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Unformatted text preview: = 2, and µ = 3, so plugging into (4.25) and
multiplying numerator and denominator by 34 = 81, we have
1 ⇡ (0) = 2/3
81
=
4
(2/3)
81 1
2
⇡ (1) = ⇡ (0) = 18/65
3
2
⇡ (2) = ⇡ (1) = 12/65
3
2
⇡ (3) = ⇡ (2) = 8/65
3 54
= 27/65
16 From the equation for the equilibrium we have that the average queue length
L=1· 18
12
8
66
+2·
+3·
=
65
65
65
65 Customers will only enter the system if there are < 3 people, so
a = 2(1 ⇡ (3)) = 114/65 and using the idle time formula (3.5)
⇡ (0) = 1 a 3 =1 114
81
=
195
195 140 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS Using Little’s formula, Theorem 3.6, we see that the average waiting time for
someone who enters the system is
W= L = a 66/65
66
=
= 0.579 hours
114/65
114 To check this we note that 1
2
1
W=
⇡ (0) + ⇡ (1) · + ⇡ (2) · 1
1 ⇡ (3)
3
3
27 1 18 2 12 3
9 + 12 + 12
33
66
=
·+
·+
·=
=
=
57 3 57 3 57 3
57
57
114
From the last computation we see that WQ = W 1/3 = 14/57. We do not
compare this result with the PollaczekKhintc...
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 Spring '10
 DURRETT
 The Land

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