a = 10 b=6 c=9 d=3 e=10 f=8 Use the example below to answer question 3:

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a = 10 b=6 c=9 d=3 e=10 f=8

A 10 B 6 C 9 D 3 E 10 F 8 Use The Example Below To Answer Question 3 1 (283.28 KiB) Viewed 44 times

Use the example below to answer question 3:

A 10 B 6 C 9 D 3 E 10 F 8 Use The Example Below To Answer Question 3 2 (167.79 KiB) Viewed 44 times

3. Use the last six digits of your Banner number 000abcdef or 001abcdef to generate weights for the edges marked a, b, c, d, e and f in the network below. In this question, if one of your digits is 0, you should use 10 instead. A 6 B 2 a 9 C b 9 6 9 D 6 9 E F 3 G 5 E C d e 2 324 с (a) Find the shortest path between A and L your resulting network. (b) Now we make the above graph a directed graph as follows (note that all the edge weights remain the same): B 2 D d е H 2 I G 5 H 6 2 5 5 6 9 2 K 9 5 K L 5 L F Find the critical path between A and L your resulting network. For full marks you should show your working out at each stage to gain full marks. (20 marks)
6 a 6 B 9 ((A) = 0 l(C) = 4 C 9 6 2 b X 6 9 6 B C 2 3 and is of weight 36. Hence to critical path is 6 9 D E 47 F Find the critical path between A and L in the resulting directed graph showing all of your working out at each stage. ((E)= max(l(B) + 3, (C) +6} = max(9, 9, 10) = 10 (20 marks) Solution: Here we choose a = 4, b = 3,c= 6, d = 2, e = 7, f = 9 so we will calculate the critical path for the following graph: l(G) = max{l(D) +3, (E) +6} = {11,16) = 16 2 (1) = 16 (K) = max((H) +6, (I)+2} = max(27, 18} = 27 6 D 9 E 47 F 8 D C d 9,9,10,10 E 4♥ e F 13,14,14 2 9 3 6 2 A 7 2 G H 7 = G We use the convention of using blue for temporary labels and red for permanent labels. The labels are as follows: 6 H 5 6 Hence we get the following graph with its critical path shown below: 2 5 11,16.16 G X I 2 12,21,21 H 5 ↑ 16 .6 (B)=6 (D) = 8 (F) = max{1(E) + 4, (C) +9} = max(14, 13} = 14 (H) = max{l(E) + 2, l(F) +7} max{10, 21} = 21 9 K 9 6 K 2 (J) = max{{(G) + 5, l(H) +5} = max(21, 26) 26 (L) = max(l(K) + 9, l(H) + 9, l(J) +5} = max(36, 30, 31} = 36 AC EF→H→K→L 5 21,26,26 9 9. 5 K 27,18,27 L 9 L 9 UNIVERSITY GREENWICH L 36,30,31,36