Question

A machine shop has received four jobs A, B, C, and D for processing on a single CNC machine. All jobs are available for processing on the first day of the production schedule calendar, and processing times and due dates as applicable on the first day are given below. Using earliest due date rule, the average tardiness (in days) is _________. 

Hint:

To minimize tardiness, use the earliest due date rule for scheduling jobs.

Answer 1

Correct Answer: 4

To calculate the average tardiness, we first use the earliest due date rule to process the jobs in the order of their due dates: B, C, A, D. The completion times for each job are:
- Job B: \( 5 \, \text{days} \),
- Job C: \( 5 + 7 = 12 \, \text{days} \),
- Job A: \( 12 + 8 = 20 \, \text{days} \),
- Job D: \( 20 + 9 = 29 \, \text{days} \).
The tardiness for each job is:
- Job B: \( \text{tardiness} = 0 \, \text{days} \) (since completion time is before due date),
- Job C: \( \text{tardiness} = 0 \, \text{days} \),
- Job A: \( \text{tardiness} = 20 - 14 = 6 \, \text{days} \),
- Job D: \( \text{tardiness} = 29 - 19 = 10 \, \text{days} \).
Thus, the average tardiness is: \[ \frac{0 + 0 + 6 + 10}{4} = 4 \, \text{days}. \] The average tardiness is 4 days.