On priority queues with impatient customers

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On priority queues with impatient customers
by Foad Iravani, Baris Balcioglu
Department of Mechanical and Industrial
Engineering University of Toronto E-mailbaris_at_mie
.utoronto.ca
29/08/2008- Can Queue Carlton University, Ottawa
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Literature on Queues with Impatient Customers

• Extensive Literature on Single Class of
  Impatient Customers
• Choi, Kim and Chung (2001) M/M/1 Queue, 2
  classes of customers,high-priority has
  deterministic patience limitand
  preemptive-resume priority over patient 2nd class
• Brandt and Brandt (2004) M/M/1 Queue, 2
  classes of customers, high-priority has i.i.d.
  general impatience time and preemptive-resume
  priority over patient 2nd class
• Argon, Ziya and Righter (2008) M/M/1 Queue, 2
  classes of finite number of customers, both
  class have exponential patience time

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Literature on Queues modeling Call Centers

• Brandt Brandt (1999) M(n)/M(m)/NGI Queue
• Armony and Maglaras (2004) M/M/N Queue
• Ahghari and Balcioglu (2008)

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Problems

• Single server queue with preemptive-resume
  priority when both classes have exponential
  times to abandon
• Single server queue with non preemptive priority
  with only high-priority class having exponential
  times to abandon
• A multi server queue with impatient customers
  with call-back option

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Problem 1
exponential time-to-abandon, rate w1
High-priority customers
Poisson(l1) Arrivals
S
Poisson(l2) Arrivals
CDF and pdf of General Service Times B1(x),
b1(x) Class 1 B2(x), b2(x) Class 2
Low-priority customers
exponential time-to-abandon, rate w2
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Solution to Problem 1

• For class 1 (that does not see class 2), we have
  the M/G/1M analysis
• For class 2, we employ the level crossing method
  due to Brill (1975, 1979), Brill and Posner
  (1977)
• fi(x) virtual waiting time density function of
  class i1,2 until the first time its service
  starts

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Analysis of the M/GI/1GI Queue

• Stanford (1978)
• Let V denote the workload seen by an arriving
  customer withF(x) and f(x) as its distribution
  and density function

where is the complementary
abandon-time distribution
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Analysis of the M/GI/1GI Queue (2)
After some manipulation
where
and
is the LT of comp. service time
Hence, by numerically inverting
one can obtain WS(x)
For numerical inversion see such as Jagerman
(1982)
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Level Crossing in the 2nd Queue
Observe that the system alternates between idle
(IP) phases with no customers and active (AP)
phases, noting
V
Down-crossing
x
time
Customer arrivals upcrossing
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Solution to Problem 1 (II)

• In the AP, we first consider the initial jump
  with CDF GI(x)
• If a class 1 customer initiates the AP, the
  workload for class 2 is the busy period generated
  by a class 1 customer
• Rao (1967) provides the busy period
  distribution (BPD) L0(x) in the M/G/1M queue
  with LST
• If a class 2 customer initiates the AP, the
  workload for class 2 is the completion time with
  CDF C2(x) (Gaver, 1962)

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Solution to Problem 1 (III)

• Denoting E(AP) as the expected length of AP, we
  equate expected no of upcrossings and
  downcrossings at level x

• We observe a recursion

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Solution to Problem 1 (IV)

• Assuming that we stop the recursion at the mth
  step

is the LT of the comp. completion
time distribution

• Denoting the E(IP)1/(l1l2) as the expected
  length of IP

• Hence, the Laplace transform of f2(x) is in hand

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Solution to Problem 1 (V)

• One can find, P1,0, probability that no class 1
  customer exists
• Following Stanford (1979), probabilities of
  abandonment

• Following Stanford, waiting time distributions
  of successful and abandoning customers in each
  class can be found too.

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Problem 2
exponential time-to-abandon, rate w1
High-priority customers
Poisson(l1) Arrivals
S
Poisson(l2) Arrivals
CDF and pdf of General Service Times B1(x),
b1(x) Class 1 B2(x), b2(x) Class 2
Low-priority customers
Non-preemptive priority rule!
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Solution to Problem 2

• For class 1, observing that

• The initial jump for class 1

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Solution to Problem 2 (II)

• We equate expected no of upcrossings and
  downcrossings at level x

• We observe another recursion, and with some
  difficulty one obtainsand E(AP) then

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Solution to Problem 2 (III)

• For class 2 customers, the initial jump

where HB2L, with L as the period generated by
patient class 1 customers to arrive during B2

• If class 2 service time is x, dist. of no of
  class 1 customers at the end of x is (Stanford,
  1979)

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Solution to Problem 2 (IV)

• With Lj-1 denoting the length of busy period
  initiated by j-1 class 1 customers and

• Only for deterministic, exponential and k-stage
  hyperexponential distributions, we can find

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Solution to Problem 2 (V)

• We can see

and
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Problem 3 Call Back Option
With q, leave system
exponential time-to-abandon, rate w1
S1
Poisson(l) Arrivals
S1
. . .
When all-servers busy, with (1-D), leave a
voice-message
With (1-q), leave a voice-message
SN
N identical Servers Exponential Service Times
Call-back queue
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Solution to Problem 3

• Define AP as the time when all servers are busy.
• For class 1 (customers who stay on line)

• Let Pj be the probability that j (lt N-1)
  servers busy

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Solution to Problem 3 (II)

• During the AP

s.t.
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Solution to Problem 3 (III)

• The rest of the analysis makes use of

• After some manipulation with K being a constant

• We can also find P0,
• probability that a customer leaves a voice
  message
• Probability that a customer on line will be
  served before leaving a message or reneging
• Probability of leaving a call-back message after
  waiting on line for a while
• And probability of abandonment

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Solution to Problem 3 (IV)

• For the call-back queue, we cannot make use of
  level-crossing technique
• An analysis similar to Brandt and Brandt (1999)
  gives the factorial moments of the call back
  queue size

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Future Work

• Announcing mean waiting time in a contact
  center, how does it affect the customer behavior?
• Can it be used to dynamically direct callers to
  call-back queue?