1692799753 Exam Solution February 2023

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Section I – Q1 5 94 B 8 123 18 213 H 18 2105 127 C 5 120 21 265 I 21 260 12 142 F 19 217 The critical path is A-C-E-H-I. The project duration is 26 weeks. 0 55 A 0 50 12 131 G 20 218 12 186 E 12 180 9 178 D 13 214 Section I – Q1 (Solution) Section I – Q1 (Solution) Activity Immediate Predecessor Duration (weeks) Duration Reduction Total Crashing Cost (€) Unitary Crashing Cost (€) A - 5 3 36 000€ 12 000€ B A 4 1 9 000€ 9 000€ C A 7 6 36 000€ 6 000€ D B 8 2 3 000€ 1 500€ E B, C 6 4 32 000€ 8 000€ F C 2 1 1 000€ 1 000€ G C 1 - - - H E 3 2 14 000€ 7 000€ I D, F, G, H 5 2 18 000€ 9 000€ ➢ Among activities of the critical path A-C-E-H-I, the lowest unitary crashing cost and highest potential benefit is associated with the crashing of activity C. However, crashing C, would lead to an overall project duration reduction of 3 weeks, due to the emergence of another critical path (A-B-E-H-I). As a consequence, crashing C is not convenient since the real economic saving (3*10 000 = 30 000€) is lower than the total crashing cost (36 000€). ➢ Consequently, the most convenient activity to crash is activity E, that leads to a real project duration reduction of 4 weeks. ➢ In this case the margin achieved is: 4*10 000 – 32 000 = 8 000€. 5 94 B 5 90 14 173 H 14 1705 127 C 5 120 17 225 I 17 220 12 142 F 15 173 After crashing E, the critical paths are: ➢ A-B-D-I ➢ A-C-E-H-I 0 55 A 0 50 12 131 G 16 174 12 142 E 12 140 9 178 D 9 170 Section I – Q1 (Solution) Section I – Q1 (Solution) Activity Immediate Predecessor Duration (weeks) Duration Reduction Total Crashing Cost (€) Unitary Crashing Cost (€) A - 5 3 36 000€ 12 000€ B A 4 1 9 000€ 9 000€ C A 7 6 36 000€ 6 000€ D B 8 2 3 000€ 1 500€ E B, C 6 4 32 000€ 8 000€ F C 2 1 1 000€ 1 000€ G C 1 - - - H E 3 2 14 000€ 7 000€ I D, F, G, H 5 2 18…