Capacity Planning Chapter 14.

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Presentation transcript:

Capacity Planning Chapter 14

Capacity Planning Capacity planning decisions involve trade-offs between the cost of providing a service (i.e., increasing the number of servers) and the cost or inconvenience to the customer Involves determining the appropriate level of service capacity by specifying the proper mix of facilities, equipment, and labor required to meet anticipated demand

Classification of Queuing Models Infinite (any number of customers) vs. finite (a specified number of customers) queues Exponential Service Times - normally used to measure a process such as waiting lines. It is a continuous probability distribution General Service Times – pre-specified Single server or multiple servers

Single Server M/M/1 λ OOOO ■ λ = arrivals per time period O = customer

M/M/1 Boat Launch Example If: λ = 6 boats arrive per hour μ = 10 boats launched per hour p = λ / μ Then: probability that the customer must wait is p = 6 / 10 = 0.6 or 60% probability that the customer does not have to wait is p0 = 1- 0.6 = 0.4 (40%)

M/M/1 Boat Launch Example Mean number of boats in the system: Ls = λ / (μ – λ) Ls = 6 / (10 – 6) = 6/4 = 1.5 boats Mean number of boats in the queue: Lq = (pλ) / ((μ – λ) Lq = (.6 x 6) / (10 – 6) = 3.6 / 4 = .9 boats

M/M/1 Boat Launch Example Mean time in the system: Ws = 1 / (μ – λ) = 1 / (10 – 6) = ¼ hour Mean time in queue: Wq = p / (μ – λ) = 0.6 / (10 – 6) = .15 hour

Multiple Servers M/M/c ■ λ OOOO λ = arrivals per time period O = customer ■ = servers

M/M/4 Secretarial Pool Example If: λ = 9 requests per hour c = 4 servers p = λ / c (used to calculate the value of p0) Then: probability that a faculty member does not have to wait for secretarial work is P0 = 1 (9/4)0 + (9/4)1 + (9/4)2 + (9/4)3 + (9/4)4 0! 1! 2! 3! 4!(1-9/16) P0 = 0.098

M/M/4 Secretarial Pool Example Mean number of jobs in the system: p c+1 Ls = (P0 + p) (c-1)!(c-p)2 (9/4)5 Ls = (0.098) + 9/4 (4-1)!(4-9/4)2 Ls = 2.56

M/M/4 Secretarial Pool Example Mean time for each secretarial job in the system: Ls Ws = λ 2.56 Ws = = 0.28 or 17 minutes per job 9