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Finding the number of rentals and purchases given the revenue

  1. Jun 6, 2017 #1
    1. The problem statement, all variables and given/known data
    "A movie is released in both rentals and purchasable formats online. The total number of views is 4 million, but this includes both rentals and purchases. Assume that each rental brings $3, and each purchase yields $10. The total revenue from the release of the movie is 19 million. Find the number of rentals and purchases."

    2. Relevant equations
    Let ##r## denote the revenue from rentals and ##p## the revenue from purchases.
    Furthermore, let ##a_1## denote the volume of rentals online and ##b_1## denote the volume of purchases online; let ##a_f## and ##b_f## denote the total volume of each for the movie.

    3. The attempt at a solution
    Then we have ##a_1r+b_1p=4_{10^6}## and ##a_fr+b_fp=19_{10^{19}}## and so...
    ##r⋅\begin{pmatrix}
    a_1 \\
    a_f\end{pmatrix}## + ##p⋅\begin{pmatrix}
    b_1 \\
    b_f\end{pmatrix}## = ##10^6⋅\begin{pmatrix}
    4 \\
    19\end{pmatrix}##

    I tried putting it into an augmented matrix. But the problem is, that there are simply too many unknowns to solve for. I don't know what I'm doing wrong.
     
    Last edited: Jun 6, 2017
  2. jcsd
  3. Jun 6, 2017 #2

    Orodruin

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    You know a lot of the variables and so they are not really unknown. Rather than trying to put it on matrix form, which you seem to struggle a bit with, write your equations on regular form first. Once you have them you can construct the corresponding matrix form.

    Note: You only have two unknowns!
     
  4. Jun 6, 2017 #3

    scottdave

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    Are you assuming that a person who purchases can view multiple times? If these count as multiple views, then yes there is not enough information. Because some people may only watch a purchased video once, while others may view it 2 or 3 or more. You just are not told this information. In fact, rentals usually are good for a certain time period (24 or 48 hours, maybe). A rental person could view more than once. My suggestion is to make an assumption statement that a purchase equals 1 view, and a rental equals 1 view, then see about solving it with less variables.
     
  5. Jun 6, 2017 #4

    Mark44

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    This didn't come out the way you planned. More likely, the following is probably a bit closer to what you were trying to say
    ##a_1r+b_1p=4 \cdot 10^6## and ##a_fr+b_fp=19 \cdot 10^6##
     
  6. Jun 6, 2017 #5

    Ray Vickson

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    Generally in such problems (where the variables have direct "managerial" significance) you should choose as your variables the things about which a direct managerial decision is to be made. In this case a manager will not decide on the "revenues", but instead will decide on the numbers sold. From there, the revenues can be calculated.

    So, if I (or my Operations Research students) were doing the problem we would choose variables R = number of rentals (in millions) and P = number of purchases (millions). You get two simple linear equations in R and P, which are essentially trivial to solve.

    Whatever you do, avoid matrices in this problem as they will just get in the way and are completely unnecessary. They hide the simplicity of the problem.
     
  7. Jun 6, 2017 #6
    Okay, so I have for total views: ##3R+10P=19##, and for online views, ##3(R-\delta R)+10(P-\delta P)=4##. I'm going to assume that ##R=3## and ##P=10## because that's the only way that the first equation will make sense (since ##3|(19-10⋅1)## and ##10|(19-3⋅3)##). So, I think I can solve for the variables in the second equation by substitution: ##3(3-\delta R)+10(1-\delta P)=4=9+10##, which implies that ##3\delta R+10\delta P=15##. Then I can say similarly, from my divisibility argument, that ##\delta R=5## and ##\delta P=0##. In total, I have ##∑R=8⋅10^6## rentals, and ##∑P=10^6## purchases?
     
  8. Jun 6, 2017 #7

    Orodruin

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    No. You cannot just randomly assume stuff about R and P. Also, you have not defined what you mean by ##\delta P##. If you put R = 3 and P = 10 your first equation reads 109 = 19 which is clearly nonsensical.
     
  9. Jun 6, 2017 #8

    scottdave

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    The problem stated that there were a total of 4 million, but your numbers adds up to 9 million?
    Also, rereading your Relevant Equations, you need to have a1 + b1 = 4 x 106, since this is just a statement about the number of items, nothing about price or revenue. But I cannot figure out what you intend for af and bf to represent. What is the relationship of a1 to af, for example?
     
    Last edited: Jun 6, 2017
  10. Jun 6, 2017 #9
    ...Wow, I realized that I completely misunderstood the problem. I misunderstood the "four million" to be four million dollars, and not four million views. Once again, I must apologize to everyone for my lack of reading comprehension. So anyway, let ##r## denote the number of rental-views in millions, and ##p## represent the number of purchase-views in millions. Then...

    ##r+p=4##
    ##3r+10p=19##

    ##p=4-r##
    ##3r+10(4-r)=19=3r+(40-10r)##
    ##-7r=-21##
    ##r=3## and ##p=1##

    So there were three million rentals and one million purchases?
     
  11. Jun 6, 2017 #10

    scottdave

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    It happens. Yes that is what I got, as well.
     
  12. Jun 7, 2017 #11
    Thanks for bearing with me, everyone, and for your help...
     
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