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µ
P1
To ﬁnd the value of ⇡ (0), we recall that when ✓ < 1, n=0 ✓n = 1/(1 ✓).
From this we see that if < µ, then
1
X n=0 ⇡ (n) = 1
X ✓ ◆n n=0 µ ⇡ (0) = 1 ⇡ (0)
( /µ) So to have the sum 1, we pick ⇡ (0) = 1 ( /µ), and the resulting stationary
distribution is the shifted geometric distribution
✓
◆ ✓ ◆n
⇡ (n) = 1
for n 0
(4.23)
µ
µ
It is comforting to note that this agrees with the idle time formula, (3.5), which
says ⇡ (0) = 1
/µ.
Having determined the stationary distribution we can now compute various
quantities of interest concerning the queue. We might be interested, for example, in the distribution of the time TQ spent waiting in the queue when the
system is in equilibrium. To do this we begin by noting that the only way to
wait 0 is for the number of people waiting in the queue Q to be 0 so
P (TQ = 0) = P (Q = 0) = 1 µ When there is at least one person in the system, the arriving customer will
spend a positive amount of time in the queue. Writing f (x) for the density
function of TQ on (0, 1), we note that if the...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
 Spring '10
 DURRETT
 The Land

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