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Very simple problem on % - explanation

  1. Nov 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello!
    For example, there is a 'cash flow' of 80 000. Discount rate is 13%. Then to find discounted cash flow I need to do the following: 80 000 / 1.13 = 70 796.46

    If, for example, the same 80 000 is a gross amount and I need to find net one with the same rate of 13%, then 80 000 * 0.87 = 69 600

    Why are these numbers different? What is the meaning behind the first and second equation? I realize that it is very simple but somehow I have blanked out on this one.

    Thank you!
    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Nov 6, 2014 #2

    LCKurtz

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    If you take the rate as a decimal ##r=.13##, in one equation you are dividing by ##1+r## and in the other you are multiplying by ##1-r##. They are not the same because$$\frac 1 {1+r} \ne 1-r$$In fact$$\frac 1 {1+r} = 1 - r + r^2 - r^3...$$which is an infinite geometric series. Notice that if ##r## is very small, the higher order terms are really small ##\frac 1 {1+r}## is close to ##1-r## in that case.
     
  4. Nov 6, 2014 #3

    DrClaude

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    In the first case, the 13% refers to the discounted cash flow. 70 796.46 plus 13% of 70 796.46 equals 80 000. In other words, the question asked is: how much cash flow do you need such that the cash flow growing by 13% will give you 80 000.

    In the second case, the 13% refers to the 80 000. 80 000 minus 13% of 80 000 equals 69 600.
     
  5. Nov 6, 2014 #4

    Ray Vickson

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    As others have indicated, the mathematical reason is that $$ \frac{1}{1.13} \neq 1 - 0.13$$
    In other words, ##0.8849557522 \neq 0.87##.
     
    Last edited by a moderator: Nov 6, 2014
  6. Nov 7, 2014 #5
    yes, exactly, I understand that, but I don't understand why is this so :) it seems that there are some basics which I miss.
     
  7. Nov 7, 2014 #6
    thank you, yes, it's true, I understand this with the use of these formulas, but I don't understand the reason
     
  8. Nov 7, 2014 #7

    DrClaude

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    Do you mean that the problem is not with the math, but with the economics?
     
  9. Nov 7, 2014 #8
    I've tried to delete this particular massage to repost it as a new thread because it refers to a bit different topic, but couldn't, so I have deleted the text
     
    Last edited: Nov 7, 2014
  10. Nov 7, 2014 #9
    no no - no problems with economics, but a huge one with basic math ;)
     
    Last edited: Nov 7, 2014
  11. Nov 7, 2014 #10

    PeroK

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    I guess your problem is as follows:

    You start with 100 and take 10% off. This gives you 90. Then you add 10% on. This gives you 99.

    Why, when you take 10% off, then add it back on again, do you not get back to where you started?

    The simple answer is that the first 10% was 10% of 100 and the second 10% was 10% of only 90. And they are different.

    Does that help?
     
  12. Nov 7, 2014 #11
    simple as it is! :) Thank you so much! Right! kindergarten issue - sometimes it happens :) Please, take a look at the following:

    1) in case 80 000 is 100%, then 80 000 less 13% is 0.87 of 80 000 and equals 69 600

    to "restore" back to 80 000 we have to multiply 69 600 by 1.1494 which is higher then initial 13% - why in this case is this one higher, if we are not changing the percentage base?

    2) if 80 000 is 113%, then to find 100% value (present value), I have to 80 000 / 1.13.
     
  13. Nov 7, 2014 #12

    Mark44

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    If you write it as an equation it might be easier to understand. I'm using PV for present value and FV for future value (= 80,000).
    80,000 = 1.13PV
    So PV = 80,000/1.13 = 70,796.46 (rounded to nearest cent)
     
  14. Nov 7, 2014 #13

    LCKurtz

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    If you have a stock with value ##V## and it goes up ##r##% expressed as a decimal its new value is ##V(1+r)##. If it then goes down ##r%## its final value is ##V(1+r)(1-r)=V(1-r^2)##. On the other hand suppose the stock starts by going down ##r%## to ##V(1-r)## then goes back up by ##r%## to ##V(1-r)(1+r) = V(1-r^2)##. Either way you end up with your original value multiplied by ##1-r^2## which is less than 1, so you have lost money. That is one problem with working with percents. As others have pointed out, you are taking the same percent but of different amounts in these calculations. If you don't use percents there is no problem. If your stock goes up or down by ##D## dollars then goes back the other way by ##D## dollars, you end up where you started. No problem if you don't use percents.
     
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