OSCM 373 Exam 2 Review

The table below shows an intermediate step of the Hungarian Method for the Assignment Model. Is a minimum cost assignment available at this step? Job\Worker A B C D 1 0 2 1 5 2 0 4 2 0 3 6 3 2 0 4 0 0 3 3

No.

Plan A has a fixed cost of $10 and a variable cost of $0.05 per minute. Plan B has a fixed cost of $25 and no variable. What is the point of indifference between the two plans?

Point of indifference is the level of volume in which total cost is the same. 25 - 10 = 15 => 15/0.05 = 300 minutes

A machine has a design capacity of 5,000 and an effective capacity of 4.000 units per year. If the expected output is 3,000 per year, what would be the Utilization?

Utilization = Actual Output / Design Capacity = 3,000 / 5,000 = 60%

A project activity C has a duration of 3 days and two predecessors A&B. A has an EF of 10 and B an EF of 12, What is the earliest C may be completed?

12>10 => 12+3 = 15

A process has a fixed cost of $20,000, variable costs per unit of $10 and outputs can be sold for $15 per unit. What is the Break-Even Point?

Break-Even Point = Fixed Cost / (Price - Variable Cost) = 20,000 / (15-10) = 4,000 units

A project's indirect costs are $1,000 per day. Currently there are 3 activities available to crash. All can be shortened by up to 1 day and none is on the Critical-Path. Crashing costs are: A for $800 per day B for $1,200 per day C for $500 per day What should you do?

Do not crash any. You only crash activities on the Critical-Path. You then crash the longest tasks (shorter tasks are typically harder to shorten the duration of).

What should be your goal when using John's Rule?

John's rule is a method that involves sequencing in two job stations to minimize total time to complete all jobs. Therefore, the answer is minimize the time it takes to complete all jobs.

Look at the project network below, which activity would you crash first? Assume they all cost the same to crash the numbers are duration in days. S-A (2) (on Critical Path) G-F (5) (not on Critical Path) F-E (3) (not on Critical Path) C-D (4) (not on Critical Path)

S-A, you always crash a task on the Critical Path, then followed by longest duration.

Using the Shortest Time Sequencing Rule and Assuming Start Time is 0, when is Job C expected to end? Job Duration Due A 3 5 B 7 8 C 5 10 D 9 12 E 6 6

Sequence based on Shortest Time Sequencing: A-C-E-B-D => 3+5 = 8

This Critical Path is never: The path with the least activities The longest path The shortest path The path with the least nodes

The shortest path