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.
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.
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.
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.