Post on 19-Dec-2015
transcript
The Theory of Queues
Models of Waiting in line
Queuing Theory
• Basic model:• Arrivals Queue Being Served Done
– Queuing theory lets you calculate:• How long will it take an average customer to
get done?• How many customers are waiting, on average?
Why this is useful
• How long will it take a customer to get done?– Time allocation
• For customers• For service
• How many customers are waiting?– Space allocation– Social dynamic
Arrivals Queue Being Served Done
– Jargon:
• Queue < System• Queue is a subset of the System• System = In queue + being served
Arrivals Queue Being Served Done
– Standard Greek letters:
• λ (Lambda) – Average arrival rate• μ (Mu) – Average service rate
• 1/λ = Average time between arrivals• 1/μ = Average time for service
Steady State -- when average performance is not changing
• λ Lambda – Average arrival rate• μ Mu – Average service rate
• If μ > λ – if the service rate is faster than the arrival rate – there is a steady state.
• If λ > μ, there is no steady state. The queue will grow and grow.
Y a Q if λ < μ?
• Why is there a queue if customers are served faster than they are arriving, on average?
Y a Q if λ<μ?
• Why is there a queue if customers are served faster than they are arriving, on average?
• Because of random variation in arrival and service times.
Randomness Modeled
– Suppose …• An event can happen at any time• Events are independent• We know the average rate of occurrence
– Then we use …• Poisson distribution
– random variation with no limit– extremes are unlikely but possible
Poisson distribution
• Limit of the binomial• Cut t amount of time into smaller and smaller
pieces• Only one arrival allowed during each piece.• The probability of an arrival during the piece is
λ/N, where N is the number of pieces.• As N∞, the probability of x arrivals during t
amount of time is …
Poisson distribution
• Prob (x arrivals during time t) =
!
)(
x
te xt
Divide time period into 2 pieces, with 1/2 probability of event in each piece.Each time period is independent of the others.
0.5 0.50 0 1/2*1/2 0.25 10 1 1/2*1/2 0.25 21 0 1/2*1/2 0.251 1 1/2*1/2 0.25 1
Divide time period into 3 pieces, with 1/3 probability of event in each piece.0.33333 0.33333 0.3333
0 0 0 2/3*2/3*2/3 0.2963 11 0 0 1/3*2/3*2/3 0.14815 30 1 0 2/3*1/3*2/3 0.148150 0 1 2/3*2/3*1/3 0.148151 1 0 1/3*1/3*2/3 0.07407 31 0 1 1/3*2/3*1/3 0.074070 1 1 2/3*1/3*1/3 0.074071 1 1 1/3*1/3*1/3 0.03704 1
Divide time period into 4 pieces, with 1/4 probability of event in each piece.0.25 0.25 0.25 0.25
0 0 0 0 3/4*3/4*3/4*3/4 0.31641 11 0 0 0 1/4*3/4*3/4*3/4 0.10547 40 1 0 0 3/4*1/4*3/4*3/4 0.105470 0 0 1 3/4*3/4*3/4*1/4 0.105470 0 1 0 3/4*3/4*1/4*3/4 0.105471 0 0 1 1/4*3/4*3/4*1/4 0.03516 61 1 0 0 1/4*1/4*3/4*3/4 0.035161 0 1 0 1/4*3/4*1/4*3/4 0.035160 1 0 1 3/4*1/4*3/4*1/4 0.035160 0 1 1 3/4*3/4*1/4*1/4 0.035160 1 1 0 3/4*1/4*1/4*3/4 0.035161 1 1 0 1/4*1/4*1/4*3/4 0.01172 41 1 0 1 1/4*1/4*3/4*1/4 0.011720 1 1 1 3/4*1/4*1/4*1/4 0.011721 0 1 1 1/4*3/4*1/4*1/4 0.011721 1 1 1 1/4*1/4*1/4*1/4 0.00391 1
10 pieces
0 1 2 3 4 5 6 7 80
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5Prob(x)
Poisson
60 pieces
A model of randomness and independence of events
– Using the Poisson distribution in your queuing model means adopting to binomial distribution’s assumptions:
• An event can happen at any time• With equal probability• Independent events
Poisson distribution application
• Running total of Poisson distribution can show how much inventory is needed.
Exponential Distribution
• Probability that the next ARRIVAL will be during the next t amount of time =
te 1
Exponential Distribution
• Probability that the SERVER will finish with the current customer during the next t amount of time =
te 1
Exponential Distribution example
• If μ = 1 per hour,• Probability the the SERVER will finish current
customer during the next hour is 1 – e-1
• = 0.632• For half of the customers, the service takes
less than 42 minutes.
Steady State
• When the distribution of probabilities of various numbers in the system stabilizes
• When startup conditions are far behind• λ<μ required for steady state to exist• ρ=λ/μ is how busy the server is• ρ is server busy time ÷ all time• ρ< 1 for steady state to be possible
Steady State with Poisson arrivals and service times
• For single-server single-stage queue with Poisson arrivals and service:
• Probability of n in the system (waiting or being served)= (1-ρ)ρn
Customers in System and in Queue
• L – customers in system =
• Lq – customers in queue = ρL
• These are averages, once the steady state is reached.
• There are usually fewer in the system than L because the prob of n in system is skewed.
1
L = ρ/(1-ρ) = λ/(u-λ)
• When the arrival rate nearly equals the service rate, the line can get very long.
• If the arrival rate nearly equals the service rate, small changes in either rate can make a big difference.
Time in System
• W – time in system = L/λ
• Wq – time waiting to be served = ρW
• These are averages, once the steady state is reached.
• Most customers get through in less time.
W = L/λ = 1/(u-λ)
• When the arrival rate and the service rate are nearly equal, the wait can be very long.
• When the arrival rate and the service rate are nearly equal, small changes in either rate can make a big difference.
Economic Analysis
Weigh• The cost of waiting
– Against -- • The cost of speeding up service or reducing
arrivals.