes or Less (1-eMinutes or Less (1-e-t-t))

0 1.00 0

0.5 0.61 0.39

1.0 0.37 0.63

1.5 0.22 0.78

2.0 0.14 0.86

Poisson DistributionPoisson Distribution• If arrival process is random, it its Poisson Poisson

distributeddistributed• A probability distribution that represents the

number of random events occurring over a fixed period of time

Poisson DistributionPoisson DistributionEx. Mean arrival rate of units into a system is three

per minute (λ=3), find the probability that exactly five units will arrive within a one-minute period (n=5, T=1)

*there is a 10.1 percent 10.1 percent that there will be five arrivals in any one-minute interval

Other Arrival CharacteristicsOther Arrival Characteristics• Arrival Patterns

• Size of Arrival Units

• Degree of Patience

ARRIVAL PATTERNSARRIVAL PATTERNS

• Controllable

• Uncontrollable

SIZE OF ARRIVAL UNITSSIZE OF ARRIVAL UNITS

• Single Arrival

• Batch Arrival

DEGREE OF PATIENCEDEGREE OF PATIENCE

• Patient Arrival

• Impatient Arrival– Balking– Reneging

QUEUING SYSTEMQUEUING SYSTEM

Consists primarily of:The Waiting Line(s)

Available Number of Servers

THE WAITING LINETHE WAITING LINEFactors to consider: Line Length

Infinite Potential LengthLimited Line Capacity

Number of LinesSingle LineMultiple Lines

- a priority rule or set of rules for determining the order of service to customers in a waiting line.

Examples:First come first serveReservations firstLargest Orders firstEmergencies First

Queue DisciplineQueue Discipline

Two major practical problems in any rule:

1.Ensuring that customers know and follow the rule.

2.Ensuring that a system exists to enable the employees to manage the line.

SERVICE TIME DISTRIBUTIONSERVICE TIME DISTRIBUTIONService Rate

- capacity of the server in number of units per time period, and not as service time.

Constant service time rule – states that each service takes exactly the same time.

LINE STRUCTURESLINE STRUCTURES Single Channel, Single Phase

Single Channel, Multiphase

Multichannel, Single Phase

Multichannel, Multiphase

Mixed

a. multiple-to-single channel structures

b. alternate path structures

Two Exit FatesTwo Exit Fates

1.The customer may return to the source population and immediately become a competing candidate for service again

2. There may be a low probability of reservice.

SAMPLE WAITING LINE SAMPLE WAITING LINE PROBLEMSPROBLEMS

EXAMPLE 1: Customers in LineEXAMPLE 1: Customers in LineWestern National Bank is considering opening a drive-through window for

customer service. Management estimates that customers will arrive at the rate of 15 per hour. The teller who will staff the window can service customers at the rate of one every three minutes.

Part 1 Assuming Poisson arrivals and exponential service, find1 Utilization of the teller.2 Average number in the waiting line.3 Average number in the system.4 Average waiting time in line.5 Average waiting time in the system, including service.

Part 1 Assuming Poisson arrivals and exponential service, find

1 Utilization of the teller.2 Average number in the waiting line.3 Average number in the system.4 Average waiting time in line.5 Average waiting time in the system, including

service.

Equations for Solving Model ProblemsEquations for Solving Model Problems

SOLUTION—Part 1

Part 2:

Because of limited space availability and a desire to provide an acceptable level of service, the bank manager would like to ensure, with 95 percent confidence, that no more than three cars will be in the system at any time. What is the present level of service for the three-car limit? What level of teller use must be attained and what must be the service rate of the teller to ensure the 95 percent level of service?

SOLUTION—Part 2

EXAMPLE.2: Equipment SelectionThe Robot Company franchises combination

gas and car wash stations throughout the United States. Robot gives a free car wash for a gasoline fill-up or, for a wash alone, charges $0.50. Past experience shows that the number of customers that have car washes following fill-ups is about the same as for a wash alone. The average profit on a gasoline fill-up is about $0.70, and the cost of the car wash to Robot is $0.10. Robot stays open 14 hours per day.

Robot has three power units and drive assemblies, and a franchisee must select the unit preferred. Unit I can wash cars at the rate of one every five minutes and is leased for $12 per day. Unit II, a larger unit, can wash cars at the rate of one every four minutes but costs $16 per day. Unit III, the largest, costs $22 per day and can wash a car in three minutes. The franchisee estimates that customers will not wait in line more than five minutes for a car wash. A longer time will cause Robot to lose the gasoline sales as well as the car wash sale.If the estimate of customer arrivals resulting in washes is 10 per hour, which wash unit should be selected?

SOLUTION:

#3. Determining the Number of Servers#3. Determining the Number of ServersLayoutLayout Service Service

PhasePhaseSource Source PopulaPopulationtion

Arrival Arrival PatternPattern

Queue Queue DisciplineDiscipline

Service Service PatternPattern

Permissible Permissible Queue Queue LengthLength

Typical Typical ExampleExample

Multichannel Single Infinite Poisson FCFS Exponential

Unlimited Parts counter in auto agency

Average number in systemAverage number in system

Average time waiting in lineAverage time waiting in line

Average total time in systemAverage total time in system

Probability of waiting in lineProbability of waiting in line

#3. Determining the Number of Servers#3. Determining the Number of ServersEx. In the service department of the Glenn-Mark Auto Agency,

mechanics requiring parts for auto repair or service present their request forms at the parts department counter. The parts clerk fills a request while the mechanic waits. Mechanics arrive in a random (Poisson) fashion at the rate of 40 per hour, and a clerk can fill requests at the rate of 20 per hour (exponential). If the cost for a parts clerk is $6 per hour and the cost for a mechanic is $12 per hour, determine the optimum number of clerks to staff the counter. (Because of the high arrival rate, an infinite source may be assumed.)

#3. Determining the Number of Servers#3. Determining the Number of Serversλ/μ

M 3

2.0 0.8888

Given:Assume:

Step 1. Determine loss.Step 1. Determine loss.

There is an average of 0.8888 mechanic waiting all day. For an 8-hour day at $12 per hour, there is a loss of machine’s time worth of:

0.8888 mechanic x $12 per hour x 8 hours0.8888 mechanic x $12 per hour x 8 hours= $85.32= $85.32

#3. Determining the Number of Servers#3. Determining the Number of ServersStep 2. Reobtain waiting time.*Compare the added cost of the additional employee

with the time saved by the mechanics.

0.1730 x $12 x 8 hours = $16.61 cost of a mechanic waiting in line

Value of mechanics’ time saved is $85.32 - $16.61 = $ 68.71

Cost of additional parts clerk is 8 hour x $6/hour = 48.00

Cost reduction by adding fourth clerk = $ 20.71

#4. Finite Population Source#4. Finite Population SourceLayoutLayout Service Service

PhasePhaseSource Source PopulationPopulation

Arrival Arrival PatternPattern

Queue Queue DisciplineDiscipline

Service Service PatternPattern

Permissible Permissible Queue Queue LengthLength

Typical Typical ExampleExample

Single Channel

Single Finite Poisson FCFS Exponential Unlimited Machine breakdown and repair in a factory

• A finite queuing situation that is most easily solved using finite tablesfinite tables, which require the manipulation of specific terms.

#4. #4. FiniteFinite Population Source Population SourceService FactorService Factor

Probability of exactly n units in Probability of exactly n units in queuing systemqueuing system

Average waiting time in lineAverage waiting time in line

Average number of Average number of units being servedunits being served

Average number Average number of units in lineof units in line

Population source less Population source less those in queuing systemthose in queuing system

Efficiency FactorEfficiency Factor

Average number of Average number of units in population units in population sourcesource

#4. Finite Population Source#4. Finite Population SourceEx. Studies of a bank of four weaving machines at the Loose Knit Textile mill have shown that, on average, each machine needs adjusting every hour and that the current serviceperson averages 7 ½ minutes per adjustment. Assuming Poisson arrivals, exponential service, and a machine idle time cost of $40 per hour, determine if a second serviceperson (who also averages 7 ½ minutes per adjustment) should be hired at a rate of $7 per hour.

#4. Finite Population Source#4. Finite Population Source*can be solved using finite queuing tables; compare

costs of machine downtime (either waiting in line or being serviced) and the cost of one repairperson, to the cost of machine downtime and two repair people.

average number of machines that are in service system x downtime cost per hour + repair people’s cost

N = Number of machines in the people

M = Number of repair people

T = Time required to service a machine

U = Average time a machine runs before requiring service

X = Service factor, or proportion of service time required for each machine (X=T/(T+U))

L = Average number of machines waiting in line to be serviced

H = Average number of machines being serviced

#4. Finite Population Source#4. Finite Population Source

#4. Finite Population Source#4. Finite Population SourceThe values to be determined from the finite tables are:

Case 1. One repairperson. From problem statement,N=4, M=1, T=7½ minutes, U=60 minutes

D = Probability that a machine needing service will have to wait

F = Efficiency factor, which is a measure of the effect of having to wait in line to be serviced

#4. Finite Population Source#4. Finite Population Source

Population 4

X M D F

0.110 2 .034 .998

1 .321 .958

N=4, F is interpolated as being approximately 0.957 at X=0.111 and M=1.

The number of machines waiting in line to be serviced is L, where:

#4. Finite Population Source#4. Finite Population SourceA comparison of downtime costs for service and repair of four machines:

The number of machines being served is H, where

Number of Repair People

Number of Machines Down (H+L)

Cost per Hour for Machines Down [(H+L)x$40/hour]

Cost of Repairpeople ($7/hour each)

Total Cost per Hour

1 0.597 $23.88 $7.00 $30.88

2 0.451 18.04 14.00 32.04

#4. Finite Population Source#4. Finite Population SourceCase II. Two repairpeople.The number of machines waiting in line, L

The number of machines being serviced, H

Population 4

X M D F

0.110 2 .034 .998

1 .321 .958

X=0.111, M=2, X=0.111, M=2, F=0998F=0998