Six parameters in shorthand

1
Queueing Theory

• Six parameters in shorthand
• First three typically used, unless specified
• Arrival Distribution
• Probability of a new packet arrives in time t
• Service Distribution
• Probability distribution packet is serviced in
  time t
• Number of servers
• Total Capacity (infinite if not specified)
• Population Size (infinite)
• Service Discipline (FCFS/FIFO)

G/G/3/20/1500/SPF General arrival and service
distributions, 3 servers, 17 queue slots (20-3),
1500 total jobs, Shortest Packet First
2
Birth-And-Death Process
3
Generalized Birth Death Process
lk
lk-1
K-1
K
K1
mk
mk1
4
Generalized Birth Death Process
5
Generalized Birth Death Process
6
M/M/1
7
M/M/1
8
M/M/1
9
M/M/1
10
M/M/1
11
M/M/1
12
M/M/1
13
Discouraged Arrival
14
Discouraged Arrival
15
Discouraged Arrival
16
Responsive Servers
17
Responsive Servers
18
M/M/m
19
M/M/m
20
M/M/m
21
M/M/1/K
22
M/M/1/K
23
M/M/1/K
24
M/M/m/m
25
M/M/1//M
26
M/M/1//M
27
M/M/a//M
28
M/M/a//M
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M/M/m/K/M
30
M/M/m/K/M
31
M/M/m/K/M
32
M/G/1 System

• Customers arrive according to a Poisson process
  with rate l, but the customer service times have
  a general distribution not necessarily
  exponential as in the M/M/1 system. Suppose that
  customers are served in the order they arrive and
  that Xi is the service time of the ith arrival.
  We assume that the random variables (X1, X2, ..)
  are identically distributed, mutually
  independent, and independent of inter-arrival
  times.
• Let EX 1/m Average service time
• EX2 Second moment of service time
• Pollaczek-Khinchin (P-K) formula
• W lEX2/(2(1-r))
• where W is the expected wait time in queue
• r l/m lEX.
• Total time in queue and service T EX
  lEX2/(2(1-r))

33
M/G/1 System

• Pollaczek-Khinchin (P-K) formula
• W lEX2/(2(1-r))
• where W is the expected wait time in queue
• r l/m lEX.
• Total time in queue and service T EX
  lEX2/(2(1-r))
• Applying Littles formula to W and T, we get the
  expected number of customers in the queue NQ and
  the expected number in the system N
• NQ l2EX2/(2(1-r)) N r l2EX2/(2(1-r))
• When service times are exponentially distributed
  (M/M/1), EX22/m2 and the P-K formula reduces
  to
• W l 2/m2 /(2(1-r)) r/m(1-r) (M/M/1)
• When service times are identical for all
  customers (M/D/1), EX21/m2 and the P-K formula
  reduces to
• W l 1/m2 /(2(1-r)) r/2m(1-r) (M/D/1)

34
M/G/1 System

• When service times are exponentially distributed
  (M/M/1), EX22/m2 and the P-K formula reduces
  to
• W l 2/m2 /(2(1-r)) r/m(1-r) (M/M/1)
• When service times are identical for all
  customers (M/D/1), EX21/m2 and the P-K formula
  reduces to
• W l 1/m2 /(2(1-r)) r/2m(1-r) (M/D/1)
• Since the M/D/1 case yields the minimum possible
  value of EX2 for given m, it follows that the
  values of W, T, NQ and N for an M/D/1 queue are
  lower bounds to the corresponding quantities for
  an M/G/1 queue of the same l and m. It is
  interesting to note that W and NQ for the M/D/1
  queue are exactly one-half their values for the
  M/M/1 queue of the same l and m. The values of T
  and N for M/D/1, on the other hand, range from
  the same as M/M/1 for small r to one half of
  M/M/1 as r approaches 1. The reason is that the
  expected service time is the same in the two
  cases, and for small r, most of the waiting
  occurs in service, whereas for large r, most of
  the waiting occurs in the queue.

35
M/G/1 System

• Pollaczek-Khinchin (P-K) formula
• W lEX2/(2(1-r))
• where W is the expected wait time in queue
• r l/m lEX.
• Total time in queue and service T EX
  lEX2/(2(1-r))
• In an M/G/1 queue, r lt 1 does not mean finite
  queueing delay rather W can be infinite if the
  second moment EX2 is infinite. What is
  happening in this case is that a small number of
  customers have incredibly long service times.
  When one of these customers is served, an
  incredible number of arrivals are queued and
  delayed by a significant fraction of that long
  service time. Thus the contribution to W is
  proportional to the square of the service time,
  leading to an infinite W if EX2 is infinite.

36
M/G/1 Example

• There are two workers competing for a job. Able
  claims an average service time which is faster
  than Bakers, but Baker claims to be more
  consistent, if not as fast. The arrivals occur
  according to a Poisson process at a rate of l 2
  per hour. (1/30 per minute). Ables statistics
  are an average service time of 24 minutes with a
  standard deviation of 20 minutes. Bakers service
  statistics are an average service time of 25
  minutes, but a standard deviation of only 2
  minutes. If the average length of the queue is
  the criterion for hiring, which worker should be
  hired?

37
M/G/1 Example (cont.)

• For Able, l 1/30 (per min), m-1 24
  (min), r l / m 24/30 4/5 s2
  202 400(min2)
• W lEX2/(2(1-r)) 1/30 m-2 s2/(2(1-4/5))
• 1/30 242 202/(2/5) 1/309765/2
  976/12244/381.33 min
• NQ l W 1/30 81.33 2.711 (customers)
• For Baker, l 1/30 (per min), m-1
  25 (min), r l / m 25/30 5/6
  s2 22 4(min2)
• W lEX2/(2(1-r)) 1/30 m-2 s2/(2(1-5/6))
• 1/30 252 22/(1/3) 1/306293 62.9
  min
• NQ l W 1/30 62.9 2.097 (customers)

38
M/G/1 Example (cont.)

• Although working faster on the average, Ables
  greater service variability results in an average
  queue length about 30 greater than Bakers. On
  the other hand, the proportion of arrivals who
  would find Able idle and thus experience no delay
  is P0 1 - r 1 / 5 20, while the proportion
  who would find Baker idle and thus experience no
  delay is P0 1 - r 1 / 6 16.7. On the basis
  of average queue length, NQ , Baker wins.

39
M/G/1 System

• r l / m ,
• L r l2 (m-2 s2) / 2 (1 - r) r r2
  (1 s2 m2) / 2 (1 - r)
• W m-1 l (m-2 s2) / 2 (1 - r)
• Wq l (m-2 s2) / 2 (1 - r)
• Lq l2 (m-2 s2) / 2 (1 - r) r2 (1
  s2 m2) / 2 (1 - r)
• P0 1 - r