pset7sol

pset7sol - 18.06 Problem Set 7 Solutions Problem 1: Do...

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Unformatted text preview: 18.06 Problem Set 7 Solutions Problem 1: Do problem 11 from section 8.3. Solution A Markov matrix must conserve probability. Hence the columns must sum to 1: A = . 7 . 1 . 2 . 1 . 6 . 3 . 2 . 3 . 5 (1) The steady state vector x satisfies Ax = x . In otherwords, x is an eigenvector of A with eigenvalue 1. Therefore x solves: ( A- I ) x = - . 3 . 1 . 2 . 1- . 4 . 3 . 2 . 3- . 5 x 1 x 2 x 3 = 0 (2) - . 11 . 11 . 1- . 4 . 3 . 11- . 11 x 1 x 2 x 3 = 0 (3) (4) Take x 3 = 1, then x 2 = 1 and . 1 x 1 = . 4- . 3 = . 1 so x 1 = 1 and x = (111) T . In general, a symmetric Markov matrix A T = A has a steady solution x = (111) T . This follows from: A is Markov columns of A sum to 1. A T = A rows of A sum to 1. The vector x = (111) T sums the row elements. Hence, Ax = x . Problem 2: Do problem 12 from section 8.3. Solution B = ( A- I ) x =- . 2 . 3 . 2- . 3 (5) 1 The eigenvalues satisfy the characteristic equation: (- . 2- )(- . 3- )- ( . 2)( . 3) = 0 (6) ( + . 5) = 0 (7) The eigenvalues are = 0 and =- . 5. In general, when A is Markov, A- I will have a = 0 eigenvalue. Specifically, this follows from the (more general) fact that if is an eigenvalue of any matrix A , then - c is an eigenvalue of A- cI . Markov matrices have an eigenvalue of 1, hence A- I must have an eigenvalue 1- 1 = 0. The eigenvectors of B are, x 1 = ( . 3 ,. 2) T corresponding to = 0, and x 2 = (1 ,- 1) T corresponding to =- . 5. A general solution to the ODE du dt = ( A- I ) u has the form: u = c 1 e t x 1 + c 2 e- . 5 t x 2 (8) Here c 1 and c 2 are integration constants (determined by initial values). As t , u becomes: u c 1 x 1 (9) Problem 3: Do problem 16 section 8.3. Solution A = . 4 . 2 . 3 . 2 . 4 . 3 . 4 . 4 . 4 (10) Since A is a Markov matrix, we know = 1 is an eigenvalue. In addition, detA = 0, so = 0 must also be an eigenvalue (ie det ( A- I ) = 0 ). The third eigenvalue = . 2 can be found by i ) inspection, ii ) using MATLAB , iii ) by direct computation. Using MATLAB, we can diagonalize A = S S- 1 , where the columns of S are eigenvectors of A: S = . 5145 . 7071- . 4082 . 5145- . 7071- . 4082 . 686 . 8165 (11) = 1 0 0 . 2 0 0 0 0 (12) 2 Now A k = S k S- 1 , so: A k u = ( S S- 1 ) k u (13) = S k S- 1 u (14) (15) Since S- 1 = . 5831 . 5831 . 5831 . 7071- . 7071- . 4899- . 4899 . 7348 (16) then...
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Fall '08 term at MIT.

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pset7sol - 18.06 Problem Set 7 Solutions Problem 1: Do...

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