Weekend Maths Brainteaser

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turb

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#317397 12-Oct-2024 06:44

There are 5 team members:

A with a WTE of 0.6

B, 0.5

C, 0.3

D, 0.3

E, 0.3

Part of their job is delivering a particular task, but no one likes doing it. 2 team members work together to deliver each task.

They have decided the fairest way is to allocate the duty proportionally according to their WTE

There are x tasks per year.

What is the formula we could use to calculate how many sessions does each person has to do, to ensure all the tasks are covered?

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floydbloke

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#3296319 12-Oct-2024 07:27

I googled it so you don't have to.

I believe WTE above stands for Whole-Time Equivalent (not Waste-To-Energy), more commonly known here in NZ as FTE I thought.

Sometimes I use big words I don't always fully understand in an effort to make myself sound more photosynthesis.

eracode

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#3296323 12-Oct-2024 08:01

I copied the post verbatim and pasted it into ChatGPT. It has given me workings resulting in a plausible answer but I won't post it here because I cheated.

I will be interested to see whether it's the same as answers posted here by people who presumably worked from first principles.

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nova

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#3296334 12-Oct-2024 08:59

Ii's not really a brain teaser, more of a straight maths problem. Total sum of everyone's WTE allocation is 2 WTE. There is 2 WTE required per task. This cancels out, so the percentage allocation for each person is A 60% of the time, B 50% of the time, C,D&E 30% of the time. Or 0.6x, 0.5x, 0.3x if you prefer. This only works properly if the number of tasks is a multiple of 10. If there were only 2 tasks for example then A+B would do the first and A+C would do the second, and D+E would do nothing. Or maybe they could shift roles part way through the task.

eracode

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#3296389 12-Oct-2024 11:35

nova:

Ii's not really a brain teaser, more of a straight maths problem. Total sum of everyone's WTE allocation is 2 WTE. There is 2 WTE required per task. This cancels out, so the percentage allocation for each person is A 60% of the time, B 50% of the time, C,D&E 30% of the time. Or 0.6x, 0.5x, 0.3x if you prefer. This only works properly if the number of tasks is a multiple of 10. If there were only 2 tasks for example then A+B would do the first and A+C would do the second, and D+E would do nothing. Or maybe they could shift roles part way through the task.

Not sure how this works - your allocation is more than 100%?

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floydbloke

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#3296391 12-Oct-2024 11:42

I'd just sack the entitled lot of them and hire a couple of people who aren't afraid to do an honest full day's work. 😝

Sometimes I use big words I don't always fully understand in an effort to make myself sound more photosynthesis.

nova

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#3296414 12-Oct-2024 12:30

eracode:

nova:

Ii's not really a brain teaser, more of a straight maths problem. Total sum of everyone's WTE allocation is 2 WTE. There is 2 WTE required per task. This cancels out, so the percentage allocation for each person is A 60% of the time, B 50% of the time, C,D&E 30% of the time. Or 0.6x, 0.5x, 0.3x if you prefer. This only works properly if the number of tasks is a multiple of 10. If there were only 2 tasks for example then A+B would do the first and A+C would do the second, and D+E would do nothing. Or maybe they could shift roles part way through the task.

Not sure how this works - your allocation is more than 100%?

The problem states that two people are required for each task. So if there were 10 tasks, you would need a total of 20 allocations. And to do it in proportion, A would do 6 tasks, B would do 5, and C,D&E would do 3 each. So the total should add up to twice the number of tasks, i.e. 200%. If the number of tasks is not a multiple of 10 it is impossible to do it fairly, unless people swap over during a task and each do part of it.

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eracode

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#3296418 12-Oct-2024 12:42

nova:

eracode:

nova:

Ii's not really a brain teaser, more of a straight maths problem. Total sum of everyone's WTE allocation is 2 WTE. There is 2 WTE required per task. This cancels out, so the percentage allocation for each person is A 60% of the time, B 50% of the time, C,D&E 30% of the time. Or 0.6x, 0.5x, 0.3x if you prefer. This only works properly if the number of tasks is a multiple of 10. If there were only 2 tasks for example then A+B would do the first and A+C would do the second, and D+E would do nothing. Or maybe they could shift roles part way through the task.

Not sure how this works - your allocation is more than 100%?

The problem states that two people are required for each task. So if there were 10 tasks, you would need a total of 20 allocations. And to do it in proportion, A would do 6 tasks, B would do 5, and C,D&E would do 3 each. So the total should add up to twice the number of tasks, i.e. 200%. If the number of tasks is not a multiple of 10 it is impossible to do it fairly, unless people swap over during a task and each do part of it.

@nova Sorry I'm thick but I still don't see how this works given that they are required to 'work together to complete each task' - working in pairs. Let's say they were required to manually fill 10 sandbags - one shoveling, one holding the bag. Based on your solution, how would that work per person - i.e. who works with who and for how many sessions?

Sometimes I just sit and think. Other times I just sit.

nova

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#3296842 13-Oct-2024 10:33

eracode:

nova:

The problem states that two people are required for each task. So if there were 10 tasks, you would need a total of 20 allocations. And to do it in proportion, A would do 6 tasks, B would do 5, and C,D&E would do 3 each. So the total should add up to twice the number of tasks, i.e. 200%. If the number of tasks is not a multiple of 10 it is impossible to do it fairly, unless people swap over during a task and each do part of it.

@nova Sorry I'm thick but I still don't see how this works given that they are required to 'work together to complete each task' - working in pairs. Let's say they were required to manually fill 10 sandbags - one shoveling, one holding the bag. Based on your solution, how would that work per person - i.e. who works with who and for how many sessions?

Say there are 10 tasks, the breakdown of the two people for each task could be:

AB, AB, AB, AC, AD, AE, BC, BD, CE, DE

So A does 6 tasks, B 5 tasks, and C,D,E do 3 each. The proportions are correct, and the formula for working how many tasks is persons_wte x num_tasks for each person ( or more generally (persons_wte/total_wte) x num_tasks x people_per_task. But since total_wte is 2 and persons_per_task is also 2, that cancels out in this specific problem.