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# hw4 - CHAPTER 4 t M MARKOVIAN QUEUES PROBLEM 4.1 Consider...

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Unformatted text preview: CHAPTER 4 t M MARKOVIAN QUEUES PROBLEM 4.1 Consider the Markovian queueing system shown below. Branch labels are birth and death rates. Node labels give the number of customers in the system. (a) Solve for pk. (b) Find the average number in the system. (c) For /\ = u, what values do we get for parts (a) and (b)? Try to interpret these results. ((1) Write down the transition rate matrix Q for this problem and give the matrix equation relating Q to the probabilities found in part (a). SOLUTION (a) Using the ﬂow conservation law for states 0 and 2 and the conservation of probability, we get the following three independent equations: M70 = MP1 + MP2 #P2 = A121 P0+P1+P2=1 97 98 CHAPTER 4 MARKOVIAN QUEUES PROBLEM 4.3 99 Solving this gives (a) Draw the state diagram showing all the transition rates. p0 = xii—ﬂ (b) Write down the equilibrium equation for (i, j) where 1 < i < k, 1 < j s n. p1 = 01% SOLUTION A2 (a) The state-transition-rate diagram is p2 = W (b) We have kt (CUSTOMER "ARRIVES") 1V=O-po+l-pl+2-p2=)zﬁf:—:’;: IV = A(2)\ + p.) (A + #02 (c) If A = u, the results in parts (a) and (b) become Po=%.p1=P2=%,N=% To interpret these results, consider a cycle from state 0 back to state 0. The rate out of state 0 is A (= it), which puts the system into state 1. The rate out of state 1 is )t + M, = 21.1., and so the .fraction of time spent in state 1 must be half that spent in state 0. From state 1 we arrive at state 2 with probability % (or return directly to state 0 With probability %) and depart state 2 at rate 11.; therefore we spend as much time, on the average, in state 2, (i.e., % ' (1 /;.t)) as in state 1 (Le, 1/2p.). (d) Equation (1.53) implies that -q,-,' is the rate at which the system departs from state i, while (1;; (i 95 j) is the rate at which it moves from state i to state j. ."=3 § 5 a a a 2 2 2 g s e Thus —z\ A 0 Q = u -(u + A) A M 0 -u From Eq. (1.56) we have directly that (b) Using Flow Out = Flow In, we obtain "Q = 0 (n- = p = [po.p1,p2]) D (M + ”#01717 = kAp,_1,,-+ "Wm—1 f0” < i < k, 1 < j S n '3 PROBLEM 4.2 PROBLEM 4.3 Consider an Ek/En/l queueing system where no queue is permitted to form. A . customer who arrives to ﬁnd the service facility busy is “lost” (he departs with no service). Let (i, j) be the system state in which the “arriving” customer is in the ith arrival stage and the customer in service is in the jth service stage (note that there is always some customer in the arrival mechanism and that if there is no customer in the service facility, then we let j = 0). Let l/kA be the average time spent in any (21) Find P 1'1 j = 0v 1,...,r. arrival stage and 1/ nu be the average time spent in any service stage. (b) Find the probability of a busy system. Consider an M/Er/l system in which no queue is allowed to form. Let j be the number of stages of service left in the system and let P ,- be the equilibrium probability of being in state (i, j). 100 CHAPTER 4 MARKOVIAN QUEUES SOLUTION The state-transition—rate diagram is (a) The ﬂow equations are APO=VMP1 j=0 rMPj=rtu+1 ISer—l NLPr=AP0 j=r Of these r + 1 equations, one is redundant; using the ﬁrst r we see that A PO—Pl—P2—"'—Pr—l—Pr ru Also 23.20 P j = 1 implies that 1=1 ”L Thus p. P = 0 )t + M and therefore A P‘ = _ ' S J rot + ,t) J ’ (b) We have P[busy system] = 1 -— P0 = 1 — IL A + u P[bus s stem] — A y y )t + IL PROBLEM 4.4 PROBLEM 4.4 Consider an M/H2/l system in which no queue is allowed to form. Service is of the hyperexponential type with m = 2pm and [L2 = 2p,(l - a1). (a) Solve for the equilibrium probability of an empty system. (b) Find the probability that stage 1 is occupied. (c) Find the probability of a busy system. SOLUTION Let 1,- represent the state when there is one customer in the system and that customer is in stage i. The state diagram for this system is as follows: As usual, we have two independent ﬂow equations and the conservation of probability: M70 = 2Ma1pl, + 2M1 ‘ (101712 M11170 = 2M0“?!l p0+plr +p12 =1 Thus pt A+p P0: A P1. =P12 = m (a) The probability of an empty system is = = M Pth system] , P9 ,, or + p, (b) The probability that stage 1 is busy is A P[stage l busy] = p1, = m 101 102 CHAPTER 4 MARKOVIAN QUEUES PROBLEM 4.5 103 (C) The probability of a busy System is I — — Mp0 I-LPO Z _ x P(z) = = ————— ”busysysteml—1*Po=Pu+mz = ,\+,L D A(1—22)+M(1‘§) Maw“) PROBLEM 4.5 (Note that the average arrival rate X = 2A, and so p = Xi = 2A/u.) Thus - . P(z) = ____2p—o.__ Con31der an M/M/l system w1th parameters A and u in which exactly two customers 2 — pz(z + 1) arrive at each amval instant. Since P( 1) = 1 = 2pc / (2 — 2p) we have p0 = 1 — p. Hence (3) Draw the state—transiti’on‘rate diagram. ' P(z) = M (b) By inspection, write down the equilibrium equations for pk (k = 0, 1,2, . . .). 2 — pz(z + 1) (c) Let p = 2A/tr. Express P(z) in terms ofp and 2. (d) By Eq. (1'82)’ ((1) Find P(z) by using the bulk arrival result given in Eq. (1.82). P _ MU _ le — 2) (e) Find the mean and variance of the number of customers in the system from (Z) — “(I _ z) _ MU — G(z)] P (z). . , . In the system under consideration, bulks have constant size 2. Thus G(z) = 22 (f) Repeat pans (a)—(e) With exactly r customers arr1v1n ' ' g at each arrlval instant = (I) = (and p = rA/p). (and p AG (1)/p, 2A/u). Therefore M1 — p)(1 — z) P(z) = ——————————— SOLUTION “(1 — Z) — A20 ’ 22) . . . This sim liﬁes as before to (a) The state-transntion-rate diagram is as follows: P 2(1 — P) P = _.___—..—_——— x i (Z) 2 - pz(z + 1) V (e) The mean and variance of the number of customers may be found from the o . a ﬁrst and second derivatives of P(z). We ﬁnd that u u dP(z) = 2(1 — p)p(22 + 1) dz [2 - pz(z + 1)]2 (b) The equilibrium equations are [7 = dP(Z) = 2(1 _ P)P(3) . dz 2:] (2 —- 2p)2 )t = = po up; k 0 N = E p (/\+IL)P1=Mp2 k=1 21—p (A + Mpk = )tpk_2 + pp”l k 2 2 After simpliﬁcation, the second derivative is 2 _ 2 (c) Multiply the kth equation by 2" and sum for k 2 0. This gives d P(z) = 4(1 ” P)P ————“—‘——‘[2 pz(z + 1)] + p(22 + 1) , , ,,,,, m 7 7 ,_ 7 , (122 [2 —- pz(z + 1)]3 m w w ., —- 2 _ , ,_, V. Azpkzk-{Hp‘zpkzk=A2pk~22k+ﬂvzpk+1zk 2—N= dP(z) =4(1._p)p Z_2_P_+_9P k=0 k=l k=2 k=0 dz2 2:1 (2 — 2p)3 xP(z) + MHz) - p0] = )tzzP(z) + %[P(z) — p0] = 271%?(2 + 7p) 104 CHAPTER 4 MAHKOVIAN QUEUES By deﬁnition, we may ﬁnd the variance of N as a§=N5—(N)2=(1‘v_2—1V)+1V—(1V)2 p 3 p 9 102 =—-— + +——————_ 211—12)”2 7") 21—p 4(1—p)2 2=P(10—P) “” 40-12)2 (1‘) The state-'transition—rate diagram is The equilibrium equations for pk are M70 = MP1 k = 0 0t + [.0121 = Mpk+1 1 5 0l + MP]: = AIM—r + Mpk+1 k 2 r Multiply the kth equation by z" and sum: A 21712" + uZka" = A :1»er + #:14sz k=0 k=1 k=r k=0 ”(2) + 1L[P(2) — p0] = Az’mz) + 3112(2) — p01 __ ”4170(2 — 1) Pm “ M2 — 1) — )tz(z’ — 1) P(Z) _ MPO IL - M 22312" As p = rA/p, (X = M and so p = X? = M/M), we may write 7170 p = _____ (Z) r _ p 22:1 2k Also P(l) = 1 = rpo/(r — rp) implies that p0 = 1 — p. Thus VW "i ”MW WWW” ‘ N {(1‘ijH’Wﬁﬁrmmw [3 _ (z) r P 22:12" To see this in another way, for the bulk arrival system with constant bulk size r, we have G(z) = 2’. Substituting this into Eq. (1.82) and simplifying gives PROBLEM 4.5 105 as before To ﬁnd N we note that (110(2): _ Ez=lkzk'1 dz r(l p)p——— so that _ dP(z) r(r + l)/2 N = :g 1 — dz z=1 r( p)p (r — rP)2 Thus N = r + 1L 2 l - p To ﬁnd of, we ﬁrst obtain (1213(2) dz2 (r — rp)22=1k(k-1) + 2142221 k)2 [W—IV = (r - rp)3 = r(l—p)p [ z=l Now recall that r :k=r(r+1)and Zk2=r(r+l)(2r+l) k=1 2 k=1 6 Therefore Zk(k— 1) = (r — l)r(r +1) k=1 3 and __ 2 — r(l-p)(r l);(r+l) +2p (r(r;1)) N2 — IV = 1— " ”)9 1r(1— p)13 (r +1)P = “60 _ p)2(2r — 2 + pr + 5p) and so 0—; =(1TI—2—1V)+1V—(1V)2 '='(r‘+‘ljﬁ'm‘_kﬂﬂ ' (r+1"')p'_(Fél¥1)2p2 6(1— 1112‘” 2 + ”r + 5") + 2(1— p) 4(1~ p)2 +1‘ aﬁ=——————1(2"(1_)pﬁ;2(4r+2—pr+p) Cl 138 CHAPTER 5 THE QUEUE MIG/1 (b) We ﬁrst simplify the expression obtained in part (a), and then ﬁnd the limit as t——)°°. _M()tt)" _ " 1 ' "‘k P[N(t)=k]= 2e n (gt)? /[1 300]de [F/OB(x)dx] n—k k A f0’[1 —B(x)]dx] [A fo‘ B(x)dx] = [M Z [ k! ' (n — k)! n=k _’e w [A fo‘n —kB(x)]dx] z: [A fo’Bocwx] [A fo’n —-B(x)]dx]k , P[N(t) = k] = (“Hex 1,8qu k! [A fo'n — B(x)]dx]k ET“ Thus, for every t, N(t) lS Poisson with parameter A f0[1 - B(x)] dx. Letting t —> 00 and noting that _ e—A fo‘n—mxndx lim/ [1 —B(x)]dx 2/”[1_B(x)]dx = X 0 we see immediately that - A‘ k pk é lim Pk(t) = e_”( x) t~—»°° k! Thus as t —» 00, the limiting distribution of number in system is Poisson with parameter A}, which is independent (except for the mean) of B(x). El PROBLEM 5.9 Consider M/Ez/l . (a) Find the polynomial for G*(s). (b) Solve, for S(y) = P[time in system S y]. ,, WVSOLLITION _ (a) For the M/Ez/I system, the Laplace transform of the service time density is 2 2 3*“) = (S +2“) PROBLEM 5.9 139 Thus Eq. (1.111) gives G*(S) = [4—]2 s + A — AG*(s) + 2p. Expanding, we get )12[G”(s)]3 — 2A(s + A + 2M)[G*(s)]2 + (s + A + 2p.)2G*(s) — 4,12 = 0 (b) Equation (1.106) gives S(1~ p) s — )\ + AB*(s) ( 2» )2 s(1—p) s+2p, 2M 2 s—A+A<S+2#) S*(s) = B*(s) Thus S *(S) 4M2(1 _ P) 32 + (4p, ~ )1)s + 4,1,(11. -- A) ll The denominator s2 +(4p.—r/\)s+4p.(p,— )t) has roots 31, 32 (where p = A/u): -M(4 - p) + m/p2 + 8p 2 ~u(4-p)-M 02+8P ———————-——2 t 11 51 52 We note that, for p < 1, we have 16p < 16 and thus (4 — p)2 > p2 + 8p. Hence 5; < s, < 0 for O < p < 1. Factoring, 4130 ~ 1)) (S — SIXS — S2) __4p,2(1—p)( 1 _ 1) ’1‘ p2+8p S_31 S”52 3*(S) = Invert to ﬁnd the pdf s(y) as 4M1 - p) (em .. em) S = (y) . .pzwiztSp Thus the PDF S(y) is 4M1—P) ——(€ s;y_ S = (y) p2 + 8p SI 1) ~ ﬁes” — 1)] a 140 CHAPTER 5 THE QUEUE MIG/1 PROBLEM 5.10 Consider an MID/1 system for which E = 2 sec. (a) Show that the residual service time pdf 13(x) is a rectangular distribution. '(b) For p = 0.25, show that the result of Eq. (1.108) with four terms may be used as a good approximation to the distribution of queueing time. SOLUTION (a) The service time distribution is given by 0' x<2 1 x22 B(x) '= { The residual service time pdf is Thus 13(x) is rectangular. th) ‘/z (b) The ﬁrst four terms of the series in Eq. (1.108) give wo) '=~ wappmxo) '3 (1 — p) [now + p13(y)+ pzémo) + p313<3)(y)] As 1 < 2 120) = {2 y 0 y 2 2 NWVW—éwgéwewtﬁa‘f WW" I V i) 0 S y S 2 b(2)(}’) = PROBLEM 5.10 141 and 2 y _. 05 \$2 16 y _ 2 3 3 b )= —Z—+— —— 25 34 mg 8 4y 4 y 2 y 3 9 ——— +— 4s 56 16 4y 4 y SW) gmﬂY) {1(3)in 1/2 1/2 1/2 3/8 ‘1 Y Y 0 1 2 3 0 1 2 3 4 0 1 2 3 4 5 6 We compare w(y) and wappmx(y) in three different ways. First, the area A under'the curve w(y) minus the area Aapprox under the curve wappm( y) is A — Aapprox = (1 _ WZPk/O 500(2))‘12’ = (1 _ p>Zpk k=4 k=4 1 = (1 ‘ PM4 (a) = P4 #l--‘ u Asp: _ l A ’ Aapprox — ﬁg Thus, in terms of area, we have a “good” approximation. Second, we note that wappmbi) = O for y 2 6. Thus the tail of the density w(y) is not approximated very well. Third, we compare the mean wait W with an approximation Wapprox calcu- lated from wappmx(y). [Note that wappmx(y) is not a pdf.) 00 7W=fiy7vrvm<§3£iyi éh — pi}: 22* f0 Siam ' " k=l We now observe that f: yl3(k)(y) dy has value k, since it represents the mean .sWQHﬁaV—Nﬁ,» .. - “214;: r w. . -~« 142 CHAPTER 5 THE OUEUE MIG/1 of a sum of k random variables each having mean 1. Thus ' °° a 1 W=l- k"=1~— ( ) ( 10),;=1 p ( Modp l_p or w = _‘.’.._ 1 - p Forp = 31, _ 1 , W ‘ § Now an 3 no WWW =/ ywappmwmy = (1 “ ”)2“ / yb<k>(y)dy 0 k=l 0 3 = <1 “ WZkP‘ =<1- pxp + 2p2 + 3:23) k=l Forp = %, W<sz>az“ or Wapprox = 031640625 Thus W - W l- (2)4 ”9““ = 3 14 30.0508 W 3 and so Wappmx is within 5% of [he mean W. ...
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