Presentation is loading. Please wait.

Presentation is loading. Please wait.

To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Chapter 16 Waiting Line Models and.

Similar presentations


Presentation on theme: "To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Chapter 16 Waiting Line Models and."— Presentation transcript:

1 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Chapter 16 Waiting Line Models and Service Improvement To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved.

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

3 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Components of Queuing System Source of customers— calling population ServerArrivals Waiting Line or“Queue”Servedcustomers Figure 16.1

4 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Elements of a Waiting Line Calling population Calling population Source of customers Source of customers Infinite - large enough that one more customer can always arrive to be served Infinite - large enough that one more customer can always arrive to be served Finite - countable number of potential customers Finite - countable number of potential customers Arrival rate ( ) Arrival rate ( ) Frequency of customer arrivals at waiting line system Frequency of customer arrivals at waiting line system Typically follows Poisson distribution Typically follows Poisson distribution

5 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Elements of a Waiting Line Service time Service time Often follows negative exponential distribution Often follows negative exponential distribution Average service rate =  Average service rate =  Arrival rate ( ) must be less than service rate  or system never clears out Arrival rate ( ) must be less than service rate  or system never clears out

6 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Elements of a Waiting Line Queue discipline Queue discipline Order in which customers are served Order in which customers are served First come, first served is most common First come, first served is most common Length can be infinite or finite Length can be infinite or finite Infinite is most common Infinite is most common Finite is limited by some physical structure Finite is limited by some physical structure

7 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Basic Waiting Line Structures Channels are the number of parallel servers Channels are the number of parallel servers Phases denote number of sequential servers the customer must go through Phases denote number of sequential servers the customer must go through

8 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Single-Channel Structures Single-channel, single-phase Waiting lineServer Single-channel, multiple phases Servers Waiting line Figure 16.2

9 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Multi-Channel Structures Servers Multiple-channel, single phase Waiting line Servers Multiple-channel, multiple-phase Figure 16.2

10 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Operating Characteristics Mathematics of queuing theory does not provide optimal or best solutions Mathematics of queuing theory does not provide optimal or best solutions Operating characteristics are computed that describe system performance 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 Steady state is constant, average value for performance characteristics that the system will reach after a long time

11 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Operating Characteristics NOTATIONOPERATING CHARACTERISTIC LAverage number of customers in the system (waiting and being served) L q Average number of customers in the waiting line WAverage time a customer spends in the system (waiting and being served) W q Average time a customer spends waiting in line Table 16.1

12 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Operating Characteristics NOTATIONOPERATING CHARACTERISTIC P 0 Probability of no (zero) customers in the system P n Probability of n customers in the system  Utilization rate; the proportion of time the system is in use Table 16.1

13 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Cost Relationship in Waiting Line Analysis Expected costs Level of service Total cost Service cost Waiting Costs Figure 16.3

14 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Waiting Line Costs and Quality Service Traditional view is that the level of service should coincide with minimum point on total cost curve 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 TQM approach is that absolute quality service will be the most cost- effective in the long run

15 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Single-Channel, Single- Phase Models All assume Poisson arrival rate All assume Poisson arrival rate Variations Variations Exponential service times Exponential service times General (or unknown) distribution of service times General (or unknown) distribution of service times Constant service times Constant service times Exponential service times with finite queue length Exponential service times with finite queue length Exponential service times with finite calling population Exponential service times with finite calling population

16 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Basic Single-Server Model Assumptions: Assumptions: Poisson arrival rate Poisson arrival rate Exponential service times Exponential service times First-come, first-served queue discipline First-come, first-served queue discipline Infinite queue length Infinite queue length Infinite calling population Infinite calling population = mean arrival rate = mean arrival rate  = mean service rate  = mean service rate

17 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Formulas for Single- Server Model L =   -   - Average number of customers in the system Probability that no customers are in the system (either in the queue or being served) P 0 = 1 -  Probability of exactly n customers in the system P n = P 0 n = 1 -  n  Average number of customers in the waiting line L q =   (  - )

18 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Formulas for Single- Server Model  =  Probability that the server is busy and the customer has to wait Average time a customer spends in the queuing system W = = 1  -  L Probability that the server is idle and a customer can be served I = 1 -   = 1 - = P 0 Average time a customer spends waiting in line to be served W q =  (  - )

19 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. A Single-Server Model Given = 24 per hour,  = 30 customers per hour Probability of no customers in the system P 0 = 1 - = 1 - = 0.20  2430 L = = = 4 Average number of customers in the system  -  24 30 - 24 Example 16.1 Average number of customers waiting in line L q = = = 3.2 (24) 2 30(30 - 24) 2  (  - )

20 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. A Single-Server Model Given = 24 per hour,  = 30 customers per hour Example 16.1 Average time in the system per customer W = = = 0.167 hour 1  -  1 30 - 24 Average time waiting in line per customer W q = = = 0.133  (  -  ) 24 30(30 - 24) Probability that the server will be busy and the customer must wait  = = = 0.80  2430 Probability the server will be idle I = 1 -  = 1 - 0.80 = 0.20

21 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. 1.Another employee to pack up purchases 2.Another checkout counter Waiting Line Cost Analysis To improve customer services management wants to test two alternatives to reduce customer waiting time: Example 16.2

22 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Waiting Line Cost Analysis Add extra employee to increase service rate from 30 to 40 customers per hour Add extra employee to increase service rate from 30 to 40 customers per hour Extra employee costs $150/week Extra employee costs $150/week Each one-minute reduction in customer waiting time avoids $75 in lost sales Each one-minute reduction in customer waiting time avoids $75 in lost sales Waiting time with one employee = 8 minutes Waiting time with one employee = 8 minutes Example 16.2 W q = 0.038 hours = 2.25 minutes 8.00 - 2.25 = 5.75 minutes reduction 5.75 x $75/minute/week = $431.25 per week New employee saves $431.25 - $150.00 = $281.25/wk

23 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Waiting Line Cost Analysis New counter costs $6000 plus $200 per week for checker New counter costs $6000 plus $200 per week for checker Customers divide themselves between two checkout lines Customers divide themselves between two checkout lines Arrival rate is reduced from  = 24 to  = 12 Arrival rate is reduced from  = 24 to  = 12 Service rate for each checker is  = 30 Service rate for each checker is  = 30 Example 16.2 W q = 0.022 hours = 1.33 minutes 8.00 - 1.33 = 6.67 minutes 6.67 x $75/minute/week = $500.00/wk - $200 = $300/wk Counter is paid off in 6000/300 = 20 weeks

24 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Waiting Line Cost Analysis Adding an employee results in savings and improved customer service Adding an employee results in savings and improved customer service Adding a new counter results in slightly greater savings and improved customer service, but only after the initial investment has been recovered Adding a new counter results in slightly greater savings and improved customer service, but only after the initial investment has been recovered A new counter results in more idle time for employees A new counter results in more idle time for employees A new counter would take up potentially valuable floor space A new counter would take up potentially valuable floor space Example 16.2

25 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Constant Service Times Constant service times occur with machinery and automated equipment 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 Constant service times are a special case of the single-server model with general or undefined service times

26 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Operating Characteristics for Constant Service Times  P 0 = 1 - Probability that no customers are in system Average number of customers in system L = L q +  Average number of customers in queue Lq =Lq =Lq =Lq = 2 2  (  - )

27 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Operating Characteristics for Constant Service Times   = Probability that the server is busy Average time customer spends in the system W = W q + 1 Average time customer spends in queue Wq =Wq =Wq =Wq =  L q

28 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. L q = = = 1.14 cars waiting (10) 2 2(13.3)(13.3 - 10) Constant Service Times Automated car wash with service time = 4.5 min Cars arrive at rate = 10/hour (Poisson)  = 60/4.5 = 13.3/hour Example 16.3 W q = = 1.14/10 =.114 hour or 6.84 minutes L q 2 2  (  - )

29 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Finite Queue Length A physical limit exists on length of queue A physical limit exists on length of queue M = maximum number in queue M = maximum number in queue Service rate does not have to exceed arrival rate (  ) to obtain steady-state conditions Service rate does not have to exceed arrival rate (  ) to obtain steady-state conditions P 0 = Probability that no customers are in system 1 - /  1 - ( /  ) M + 1 Probability of exactly n customers in system  P n = (P 0 ) for n ≤ M n L = - Average number of customers in system /  /  1 - /  (M + 1)( /  ) M + 1 1 - ( /  ) M + 1

30 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Finite Queue Length Let P M = probability a customer will not join system Average time customer spends in system W =W =W =W = L  (1 - P M ) L q = L -  (1- P M )  Average number of customers in queue Average time customer spends in queue W q = W - 1

31 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Finite Queue Quick Lube has waiting space for only 3 cars. = 20,  = 30, M = 4 cars (1 in service + 3 waiting) = 20,  = 30, M = 4 cars (1 in service + 3 waiting) Example 16.4 Probability that no cars are in the system P 0 = = = 0.38 1 - 20/30 1 - (20/30) 5 1 - /  1 - ( /  ) M + 1 P n = (P 0 ) = (0.38) = 0.076 Probability of exactly 4 cars in the system 2030 4  n=Mn=Mn=Mn=M L = - = 1.24 Average number of cars in the system /  /  1 - /  (M + 1)( /  ) M + 1 1 - ( /  ) M + 1

32 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Finite Queue Example 16.4 Average time a car spends in the system W = = 0.067 hr L  (1 - P M ) L q = L - = 0.62  (1- P M )  Average number of cars in the queue Quick Lube has waiting space for only 3 cars. = 20,  = 30, M = 4 cars (1 in service + 3 waiting) = 20,  = 30, M = 4 cars (1 in service + 3 waiting) Average time a car spends in the queue W q = W - = 0.033 hr 1

33 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Finite Calling Population Arrivals originate from a finite (countable) population Arrivals originate from a finite (countable) population N = population size N = population size Probability of exactly n customers in system P n = P 0 where n = 1, 2,..., N n  N! (N - n)! Average number of customers in queue L q = N - (1- P 0 )  +  Probability that no customers are in system P 0 = n = 0 n = 0 N! (N - n)!  N n 1

34 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Finite Calling Population Arrivals originate from a finite (countable) population Arrivals originate from a finite (countable) population N = population size N = population size Wq =Wq =Wq =Wq = L q L q (N - L) (N - L) Average time customer spends in queue L = L q + (1 - P 0 ) Average number of customers in system W = W q + Average time customer spends in system 1

35 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. 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) Probability that no machines are in the system P 0 = 0.652 Average number of machines in the queue L q = 0.169 Average number of machines in system L = 0.169 + (1 - 0.652) =.520 Example 16.5 Finite Calling Population

36 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. 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) Example 16.5 Finite Calling Population Average time machine spends in queue W q = 1.74 hrs Average time machine spends in system W = 5.33 hrs

37 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Multiple-Channel, Single-Phase Models Two or more independent servers serve a single waiting line Two or more independent servers serve a single waiting line Poisson arrivals, exponential service, infinite calling population Poisson arrivals, exponential service, infinite calling population s  >  s  >  P 0 = 11s!  s s  s  - s  -  n=s-1 n=0 1 11n!n!11n!n!  n +

38 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Multiple-Channel, Single-Phase Models Two or more independent servers serve a single waiting line Two or more independent servers serve a single waiting line Poisson arrivals, exponential service, infinite calling population Poisson arrivals, exponential service, infinite calling population s  >  s  >  P 0 = 11s!  s s  s  - s  -  n=s-1 n=0 1 11n!n!11n!n!  n + Computing P 0 can be time-consuming. Tables can used to find P 0 for selected values of  and s.

39 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Multiple-Channel, Single-Phase Models Probability of exactly n customers in the system P n = P 0, for n > s 1 s! s n-s n P 0, for n > s 1 11n!n!11n!n!  n Probability an arriving customer must wait P w = P 0 1 11s!s!11s!s! s  s  - s  - s Average number of customers in system L = P 0 +  ( /  ) s (s - 1)!(s  - ) 2 

40 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Multiple-Channel, Single-Phase Models W = L Average time customer spends in system   =  /s  Utilization factor Average time customer spends in queue W q = W - = 1 L q L q = L -  Average number of customers in queue

41 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Multiple-Server System Customer service area = 10 customers/area = 10 customers/area  = 4 customers/hour per service rep s  = (3)(4) = 12 P 0 = 0.045 Probability no customers are in the system Number of customers in the service department L = 6 Waiting time in the service department W = L / = 0.60 Example 16.6

42 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Multiple-Server System Customer service area = 10 customers/area = 10 customers/area  = 4 customers/hour per service rep s  = (3)(4) = 12 L q = L - /  = 3.5 Number of customers waiting to be served Average time customers will wait in line W q = L q / = 0.35 hours Probability that customers must wait P w = 0.703 Example 16.6

43 To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Add a 4th server to improve service Add a 4th server to improve service Recompute operating characteristics Recompute operating characteristics P 0 = 0.073 prob of no customers P 0 = 0.073 prob of no customers L = 3.0 customers L = 3.0 customers W = 0.30 hour, 18 min in service W = 0.30 hour, 18 min in service L q = 0.5 customers waiting L q = 0.5 customers waiting W q = 0.05 hours, 3 min waiting, versus 21 earlier W q = 0.05 hours, 3 min waiting, versus 21 earlier P w = 0.31 prob that customer must wait P w = 0.31 prob that customer must wait Improving Service


Download ppt "To Accompany Russell and Taylor, Operations Management, 4th Edition,  2003 Prentice-Hall, Inc. All rights reserved. Chapter 16 Waiting Line Models and."

Similar presentations


Ads by Google