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02q1soln - Massachusetts Institute of Technology...

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Massachusetts Institute of Technology 1.203J/6.281J/13.665J/15.073J/16.76J/ESD.216J Logistical and Transportation P lanning Methods Fall 2002 Quiz 1 Solutions 1 (a)(i) Without loss of generality we can pin down X 1 at any fixed point. X 2 is still uniformly distributed over the square. Assuming that the police car will always follow a shortest route to the emergency incident, the max possible distance between X 1 and X 2 is 2 km. The travel distance is thus uniformly distributed between 0 and 2 km. (a)(ii) Following similar logic, the max possible distance is now 4 km. The travel distance is thus uniformly distributed between 0 and 4 km. (b) Let’s number the links as shown in Figure 1. There is a 1 chance that the emergency 12 incident will be on any one of the 12 links. Thus if we can determine the conditional pdf for the travel distance from X 1 (conditioned to be uniformly distributed on link 7) to X 2 for each possible link for X 2 , we are done. All we do then is add the resulting conditional pdfs, multiplying each 1 by 12 , the probability of occurrence. Careful inspection of the problem reveals that with regard to computing the conditional travel distance pdf between X 1 and X 2 there are three sets of links 1 km 3 8 1 4 police car here 5 2 6 9 10 7 11 12 Figure 1: Link Numbering Set 1: 1, 3, 4, 5, 6, 8, 9, 10, 11 Page 1 of 7
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Massachusetts Institute of Technology 1.203J/6.281J/13.665J/15.073J/16.76J/ESD.216J Logistical and Transportation P lanning Methods Fall 2002 Consider link 1. Define X 2 to be the distance from the right most point of link 1. Then the conditional travel distance, given that X 1 is defined to be the distance on link 7 from its left most point and that X 2 is on link 1, is ( D | 1 , 7) = X 1 + X 2 + 1. Let’s define V = X 1 + X 2 . Note that X 1 , X 2 U(0 , 1) i.i.d. Either by convolution, or by the “never fail” sample space method using cumulative distribution functions, or by recalling problem 2(e)(i) of HW1, we find d, d [0 , 1] f V ( d ) = 2 d, d [1 , 2] 0 , otherwise Now the conditional pdf we want for link 1 is f V ( d ) “shifted to the right” by one unit of distance. Call this conditional pdf f ( D | 1 , 7)( d ). Then we have for link 1, f ( D | 1 , 7)( d ) = f V ( d 1). By inspection we also have f ( D | 3 , 7)( d ) = f ( D | 8 , 7)( d ) = f ( D | 11 , 7)( d ) = f ( D | 1 , 7)( d ) = f V ( d 1). For the remaining links in Set 1, links 4, 5, 6, 9 and 10 “touch” link 7, so there is no shifting of the pdf by one. That is, there is no intermediate link between them that would add 1.0 km to the travel distance. Hence, f ( D | 4 , 7)( d ) = f ( D | 5 , 7)( d ) = f ( D | 6 , 7)( d ) = f ( D | 9 , 7)( d ) = f ( D | 10 , 7)( d ) = f V ( d ).
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  • Fall '06
  • hansman
  • Probability theory, Massachusetts Institute of Technology, Cumulative distribution function, Massachusetts Institute, busy period

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