Dear Bimal,
 
This is a good question, because we never think that a job should be treated as a product too.
 
May be we could assume some task like this :

Task 1  : Type invitation letter for Local Manager. (Target : Wednesday, 2 pm)
Task 2  : Make daily report to Direct Supervisor. (Target : Wednesday, 2 pm)
Task 3  : Prepare Power Point presentation for all Directors. (Target : Wednesday, 2 pm)
 
Since the target is same for all task, so that we have make some priority by giving weight for each of task.

Task 1  : Type invitation letter for Local Manager. (Target : Wednesday, 2 pm) – Weight : 10
Task 2  : Make daily report to Direct Supervisor. (Target : Wednesday, 2 pm) – Weight : 20
Task 3  : Prepare Power Point presentation for all Directors. (Target : Wednesday, 2 pm) – Weight : 40
 
From this point, we can count the maximum score for each task by its fulfillment.

Task 1 : Start Monday (2 days before). So, the daily value is [(x+1) / 2] x Weight, that is (3 / 2) x 10 = 15.
Task 2 : Start Monday (2 days before). So, the daily value is [(x+1) / 2] x Weight, that is (3 / 2) x 20 = 30.
Task 3 : Start Monday (2 days before). So, the daily value is [(x+1) / 2] x Weight, that is (3 / 2) x 30 = 45.
 
If Task 1 is delay some days, so the value estimation is like this :

Days delay : 1, convert to negatif value (-1)
Weight : 10
Value = (0 / 2) x 10 = 0
 
Days delay : 2, convert to negatif value (-2)
Weight : 10
Value = (-1 / 2) x 10 = -5
 
Days delay : 3, convert to negatif value (-3)
Weight : 10
Value = (-2 / 2) x 10 = -10
 
The same formula is working for the other task, so that we can count the value of job completion.
 
I think that’s my answer, while I’m still considering another formula for it.

Thanks.

Negi