ISyE 6650 Test 2 Solutions

Transcription

1 1 NAME ISyE 665 Test 2 Solutions Summer 2 This test is open notes, open books. The test is two hours long. 1. Consider an M/M/3/4 queueing system in steady state with arrival rate λ = 3 and individual service rate µ = 1. (a) Find the probability that the system is empty. Solution: First of all, we see that λ = λ 1 = λ 2 = λ 3 = λ = 3 and Since we have µ 1 = µ = 1, µ 2 = 2µ = 2, µ 3 = µ 4 = 3µ = 3. c n = λ n 1 λ 1 λ µ n µ 2 µ 1, c = 1, c 1 = 3, c 2 = c 3 = c 4 = 9 2. Further, P n = c n P, where P = ( ) 1 c n =.571. n= (b) What is the expected number of customers in the system (queue + servers)? Solution: L = np n 4 = P nc n = n= n=1

2 2 2. Short Questions. Just write your answer. (a) What is L = λw known as? Law). Solution: Little s Law (or the Conservation (b) TRUE or FALSE? A Poisson process is a renewal process. Solution: TRUE. (c) TRUE or FALSE? The steady-state probability distribution of the number of customers in an M/G/ system is Pois(λ/µ). Solution: TRUE. (d) What is the traffic intensity of an M/M/k queueing system with arrival rate λ and individual service rate µ? Solution: λ/kµ. (e) TRUE or FALSE? The Pollaczek-Khintchine formula can be used to analyze the steady-state expected queue size for an M/M/1 system. Solution: TRUE. (f) Which of the following would a queueing theorist most likely be interested in PASTA, PESTO, Pr{ESTO}, or ANOVA? Solution: PASTA. (g) TRUE or FALSE? CTMC s are semi-markov processes. Solution: TRUE. (h) Suppose one day I place a Pois(1) number of bets at a casino; then I get tired and go home (presumedly to take a nap, as described in the next problem). Each time I place a bet, I win $X, where X Nor( 1, 9) (so I tend to lose a little money, but not always). Assume the outcomes of all bets are i.i.d. By the time I go home, what are my expected winnings? Solution: E[ N i=1 X i ] = E[N]E[X i ] = 1.

3 3 3. Suppose that I write papers for an Exp(3/day) amount of time before taking a nap for an Exp(1/day) amount of time; then I repeat the process over and over (and all research times and nap times are independent of each other). Suppose I start doing research at midnight on Monday morning. (a) What s the probability that I take my first nap before 6 am on Monday morning? Solution: (Wow! These are long naps, eh?) Let X 1 denote the time that the first nap begins; clearly, X 1 Exp(3). This implies that Pr(X 1 < 1/4) = 1 e 3/4 =.528. (b) What s the probability that I ll be sleeping at 6 am on Monday morning? Solution: Let state denote working; and let state 1 denote napping. We re interested in P 1 (t), where t = 1/4, λ = 3, and µ = 1. Then, from class notes, we have ] λ P 1 (t) = [1 e (λ+µ)t =.474 λ + µ (c) What s the probability that I ll be sleeping at 6 am next Monday morning? Solution: Let t get big in your solution above. then P 1 = λ = 3/4. λ+µ

4 4 4. Suppose customers arrive at a service station according to a PP(λ = 4). Would you rather have (i) one fast server at the station, i.e., one guy who could process customers in an Exp(µ = 6) amount of time, or (ii) two slower servers, i.e., two parallel servers, each of whom could process customers in an Exp(µ = 3) amount of time? Solution: Let s base the comparison on w, the steady-state expected amount of time in the system. (i) M/M/1: Here λ = 4, µ = 6, so that ρ = 2/3. Then L = ρ/(1 ρ) = 2, and by Little s Law, w = L/λ = 1/2. (ii) M/M/2: Here λ = 4, µ = 3, and k = 2, so again ρ = 2/3. Now we have (from class notes) [ k 1 (kρ) n ] P = + (kρ)k ) 1 =.2. n! k!(1 ρ) This implies that n= L = kρ + (kρ)k+1 P k(k!)(1 ρ) 2 = 2.4, yielding w = L/λ =.6. Thus, k = 1 is better! Actually, there might be a better way to do this problem... (i) M/M/1: λ = λ 1 = λ 2 = = λ = 4 and µ 1 = µ 2 = µ 3 = = 6. (ii) M/M/2: We still have λ = λ 1 = λ 2 = = λ = 4, but now µ 1 = 3 and µ 2 = µ 3 = = 6. Comparing the two alternatives reveals that everything is the same except for µ 1, which is greater for the M/M/1 case.

5 5 5. Renewal Theory Questions. (a) TRUE or FALSE? The event {N(t) < n} implies that the nth arrival occurs after time t. Solution: TRUE. (b) TRUE or FALSE? The event {N(t) < n} implies S n t. Solution: TRUE. (c) If the renewal function m(t) = 7t, what is the interarrival-time distribution? Solution: Exp(7). (d) TRUE or FALSE? S N(t) t < S N(t)+1 for all t. Solution: TRUE. (Actually, this could be FALSE if we allow interarrival times why?) (e) Consider a renewal process having i.i.d. interarrivals occuring at the rate of 5/day. What does N(t)/t converge to as t gets big? Solution: N(t)/t 1/µ = 5. (f) TRUE or FALSE? The Elementary Renewal Theorem implies that t/n(t) µ (where µ = E[X 1 ] is the mean renewal time). Solution: FALSE. (It s usually true, but the ERT deals with m(t), not N(t).) (g) What do you get when you add up the age at time t with the excess at time t? Solution: X N(t)+1. (h) Suppose arrivals occur according to a PP(5/hr). What s the approximate probability that we observe between 95 and 15 arrivals in a 2-hour period? Solution: Since λ = 5, we have µ = 1/λ = 1/5 and σ 2 = 1/λ 2 = 1/25. Now, by the Central Limit Theorem, N(t) t µ tσ 2 /µ 3 = N(t) 5t 5t Nor(, 1). This implies that Pr(95 N(t) 15) ( 95 1 Pr Nor(, 1) 1 ) = Pr(.5 Z.5) = 2Φ(.5) 1 =.383.

6 6 6. Suppose that W(t) is a standard Brownian motion process. (a) What is the mean of W2 (t) dt? Solution: [ ] E W 2 (t) dt = E[W 2 (t)] dt = Var(W(t)) dt = t dt = 1/2. (b) What is the variance of W2 (t) dt? Solution: (OK, I admit that this was a pretty difficult problem! Nevertheless... ) Suppose s < t, and let s calculate E[W 2 (t)w 2 (s)] = E[(W(t) W(s) + W(s)) 2 W 2 (s)] = E[(W(t) W(s)) 2 W 2 (s)] + 2E[(W(t) W(s))W 3 (s)] + E[W 4 (s)] = E[(W(t) W(s)) 2 ]E[W 2 (s)] + 2E[(W(t) W(s))]E[W 3 (s)] + E[W 4 (s)] (by independent increments of Brownian motion) = E[(W 2 (t s))]e[w 2 (s)] + E[W 4 (s)] (by stationary increments and the fact that E[W(t)] = ) { } 2 = Var(W(t s))var(w(s)) + Var(W 2 (s)) + E[W 2 (s)] = (t s)s + s 2 Var(χ 2 (1)) + s 2 {E[χ 2 (1)]} 2 (since W(s) Nor(, s)) = (t s)s + 2s 2 + s 2 = ts + 2s 2. Thus, if s < t, Cov(W 2 (t), W 2 (s)) = E[W 2 (t)w 2 (s)] E[W 2 (t)]e[w 2 (s)] = E[W 2 (t)w 2 (s)] Var(W(t))Var(W(s)) = ts + 2s 2 ts = 2s 2. So finally we have ( Var ) W 2 (t) dt ( = Cov W 2 (t) dt, ) W 2 (s) ds

7 7 = = 2 = 2 t Cov(W 2 (t), W 2 (s)) ds dt (by symmetry) t Cov(W 2 (t), W 2 (s)) ds dt 2s 2 ds dt = 1/3. (c) Suppose that B(t) = W(t) tw(1) is the associated Brownian bridge process. What is the mean of B2 (t) dt? Solution: [ ] E B 2 (t) dt = E[B 2 (t)] dt = Var(B(t)) dt = t(1 t) dt = 1/6.

8 8 7. (Only if you have time!) Consider a stationary, ergodic DTMC X 1, X 2,... (i.e., one that s been running long enough so that it s in steady state). Denote the transition probabilities by P ij and the stationary probabilities by π i. Suppose that, starting at some point n, we trace the sequence of states X n, X n 1,... backwards. (a) Prove that the backwards transition probabilities satisfy Pr{X m = j X m+1 = i} = π jp ji π i, for all i, j. Solution: Hey, this is all reversibility stuff! Pr{X m = j X m+1 = i} = Pr{X m = j, X m+1 = i} Pr{X m+1 = i} = Pr{X m = j} Pr{X m+1 = i X m = j}. Pr{X m+1 = i} (b) Prove that the backwards process is a DTMC. Solution: See text.