HW9_Sol

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online