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HW3-Sol

# HW3-Sol - ENGR62/MS&E111 Introduction to Optimization Prof...

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ENGR62/MS&E111 Spring 2006 Introduction to Optimization June 5, 2006 Prof. Amin Saberi Homework Assignment 3: Solutions Problem 1 (a) The given data correspond to the following graph (not to scale): o A B C D E d 40 50 60 10 20 50 70 10 80 60 55 40 Figure 1: Graph for problem 1 where each line represents two arcs 1 in opposite directions (note, however, that not all of these arcs are necessary to find the shortest path since, for instance, we will never come back to the origin after leaving it). Thus, we have a graph where: V = { o,A,B,C,D,E,d } (1) E = { ( o,A ) , ( o,B ) , ( o,C ) , ( A,B ) , ( B,C ) ,... } (2) (b) The shortest path can be found by solving the following LP: min ( i,j ) E c ij f ij subject to { k | ( j,k ) E } f jk { i | ( i,j ) E } f ij = 0 j V \{ o,d } { o | ( j,o ) E } f ok { i | ( i,o ) E } f io = 1 { d | ( j,d ) E } f dk { i | ( i,d ) E } f id = 1 0 f ij 1 ( i,j ) E (3) where the c ij correspond to the distances given in the graph above. (c) The shortest path, as computed in Excel, is given by ( o,A,B,E,D,d ) with a corresponding distance of 160 miles (see the spreadsheet on the course webpage 2 ). 1 Note that the roads can alternatively be interpreted as being unidirectional. This is ok, and results in a shortest path of 165. 2 Note: This spreadsheet shows how formulate network flow problems in Excel using a different method than you may be used to. It is simply another way you can formulate these problems in Excel. 1

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(d) If the numbers in the table are interpreted as times rather than distances then the process of finding the minimum-time-path is identical to the one shown above. Thus, the shortest time path is ( o,A,B,E,D,d ) with a corresponding time of 160. (e) If the numbers in the graph instead correspond to flow capacities, then we can find the maximum flow between o and d by appending an edge, ( d,o ), to the graph above and solving the following LP: max f do subject to { k | ( j,k ) E } f jk { i | ( i,j ) E } f ij = 0 j V 0 f ij u ij ( i,j ) E (4) where the u ij are the “capacities” given in the graph above and E = E ∪ { ( d,o ) } as discussed in the notes.
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HW3-Sol - ENGR62/MS&E111 Introduction to Optimization Prof...

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