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Unformatted text preview: 8/13/2010 Waiting Line Analysis Part 2
General notation (also called Kendall notation) for classifying waiting lines A/B/k or A/B/k/x/y
A‐ probability distribution of arrivals
B‐ probability distribution of service times
k‐number of channels(servers)
x‐ number of units allowed in the system at one time
x‐potential size of calling population
“M” used for A & B when arrivals are Poisson and service times are exponential
“D" used for A & B when arrivals are deterministic
“G” used for A & B when arrivals are and service times are general distributions Behavior of Arrivals
• balk‐ customers refuse to enter the system because the line is too long
• reneg‐customers enter but decide to leave without service because the become impatient
• blocked ‐ no waiting allowed; when all servers are busy customers are not allow to enter the system Transient Interval
ˆ
λ= λˆ number of arrivals in time interval t
t ˆ
λ = λ = constant λ Transient Interval Steady State Operation 1 8/13/2010 Single Channel System with Poisson Arrivals and General Service Distribution
The time to copy documents follows approximately a normal distribution with a mean of 2.5 minutes and a standard deviation of 0.5 minutes. Determine
1)the average number of customers
waiting or using the copier.
2) the probability an arriving
customer must wait in line.
3) the proportion of time the copier is idle.
4) the average time a customer must waiting line before using the copier. Model as an M/G/1 system with λ = 12 per hour = 12/60 = .2/minute
1/µ = 2.5 minutes
µ = 1/(2.5) = .4 per minute
σ = 30 seconds = .5 minutes 1) Average number of customers waiting to use the copier L = Lq + λ
μ Lq = λ 2σ 2 + (λ μ )2
2(1 − λ μ ) Substituting numbers L = .26 + (.2/.4) = .76 customers
2) Probability an arriving customer must wait in line P = λ μ = .2 .4 = 0.5
W M/G/1 Copy‐System Analysis Cont’d
3) Proportion of time the copier is idle P0 = 1 − λ μ = 1 − .2
= 0.5
.4 4) Average time a customer must wait in line before using the copier
Wq = Lq λ = .26
= 0.13minutes
.
.2 u = 1 − P0 = 0.5 5) What is the system utilization?
The system is busy on average 50% of the time. 2 8/13/2010 3 8/13/2010 4 8/13/2010 5 8/13/2010 6 8/13/2010 7 ...
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 Spring '10
 J.LOUTEN

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