Recitation 10

Example: Consider the CPM project network in the figure, where time is measured in days.

Activity Usual time Crash time Cost per day
(0,1) 4 2 $500 (2,3) 10 6$700
(2,4) 7 4 $600 (5,6) 5 3$300
1. Meeting a project deadline: If the durations on the edges of the network are the usual times, and the crash times and the costs of a day’s reduction for four of the activities are given in the table, set up a linear program to determine how to reduce the length of the project to 19 days as cheaply as possible.
2. Minimizing total project cost: If there is a fixed cost involved with the project of $1,000 a day, set up the LP to minimize total project cost. 3. Staying within budget: If a budget of$3,000 is available for reduction in time, set up the LP to determine the shortest time.

Solution: Let t_0, t_1, \dots, t_6 indicate the time of the event of the nodes, and let t_{01}, t_{23}, t_{24}, t_{56} indicate the time of the required to complete activity (0,1), (2,3), (2,4), (5,6).

Meeting a project deadline:

Minimize 500(4-t_{01}) + 700(10-t_{23})+600(7-t_{24})+300(5-t_{56}) t_1 - t_0 \ge t_{01} t_2 - t_0 \ge 5 t_3 - t_1 \ge 8 t_3 - t_2 \ge t_{23} t_4 - t_2 \ge t_{24} t_5 - t_3 \ge 3 t_5 - t_4 \ge 2 t_6 - t_5 \ge t_{56} 4 \ge t_{01} \ge 2 10 \ge t_{23} \ge 6 7 \ge t_{24} \ge 4 5 \ge t_{56} \ge 3 t_{6} - t_0 \le 19

Minimizing total project cost:

Minimize 500(4-t_{01}) + 700(10-t_{23})+600(7-t_{24})+300(5-t_{56})+1000(t_6-t_0) t_1 - t_0 \ge t_{01} t_2 - t_0 \ge 5 t_3 - t_1 \ge 8 t_3 - t_2 \ge t_{23} t_4 - t_2 \ge t_{24} t_5 - t_3 \ge 3 t_5 - t_4 \ge 2 t_6 - t_5 \ge t_{56} 4 \ge t_{01} \ge 2 10 \ge t_{23} \ge 6 7 \ge t_{24} \ge 4 5 \ge t_{56} \ge 3

Staying within budget:

Minimize t_6-t_0 t_1 - t_0 \ge t_{01} t_2 - t_0 \ge 5 t_3 - t_1 \ge 8 t_3 - t_2 \ge t_{23} t_4 - t_2 \ge t_{24} t_5 - t_3 \ge 3 t_5 - t_4 \ge 2 t_6 - t_5 \ge t_{56} 4 \ge t_{01} \ge 2 10 \ge t_{23} \ge 6 7 \ge t_{24} \ge 4 5 \ge t_{56} \ge 3 500(4-t_{01}) + 700(10-t_{23})+600(7-t_{24})+300(5-t_{56})\le 3000