Exercise 3 Stochastic Models of Manufacturing Systems 4T400, 6 May

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1 Exercise 3 Sochasic Models of Maufacurig Sysems 4T4, 6 May. Each week a very popular loery i Adorra pris 4 ickes. Each ickes has wo 4-digi umbers o i, oe visible ad he oher covered. The umbers are radomly disribued over he ickes. If someoe, afer ucoverig he hidde umber, fids wo ideical umbers, he wis a large amou of moey. Wha is he average umber of wiers per week? Aswer: Le X i = if icke i is a wiig icke, ad X i = oherwise. Le deoe he umber of ickes (i his case = 4. The P (X i = = P (X i =, ad hus E(X i =. Le X be oal umber of wiers, so X = X + + X ad E(X = E(X + + X = E(X + + E(X = =. So he average umber of wiers does o deped o he size of he loery. 2. Two people, sragers o oe aoher, boh livig i Eidhove, mee each oher. Each has approximaely 2 acquaiaces i Eidhove. Wha is he probabiliy of he wo people havig a acquaiace i commo? Aswer: Le w be he umber of acquaiaces (so w = 2 ad r deoe he umber of ihabias of Eidhove mius w + 2 (so assumig Eidhove has 2. ihabias, he r = Le p be he probabiliy of havig a leas oe acquaiace i commo, he ( r p = ( w r(r (r w + r+w = (r + w(r + w (r +. w So for he specific values of r ad w we ge p X is he disace o of a radom poi i a disk of radius r. Wha is he desiy of X? Aswer: We have F (x = P (X x = πx2 πr 2 = x2 r 2. Hece, for he desiy we ge Calculae E(X ad var(x. Aswer: Similarly E(X 2 = 2 r2, so E(X = f(x = d 2x F (x = dx r 2. r xf(xdx = r 2x 2 r 2 dx = 2 3 r. var(x = E(X 2 (E(X 2 = 2 r2 4 9 r2 = 8 r2. 4. Le X be radom o (, ad Y be radom o (, X.

2 Wha is he desiy of he area of he recagle wih sides X ad Y? Aswer: The joi desiy of X ad Y is give by f(x, y = x, < y < x <. Le Z = XY. The F (z = P (Z z = = 2 z z. z x x dydx + z x z x dydx Hece f(z = d dz F (z = z. Calculae he expeced value of he area of he recagle wih sides X ad Y. Aswer: E(Z = zf(zdz = ( z z dz = A bach cosiss of iems wih probabiliy ( pp,. The producio ime of a sigle iem is uiform bewee 4 ad miues. Wha is he mea producio ime of a bach? Aswer: Le N deoe he umber i he bach, so ad P (N = = ( pp,, E(N = P (N = = ( pp = p. Le X i deoe he producio ime of iem i i he bach, hus E(X i = 4+ 2 = 7 miues. Furher, le X deoe he producio ime of he whole bach, so X = X + + X N, ad ( N ( N E(X = E X i = E X i N = P (N = = = ( E X i N = P (N = = ( E X i P (N = E(X P (N = = E(NE(X = 7 p miues. 6. Le X have a mixed biomial disribuio: wih probabiliy.5, he radom variable X is biomial wih parameers = 2 ad p =., ad oherwise, X is biomial wih parameers = 2 ad p =.9. Le X i, i =, 2,... be idepede radom variables, wih he same disribuio as X, ad cosider he sum S = X X. For each =, 5,, 2, geerae may samples of S (for example, by usig χ ad plo a hisogram (for example i R. Wha is your coclusio? Aswer: Below he hisograms are ploed of S, S 5, S ad S 2. Each hisogram is based o 5 samples, geeraed by a χ program. The coclusio is ha, alhough he mixed biomial disribuio does o resemble he Guassia disribuio a all, he disribuio of

3 he sum of a few mixed biomial radom variables already closely maches ha of a Gaussia oe. Hisogram of S Hisogram of S Hisogram of S Hisogram of S Cosider a wo-machie producio lie producig fluid. The producio rae of machie is 5, he rae of machie 2 is 4. Machie 2 ever fails, bu machie is subjec o breakdows. The up imes ad dow imes have bee colleced durig a log period. The size of he fluid buffer is K. Fi some simple disribuios o he sample mea ad sample variace of he up ad dow imes, ad graphically compare he fied ad he empirical disribuios. Aswer: Le he radom variable U be he upime, ad D he dowime. The sample mea of he upime U( wih = samples is equal o U( = ad he sample variace S 2 ( is equal o S 2 ( = U i = 8.9 (U i U( 2 = 63

4 Accordigly, he sample mea ad sample variace of he samples of he dowime are D( =.7 ad S 2 ( = The hisograms of he samples of he upimes (lef ad he dowimes (righ are show below. Hisogram of upimes Hisogram of dowimes We ca fi a Gamma disribuio wih desiy f( = ( a Γ(a b b e /b,, o he mea µ ad variace σ 2 by seig a = σ2 µ 2, b = µ a. This meas ha we ge a =.49 ad b = 8.3 for he upimes, ad a =.5 ad b = 7.64 for he dowimes. The desiies of he Gamma disribuio fied o he upimes (lef ad he dowimes (righ are show below. desiy Gamma desiy Gamma Esimae, by simulaio, he hroughpu for K = 2 over a log simulaio horizo T, say T = 4 ime uis. Repea his may (say 3 imes, ad plo a hisogram of he esimaes for he hroughpu. Wha is your coclusio abou he disribuio of hese esimaes ad how ca his be explaied? Aswer: Below we lis a hisogram of he esimaes, showig ha i is approximaely ormal disribued. The explaaio is ha he hroughpu is esimaed by 4( I/T

5 K Empirical Gamma Expoeial Table : Throughpu for up- ad dow imes sampled from empirical, Gamma ad expoeial disribuios. where I is he sum of all idle imes of machie 2 durig he ierval [, T ]. Hece, by he ceral limi heorem, oe ca expec ha 4( I/T is approximaely ormal disribued as T (ad hus he sum of idle imes is large. Hisogram of Throughpu Throughpu Esimae he log-ru hroughpu of he producio lie for various values of he buffer size K, by usig boh he empirical ad fied disribuios. Wha are your coclusios? Aswer: Below we lis he hroughpu for various values of he buffer size K for up- ad dow imes sampled from heir empirical disribuio ad fied Gamma disribuio; he hroughpu has bee esimaed by a simulaio ru of legh 6, usig a χ model. The resuls show ha boh disribuios yield similar performace. If a expoeial disribuio is fied o he mea oly, he he performace is sigificaly differe from he oe for he empirical disribuio; he hroughpu is larger, which migh be expeced, sice he variabiliy of he expoeial (i.e. coefficie of variaio is less ha ha of he empirical.