1. Capacity Planning
Chapter 14

2. 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

3. 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

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

5. 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 = = 0.4 (40%)

6. 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

7. 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

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

9. 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 =
(9/4)0 + (9/4)1 + (9/4)2 + (9/4)3 + (9/4)4
0! ! ! ! !(1-9/16)
P0 = 0.098

10. 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

11. 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