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ADMS 3351 (2)
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Department
Administrative Studies
Course
ADMS 3351
Professor
Humayun Chaudhary
Semester
Summer

Description
Chapter 3 1. a. 6 9 15 19 19 26 B (3) E (4) G (7) 12 15 15 19 19 26 0 6 A (6) 0 6 6 13 13 15 15 18 C (7) D (2) F (3) 6 13 13 15 16 19 b. A-C-D-E-G, also shown in the network above as the bold path. c. 26 weeks, 6+7+2+4+7. d. 6 weeks, 15-9. 2. a. 1 5 5 7 7 9 9 11 B (4) D (2) F (2) G (2) 1 5 5 7 8 10 10 12 12 15 0 1 H (3) A (1) 12 15 0 1 1 4 7 12 C (3) E (5) 4 7 7 12 b. A-B-D-E-H, also shown in the network above as the bold path. c. 15 weeks, 1+4+2+5+3. 1 d. C, 3 weeks; F, 1 week; and G, 1 week. 3. a. 3 5 5 11 B (2) E (6) 7 9 9 15 0 3 3 7 7 13 13 15 15 18 A (3) C (4) F (6) G (2) I (3) 0 3 3 7 7 13 13 15 15 18 3 7 7 10 D (4) H (3) 3 7 12 15 Note that G has both D and F as immediate predecessors. However, D is redundant because F has D as an immediate predecessor. b. A-C-F-G-I, and A-D-F-G-I. c. B is not on a critical path and has slack of 4; therefore, do not shorten as it will not change the project completion time. Shorten C, D, and G one week each. C and D are on parallel critical paths, reducing them will only reduce project completion time by 1 week. d. A-C-F-G-I; and A-D-F-G-I. Project completion time is reduced from 18 to 16 weeks. 2 4 a. 4 6 8 13 13 16 2 (2) 5 (5) 8 (3) 9 11 11 16 16 19 0 4 4 8 8 14 14 19 19 26 1 (4) 3 (4) 6 (6) 9 (5) 10 (7) 0 4 4 8 8 14 14 19 19 26 4 7 7 9 4 (3) 7 (2) 9 12 12 14 b. 1-3-6-9-10. c. The most logical option would be to cut activity 3 by 3 weeks, and then reduce activity 6 or 9 by one week. This is the lowest cost option, and does not create an additional critical path. Another option would be to reduce activities 6 and 9 by a total of 3 weeks, and then reduce activity 3 by one week. This also has the same cost, but creates an additional critical path (1- 3-5-8-10). 5. 7 12 14 20 B (5) C (6) 9 14 14 20 0 7 11 14 20 26 A (7) G (3) D (6) 0 7 11 14 20 26 7 11 11 19 E(4) F (8) 7 11 12 20 3 a. A-E-G-C-D. b. 26 weeks. c. No difference in completion date. B already has slack time of 2 and F has slack time of 1. 6. 7 10 16 20 B (3) F (4) 8 11 16 20 0 7 11 16 20 25 A (7) D (5) G (5) 0 7 11 16 20 25 7 11 16 18 C (4) E (2) 7 11 18 20 a. A-C-D-F-G b. Activity Normal Crash Time Normal Cost Crash Cost NT-CT Cost/day to Time (NT) (CT) (NC) (CC) expedite A 7 6 $7,000 $8,000 1 $1,000 B 3 2 5,000 7,000 1 2,000 C 4 3 9,000 10,200 1 1,200 D 5 4 3,000 4,500 1 1,500 E 2 1 2,000 3,000 1 1,000 F 4 2 4,000 7,000 2 1,500 G 5 4 5,000 8,000 1 3,000 First, lowest cost activities to crash are A and E at $1,000 per day. E is not on the critical path, therefore select A. Critical path remains the same. Second, lowest cost activity on the critical path is C. Crash activity C. The critical path remains the same. Third, D and F are next lowest cost activities on the critical path. Both have a cost of $1,500 per day. Select D then F or reverse the order (F then D). F can not be reduced by two day because it would cause E to become part of a critical path. 4 Summary of steps to reduce project by four days: Step Activity to crash Cost to crash Days saved 1 A $1,000 1 2 C 1,200 1 3 D (or F) 1,500 1 4 F (or D) 1,500 1 7. 0 10 20 50 b (10) d (30) 10 20 50 80 0 0 80 100 100 100 a (0) g (20) h (0) 0 0 80 100 100 100 0 20 20 40 40 80 c (20) e (20) f (40) 0 20 20 40 40 80 a. 100 Hours. b. b and d are not on the critical path. Their start can be delayed without delaying the start of any subsequent activities. 5 8. 15 21 D (6) 5 15 15 21 B (10) 5 15 0 5 13 20 21 25 A (5) E (7) G (4) 0 5 14 21 21 25 5 13 C (8) 6 14 13 17 F (4) 17 21 a. A-B-D-G, 25 weeks, 5+10+6+4. b. Activity Normal Normal Cost Crash Time Crash Cost NT-CT Cost/week Time (NT) (NC) (CT) (CC) to expedite A 5 $7,000 3 $13,000 2 $3,000 B 10 12,000 7 18,000 3 2,000 C 8 5,000 7 7,000 1 2,000 D 6 4,000 5 5,000 1 1,000 E 7 3,000 6 6,000 1 3,000 F 4 6,000 3 7,000 1 1,000 G 4 7,000 3 9,000 1 2,000 First, reduce D (lowest cost activity on the critical path) by one week. This adds an additional critical path with activities C and E in it. Second, crash activity G by one week. Critical paths remain the same. Third, crash activity A by one week at a cost of $3,000, which is the least expensive. 6 Summary of activities crashed: Step Activity Cost to crash Weeks reduced 1 D $1,000 1 2 G 2,000 1 3 A 3,000 1 Total cost $6,000 9. 7 9 16 20 B (2) F (4) 9 11 16 20 0 7 11 16 20 25 A (7) D (5) G (5) 0 7 11 16 20 25 7 11 16 18 C (4) E (2) 7 11 18 20 a. A-C-D-F-G. b. 25 weeks, 7+4+5+4+5. c. B, 2weeks; E, 2 weeks. Activity Normal Crash Time Normal Cost Crash Cost NT-CT Cost/week Time (NT) (CT) (NC) (CC) to expedite A 7 6 $7,000 $8,000 1 1,000 B 2 1 5,000 7,000 1 2,000 C 4 3 9,000 10,200 1 1,200 D 5 4 3,000 4,500 1 1,500 E 2 1 2,000 3,000 1 1,000 F 4 2 4,000 7,000 2 1,500 G 5 4 5,000 8,000 1 3,000 7 First, shorten activity A by one week at a cost of $1,000. This is the lowest cost/week activity on the critical path. Second, shorten activity C by one week at a cost of $1,200. This is the next lowest cost/week activity on the critical path. 10. a. 9 17 24 29 B (8) D (5) 21 29 29 0 9 34 34 36 A (9) F (2) 0 9 34 9 24 34 36 24 C (15) E (10) 9 24 24 34 b. and c. Path Normal Length Crashed Length ABDF 24 15 ACDF 31 20 ACEF 36 23 d. We would only crash the project until 31 weeks since the cost of crashing C is $4000 which is greater than the $3500 in additional profit. 8 LENGTH ABDF - 24 24 24 24 23 22 22 22 22 ACDF - 31 31 31 31 30 29 28 27 26 ACEF - 36 35 34 33 32 31 30 29 28 Activity CrashedE E E E A A A C Crash cost 2500 2500 2500 2500 3000 3000 3000 4000 Cumulative Cost 2500 5000 7500 10000 13000 16000 19000 23000 9 11. Activity Normal Crash Time Normal Cost Crash Cost NT-CT Cost/week Time (NT) (CT) Weeks (NC) (CC) to expedite Weeks A 2 1 $50 $70 1 20 B 4 2 80 160 2 40 C 8 4 70 110 4 10 D 6 5 60 80 1 20 E 7 6 100 130 1 30 F 4 3 40 100 1 60 G 5 4 100 150 1 50 a. 2 6 6 13 B (4) E (7) 2 6 6 13 0 2 2 8 8 12 13 18 A (2) D (6) F (4) G (5) 18 18 0 2 4 9 9 13 13 18 End 2 10 C (8) 18 18 10 18 b. Critical path is A-B-E-G c. Duration Activity to crash Cost to crash Total direct costs 18 None $0 $500 17 A 20 520 16 E 30 550 15 G 50 600 14 B & D 60 660 13 B & F 100 760 10 d. Duration Total direct costs Total indirect costs Total Costs 18 $500 $400 $900 17 520 350 870 16 550 300 850 15 600 250 850 14 660 200 860 13 760 150 910 Select either 15 or 16 weeks at a total cost of $850. An alternative solution could be developed for this problem. When crashing an activity, there are two assumptions that can be made with in regards to the crash time. One, which was used above, is that any time between the normal time and the crash time can be used. In this problem, activity B could be reduced to either 3 weeks or to 2 weeks. In many cases, this is a realistic approach. However, in some cases it is an either/or situation. For example, activity B could be either the normal time of 4 weeks, or the crash time of 2 week, but nothing in-between. If we calculate the cost per time period to expedite, we generally assume that a linear approximation for the cost is appropriate, and any time in-between normal time and crash time is acceptable. This is the case in the above solution. If you do not make this assumption, the following solution is what you would develop. 11 c. Alternative solution, see explanation above. Duration Activity to crashCost to crash Total direct costs 18 None $0 $500 17 A 20 520 16 E 30 550 15 G 50 600 14 B(2 weeks) & D 100 700 13 F 60 760 d. Alternative solution, see explanation above. Duration Total Total Total Costs direct costs indirect costs 18 $500 $400 $900 17 520 350 870 16 550 300 850 15 600
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