HW7_Solutions - IE:3610 Stochastic Modeling(Fall 2015...

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IE:3610 Stochastic Modeling (Fall 2015) Homework 7 (due Thursday, Oct 15) Solutions HW7 (Due Thu, Oct 15): 29.7-1(a,c,d); 29.7-2, 29.8-1; 29.8-2. 29.7-1(a,c,d) Solution: (a) P 00 = P TT = 1; P i,i - 1 = q ; P i,i +1 = p ; P i,k = 0 else P (1) = 1 0 0 0 . . . q 0 p 0 . . . . . . . . . q 0 p 0 0 q 0 p 0 0 0 1 (b) Class 1: { 0 } absorbing Class 2: { T } absorbing Class 3: { 1,2,. . . , T-1 } transient (c) When T = 3 and p = 0 . 3: P (1) = 1 0 0 0 0 . 7 0 0 . 3 0 0 0 . 7 0 0 . 3 0 0 0 1 We have two absorbing states: R 1 = { 0 } and R 2 = { 3 } , set of tran- sient states T = { 1 , 2 } . Absorption probabilities f TR = ( I - P TT ) - 1 p TR . f 10 f 20 = 1 0 0 1 - 0 0 . 3 0 . 7 0 - 1 0 . 7 0 = 1 - 0 . 3 - 0 . 7 1 - 1 0 . 7 0 = f 10 f 20 = 10 / 79 10 3 7 10 0 . 7 0 = 1 / 79 70 49 = 0 . 886 0 . 62 f 13 f 23 = 1 0 0 1 - 0 0 . 3 0 . 7 0 - 1 0 0 . 3 = 10 / 79 10 3 7 10 0 0 . 3 f 13 f 23 = 1 / 79 9 30 = 0 . 1139 0 . 3797 1
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(d) When T = 3 and p = 0 . 7: P (1) = 1 0 0 0 0 . 3 0 0 . 7 0 0 0 . 3 0 0 . 7 0 0 0 1 We have two absorbing states: R 1 = { 0 } and R 2 = { 3 } , set of tran- sient states T = { 1 , 2 } . Absorption probabilities f TR = ( I - P TT ) - 1 p TR . f 10 f 20 = 1 0 0 1 - 0 0 . 7 0 . 3 0 - 1 0 . 3 0 = 1 / 79 30 9 = 0 . 3797 0 . 1139 f 13 f 23 = 1 0 0 1 - 0 0 . 3 0 . 7 0 - 1 0 0 . 3 = 1 / 79 70 49 = 0 . 62 0 . 886 29.7-2 Solution: (a) 0 = Have to honor warranty 1 = Reorder in 1st year 2 = Reorder in 2nd year 3 = Reorder in 3rd year P (1) =
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