Just a bit of procrastination here but at the bus stop I was wondering how should I study before multiple tests given the percentage of the grade each one was worth, the days until each test and the amount of hours you want to study that day.

Here's my attempt

I have come to the conclusion that it would be an inverse relationship between the percentage of time you should allocate (S) and the time until the test (t). Multiplied this by the grade percentage the test was worth (G).

$$S\propto \frac{G}{t}$$

Given that you have multiple tests and 8 hours (S=8) a day to study I sum the above relationship to and equate the hours to yield the following: (k= 8/sum)

$$8=k \sum^n_{i=1} S_i$$ $$8=k \sum^n_{i=1} \frac{G_i}{t_i}$$

For instance let's say that I have 6 tests: English($G_1=0.25$, $t=58$ days) Maths Methods($G_2=0.5$, $t=60$ days), Specialist Maths($G_3=0.5$, $t=65$ days), Chemistry($G_4=0.5$, $t=72$ days), Modern History($G_5=0.25$, $t=74$ days), Physics($G_6=0.5$, $t=76$ days). Today I have 8 hours to study (more like 7 after I'm done with this), using my formula I can calculate how long I should study for each test today ($S_i$).

$$8=k \sum^n_{i=1} \frac{G_i}{t_i}= k\Big(\frac{G_1}{D_1}+\frac{G_2}{D_2}+\frac{G_3}{D_3}+\frac{G_4}{t_4}+\frac{G_5}{t_5}+\frac{G_6}{t_6}\Big)$$


$k\approx 214.84$ Therefore,

  • English $S= 0.93$ hours
  • Maths Methods $S= 1.79$ hours
  • Specialist Maths $S= 1.65$ hours
  • Chemistry $S= 1.49$ hours
  • Modern History $S= 0.73$ hours
  • Physics $S= 1.41$ hours

Now given that the time approaches the day before the first test (English) I want to just study the content of the first test so I am ready for the test. So let $t_1=1$


$k\approx 14.45 $ Therefore

  • English $S= 3.61$ hours
  • Maths Methods $S= 2.4$ hours
  • Specialist Maths $S=0.90 $ hours
  • Chemistry $S= 0.48$ hours
  • Modern History $S= 0.21$ hours
  • Physics $S= 0.38$ hours

This is where my model somewhat fails as I want most of the day to be studying the most urgent subject. If the test dates were more 'spaced out' or English had a 50% of the marks test then I would spend most of the day before the test studying the subject.

Can any of you 'test anxious' mathematicians suggest any improvements of this crude model (I am a high school student)? I was thinking about throwing in some differential equations ($\frac{\partial S}{\partial t_1} )$ but I'm not smart enough for that yet. Thank you


I have more of a think about it and have come up with the following conditions. Ignoring $S\propto \frac G t$

$$1=k \sum^n_{i=1} S_i $$

$$0=\sum^n_{i=1} \frac{dS_i}{dt} $$

$$ S_i(t_i=1)=1 $$

$$ S_i(t_i<1)=0$$

$t$ is the days until the exam. I have come up with the following model for 3 different tests with test dates t=10, t=20, t=30 https://www.desmos.com/calculator/jrrawaofki

This application can be extended to all tasks you pursue which have a deadline. This isn't a serious model and there is plenty of space to play with.

How could I extend and improve this model using differential equations?

This is the second bounty

  • 3
    $\begingroup$ Given that you are smart enough to even pose this question, and given that you are in high school rather than college, I would be surprised if 4 hours a day (or significantly less) of homework, wasn't sufficient for you to make straight A's in all of your courses. The approach that I would advise is to forgo short term memory gain, study to learn the material, reviewing what you have learned daily, and when test time comes around, just reveal what you have learned. Often, daily review of previous lessons will very rewardingly require less than 3 minutes. $\endgroup$ Aug 29 at 0:26
  • 1
    $\begingroup$ It seems like time needs to play more of a factor. What happens if you have $t^2$ instead of $t$ in the denominator? $\endgroup$
    – Kman3
    Aug 29 at 0:27
  • $\begingroup$ @user2661923 thank you for your advice. I split my study to 2 hours/day on weekdays and 8 hours on a Sunday. However my formula works for any given study periods you want, I just chose 8 hours arbitrarily and it is a Sunday after all. $\endgroup$
    – hwood87
    Aug 29 at 0:34
  • 3
    $\begingroup$ You may want to indicate whether this is a serious model or just something to play with. That said you should take into account negative effects of overworking yourself. I would be tempted to model it as a linear optimization problem (en.m.wikipedia.org/wiki/Linear_programming) with constraints that represent things like sleeping, play time, etc. Studying is only beneficial if it's high quality and your brain needs those other things to be at its best. Personally I found short sessions with breaks between, sometimes with a nap, helped me internalize new information the fastest. $\endgroup$ Sep 5 at 6:19
  • 1
    $\begingroup$ You could have used this time of making this problem to actually study for the test.. $\endgroup$
    – Buraian
    Sep 11 at 13:36

Nice question.

I first of all disagree with your 'conclusion' that the percentage of time allocated should be inversely proportional to the time until the test. My reason is pretty basic. Say you have two tests, one in 80 days and one in 81 days, worth the same amount. Then your 'conclusion' would mean that, on the first day, roughly the same amount of time should be spent studying for both tests, and after 79 days, twice as much time should be spent on one than the other. However, the days $1,2,\dots,79$ are indistinguishable from each other, so you can just interchange what you do on days $1$ and $79$.

In any event, here is my model. Say there are $n$ tests, with test $j$ being a proportion/weight $w_j$ of your final grade. For ease, I will normalize so that you have one hour, or one unit of time, to study each day (you can easily change this to $8$). You're given days $d_1,\dots,d_n \in \mathbb{N}$. To have a more advanced model, I will also introduce learning parameters, $(\lambda_j)_{j=1}^n$, measuring how quickly you can learn the subject that test $j$ is testing on. The smaller $\lambda_j$, the more slowly you learn.

Finally, for a given $\lambda$, we'll have a function $f_\lambda : \mathbb{R}^{\ge 0} \to [0,1]$, where $f_\lambda(t)$ reflects how much you have learned after spending time $t$ studying, if your learning parameter (for the given material) is $\lambda$. The expected grade you get on the test after studying for time $t$ is $100f_\lambda(t)\%$, which I'll just call $f_\lambda(t)$ for ease.

What should $f_\lambda$ be? We assume $f_\lambda(0) = 0$; that is, you start off knowing nothing (it is easy to change this if you wish). The function $f_\lambda$ should of course be increasing and reflect the fact that learning is most rapid at the beginning and then slows down and asymptotically approaches a limit (of perfect knowledge). It makes sense to have $$\frac{\partial f_\lambda}{\partial t}(t) = \lambda (1-f_\lambda(t));$$ intuitively, starting at $t=0$, $f_\lambda$ starts at $0$, starts out increasing with rate $\lambda$, and then increases at a slower (and slower) rate as $f_\lambda$ increases. Solving the differential equation yields $$f_\lambda(t) = 1-e^{-\lambda t}.$$

With all these parameters (and given context), what is the appropriate utility function? It's pretty clear to me that it is the sum of the expected grades, weighted by the worth of the test: $$U = \sum_{j=1}^n \left(1-e^{-\lambda_j t_j}\right)w_j,$$ where we studied time $t_j$ for test $j$.

Therefore, we come to the following optimization problem. First, let us write $\vec{t}^{(d)} = (t^{(d)}_1,\dots,t^{(d)}_n)$ to denote the times spent studying for each test on day $d$.

$\textbf{Optimization Problem}$: Given $n \in \mathbb{N}$, learning parameters $\lambda_1,\dots,\lambda_n > 0$, weights $w_1,\dots,w_n \in [0,1]$, and days $d_1,\dots,d_n \in \mathbb{N}$, choose times $(\vec{t}^{(d)})_{d \ge 0}$ satisfying $\sum_{j=1}^n t^{(d)}_j = 1$ for each $d \ge 0$ in order to maximize $\sum_{j=1}^n \left(1-e^{-\lambda_j t_j}\right)w_j$, where $t_j := \sum_{d=0}^{d_j-1} t^{(d)}_j$.

One can solve this optimization problem for any inputs. We will do so for $n=2$. Let me first remark though that the main regime of interest is when, roughly, $\lambda_j \approx 1/d_j$ (the point is that $e^{-y}$ is close to $1$ when $y$ is very small and is close to $0$ when $y$ is large).

$\textbf{Solution for $2$ Tests}$ Say $d_2 = d_1+\Delta$, where $\Delta \ge 0$. Of course, on days $d_1,\dots,d_1+\Delta-1$ (if there are any), we will only study for test $j=2$. So write $t_2 = \Delta+\overline{t_2}$. We wish to maximize $(1-e^{-\lambda_1 t_1})w_1+(1-e^{-\lambda_2 t_2})w_2$, which is of course equivalent to minimizing $e^{-\lambda_1 t_1}w_1+e^{-\lambda_2 t_2}w_2$ (since $w_1,w_2$ are fixed). We may write this as $$e^{-\lambda_1 t_1}w_1+e^{-\lambda_2\Delta}e^{-\lambda_2 \overline{t_2}}w_2.$$ We want $t_1+\overline{t_2} = d_1$ (we can then do $t^{(d)}_1 := t_1/d_1$ and $t^{(d)}_2 := \overline{t_2}/d_1$ for each $0 \le d \le d_1-1$). So we wish to minimize $$e^{-\lambda_1 (d_1-\overline{t_2})}w_1+e^{-\lambda_2\Delta}e^{-\lambda_2 \overline{t_2}}w_2.$$ Differentiating with respect to $\overline{t_2}$ to find the minimum yields $$\overline{t_2} = \frac{\ln\left(\frac{\lambda_2 w_2}{\lambda_1 w_1}\right)+(\lambda_1 d_1 - \lambda_2 \Delta)}{\lambda_1+\lambda_2}.$$ You can now obtain $t_1$ and thus how much you study on each day.

  • $\begingroup$ Thank you that was an excellent answer I really appreciate it. When using the differential equation to find the function $f_{\lambda} (t)$ why do you use $2$ in the exponent? Thank you for your help $\endgroup$
    – hwood87
    Sep 15 at 1:41
  • $\begingroup$ @hwood87 I have no idea. The $2$ was actually bothering me, so I'm happy I can remove it. Hopefully all the unnecessary $2$'s are now removed. $\endgroup$ Sep 15 at 1:51

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