CS 70

Amortized Time vs. …

Contrasting Amortized Time

Suppose you're thinking about doing some volunteer work in the community for ten weeks over the summer. There are several projects you could work on, and they each one has different time commitments:

  1. Worst-Case Kitchen: Maximum of 5 hours per week, every week.
  2. Amortization Park: At the end of week \( m \) you'll have done at most \( 4m \) hours of work, with a maximum of 40 hours total over the ten-week period.
  3. Expectation Center: Each week varies, but it averages out to be 4.5 hours per week, with a 1-hour standard deviation.

What's the amortized time per week at Amortization Park?

A friend tells you that most weeks at Amortization Park only require two hours of work, but the penultimate week (week 9) can be a doozie. Assuming your friend is right, how long might you have to work on week 9?

Amortization Park's promise is that by the end of week 9, you will have done at most 36 hours of work. If the previous weeks each only asked you to work two hours, you'll only have done 16 hours by week 9, so during week 9 you could be asked to work for 20 hours (i.e., \( 36 - 16 \) hours).

Suppose that you're flexible about how you arrange your other commitments, but you don't have enough time to volunteer for more than 48 hours over the whole summer, which option(s) should you avoid? (Briefly explain why.)

Both worst-case time and amortized time provide a hard guarantee about the total amount of time you'll spend working over the summer: 50 hours at the Worst-Case Kitchen and 40 hours at Amortization Park. Unfortunately, 50 hours is too many.

Assuming a normal distribution, there is a 82.9% chance that volunteering at the Expectation Center will take less time than your 48-hour limit, but there's also a 17.1% chance that it will take longer.

So it seems like your best choice is Amortization Park, followed by the Expectation Center, and then the Worst-Case Kitchen.

What is the chance that if you picked Expectation Center, you'd end up working more than 60 hours over the whole summer? (Assuming a normal distribution. Feel free to just guess.)

About

People often overestimate the chance of unlikely events. Assuming a normal distribution with a mean of 4.5 and a standard deviation of 1 for each week's work, the mean for the sum of ten samples is 45, with a standard deviation of \( \sqrt{10} \). So 60 hours would be 4.75 standard deviations away from the mean. That's really unlikely, with a chance of about 1 in a million.

Is any option definitively the best for everyone?

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