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Unformatted text preview: -1-QUIZ 2 SOLUTIONSLogistical and Transportation Planning MethodsMassachusetts Institute of TechnologyDecember 5, 1999(1)Nodes 4 & 6 as well as nodes 3 &4 comprise Hakimi 2-median(2)Nodes 2, 4 or 6 are all one-vertex centersThe maximum distance vector is:Node, kMax, m(k)182638465766(3)The absolute center is easy to compute if one remarks the symmetry of theproblem. Only, arc (2,3), (2,6), (3,6), (5,6) have to be inspected. Moreover, ifone starts looking at arc (2,6) the local center is at 1 distance unit from node 2(middle of the arc) and the maximum distance is 5. Looking at the maximumdistance vector above, we conclude:m(2)+m(3)-d(2,3)2=6+8-42=5 . Therefore, we can discard the candidacy ofarc (2,3).Similarly for arc (3,6) because:m(3)+m(6)-d(3,6)2=8+6-32=5.5We can discard arc (3,6).For arc (5,6):m(5)+m(6)-d(5,6)2=7+6-42=4.5We have to look at (5,6).Searching for the local center on arc (5,6) we find:-2-We find that the local center of arc (5,6) is in node 6, with a maximum distance of...
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- Fall '06