# sol12 - \$32 1)Mn “{pa =[X(n — 1)M_1 n i=1 1 l...

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Unformatted text preview: \$32 1 )Mn “ {pa- = %[X,, + (n — 1)M,,_1] n i=1 1 l = _Xn 'i' - _')M'n--1 TL in: Clearly if Mad is given then 1Mn depends only on X” and is independent of Mn_2, MP3, M’ﬂ is a. Markov process. b) an(\$ ] Mn—l = y)da: P[m<MnS\$+dIIMn—1=y] l l . — au— < P[x<n+(l n)y_a:+da:] = P[nw—(n—1)y <annx—(n—l)y+dx] fr(n\$ -— (n — 1)y)dz H P‘D-U K1 2 TYn—l + X71. YE] : 0 Yn w (rm—1 = Xn qu(ylyn—1 : yl Yn—2 : = fxnﬁl _ ryl) fYnC‘JIY «1 = \$11) : Yn is a Markov process. ® ‘ u mud/26’- ’? (00: “:6, ‘1 X 331’“ M le '&4}%\W:%343i1ﬂ,s:muéxr%\Mﬁ¥i . ‘1 1P11m<“m*ﬁuiiwf%_g :_?Lmaa+r r r, < ‘Xﬂﬁxf‘. ‘3 ~ 1" A+Cgaig VH‘I yAssume Xn is discrete, then P[X.n = xanA = :51,Xn_2 = :172, = .P[Xn = 1:] since Xn is iid = PIX’n : Ian_1 = :81} X“ is a. Markov process with transition probabilities: P[Xn = \$‘Xﬁ_1 = \$1] = = w] all .171 «13.3 a) ATé5 b)).-m=2 5 m 5 “)T2xz5(§)—§m @a) 5m1= AliEINi] b) gm = §£[N] = 2?sz c) gm = g 2 mm] = 2%[11-1 Let A;(f) = # type 2' arrivals during [0,t] AGE) = Z A;(t) : total # arrivals 1 Am < T > = m 2 1";- avemge time in system 1 A1(t) A2 (t) Ana) _ 1 Ana) A1“) Ana) M” ‘" Am [Am E3 12*” Ana) E T'" _ A10) 1 A1“) Ana) 1 An“) Am Am) 2 T“ + + Am Ana) {2 T‘" W W " 1} €[T11 a“ sin} as t —r 00 => same result as above. 9 a)P[N2n1=(1—p)§pj=(1ﬂp>fp=p” 1 b) P[N2 10] = p10 : 10—3 =>p=lO"0'3m 5 1 Am— Z’” 2 MD A net proﬁt is made if ; M u 5>E[T}—1_p A 1 =>1——>-—=>5 —5A>1 H 5w ‘1 1 => 0<A<p—— 5 ...
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## This note was uploaded on 01/15/2011 for the course ECE 616 taught by Professor Khkjk during the Winter '10 term at Concordia Canada.

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sol12 - \$32 1)Mn “{pa =[X(n — 1)M_1 n i=1 1 l...

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