The order takes an exponentially distributed amount

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Unformatted text preview: 4.28), we investigate stability by solving the system of equations for the rj that represent the arrival rate at station j . As remarked earlier, the departure rate from station j must equal the arrival rate, or a linearly growing queue would develop. Thinking about the arrival rate at j in two di↵erent ways, it follows that K X rj = j + ri p(i, j ) (4.30) i=1 This equation can be rewritten in matrix form as r = r = (I p) + rp and solved as (4.31) 1 By reasoning in Section 1.9, where unfortunately r is what we are calling p here: r= 1 X n=0 pn = 1K XX ip n (i, j ) n=0 i=1 The answer is reasonable: pn (i, j ) is the probability a customer entering at i is at j after he has completed n services. The sum then adds the rates for all the ways of arriving at j . 148 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS Having found the arrival rates at each station, we can again be brave and guess that if rj < µj , then the stationary distribution is given by ⇡ (n1 , . . . , nK ) = K Y ✓ rj ◆nj ✓ µj j =1 1 rj µj ◆ (4.32) This is true, but the proof is more complicated than for the two s...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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