Waiting Line Models And Service Improvement Chapter 16 Waiting Line Models And Service Improvement © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e

Why do waiting lines form at a service facility, even though there may be more than enough service capacity to meet normal demand in the long run? 2000 by Prentice-Hall, Inc

Elements Of Waiting Line Analysis Queue a single waiting line Waiting line system consists of arrivals servers waiting line structures © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 2

Components Of Queuing System Source of customers Arrivals Waiting Line or queue Server Served customers © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 5

Russell/Taylor Oper Mgt 3/e Calling population source of customers infinite - large enough that one more customer can always arrive to be served finite - countable number of potential customers – examples? Arrival rate () frequency of customer arrivals at waiting line system typically follows Poisson distribution © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 3

Russell/Taylor Oper Mgt 3/e Service time often follows negative exponential distribution average service rate =  Arrival rate () must be less than service rate or system never clears out © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 4

Queue Discipline And Length order in which customers are served first come, first served is most common When would FCFS not be appropriate? Length can be infinite or finite infinite is most common finite is limited by some physical structure – examples? © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 6

Basic Waiting Line Structures Channels are the number of parallel servers Phases denote number of sequential servers the customer must go through © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 7

Single-Channel Structures Single-channel, single-phase Waiting line Server Single-channel, multiple phases Waiting line Servers © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 8

Multi-Channel Structures Multiple-channel, single phase Servers Multiple-channel, multiple-phase Waiting line Servers © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 9

Operating Characteristics Mathematics of queuing theory does not provide optimal or best solutions Operating characteristics are computed that describe system performance Steady state is constant, average value for performance characteristics that the system will reach after a long time © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 10

Operating Characteristics Notation Description L Average number of customers in the system (waiting to be served) Lq Average number of customers in the waiting line W Average time a customer spends in the system (waiting and being served) Wq Average time a customer spends waiting in line © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 11

Russell/Taylor Oper Mgt 3/e P0 Probability of no (zero) customers in the system Pn Probability of n customers in the system  Utilization rate; the proportion of time the system is in use © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 12

Cost Relationship In Waiting Line Analysis Total cost Expected costs Service cost Waiting Costs Level of service © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 13

Waiting Line Analysis And Quality Traditional view is that the level of service should coincide with minimum point on total cost curve TQM approach is that absolute quality service will be the most cost-effective in the long run © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 14

Single-Channel, Single-Phase Models All assume Poisson arrival rate Variations exponential service times general (or unknown) distribution of service times constant service times exponential service times with finite queue length exponential service times with finite calling population © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 15

Basic Single-Server Model Assumptions: Poisson arrival rate exponential service times first-come, first-served queue discipline infinite queue length infinite calling population  = mean arrival rate  = mean service rate © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 16

Formulas For Single-Server Model P0 =   (1 - ) Probability that no customers are in system   ( ) n Probability of exactly n customers in system Pn = P0 =   ( ) n   (1 - )   Average number of customers in system L =   Average number of customers in queue Lq = © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 17

Russell/Taylor Oper Mgt 3/e Average time customer spends in system   L  W = = Wq =   Average time customer spends in queue   Probability that server is busy, utilization factor =   (1 - )  =  = Probability that server is idle & customer can be served = P0 © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 18

Single-Server Example Given  = 24 per hour,  = 30 customers per hour, calculate   (1 - ) = 1 - (24/30) = 0.20 Probability that no customers are in system P0 =   Average number of customers in system L = = 24/(30-24) = 4   Average number of customers in queue Lq = = 242/30(30-24) = 3.2 © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 19

Russell/Taylor Oper Mgt 3/e Average time customer spends in system   W = = 1(30-24) = 0.167 hr = 10 min Average time customer spends in queue   Wq= = 24/30(30-24) = 0.133 hr = 8 min   Probability that server is busy, utilization factor = = 24/30 = 0.80 Probability that server is idle & customer can be served I =  = 1 - 0.80 = 0.20 © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 20

Waiting Line Cost Analysis To improve customer services management wants to test two alternatives to reduce customer waiting time: 1. Another employee to pack up purchases 2. Another checkout counter © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 21

Russell/Taylor Oper Mgt 3/e Alternative 1 Extra employee costs $150/week Each one-minute reduction in customer waiting time avoids $75 in lost sales Extra employee will increase service rate to 40 customers per hour Recompute operating characteristics Wq = 0.038 hours = 2.25 minutes, originally was 8 minutes 8.00-2.25 = 5.75 minutes 5.75 x $75/minute/week = $431.25 per week New employee saves $431.25 - 150.00 = $281.25/wk © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 22

Russell/Taylor Oper Mgt 3/e Alternative II New counter costs $6000 plus $200 per week for checker Customers divide themselves between two checkout lines Arrival rate is reduced from = 24 to = 12 Service rate for each checker is  = 30 Recompute operating characteristics Wq = 0.022 hours = 1.33 minutes, originally was 8 minutes 8.00-1.33 = 6.67 minutes 6.67 x $75/minute/week = $500.00/wk - 200.00 = $300/wk Counter is paid off in 6000/300 = 20 weeks Counter saves $300/wk; choose alternative II © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 23

Constant Service Times Constant service times occur with machinery and automated equipment Constant service times are a special case of the single-server model with general or undefined service times © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 24

Operating Characteristics For Constant Service Times Probability that no customers are in system   (1 - ) P0 = Average number of customers in queue Lq =   Average number of customers in system L = Lq +   Average time customer spends in queue Wq = Lq  © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 25

Russell/Taylor Oper Mgt 3/e   W = Wq + Average time customer spends in system   = Probability that server is busy, utilization factor   When service time is constant,  = 0 Formula can be simplified Lq = =     = =   © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 26

Constant Service Time Example Automated car wash with service time = 4.5 min Cars arrive at rate  = 10/hour (Poisson)  = 60/4.5 = 13.3/hour  2 (10)2 Lq = = = 1.14 cars waiting 2(13.3)(13.3-10) Wq = Lq  =1.14/10 = .114 hour or 6.84 minutes © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 27

Russell/Taylor Oper Mgt 3/e Finite Queue Length A physical limit exists on length of queue M = maximum number in queue Service rate does not have to exceed arrival rate () to obtain steady-state conditions  M Probability that no customers are in system P0 =   Probability of exactly n customers in system ( ) n Pn = (P0 ) for n M   (M + 1() M + 1 1 - ( )M+1 Average number of customers in system L = © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 28

Russell/Taylor Oper Mgt 3/e Let PM = probability a customer will not join system Lq = (1- PM)  L Average number of customers in queue L Average time customer spends in system W = (1 - PM) Average time customer spends in queue   W Wq = © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 29

Russell/Taylor Oper Mgt 3/e Finite Queue Example Quick Lube has waiting space for only 3 cars  = 20,  = 30, M = 4 cars (1 in service + 3 waiting) Probability that no cars are in system P0 =  M 1 - 20/30 20/305 = = 0.38 Pm = (P0 )   ( ) n=M = (0.38) 4 20 30 = 0.076 Probability of exactly n cars in system L =   (M + 1() M + 1 1 - ( )M+1 = 20/30 1 -20/30 (5(20/30) 5 1 - (20/30)5 = 1.24 Average number of cars in system © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 30

Russell/Taylor Oper Mgt 3/e Average number of cars in queue (1- PM)   20(1-0.076) 30 Lq= L - = 1.24 - = 0.62 L 1.24 Average time car spends in system = = 0.67 hours = 4.03 min W = (1 - PM) 20 (1-0.076)   W  = 0.033 hours = 2.03 min Average time car spends in queue Wq = = 0.067 - 30 © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 31

Finite Calling Population Arrivals originate from a finite (countable) population N = population size P0 =  n N! (N - n)!  N n = 0 Probability that no customers are in system Probability of exactly n customers in system N!   ( ) n Pn = P0 where n = 1, 2, ..., N (N - n)! Average number of customers in queue + Lq = N (1- P0)  © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 32

Russell/Taylor Oper Mgt 3/e Average number of customers in system Lq + L = (1- P0) Wq = (N - L)  Lq Average time customer spends in queue Average time customer spends in system W = Wq +   © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 33

Finite Calling Population Example 20 machines which operate an average of 200 hrs before breaking down  = 1/200 hr = 0.005/hr Mean repair time = 3.6 hrs  = 1/3.6 hr = 0.2778/hr P0 = 1 n N! (N - n)!  N n = 0 Probability that no machines are in system 1 = = 0.652 20 20!  (0.005/0.2778)n (20 - n)! n = 0 © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 34

Russell/Taylor Oper Mgt 3/e + Average number of machines in queue Lq = N (1- P0)  0.005 + 0.2778 = 20 (1- 0.652) = 0.169 0.005 Average number of machines in system L = Lq + (1-P0) = 0.169 + (1-0.62) = 0.520 Lq 0.169 Average time machine spends in queue Wq = = = 1.74 (N - L)  (20 - 0.520) 0.005 1 Average time machine spends in system 1  = W = Wq + 1.74 + = 5.33 hrs 0.278 © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 35

Multiple-Channel, Single-Phase Models Two or more independent servers serve a single waiting line Poisson arrivals, exponential service, infinite calling population s> P0 = 1  n = s - 1 n = 0 ] +   ( ) n n! s! s s s -  © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 36

Russell/Taylor Oper Mgt 3/e 1   ( ) n Probability of exactly n customers in system Pn = P0, for n > s s! sn-s 1   ( ) n Pn = P0, for n <= s n! Probability an arriving customer must wait 1   ( ) s ( s ) Pw = P0 s -  s!   ( ) s (s - 1 ! (s -  Average number of customers in system   ( ) L = P0 + © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 37

Russell/Taylor Oper Mgt 3/e Average time customer spends in system W = L   Average number of customers in queue Lq = L  1  Lq Average time customer spends in queue Wq = W =  Utilization factor  = /s © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 38

Multiple-Server Example Customer service area  = 10 customers/area  = 4 customers/hour per service rep s = (3)(4) = 12 P0 = 1  n = s - 1 n = 0 ] +   ( ) n n! s! s s s -  1 = = 0.045 ]  ( ) 1 0!    ( ) 1 1! 1  ( ) 1 2!  2 1 3!   ( ) 3 3(4) 3(4)-10 + + + © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 39

Russell/Taylor Oper Mgt 3/e   ( ) s (s - 1 ! (s -    ( ) Average number of customers in system L = P0 + (10)(4) (10/4) 3 = (3-1)! [3(4)-10] 2 (0.045) + (10/4) = 6 Average time customer spends in system W = L  = 6/10 = 0.60 hr = 36 min © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 40

Russell/Taylor Oper Mgt 3/e  Lq =  L = 6 - 10/4 = 3.5 Average number of customers in queue Wq =  = 3.5/10 = 0.35 hrs = 21 min Lq Average time customer spends in queue 1   ( ) s ( s ) s! Pw = P0 s -  Probability an arriving customer must wait 3 1 10 3(4) ( ) ( ) = (0.45) = 0.703 3(4)-10 3! 4 © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 41

Russell/Taylor Oper Mgt 3/e Improving Service Add a 4th server to improve service Recompute operating characteristics Po = 0.073 prob of no customers L = 3.0 customers W = 0.30 hour, 18 min in service Lq = 0.5 customers waiting Wq = 0.05 hours, 3 min waiting, versus 21 earlier Pw = 0.31 prob that customer must wait © 2000 by Prentice-Hall Inc Russell/Taylor Oper Mgt 3/e 2000 by Prentice-Hall, Inc Ch 16 - 42

We can have some control over the system using a variety of methods: 1. Do not overlook the effects of perception management. 2. Determine the acceptable waiting time for your customers. 3. Install distractions that entertain and physically involve the customer. 4. Get customers out of the line. 5. Only make people conscious of the time if they grossly overestimate waiting times. 6. Modify customer arrival behavior. 7. Keep resources not serving customers out of sight. 8. Segment customers by personality types. 9. Adopt a long-term perspective. 10. Never underestimate the power of a friendly server. 2000 by Prentice-Hall, Inc

Question: What changes can be made to improve service? alter arrival patterns or rates change number of service channels change number of sequential phases of service change number of service facilities (add new stores) increase server efficiency through improved methods or incentives priority rules line arrangements 2000 by Prentice-Hall, Inc