this post was submitted on 08 Jun 2024
1205 points (98.4% liked)

memes

9688 readers
4696 users here now

Community rules

1. Be civilNo trolling, bigotry or other insulting / annoying behaviour

2. No politicsThis is non-politics community. For political memes please go to [email protected]

3. No recent repostsCheck for reposts when posting a meme, you can only repost after 1 month

4. No botsNo bots without the express approval of the mods or the admins

5. No Spam/AdsNo advertisements or spam. This is an instance rule and the only way to live.

Sister communities

founded 1 year ago
MODERATORS
 
you are viewing a single comment's thread
view the rest of the comments
[โ€“] [email protected] 2 points 3 months ago (1 children)

Well, arguably still "incorrect" in real world terms since it fails to have an adjustment for divisibility of D as a function of how many people. If theoretically a task is "perfectly divisible" at two people and halves the time, it will not be the case that a million people will cause it to happen in one millionth of the time. Improvement by expressly pointing out "C" and declaring your assumption of zero for math to work. Also assumption than for any increment of X, the time impact is equal.

In math this is pedantic, but it sure impact project planning in very disastrous ways, and business people love to assume C is zero, any change to X is linear and with linear impact, and make embarrassingly bad calls as a result.

[โ€“] [email protected] 1 points 3 months ago* (last edited 3 months ago)

yup, but that answer was based entirely on the assumptions present in the question. D is all divisible work, and C is everything else, because that's literally all you can assume to make the math work. D has to therefore be 12 months worth of divisible work minus C. C could very well be 12 months of work, meaning D is zero and adding more workers won't matter.