Question

3. The network for shooting a TV commercial as shown in the table has a fixed cost of $90 per day but money can be saved by shortening the project duration. Find the least-cost schedule. Normal Time 7 Crash Time 4 Activity Description Cost Increase 1st, 2nd 3"d da $30, 50, 70 40, 45, 65 60, 60 35, 60 Contract personnel Obtain stage props Rent equipment Contract studio Set time and date 12 10 9

Answer 1

ANSWER:

• Taking place the base of the image in the table, we will primary try to locate the serious path.
• Within the image below, green represent regular time and red represent crash time.
• Because for each the image of the representation in the table, the critical pathway for conclusion of the development is 1-2-3-4 and duration is 19 days.

1-2-4=18 days

1-3-4 = 15 days

• Total possible cost as per plan duration = $1710
• On analyze the serious path and picture options for incoming at the minimum cost schedule; Step 4 is the most favorable solution and schedule.

Step	Critical path (in days)	activities to crash by day 1 Cost of the crash)	Cost change	Final total Cost ($)
1	1-2-3-4 (19) 1-2-4 (18) 1-3-4 (15)	1-2 (30)	30-90= -60	1710-60= 1650
2	1-2-3-4 (18) 1-2-4 (17) 1-3-4 (15)	2-3 (40)	40-90= -50	1650-50 = 1600
3	1-2-3-4 (17) 1-2-4 (17) 1-3-4 (15)	1-2 (50) 2-3 (45), 2-4 (35)	50-90 = -40	1560
4	1-2-3-4 (16) 1-2-4 (16) 1-3-4 (15)	1-2 (70) 2-3 (45) 2-4 (35)	70-90=-20	1540
5	1-2-3-4 (15) 1-2-4 (15) 1-3-4 (15)	2-3 (45) 2-4 (35) 1-3 (60)	45+35+60-90=50	1590
6	1-2-3-4 (14) 1-2-4 (14) 1-3-4 (14)	2-3 (65) 2-4 (60) 1-3 (60)	65+60+60-90=95	1,685
7	At 13 weeks system is fully crashed	Crash time = 4+6+3 = 13 weeks