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In a book by Wayne Winston for operations research I found this question.

enter image description here

Here's how I did it:
Let $t$ be the no.of subjects to pass and let h be the no.of hours she has in hand for studying.
Then $f_t(h)$ be the probability of passing subjects t,t+1,..3 when h hours are in hand. Let $x_t$ be the no.of hours spent on subject t.
$0<=x_t<=h; h\in {0,1,2,3,4}$

Then $f_3(4)=0.5 \\ f_3(3)=0.44 \\f_3(2)=0.4\\f_3(1)=0.3\\f_3(0)=0.1$
Let $P_t(x_t) $ be the probability of passing subject t when $x_t$ hours are spent on it.

$f_t(h)=max ${${P_t(x_t)}+f_{t+1}(h-x_t)$}
According to this I get
$f_(4)=$max$P_1(0)+f_2(4)=0.98\\P_1(1)+f_2(3)=1\\P_1(2)+f_2(2)=1\\P_1(3)+f_2(1)=0.93\\P_1(0)+f_2(4)=0.425$

The answer I get is she should study French for 1 hour, German for 1 hour and Stat for two hours. But the answer given in the book says French 1 hour, German 0 hours, Stat 3 hours.

What's wrong with my method? And what is the probability of passing at least one subject ?

callculus42's user avatar
callculus42
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askedJul 22, 2015 at 14:26
sam_rox's user avatar
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  • $\begingroup$have you worked with routing algorithms?Dijkstra algorithm? As this solves the problem of finding the route by imagining the matrix as a nodes then iterating through would be my first guess$\endgroup$CommentedJul 22, 2015 at 14:50

1 Answer1

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I have a different approach.

indices:

$ij$: The amount of hoursi Angie is learning for coursej.

$i \in \{0,1,2,3,4\}$

$j \in \{1,2,3\}$

variables:

$$x_{ij}=\begin{cases} 1, \ \texttt{if Angie learns i hours for course j} \\ 0, \ \texttt{else}\end{cases}$$

$$f_{ij}=\texttt{probability, that Angie passes course j, if she is learning i hours}$$

constraint 1: Angie is learning 4 hours a week

$$\sum_{i=0}^4 \sum_{j=1}^3 i \cdot x_{ij}=4$$

constraint 2: Only one hours/course combination for each course

$$ \sum_{i=0}^4 x_{ij}=1 \ \ \ \ \forall j \in \{1,2,3\}$$

objective function:

Because of the comments ofjoriki the objective function is

$$\texttt{max} \ \ \ 1-\prod_{j=1}^3 \left(1- \sum_{i=0}^4 f_{ij} \cdot x_{ij} \right) $$

answeredJul 22, 2015 at 15:23
callculus42's user avatar
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    $\begingroup$Your objective function is the sum of the probabilities of passing the individual tests. The intended objective function is the probability of passing at least one test, i.e. the complement of the product of the complements of the probabilities.$\endgroup$CommentedJul 22, 2015 at 15:28
  • $\begingroup$On one hand I understand your argument (more or less). On the other hand I do not really know how to change the objective function. Could you show me (us), how you would write the objective function ?$\endgroup$CommentedJul 22, 2015 at 15:42
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    $\begingroup$$$1-\prod_{j=1}^3\left(1-\sum_{i=0}^4f_{ij}x_{ij}\right)$$$\endgroup$CommentedJul 22, 2015 at 15:44
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    $\begingroup$This looks fine. I will edit my answer. Many thanks for this helpful hint.$\endgroup$CommentedJul 22, 2015 at 15:51
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    $\begingroup$@sam_rox: Yes, exactly. The way things are set up in this answer, $\sum_{i=0}^4f_{ij}x_{ij}$ is the probability of passing test $j$. (I wouldn't have set it up like that, but it's correct.) Then one minus that is the probability of not passing, i.e. of failing test $j$. The product of those probabilities for all three tests is the probability of failing all of them. And one minus that is the probability of not failing all of them, i.e. of passing at least one.$\endgroup$CommentedJul 22, 2015 at 17:34

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