Shanghai Jiao Tong University

• ECE operating

• Solutions to Homework Six 1. For an M/M/1 queue in steady state, let Na be the number of customers the next arrival will find in the system. What is the probability P{Na = n} and E[Na] ? Solution: Let N a denote the number found by the next arrival, and let X be the number currently in the system. Conditioning on X yields P{Na = n} = X∞ k=0 P{Na = n|X = k}P{X = k} = X∞ k=0 P{Na = n|X = k}(λ/µ) k (1 − λ/µ) = X∞ k=n P{Na = n|X = k}(λ/µ) k (1 − λ/µ) = X∞ i=0 P{Na = n|X = n + i}(λ/µ) n+i (1 − λ/µ) Now, for n > 0, given there are currently n+i in the system, the next arrival will find n if we have i services before an arrival and then an arrival before the next service completion. By the lack of memory property of exponential interarrival random variables, this gives P{Na = n|X = n + i} =  µ λ + µ i λ λ + µ , n > 0 Consequently, for n > 0, P{Na = n} = X∞ i=0  µ λ + µ i λ λ + µ (λ/µ) n+i (1 − λ/µ) = λ λ + µ (λ/µ) n (1 − λ/µ) X