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Unformatted text preview: Fall 2010 IEOR 160 Industrial Engineering & Operations Research September 21, 2010 Page 1 of 3 HOMEWORK 3 SOLUTIONS Chapter 12.4 4. The probability of x hits in n at bats when probability of success on each at bat is p is given by f ( p ) = n ! x !( n- x )! p x (1- p ) n- x . It is easier to maximize g ( p ) = ln( f ( p )), then g ( p ) = ln n ! x !( n- x )! + x ln p + ( n- x ) ln(1- p ) and g ( p ) = x- np p (1- p ) = 0 for p = x n . Since g ( p ) > 0 for p < x n and g ( p ) < 0 for p > x n , we know that p = x n maximizes the probability of x hits in n at bats. 9. (a) Let x be the number of people (in thousands) who take car daily. Equilibrium occurs when 20 + 5 x = 40 or x = 4(thousand) people. (b) Let TT ( x ) be the average travel time per day when x thousand people take car. Then TT ( x ) = x (20 + 5 x ) 10 + (1- x )40 10 = 4- 2 x + x 2 2 . TT ( x ) = x- 2, TT 00 ( x ) = 1 > 0. Thus x = 2(thousand) people will minimize average travel time. The moral of the problem is that when individuals are left to their own devices, they will produce suboptimal waiting times!...
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This note was uploaded on 10/31/2010 for the course IEOR 41027 taught by Professor Glassey during the Fall '10 term at Berkeley.
- Fall '10
- Operations Research