HW9_solution

HW9_solution - IEOR 4004 HW#9 solution 1 P 418#1 Find the...

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Unformatted text preview: IEOR 4004 HW #9 solution 1. P 418 #1 Find the shortest path from node 1 to node 6 in Fig. below: 1 0* 0* From 6 31 24 = 7 6 > 5 31 19 = 12 6 > 4 31 12 = 19 31 7 = 24 31 0 = 31 From 5 24 19 = 5 24 12 = 12 5 > 3 24 7 = 17 24 0 = 24 From 4 19 12 = 7 4 > 3 19 7 = 12 4 > 2 19 0 = 19 2 7 7* 7* 3 12 14 12* 12* 4 21 19 19 19* 19* 5 31 28 24 26 24* 24* 6 44 38 33 31 31 31* 31* From 3 From 2 12 7 = 5 7 0 = 7 2 > 1 12 0 = 12 3 > 1 Shortest Paths: 1 2 4 6 / 1 3 4 6 / 1 3 5 6, with length equal to 31. 2. P 418 #2 Find the shortest path from node 1 to node 5 in Fig. below: 1 0* 0* 2 2 2* 2* 3 8 7 7* 7* 4 Inf 6 6* 13 6* 5 Inf 14 16 Inf 14* 14* From 5: From 2: 14 6 = 8 2 0 = 2 2 > 1 14 2 = 12 5 >2 Shortest Path: 1 2 5, with length equal 14. 3. P 418 #4 Use Dijkstra's algorithm to find the shortest path from node 1 to node 4 in Fig. below. Why does Dijkstra's algorithm fail to obtain the correct answer? 1 0* 0* From 4: 2 1 = 1 4 >3 2 2 = 0 Shortest Path according to Dijkstra's: 1 3 4 From 2: 2 0 = 2 3 > 1 2 2 2* 2* 3 1 1* 1* 4 2 2* 2* Obviously, Dijkstra's algorithm fails to give the actual shortest path, which is 1 2 3 4, since it makes the assumption that all arc lengths are nonnegative. In this case, the arc length of (2,3) is 2, thus making the Dijkstra's algorithm fails. 4. P419 #7 Minimize the total cost of having a machine for five years. Total cost = purchase cost + maintain cost Let node i = the beginning of year i Cij = the total cost of purchasing and maintaining a machine from beginning of year I to the beginning of year j, where i < j, C12 = 170 + 38 = 208 C34 = 210 + 38 = 248 C23 = 190 + 38 = 228 C13 = C12 + 50 = 258 C35 = C34 + 50 = 298 C24 = C23 + 50 = 278 C25 = C24 + 97 = 375 C14 = C13 + 97 = 355 C36 = C35 + 97 = 395 C26 = C25 + 182 = 557 C15 = C14 + 182 = 537 C16 = C16 + 304 = 841 C56 = 300 + 38 = 338 C45 = 250 + 38 = 288 C46 = C45 + 50 = 338 6 841 765 653 693 875 653* 653* 1 2 3 4 5 0* 208 258 355 537 208* 436 486 583 258* 506 556 355* 543 537* 0* 208* 258* 355* 537* From 6: From 3: 653 537 = 116 258 208= 50 653 355 = 298 258 0 = 258 3 > 1 653 258 = 395 6 > 3 653 208 = 445 653 0 = 653 Shortest Path: 1 3 6, with length equal to 653 Therefore, a new machine should be purchased at year 1 and year 3. ...
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This note was uploaded on 02/10/2010 for the course IEOR 3600 taught by Professor Chudnovsky during the Spring '09 term at Columbia.

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