IFSM 300 – Class 25
April 29, 2002
Preliminaries
 Assignment 25: work on homework 6.
 Homework 6 (problems in assignment 25) is due Monday, May 6.
 Class Web site reminder: http://www.gl.umbc.edu/~rrobinso/
Finite Source of Customers
 Complete notes below. Summary here.
 Refer to Processing Office Work, situation 15.4, page 773.
 Notation.
 Assumptions.
 Table of performance measures.
 Numerical examples.
 Little’s Law.
Finite Capacity in the System
 Complete notes below. Summary here.
 Refer to Furniture Refinishing, situation 15.5, page 776.
 Notation.
 Assumptions.
 Table of performance measures.
 Numerical examples.
 Little’s Law.
Multiple Servers
 Complete notes below. Summary here.
 Refer to Expanding Service Facilities, situation 15.6, page 778.
 Notation.
 Assumptions.
 Table of performance measures.
 Numerical examples.
 Little’s Law.
Millionaire Q1
In a multiple-server model, the “number of channels” means the:
A. Number of lines.
B. Number of arrival positions.
C. Number of service positions. (!!)
D. Number of English Channels.
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Millionaire Q2
Steady state is achievable in a multiple-server model only when:
A. M  >  .
B.  > M .
C. S  >  . (!!)
D.  > S .
Millionaire Q3
In the finite-queue-length, single-server model (M/M/1/K model), PK is the
symbol for:
A. The fraction of arriving customers who will leave without waiting.
B. The probability that K customers are in the system.
C. The probability that the system is full.
D. All of the above. (!!)
Millionaire Q4
In an M/M/1/K system, the average rate at which customers are leaving after
being served is the:
A. Service rate.
B. (Service rate) times (fraction of waiting customers who will be served).
C. Arrival rate.
D. (Arrival rate) times (fraction of arriving customers who stay). (!!)
Millionaire Q5
In a finite-source model, the letter M denotes:
A. Markov.
B. A large cost.
C. Number of customers. (!!)
D. Maximum line length.
Millionaire Q6
When working with the finite-source model, we represent the arrival rate of
a single customer as:
A. M .
B. (M – n) .
C. (M – L) .
D.  . (!!)
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Notes: Other Queueing Models
Finite Source of Customers. In this adaptation of the basic, single-server model (M/M/1
model), we consider the case where the population of prospective customers is a fixed
number M. Try not to confuse this M with the M meaning Markov in M/M/1, or the M
meaning large cost in transportation and assignment models.
In other words, with a total of M customers available, when n customers are in the
system, the number of customers remaining to possibly arrive at the system will be
M – n.
You could call this model a finite-source M/M/1 model. Again, the “M” in M/M/1 means
Markov, not the finite number of customers in the population.
Notation. Our notation is the same as before, except for these revisions:
M – fixed number of possible customers
 – mean arrival rate of each individual customer.
Before,  was the mean total rate at which customers arrive at the system. Now,  is the
mean rate at which each customer outside the system will arrive. In this finite-source
model, therefore, the mean total arrival rate varies as the number n of customers in the
system varies. When n customers happen to be in the system, the mean total arrival rate
of customers from outside will be (M – n)  .
Assumptions. The crucial assumptions behind the finite-source model are:
1. Each customer’s time outside the system is exponentially distributed with
mean 1 / . Hence the mean arrival rate of each outside customer is  .
2. Service time is exponentially distributed with mean 1 /  . So the mean service
rate is  .
Thus Poisson-process models apply to arrivals and service but the details of the system
model are changed now because of the fixed customer population M.
Note that in this model we do not worry about achieving steady state. The number of
customers in the system cannot exceed M and therefore indefinite growth can’t occur.
Table for the Finite-Source Model. The formulas of this model are summarized in table
15.10, page 774 in the book, shown below at the top of the next page.
Except for reconsidering Little’s Law, we will not review any of the derivations, which
have moved up to a higher level of complexity. Instead, we’ll focus on numerical
examples.
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Numerical Examples. The data in these examples are M = 3 customers,  = 0.25
customers per hour, and  = 0.75 customers per hour. Start with P0:
M
P0 = 1 /
3
 [(M!/(M – n)!)]( /)n
n=0
= 1 / [  [(3!/(3 – n)!)]( 0.25/0.75)n.
n=0
Evaluating the denominator, we find
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 [(3!/(3 – n)!)]( 0.25/0.75)n = (3!/3!)(0.25/0.75)0 + (3!/2!)(0.25/0.75)1
+ (3!/1!)(0.25/0.75)2 + (3!/0!)(0.25/0.75)3
n=0
= 1.0 + 1.0 + 0.6667 + 0.2222 = 2.8889
So the result for P0 is:
P0 = 1 / 2.8889 = 0.3462.
Next, examine Pn in general:
Pn = [M!/(M – n)!]( /)n P0 , n = 0, 1, 2, …
Examples of Pn with n = 1 and n = 2 are:
P1 = (3!/2!)(0.25/0.75)1(0.3462) = 0.3462
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P2 = (3!/1!)(0.25/0.75)2(0.3462) = 0.2308
The third line in the table is LQ:
LQ = M – [( + ) / ](1 – P0) = 3 – [(0.25 + 0.75) / 0.25](1 – 0.3462)
= 0.3848 customers in the queue.
Next is L:
L = LQ + (1 – P0) = 0.3848 + (1 – 0.3462) = 1.0386 customers in the system.
Then WQ:
WQ = LQ /  (M – L) = 0.3848 / (0.25)(3 – 1.0386) = 0.7847 hours in the queue.
Last, we have W:
W = WQ + (1 / ) = 0.7847 + (1 / 0.75) = 2.118 hours in the system.
Little’s Law for a Finite Source. Because the rate at which customers in total enter the
system and the queue, equal to the mean total arrival rate (M – n) , varies as n varies, we
need the mean :
_ M
M
M
M
 =  n Pn =  (M – n) Pn =  (M  Pn –  n Pn) =  (M- L).
n=0
n=0
n=0
n=0
Using the mean  thus determined, Little’s Law in this model gives us:
_
L =  W =  (M – L) W =  MW –  LW
L + WL = MW
L = MW / (1 + W) = (0.25)(3)(2.118) / [1 + (0.25)(2.118)]
= 1.0386 customers in the system.
_
LQ =  WQ =  (M – L) WQ = (0.25)(3 – 1.0386)(0.7847)
= 0.3848 customers in the queue.
Finite Capacity in the System. The queueing model now contemplated applies to a
queueing system that has room for no more than K customers. Otherwise, the system is
basic, single-server.
The official classification of this model is M/M/1/K. With such expanded descriptors, we
would call the models studied so far M/M/1/ and finite-source M/M/1/ .
Notation. We employ the basic-single-server-model notation with one addition:
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K – maximum number of customers who can be accommodated in the system.
In other words, n  K.
Assumptions. The new assumption is that when customers have filled the system to
capacity (total number in the queue plus being served equals K) any arriving customer
cannot enter the system and therefore departs.
Once again we have Poisson arrivals and service. But the capacity limit produces
different formulas, as did the source limit we just reviewed.
Because the number of customers in the system can’t be larger than K, indefinite growth
cannot take place and so we need not question whether steady state may be achieved.
Table for the Finite-Capacity Model. The table of formulas – table 15.11 from page
777 in the book – is shown below. Once again we will review numerical examples but
skip derivations, except for the derivation of Little’s Law.
Numerical Examples. The data for these examples are K = 7 customers,  = 30
customers per week, and  = 33 customers per week. Begin with P0.
P0 = [1 – (/)] / [1 – (/)K+1] = [1 – (30/33)] / [1 – (30/33)7+1] = 0.1704 .
Observe that the fundamental time unit in this example is one week. That’s fine so long
as we are consistent, by expressing both  and  “per week.”
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In general, Pn is:
Pn = ( / )n P0 , n = 0, 1, 2, …
This is the same as Pn for the unrestricted basic, single-server model. Examples of P1 and
P2 are:
P1 = ( / )1 P0 = (30/33)( 0.1704) = 0.1549
P2 = ( / )2 P0 = ( / ) P1 = (30/33)(0.1549) = 0.1408.
As we saw when reviewing the M/M/1 model, the easy way to move from P0 to P1 to P2
and so on is to keep multiplying by ( / ).
Next we come to a formula characteristic of this finite-capacity model. When the system
becomes full, arriving customers are turned away. The question asked now is: What’s the
probability that a newly arrived customer will be turned away? The answer is:
Pr {arriving customers are turned away} = Pr {the system is full} = PK.
We already know the formula for PK since it’s Pn when n = K:
PK = ( / )K P0 = (30/33)7(0.1704) = 0.08744 .
The next formula in the table is that for L:
L = [( / ) / (1 - ( / ))] – [(K + 1) ( / )k+1 / (1 - ( / )k+1)]
= [(30/33) / (1 - (30/33))] – [(7 + 1) (30/33)7+1 / (1 - (30/33)7+1)]
= 3.004 customers in the system.
Now LQ:
Lq = L – [(1 – Pk) / ] = 3.004 – [30(1 – 0.08744) / 33]
= 2.174 customers in the queue.
Then WQ:
WQ = Lq / (1 – Pk) = 2.174 / 30(1 – 0.08744)
= 0.0794 weeks in the queue.
Finally W:
W = WQ + (1 / ) = 0.0794 + (1/33) = 0.1097 weeks in the system.
Little’s Law for Finite Capacity. To employ Little’s Law, we need, as always, the mean
rate at which customers actually enter the system:
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
K-1
mean  =  n Pn =   Pn + (0) PK = (1 – Pk) + (0) PK = (1 – Pk).
n=0
n=0
Therefore,
L = (1 – Pk) W = (30)(1 – 0.08744)(0.1097) = 3.004 customers in the system.
LQ = (1 – Pk) WQ = (30)(1 – 0.08744)(0.0794) = 2.174 customers in the queue.
Multiple Servers. The fourth of the four queueing models we are reviewing is the basic
model with two or more servers. In the classification notation, the number of servers is
designated “c” meaning “channels.” This model is referred to as having “parallel
channels.” The description of it is M/M/c/ , or in the shortened version M/M/c.
Assumptions. The key idea is that instead of being limited to a single server we have S
identical servers, each serving at a rate  when busy. There continues to be a single
queue. When a customer reaches the head of the line, the customer goes to the next
available server.
In this case – as in the basic, single-server model – being able to reach steady state is not
assured. The test now, clearly, is S  >  . Observe that the total service rate is n  for n
between 1 and S, and is S  for n from S upward.
Table for the Multiple-Server Model. In the book, see table 15.12, page 780. This table
is given below, at the top of the next page.
One more time we will work examples and bypass derivations except to consider Little’s
Law.
Numerical Examples. In these examples we assume  = 4 customers an hour,  = 5
customers an hour, and S = 2 servers.
The check to confirm that the system will be able to achieve steady state is: S  > 
(10 > 4).
Starting on line 1 of the table, for P0 we have:
S-1
P0 = 1 / [  (1 / n!)( / )n + (1 / S!)( / )S(S / ( S - ))]
n=0
= 1 / [(1/0!)(4/5)0 + (1/1!)(4/5)1 + (1/2!)(4/5)2(10/(10 – 4))]
= 1/ [1 + (4/5) + (1/2)(16/25)(10/6)]
= 0.4286 .
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Then we come to the probability an arriving customer has to wait PW:
PW = (1 / S!)( / )S[S / ( S - )] P0
= (1/2!)(4/5)2(10/(10 – 4)(0.4286)
= 0.2286 .
The general Pn depends upon whether or not all servers are busy:
[( / )n / n!] P0
if n  S
Pn =
[( / )n / S!(Sn-S)] P0 if n  S .
For example,
P1 = [(4 / 5)1 / 1!] (0.4286) = 0.3429
P2 = [(4 / 5)2 / 2!] (0.4286) = 0.1372
P3 = [(4 / 5)3 / 2!(21)] (0.4286) = 0.0549.
Returning to PW, notice that with 2 servers the probability of waiting is the probability
that both servers are busy. In other words, in this numerical example,
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PW = 1 – (P0 + P1) = 1 – (0.4286 + 0.3429) = 0.2285 (same as above except for
rounding).
Next in the table comes LQ:
Lq = [()( / )S / (S – 1)!( S - )2] P0
= [(20)(4/5)2 / (1)!(10 - 4)2] (0.4286)
= 0.1524 customers in the queue.
Then we have L:
L = LQ + ( / ) = 0.1524 + (4/5) = 0.9524 customers in the system.
The next formula in the table is for W. Note that it comes directly from Little’s Law:
W = L /  = 0.9524 / 4 = 0.2381 hours in the system.
Finally, we have WQ:
WQ = W – 1 /  = 0.2381 – (1 / 5) = 0.0381 hours in the queue.
Little’s Law for Multiple Servers. The presence of more than one server doesn’t affect
the mean rate at which arriving customers enter the system. Thus that rate is  and the
formulas of the basic, single-server model apply again:
L =  W = (4)(0.2381) = 0.9524 customers in the system.
LQ =  WQ = (4)(0.0381) = 0.1524 customers in the queue.
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