Interpret success-rate/time * $

[McFluffy]

Homework Statement

You are applying for a $\$1000$ scholarship and your time is worth $\$10$ an hour. If the chance of success is $1 -(1/x)$ from $x$ hours of writing, when should you stop?

Homework Equations

Let $p(x)=1 -(1/x)$ be the rate of success as a function of time, $x$.

The Attempt at a Solution

[/B]
My way if thinking eventually led to the correct answer which is $\frac{1}{x^2}1000=10$. Solving for $x$ gives you the solution.

I was stuck at this problem and didn't know how to proceed and I tried to find out if I could find the answer just by matching up the units on both sides. I don't know how they calculated $p(x)$ but I do know that it is dimensionless. Thus, $p'(x)$ will give me the rate of success per unit time or just the unit, $1/h$. I know that $\$1000$ has dollar units and that $10\frac{}{h}$ has dollar per hour unit, so if I multiply $p'(x)=\frac{1}{x^2}$ with $\$1000$ I should get the same units as $10\frac{}{h}$. So I set $\frac{1}{x^2}1000=10$ and solved for $x$. Feeling doubtful, I checked the solution and was surprised how I got it right.

My question is how do you interpret the solution, $\frac{1}{x^2}1000=10$? Like since $p'(x)$ is defined as the success rate per unit time, how come if I multiplied it by $\$1000$, it got me the solution? I just don't understand it, $$\frac{\text{success rate}}{\text{time}}\cdot \text{currency}$$

How do you interpret this? and yes, I'm aware that the success rate is dimensionless but still, I don't understand the reasoning behind the answer. I just want someone to solve the problem with also providing some commentary on his/her methods of solving it as I felt that my reasoning is inadequate.

 

[kuruman]

The more hours you work on the application the better your chance in getting the $1000 scholarship. What is the extra dollars you expect to gain when you spend an extra hour on that application? When those expected extra dollars per hour are equal to your hourly wage of $10, then you must stop working on the application and go flip burgers or whatever. That's what the equation you discovered is saying. BTW, $p(x)$ as given is not dimensionless if $x$ in the denominator has units of hours.

 

[McFluffy]

BTW, $p(x)$ as given is not dimensionless if $x$ in the denominator has units of hours.

$p(x)$ is the rate of success function, which calculates the probability of being successful, you're referring to $p'(x)$.

 

[McFluffy]

The more hours you work on the application the better your chance in getting the $1000 scholarship.

This one I understand as this is how $p(x)$ was defined.

What is the extra dollars you expect to gain when you spend an extra hour on that application? When those expected extra dollars per hour are equal to your hourly wage of $10, then you must stop working on the application and go flip burgers or whatever.

I don't understand this one. I understand since my time is worth 10$/hour and that I want to get the $1000 scholarship, I wouldn't want to spend more than 100 hour writing for the scholarship. If I spend an extra hour on that application, it would not be worth it.

 

[kuruman]

I am referring to your statements

Let $p(x)=1 -(1/x)$ be the rate of success as a function of time, $x$.

I don't know how they calculated $p(x)$ but I do know that it is dimensionless.

If $x$ has dimensions of time, $1/x$ has dimensions of inverse time in which case $p(x)$ is not dimensionless; it has mixed dimensions because while $1$ is dimensionless, $1/x$ is not.

 

[McFluffy]

I am referring to your statementsIf $x$ has dimensions of time, $1/x$ has dimensions of inverse time in which case $p(x)$ is not dimensionless; it has mixed dimensions because while $1$ is dimensionless, $1/x$ is not.

Sorry, I feel confused a bit, isn't $p(x)$ a probability function? Aren't they supposed to be dimensionless?

 

[kuruman]

Sorry, I feel confused a bit, isn't $p(x)$ a probability function? Aren't they supposed to be dimensionless?

Of course they are. This is a minor point that has no effect on the answer. You were a bit careless in defining "$p(x)$ as a function of time $x$." Actually, $x$ is dimensionless and its definition should be "number of hours spent on the application". As a number, it is dimensionless and so is $p(x)$ in that case.

 

[McFluffy]

Of course they are. This is a minor point that has no effect on the answer. You were a bit careless in defining "$p(x)$ as a function of time $x$." Actually, $x$ is dimensionless and its definition should be "number of hours spent on the application". As a number, it is dimensionless and so is $p(x)$ in that case.

I just want to ask, even though my solution in the end was valid, was the way of thinking to obtain the solution justified?

 

[StoneTemplePython]

I just want someone to solve the problem with also providing some commentary on his/her methods of solving it as I felt that my reasoning is inadequate.

Note: in general following the units is a good idea, but this problem isn't great for that -- still worth a try though.

Two basic approaches for this problem:

they both use the fact that you have

$\text{Expected Payoff} = 1000 (1-\frac{1}{x}) - 10x$

and you want to maximize that by finding some $x \gt 0$

Approach 1:
Use calculus. This is close to what you did, I think.

Approach 2:
Why not use $GM \leq AM$?

steps:

re-write the expected payoff slightly

$\text{Expected Payoff} = 1000 \big(1-\frac{1}{x}\big) - 10x = 2000\Big(\frac{1}{2} - \frac{1}{2}\big(\frac{1}{x} + \frac{x}{100}\big)\Big)$

the goal is maximization, so we want to minimize the thing that's being subtracted i.e. where the variable $x$ is, i.e.

$\text{minimize } \frac{1}{2}\big(\frac{1}{x} + \frac{x}{100}\big)$

a nice way to do this is apply $AM \geq GM$

$\frac{1}{2}\big(\frac{1}{x} + \frac{x}{100}\big) \geq \big(\frac{1}{x} \frac{x}{100}\big)^\frac{1}{2} = \frac{1}{10}$

with equality iff $\frac{1}{x}= \frac{x}{100} \to x = 10$ for maximization to occur

 

[kuruman]

I just want to ask, even though my solution in the end was valid, was the way of thinking to obtain the solution justified?

I alluded to the way to think about it in #2, but you wrote you didn't understand it in #4. So here is what I meant stated differently. The question asks you to find the point of diminishing returns. You should keep spending time on the application as long as your expected hourly wage doing this is more than your guaranteed wage doing something else. In terms of an inequality, $1000\times(dp/dx) \ge 10$. The point of diminishing returns is at the equality and gives the answer to the question.