Capacity Planning

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# Capacity Planning - PowerPoint PPT Presentation

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

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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
Single Server M/M/1

λ OOOO ■

λ = arrivals per time period

O = customer

■ = server

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

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