Question

Four jobs are on order in a factory. As on day 20 of the production calendar, the corresponding due date and work remaining to complete these jobs (in days) are given in the table below.

Job	Due date	Work remaining (in days)
M	26	5
N	24	10
O	28	10
P	31	5

Which job(s) has/have critical ratio less than unity?

• A. Job N and Job O
• B. Job M and Job N
• C. Job P
• D. Job O and Job P

Hint:

Critical ratio is \((\text{due date}-\text{today})/\text{work remaining}\); a value under 1 means the job is running behind.

Answer 1

The Correct Option is A

Step 1: Rephrase the rule as a race between two clocks.
Each job has a time left before due date clock, $T = \text{due date} - 20$, and a days of work still needed clock, $W$. The critical ratio is just $T/W$. If $T$ runs out faster than $W$ (that is, $T < W$), the ratio is under 1 and the job is in trouble.

Step 2: List T and W for every job.
Job M: $T = 6$, $W = 5$. Job N: $T = 4$, $W = 10$. Job O: $T = 8$, $W = 10$. Job P: $T = 11$, $W = 5$.

Step 3: Compare T against W directly for each job.
Job M has $T = 6 > W = 5$, so it is ahead, $CR = 1.2$. Job N has $T = 4 < W = 10$, clearly behind, $CR = 0.4$. Job O has $T = 8 < W = 10$, also behind, $CR = 0.8$. Job P has $T = 11 > W = 5$, comfortably ahead, $CR = 2.2$.

Step 4: Collect the jobs where T falls short of W.
Only Jobs N and O have less time remaining than work remaining, so only these two give a critical ratio under 1.

Final Answer:
Job N and Job O are the ones running behind schedule on day 20. \[ \boxed{\text{Job N and Job O}} \]