Waiting Line Models Production and Operations Management Reporter: Noel M. Arellano Text Authors: Russell & Taylor

Waiting Line Models Characteristics of waiting line models Elements of waiting line analysis Waiting line analysis and quality Single-channel, single-phase models Multiple-channel, single-phase models 2

Characteristics of Waiting Line Waiting is a fact of life Americans wait up to 30 minutes daily or about 37 billion hours in line yearly US leisure time has shrunk by more than 35% since 1973 An average part spends more than 95% of its time waiting Waiting is bad for business and occurs in every arrival 3

Elements Of Waiting Line Analysis Queue A single waiting line Waiting line system consists of Arrivals Servers Waiting line structures 4

Elements Of Waiting Line Analysis Source of customers Arrivals Waiting Line or queue Server Served customers Queue System 5

Elements of Waiting Line Analysis Calling population Source of customers Infinite - large enough that one more customer can always arrive to be served Finite - countable number of potential customers Arrival rate () Frequency of customer arrivals at waiting line system Typically follows Poisson distribution 6 1

Elements of Waiting Line Models Service time Often follows negative exponential distribution Average service rate =  Arrival rate () must be less than service rate ()or system never clears out 7

Queue Discipline And Length Order in which customers are served First come, first served is most common Service in random order Length can be infinite or finite Infinite is most common Finite is limited by some physical structure 8

Basic Waiting Line Structures Channels are the number of parallel servers for serving arriving customers Phases denote number of sequential servers the customer must go through 9

Single-Channel Structures Single-channel, single-phase Waiting line Server Single-channel, multiple phases Waiting line Servers 10

Multiple-Channel Structures Multiple-channel, single phase Waiting line Servers Multiple-channel, multiple-phase Waiting line Servers 11

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 12

Operating Characteristics Notation Description L Average number of customers in the system (waiting and being served) Lq Average number of customers in the waiting line W Average time a customer spends in the Wq Average time a customer spends waiting in line 13

Operating Characteristics Notation Description 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 14

Cost Relationships In Waiting Line Total cost Expected costs Service cost Waiting Costs Level of service 15

Waiting Line Analysis & 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 16

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 17

Basic Single Server Models 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 18

Formulas For Single-Server Models 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 = 19

Formulas For Single-Server Models 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 20

Single-Server Model: Example Given  = 24 per hour,  = 30 customers per hour, calculate   (1 - ) Probability that no customers are in system = 1 - (24/30) = 0.20 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 21

Single-Server Model: Example 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 I =  Probability that server is idle & customer can be served = 1 - 0.80 = 0.20 22

Waiting Line Cost Analysis To improve customer services management wants to test two alternatives to reduce customer waiting time: Another employee to pack up purchases Another checkout counter 23

Solution For Alternative I 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 24

Solution For Alternative II New counter costs $6000 + $200/week for checker Customers divide themselves b/w 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 8min. 8.00-1.33 = 6.67 minutes 6.67x$75/min/week=$500.00/wk - 200.00 = $300/wk Counter is paid off in 6000/300 = 20 weeks Counter saves $300/wk; choose alternative II 25

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 26

OC 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  27

OC For Constant Service Times Average time customer spends in system   W = Wq +   = Probability that server is busy, utilization factor   When service time is constant,  = 0 Formula can be simplified Lq = =     = =   28

Constant Service Times: 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) = 1.14/10 = .114 hour or 6.84 minutes 29

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 Probability that no customers are in system  M P0 = Probability of exactly n customers in system   ( ) n Pn = (P0 ) for n M Average number of customers in system   (M + 1() M + 1 1 - ( )M+1 L = 30

Finite Queue Length Let PM = probability a customer will not join system Average number of customers in queue Lq = (1- PM)  L L Average time customer spends in system W = (1 - PM) Average time customer spends in queue   W Wq = 31

Finite Queue Length: 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 Probability of exactly n cars in system Pm = (P0 )   ( ) n=M = (0.38) 4 20 30 = 0.076 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 32

Finite Queue Length: Example (1- PM)  Average number of cars in queue  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)  Average time car spends in queue   W = 0.033 hours = 2.03 min Wq = = 0.067 - 30 33

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 (N - n)! where n = 1, 2, ..., N Average number of customers in queue + Lq = N (1- P0)  34

Finite Calling Population 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 +   35

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 = 20 20!  (0.005/0.2778)n (20 - n)! n = 0 = 0.652 36

Finite Calling Population: Example + 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  1 Average time machine spends in system = W = Wq + 1.74 + = 5.33 hrs 0.278 37

] + Multi-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 -  38

Multi-Channel, Single-Phase Models 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 + 39

Multi-Channel, Single-Phase Models Average time customer spends in system W = L   Average number of customers in queue Lq = L  Average time customer spends in queue 1 Wq = W  =  Lq Utilization factor  = /s 40

] + ] 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 + + + 41

Multiple Server: Example   ( ) s (s - 1 ! (s -    ( ) Average number of customers in system P0 + L = (10)(4) (10/4) 3 = (3-1)! [3(4)-10] 2 (0.045) + (10/4) = 6 Average time customer spends in the system W = L  = 6/10 = 0.60 hr = 36 min 42

Multiple Server: Example Average number of customers in queue  Lq =  L = 6 - 10/4 = 3.5 Average time customer spends in queue Wq =  = 3.5/10 = 0.35 hrs = 21 min Lq 1   ( ) s ( s ) Pw = P0 s -  s! Probability an arriving customer must wait 1 10 3(4) ( ) 3 ( ) = (0.45) = 0.703 3(4)-10 3! 4 43

Improving Service Add a 4th server to improve service Recompute operating characteristics Po = 0.073 probability of no customers L = 3.0 customers W = 0.30 hour, 18 min in service Lq = 0.5 customers waiting Wq = 0.05 hrs, 3 min waiting, vs. 21 earlier Pw = 0.31 probability that customer must wait 44