© 2007 Pearson Education Waiting Lines Supplement C

© 2007 Pearson Education Waiting Lines Waiting line: One or more “customers” waiting for service. Customer population: An input that generates potential customers. Service facility: A person (or crew), a machine (or group of machines), or both, necessary to perform the service for the customer. Priority rule: A rule that selects the next customer to be served by the service facility. Service system: The number of lines and the arrangement of the facilities.

© 2007 Pearson Education Waiting Line Models Basic Elements Service system Customer population Waiting line Priority rule Service facilities Served customers

© 2007 Pearson Education Waiting Line Arrangements Single line Service facilities Multiple lines Service facilities

© 2007 Pearson Education Service Facility Arrangements Channel: One or more facilities required to perform a given service. Phase: A single step in providing a service. Priority rule: The policy that determines which customer to serve next.

© 2007 Pearson Education Service Facility Arrangements Single channel, single phase Service facility

© 2007 Pearson Education Single channel, multiple phase Service Facility Arrangements Service facility 1 Service facility 2

© 2007 Pearson Education Multiple channel, single phase Service Facility Arrangements Service facility 1 Service facility 2

© 2007 Pearson Education Multiple channel, multiple phase Service Facility Arrangements Service facility 3 Service facility 4 Service facility 1 Service facility 2

© 2007 Pearson Education Service facility 3 Service facility 4 Service facility 1 Service facility 2 Routing for : 1–2–4 Routing for : 2–4–3 Routing for : 3–2–1–4 Service Facility Arrangements Mixed Arrangement

© 2007 Pearson Education Priority Rule  The priority rule determines which customer to serve next.  Most service systems use the first-come, first- serve (FCFS) rule. Other priority rules include:  Earliest promised due date (EDD)  Customer with the shortest expected processing time (SPT)  Preemptive discipline: A rule that allows a customer of higher priority to interrupt the service or another customer.

© 2007 Pearson Education Probability Distributions Arrival Times Example C.1 Arrival rate = 2/hour Customer Arrivals are usually random and can be described by a Poisson distribution. Probability that n customers will arrive… P n = e - T ( T) n n! P 4 = e -2(1) [2(1)] 4 4! P 4 = e -2 = Interarrival times: The time between customer arrivals. Probability that 4 customers will arrive… Mean = T Variance = T

© 2007 Pearson Education  The exponential distribution describes the probability that the service time will be no more than T time periods. Probability Distributions Service time If the customer service rate is three per hour, what is the probability that a customer requires less than 10 minutes of service? P(t ≤T) = 1 – e -  T μ = average number of customers completing service per period t = service time of the customer T = target service time P(t ≤ hr) = 1 – e -3(0.167) = 1 – 0.61 = 0.39 Example C.2 Mean = 1/  Variance = (1/  ) 2

© 2007 Pearson Education Operating Characteristics  Line Length: Number of customers in line.  Number of Customers in System: Includes customers in line and being serviced.  Waiting Time in Line: Waiting for service to begin.  Total Time in System: Elapsed time between entering the line and exiting the system.  Service Facility Utilization: Reflects the percentage of time servers are busy.

© 2007 Pearson Education Single-Server Model  The simplest waiting line model involves a single server and a single line of customers.  Assumptions:  The customer population is infinite and patient.   The customers arrive according to a Poisson distribution, with a mean arrival rate of      The service distribution is exponential with a mean service rate of    The mean service rate exceeds the mean arrival rate.  Customers are served on a first-come, first-served basis.  The length of the waiting line is unlimited.

© 2007 Pearson Education  = Average utilization of the system =  L = Average number of customers in the service system =  –  L q = Average number of customers in the waiting line =  L W = Average time spent in the system, including service = 1  –  W q = Average waiting time in line =  W  n = Probability that n customers are in the system = (1 –  )  n Single-Server Model

© 2007 Pearson Education Single-Channel, Single-Phase System Arrival rate (  = 30/hour, Service rate (  = 35/hour Average time in line = W q = 0.857(0.20) = 0.17 hour, or minutes Average time in system = W = = 0.20 hour, or 12 minutes 1 35 – 30 Average number in line = L q = 0.857(6) = 5.14 customers Average number in system = L = = 6 customers – 30 Utilization =  = = = 0.857, or 85.7% 3035 Example C.3

© 2007 Pearson Education Single-Channel, Single-Phase System /hour Arrival rate (  = 30/hour Service rate (  = 35 /hour

© 2007 Pearson Education Application C.1

© 2007 Pearson Education Example C.4

© 2007 Pearson Education Example C.4

© 2007 Pearson Education Example C.4

© 2007 Pearson Education Application C.2

© 2007 Pearson Education Multiple-Channel, Single-Phase System  With the multiple-server model, customers form a single line and choose one of s servers when one is available.  The service system has only one phase.  There are s identical servers.  The service distribution for each is exponential.   Mean service time is 1/   The service rate (s  exceeds the arrival rate ( ).

© 2007 Pearson Education American Parcel Service is concerned about the amount of time the company’s trucks are idle, waiting to be unloaded. The terminal operates with four unloading bays. Each bay requires a crew of two employees, and each crew costs $30/hr. The estimated cost of an idle truck is $50/hr. Trucks arrive at an average rate of three per hour, according to a Poisson distribution. Unloading a truck averages one hour with exponential service times. 4 Unloading baysCrew costs $30/hour 2 Employees/crewIdle truck costs $50/hour Arrival rate = 3/hourService time = 1 hour Multiple-Server Model Multiple-Server Model Example C.5

© 2007 Pearson Education Multiple-Server Model 4 Unloading baysCrew costs $30/hour 2 Employees/crewIdle truck costs $50/hour Arrival rate = 3/hourService time = 1 hour Utilization =  = 31(4)31(4) = 0.75  0 = [∑ + ( )] - 1 (3/1) n n! (3/1) 4 4! 1 1 – 0.75 = Average trucks in line = L q =  0 ( ) s  s!(1 –  ) (3/1) 4 (0.75) 4!(1 – 0.75) 2 = = 1.53 trucks Average time in line = W q = L q = 0.51 hours= Average time in system = W = W q + 11 = = 1.51 hours Average trucks in system = L = W = 3(1.51) = 4.53 trucks

© 2007 Pearson Education Multiple-Server Model Labor costs:$30(s)=$30(4)=$ Idle truck cost:$50(L)=$50(4.53)= Total hourly cost=$ Unloading baysCrew costs $30/hour 2 Employees/crewIdle truck costs $50/hour Arrival rate = 3/hourService time = 1 hour

© 2007 Pearson Education Application C.3

© 2007 Pearson Education Application C.3 hrs. (or minutes)

© 2007 Pearson Education Little’s Law  Little’s Law relates the number of customers in a waiting-line system to the waiting time of customers. L = W L is the average number of customers in the system. is the customer arrival rate. W is the average time spent in system, including service.

© 2007 Pearson Education  In the finite-source model, the single-server model assumptions are changed so that the customer population is finite, with N potential customers.  If N is greater than 30 customers, then the single-server model with an infinite customer population is adequate. Finite-Source Model

© 2007 Pearson Education Finite-Source Model  = Average utilization of the server = 1 –  0 L q = Average number of customers in line = N – (1 –  0 ) +   L = Average number of customers in the system = N – (1 –  0 ) W q = Average waiting time in line = L q [(N – L) ] –1 W = Average time in the system = L[(N – L) ] –1  0 = probability of zero customers [ ∑ ( ) n ] –1 N! (N – n)!  N n=0

© 2007 Pearson Education Number of robots = 10Loss/machine hour = $30 Service time = 10 hrsMaintenance cost = $10/hr Time between failures = 200 hrs Example C.6

© 2007 Pearson Education Finite-Source Model Example C.6 Solution  = 1 – = L q = 0.30 robots L = 0.76 robot W = hours Number of robots = 10Loss/machine hour = $30 Service time = 10 hrsMaintenance cost = $10/hr Time between failures = 200 hrs  0 = W q = 6.43 hours Labor cost:($10/hr)(8 hrs/day)(0.462 utilization)=$ Idle robot cost:(0.76 robot)($30/robot hr)(8 hrs/day)= Total daily cost=$219.36

© 2007 Pearson Education Decision Areas for Management  Using waiting-line analysis, management can improve the service system in one or more of the following areas.  Arrival Rates  Number of Service Facilities  Number of Phases  Number of Servers Per Facility  Server Efficiency  Priority Rule  Line Arrangement

© 2007 Pearson Education Application C.4

© 2007 Pearson Education Application C.5

© 2007 Pearson Education Application C.5