HW9_Sol - ECEN 646 Homework 9 Solutions Qiniar Exercise(4.6...

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Unformatted text preview: ECEN 646 , Homework 9 Solutions Qiniar Exercise (4.6), Chapter 5 We are given P(j,j) > 0 for some 3'. Let A be the set of all n 2 1 such that P"(j,j) > 0. In that case, 1 E A because P1(j, j) = PU, j) > 0. Thus the greatest common divisor of set A is 1 (the integer 1 has no higher divisor than i) so by Criterion 3.4, state 3' is aperiodic. Since the Markov chain is irreducibie, all states must be aperiodic. Qinlar Exercise (4.7), Chapter 5 We are given .9 m 3, S m 8, and A fi 4. Z;r1 is the total demand during the nth week and X” is the stock on hand just before the nth week. Since the demand process is Poisson, its distribution is A" _4 4’6 ma. :k}=eflkfime k! (a) From Exampie (1.16), X72” fl+1zj1if3<Xfl_<—8$ Zn+l§Xn Xn+i = 8 r" Zn+1= j, if Xn S 3, Zn-H 5 8 U, I o.w. The value of XHH is a. function of X71, and Zn“. Since ZR“ is independent of X0,. . . ,Xn, it is true that P{Xn+1=j|Xo,...,Xfl}= P{Xn+1 r”— len} is a Markov chain. Since 0 S Xn 5 8, the state space E = {0,1, . . . ,8}. Consider P{Xfi+1 = 3'an = t}, and let pn = 54% and rn = 2;”an 6—4417: as defined in the problem statement. For t g 3 and j aé 0, we will fill the inventory to S m S and then subtract the demand, so I . I _ . M 4034} PinH—l =J!Xn ziiflpis’znntl$3}=P{Zn+1=8"3}=e 4(8 Wj): :psvj If i S 3 and j = 0, then demand must be greater than S :4 8: 00 4k P{Xn+r — [lan = t} = P{Zn+1 > 3} — 28—4? m r7 k=8 ‘ If 4 Si 3 8 and j # 0, the inventory is not filled, so P(i,j) is only non-zero when j < i. In this case, 46—1”) (i — j): T" “‘9' If 4 S i g 8 and j m 0, the inventory is not fiiied and the demand meets or exceeds the supply. In this case, P{Xn+1=jan : 3 P{Zn+1=i—j}= 6—4 “iii 00 Pm“ : OEXR m i} = Pizn+12il= Ze k, kmi — Tiwl Thus we have the transition probabilities given in P in the problem statement. (b) Here I assume that a shortage is when demand? exceeds inventory. Since the problem didn’t explicitly define a shortage, if you made and stated a different assumption, that is okay also. With my assumption, we need the probability that, starting with 8, that demand Zn“ exceeds 8. This is 30 44k 8 44k P{Z,,+1> s} : 2e“ 76—! =r8 m 1mg}; 75 56:9 I 1:20 I If you solve for the numerical value, this is 2 54 256 194 4096 16384 65536)_07023L P z :1— ‘4 {n+1>8} 8 (1+4+8+fi+24+120+720+5040+40320 (c) i read this as, starting at Xe : 8, no replacement is needed at either t1 or t2. This is the event that X1 > s and X2 > s, or 8 — Z1 — Z2 > 3. Equivalently, Z1 + Z2 < 5. Since 2'1 + 22 are Poisson arrivals in the time interval ((3, 2}, this is also a Poisson distribution, with time interval length 2, W8 8; I m 0&0 z! 4 P{X} >3,X2>3!XG=8}mP{Zl—i-Zg <5}=Ze 120 (d) Starting with 4 units, any positive demand in a week will cause replenishment. Thus each week is a Bernoulli trial, where replenishment is not needed with probability q :- P {2,- = 0} = e” , and replenishment is needed with probability p = 1 — q = 1 — ew‘l. Then the time to first replenishment, T, is geometric, ie., P{T : n} 2 pqn—l Z (1 __ er4)e—4(nel) : (e4 _1)ee4n Qinlar Exercise {4.10), Chapter 5 (all We’re given that Y11 is the age of the equipment in use at time n, and q, is defined as P {1,114.1 = i + 1§Yn 2 Fort : O, P {YHH = llY = O} is just the probability that the equipment doesn‘t fail at time 1, ie., P {YWH : 111’}, z 0} = l -— P {equipment doesn’t fail in first interval} = 1—191. Fort: 1, . _ PY mi+i,Yn:i PYn mi+1 Manet-trivia} Maggi} }: lpfifififll H P {equipment doesn’t fail in i + 1 intervals} P {equipment doesn’t fail in 1 intervals} ‘ 1 Zgozwzpk z: 1 "“ Pk Big—4+1“ 1 “' Zlczl Pk i! (b) {Yn} is Markov because P{Yn+1:jlYgziG,...,Yn:ln} '_ P{Y mi YD—ig} n# 111"'7 Ifj 75 0 and j # in + 1, then P{Yn+1 :2 jIYg = i0,...,Yn w in} 2 0 since the lifetime can only decrease to O or increase by 1 in one time intervai. So Pfiaj) = 1-q2' i=0 0 am This corresponds to the transition matrix given in the problem statement. (C) The chain is irreducible, since state i communicates with state i + 1 for all 2', and with state 0. Since P(O, 0) > 0, the chain is aperiodic. To check whether the chain is recurrent or transient1 we solve the system of equations, Mt) ~_w Z Q(i,j)h(j), 0 3 Mi) 5 1, w 6 so 2 E {a} jEEo You can choose any is, but I choose I: = E), and Q becomes 0 (11 0 Q = 0 0 (12 So, It 2 Qh implies that M1) : «1111(3) am} 1 1 =1 1*” =0 sell 2 its Thus if h(1) > 0, then Mi) —-> 00 as i ——> 00. Since S l is required for all 2‘, clearly there is no other soiution besides h(t) = 0%. Thus aii states are recurrent. New, check whether the states are positive (non-null) recurrent or null recurrent by finding a solution to v = vP where o is a row vector. “0(1) : 9011(0) “(2) = (11141} = (11%?!(0) WU} fl Um) [€192 - (ii—«ll Applying the condition that the sum of is 1, 1 = Zeb") : 11(0) + v(0) Ego-“Gel kzl i=0 Which implies that v((}) m 1/(i + 2:021:10- - -qkwl). We aren’t given any more information about the sequence {qi}, so we dOn’t know if the sum converges. If 22:1 go - - - qk_1 < so, then there is a solution to o = ’UP, and the chain is positive recurrent. Otherwise, if 2:11:10 -- -qk_1 z 00, then the chain is null recurrent. Qiniar Exercise (8.7], Chapter 6 We are given that X is aperiodic and irreducible with a finite number of states m. Thus by Corollary (2.11), there is a unique sointion for the limiting probabilities 7r such that 7r:irP, Zia-:1 éEE That is, if we show that a particular vector it satisfies these equations, then there is no other different solution. To show that the solution is m i/mVi, we use the fact that P is doubly Matrkov, ie., 216E P('i,j) = le: «(3') = ZvrsiPm) = 352mg) = Sn—vi‘ iEE iEE and Egg}; 2 x 1. Thus the solution = l/m W is the unique solution. [:3] Qinlar Exercise (8.10), Chapter 6 (3) We’re given it : (0.2,0.1,0.3,0.2,0.2) and f m (3,—1,2,—5, 3). We’re given X is irreducible, and clearly X is recurrent. Then by Corollary (2.23), “1320 Hui—1 Z E,[f(Xmll z Ff = (0.2)(3)+{0.1)(—1)+(0.3)(2)+(e.2)(e5)+(0.2){3) : 0.7 771:0 (b) This limit is expressed in {2.24), thus ago ’rrf _ 0.7 _ 0.? m 7/3 ? (0.2)(2) + {0.i}(w1} + (0.3)(2) + (0.2)(—3) 0.3 lim : MOO 221:9 90%) N9 @ Additional Problem We are given, PM = 51-“? where to make the notation simpler I have defined 51' 3 Educ 163% Note that this notation for the sum means that we sum from is equal 1 to it, but not including i. The problem is to find the limiting distribution, 7r 2 [7r{1), . . . ,7r(n)}. With this notation, the 4 equation it = WP implies that, for any It E {1, . . . ,n}, was) 2 waek) : Emit (3) in} £95.76 By the computational hint in (6.2.14), if we can find a solution to this equation1 then we can scale it to find a solution that aiso satisfies 227:1 1r(z') = 1. One solution that satisfies Eq. 3 is 2 31‘. This is a solution by guessing - but it makes sense that the limiting probability of being at state 2' should be proportionai to the dji leading to state i. Also, you can see that letting 7r(i) = s; cancels the denominator in Eq. 3. That is7 n d. filki=E 5i_:€ $2 dik=E dkimSk iy—‘k ‘ we 29% Now that we have a solution to 7r = ‘fl'P, we normalize, ie., find a constant c such that the solution 7r(k) = csk so that Mi} : I. That is, 2&1 csi m 1 or 1. c 2 2?:1 87: Thus for a final, normalized sointion, we have Sic Ejgfi-i dkj Tr k : n : ——-————m ( ) Zizi 31' 2?:1 Ej¢édij ...
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HW9_Sol - ECEN 646 Homework 9 Solutions Qiniar Exercise(4.6...

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