Klausur ss09 solution engl PDF

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Klausur Modellbildung und Simulation (Prof. Bungartz) SS 2009 Name, Vorname:

Seite 1/6 Matrikelnummer:

1 Zufallsvariablen erzeugen (ca. 3 + 4 + 6 = 13 Punkte) Given n real numbers xi , i = 1, . . . , n from the interval (0, 1). The xi are ordered differently in pairs and in ascending order: 0 < x1 < x2 < . . . < xn < 1. As an example, we take n = 4 and the numbers 0.1, 0.2, 0.4, 0.8. Further, a random number generator R() is available which provides realizations of a random variable equally distributed to [0, 1]. a) For the sample data, sketch the emph empirical distribution function H(x), which specifies the proportion of data points that are at most as large as x: H(x) =

1 |{xi : xi ≤ x}| . n

b) Provide a method to produce realizations of a random variable X, which as distribution function FX has just the empirical distribution function H(x) (for general data records xi ).

We just need to pick one of the xi , each with a probability of 1/n: X := xi mit i := ⌈R() · n⌉ (i = 1 f¨ ur R() = 0)

Klausur Modellbildung und Simulation (Prof. Bungartz) SS 2009 Name, Vorname:

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c) Now create a random variable Y whose distribution function FY has the following properties: – FY is steady. – FY (xi ) = H(xi ) for all data points xi , i = 1, . . . , n and let FY (0) = 0. – Between every two data points xi and xi+1 and between 0 and x1 , FY should be linear. Sketch FY and its associated density fY for the sample dataset and specify (for generic datasets xi ) a method to produce Y implementations.

Wir bestimmen wie gehabt i := ⌈R() · n⌉ (i = 1 f¨ur R() = 0), und sorgen daf¨ur, dass Y im Intervall [xi−1, xi ] landet (ab jetzt sei x0 := 0). Innerhalb des Intervalls muss es gleichverteilt sein: Y := xi−1 + R() · (xi − xi−1)

Klausur Modellbildung und Simulation (Prof. Bungartz) SS 2009 Name, Vorname:

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2 Markovprozess, Systeme gew¨ohnlicher DGLn (ca. 2 + 5 + 2 + 6 = 15 Punkte) We consider a discrete homogeneous Markov process in continuous time with two states. The transition rate from state 1 to state 2 and vice versa is λ. a) Draw the transition diagram:

b) Now the system is in state 1 at the time t with a probability p1 (t) (0 leqp1 leq 1), that is with probability p2 (t) = 1 − p1 (t) in Condition 2. As in Sheet 7 ’Markov Chain Population Growth’, use the failure rate formula to specify the p1 (t + ∆t) and p2 (t + ∆t) probabilities in the Deltat tolimit0 depending on p1 (t), p2 (t) and ∆t.

ur den Wechsel des Zustands innerhalb von Die Wahrscheinlichkeit f¨ ∆t ergibt sich aus der Ausfallrate (die hier λ ist): p(T ≤ t + ∆t|T > t) = λ. ∆t→0 ∆t

hT (t) = lim

Somit ist im Grenzwert ∆t → 0 . p(1 → 2) = λp1∆t,

. p(2 → 1) = λp2∆t,

und weiter . p1(t + ∆t) = p1(t) − λp1(t)∆t + λp2(t)∆t, . p2(t + ∆t) = p2(t) + λp1(t)∆t − λp2(t)∆t.

Klausur Modellbildung und Simulation (Prof. Bungartz) SS 2009 Name, Vorname:

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c) Give a system of differential equations describing the evolution of p1 (t), p2 (t). Note: if you have lost the thread in between, just continue to count up as a substitute      −1 1 p1 p˙1 = . p˙2 1 −1 p2

Teil b) mit ∆t → 0 gibt      −λ λ p1 p˙1 . = p2 λ −λ p˙2 d) Determine eigenvectors and eigenvalues of the system matrix of c), all equilibrium points and sketch the direction field in the bottom stationary coordinate system (very rough sketch for any lambda > 0). How do the solutions in the equilibrium points deal with small disturbances? What does this mean for the considered Markov process?

Eigenwerte (am einfachsten durch Ansehen der Matrix):     1 1 0 zu , und − 2λ zu , 1 −1 Gleichgewichtspunkte: p˙1 = p˙2 = 0 ⇔ p1 = p2. Das System ist orungen indifferent gegen¨uber St¨orungen in Richtung (1, 1)T , aber St¨ T in Richtung (1, −1) werden abgebaut. Markovprozess: hier bewegen wir uns auf einem Abschnitt der Geraden p1+p2 = 1, also f¨ur beliebige Startwerte auf den Punkt (1/2, 1/2)T zu (exponentieller Abfall der Entfernung zu (1/2, 1/2)T ).

Klausur Modellbildung und Simulation (Prof. Bungartz) SS 2009 Name, Vorname:

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3 Numerik gew¨ohnlicher Differentialgleichungen (ca. 3 + 2 = 5 Pkt.) To get a numeric value for e = 2.71828 ldots, one can get out of the differential equation y(t) ˙ = y(t),

y(0) = 1

(1)

determine an approximation for y(1); since the solution to the equation is y(t) = et , we have an approximation for e. a) Give an explicit formula for the approximation etaN for y(1) depending on a N in N N, if Equation ( ref eq: ode) is numeric from t = 0 to t = 1 with the Euler method at a step size δt = 1/N is integrated.

b) How does the discretization error ηN − e behave in this method depending on δt? (O(. . .), just write down, do not deduce!) What does that mean for the ratio of computational effort to achieved accuracy?

Klausur Modellbildung und Simulation (Prof. Bungartz) SS 2009 Name, Vorname:

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4 Wahlversprecher (about 4 + 3 = 7 points) In a poll for an electoral system that complies with the Democratic Basic Laws 1–4 (ie, the collective selection function is a figure PAn toPA that satisfies the Pareto condition and the independence of irrelevant alternatives), candidates will emerge x, y and z on. One week before the election, the result would be xyz, so x would win before y and z would land at the very end. Well, x forgets in an interview to categorically rule out tax hikes. Then all voters put x at the end of their new preference i′: Now y′i x and z′i x apply to all voters i. The preferences between y and z on the other hand remain the same for all voters. a) Which optional exits varrho′ are still possible? Justify your answer!

b) It may be that in the original opinion image (i or  with xyz) x was not a voter favorite, ie v was foreachvoteryi x or zi x?

Nein: der Satz von Arrow besagt, dass es einen Diktator j geben muss. Dann gilt yj x ⇒ yx und zj x ⇒ zx, in keinem Fall w¨urde sich also das beobachtete Wahlergebnis einstellen....