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An optimisation problem

  1. Dec 14, 2013 #1
    This is a problem I got from my friend. I think this can be solved linear algebra. Please help me to transform this problem to mathematical equations at least.

    Raju has an exam after 1 month , and daily he allots 4 hours for study. He has 3 subjects to study (A, B and C) for the exam. Subject A covers 10% of total contents in the syllabus for exam, B 40% and C the rest 50%. The marks weightage for subject A in the exam is 30%, for B is 10% and for C is 60%. Raju rates his subjects according to level of toughness , like A-3/3, B-1/3, C-2/3. (3/3 - very easy, 2/3-medium, 1/3-hard) .Now how Raju must allot his timetable for effective learning considering the above given parameters. That is how much time in hours must he spent for A, B and C subjects?
     
  2. jcsd
  3. Dec 14, 2013 #2

    Ray Vickson

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    Show your work.
     
  4. Dec 14, 2013 #3
    I don't know if this is right. I am going to use some ideas from ratio and proportions here:

    hour allotted for A, B, C be x, y and z respectively.

    x + y + z = 30 * 4 = 120 - (1)

    now from ratio and proportions I think (see the attachment image)
    10:30:3:x = 40:10:1:y = 50:60:2:z - (2)

    From (1) and (2) I got x= 92.7 hrs, y= 2.57 hrs and z= 24.72 hrs. I dont know if this is right.
     

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    Last edited: Dec 14, 2013
  5. Dec 16, 2013 #4

    haruspex

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    I don't know what that notation means.

    In order to answer the question, you need to make some assumption about how much the marks will improve for a subject by spending one hour on it, marks = f(syllabus_fraction, weight, ease). (The marginal gain.)
    The way it depends on weight seems straightforward, but the other two are less clear. If a subject is easy, maybe you don't need to spend any time on it; or maybe a little time is well rewarded.
     
  6. Dec 16, 2013 #5
    10:30:3:x = 40:10:1:y = 50:60:2:z means 10/30/3/x = 40/10/1/y = 50/60/2/z

    / for division

    Can you help me here with stating your own assumptions.
     
  7. Dec 16, 2013 #6

    Ray Vickson

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    You have not helped with understanding: the notation "10:30:3:x = 40:10:1:y = 50:60:2:z" was bad enough, but "10/30/3/x = 40/10/1/y = 50/60/2/z" is even worse. Instead of trying to summarize in notation that nobody else in the world understands, why not just say in plain English what you actually mean? Are you saying that 10/x = 40/y = 50/z (one statement) and 30/x = 10/y = 60/z (another statement), etc? Or, are you trying to say something else?
     
  8. Dec 16, 2013 #7

    haruspex

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    As Ray posted, that does not explain anything. What is 10/30/3/x? As algebra it's ambiguous: a/b/c could mean (a/b)/c or a/(b/c), which are different.
    I'm sure you can think of something. I'm happy to comment on your ideas.
     
  9. Dec 16, 2013 #8
    I was trying to apply some theory from RATIO and PROPORTIONS.

    What I was trying to convey was this


    (10/30) / (3/x) = (40/10) / (1/y) = (50/60) / (2/z)

    where (a/b) / (c/d) = (a d) / (b c)

    also I don't have any justification for this. This was a try. I wanted to understand this problem. I am trying to state this problem mathematically first.

    Regarding the assumptions Raju is aiming for a 70% marks in the exam.
     
  10. Dec 17, 2013 #9

    haruspex

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    I don't think you can hope to answer the question sensibly without considering what the numbers represent in the real world. All else is numerology.
    There's another point I should have made before. You will need to allow for the law of diminishing returns. As Raju spends time on one topic, the value of spending further time on it declines. You must account for this, otherwise the answer will be to spend all the time on a single topic.
    How about this...
    - first, simplify the problem by subdividing the subjects into topics of equal content; then you only have to worry about two attributes: marks per topic and difficulty.
    - for each topic, there's a maximum mark Raju can hope to get regardless of time spent; this depends on the marks available and the degree of difficulty.
    - you want a function of time spent that starts at 0 and tends to that maximum value at infinity. Pick one.
    - you could throw in a parameter (a multiplier on the time) which also depends on the difficulty.
    You will notice that it is perhaps no longer a linear problem, but so be it.
     
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