If we let t1 and t2 be the waiting 142 chapter 4

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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 Pollaczek-Khintc...
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