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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, ZnH 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=jXo,...,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 nonzero 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—iZg <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
Manettrivia} 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) = 1q2' 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 (nonnull) 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 Ejgﬁi dkj Tr k : n : ——————m
( ) Zizi 31' 2?:1 Ej¢édij ...
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 Fall '11
 SerapSavari
 Poisson Distribution, Probability theory, Problem Statement, Poisson process, Markov chain, Qinlar Exercise

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