I cant conceptualize this equation

[KgBrother]

This math problem is related to my business. Any help is greatly appreciated:


'a' is my investment, 'b' is the multiple that my investment will grow, and 'c' is the total cost of assets+investment.

A•B=C

'A' is unknown
'B' is 10
'C' asset is 100, but the investment side depends on 'A'

Everytime I change A to solve for C, then C changes which requires A to change to resolve again and again and again infinitely.

My questions are:
1- How do I write this infinite-like equation regardless of the numbers used?

2- How can I also write this equation so that I can see what 'A' would be after the 5th cycle of resolving? Meaning after I did the equation 5 times, the 6th 'A' used to resolve the equation would be the number Id get.

Id really appreciate any help!

 

[Dr.Peterson]

I suspect that you didn't really express what you mean.

Your equation, as written, says that A times 10 is 100, so clearly A is 10. There is no infinite process, just a division by 10.

But I have known people to try solving more interesting equations in the way you describe, and ending up with an infinite process. The answer is to learn algebra. It is not always as easy as just dividing by 10, but will often (not always) provide a direct solution.

If you still need help, you might want to state the problem more fully, and then show us what you actually did, so we can see just what you are asking about.

 

[JeffM]

This math problem is related to my business. Any help is greatly appreciated:


'a' is my investment, 'b' is the multiple that my investment will grow, and 'c' is the total cost of assets+investment.

A•B=C

'A' is unknown
'B' is 10
'C' asset is 100, but the investment side depends on 'A'

Everytime I change A to solve for C, then C changes which requires A to change to resolve again and again and again infinitely.

My questions are:
1- How do I write this infinite-like equation regardless of the numbers used?

2- How can I also write this equation so that I can see what 'A' would be after the 5th cycle of resolving? Meaning after I did the equation 5 times, the 6th 'A' used to resolve the equation would be the number Id get.

Id really appreciate any help!

I think part of your problem is language. I suspect that what you mean is that a is the initial cost of the investment, b is the percentage rate at which it will grow in value each period, and c is the value of the investment after n periods.

If that is correct, that is a basic present value problem.

[MATH]a * \left (1 + \dfrac{b}{100} \right )^n = c \implies a = x \div \left (1 + \dfrac{b}{100} \right )^n.[/MATH]
You can solve that using a decent hand calculator in scientific mode.

Here is a super-easy example.

Percentage increase per year = 10

final value = 100

number of years = 2

initial investment required =

[MATH]100 \div \left (1 + \dfrac{10}{100} \right)^2 = 100 \div 1.21 \approx 82.64.[/MATH]
Let's see why this is a good approximation.

[MATH]82.64 * 0.1 = 8.26 \text { and } 82.64 + 8.26 = 90.90[/MATH]
[MATH]90.90 * 0.1 = 9.09 \text { and } 90.90 + 9.09 = 99.99. \ \checkmark[/MATH]

 

[KgBrother]

I think part of your problem is language. I suspect that what you mean is that a is the initial cost of the investment, b is the percentage rate at which it will grow in value each period, and c is the value of the investment after n periods.

If that is correct, that is a basic present value problem.

[MATH]a * \left (1 + \dfrac{b}{100} \right )^n = c \implies a = x \div \left (1 + \dfrac{b}{100} \right )^n.[/MATH]
You can solve that using a decent hand calculator in scientific mode.

Here is a super-easy example.

Percentage increase per year = 10

final value = 100

number of years = 2

initial investment required =

[MATH]100 \div \left (1 + \dfrac{10}{100} \right)^2 = 100 \div 1.21 \approx 82.64.[/MATH]
Let's see why this is a good approximation.

[MATH]82.64 * 0.1 = 8.26 \text { and } 82.64 + 8.26 = 90.90[/MATH]
[MATH]90.90 * 0.1 = 9.09 \text { and } 90.90 + 9.09 = 99.99. \ \checkmark[/MATH]

I dont think I made the question clear enough. Im not trying to solve what you have in mind.

I want Investment 'A' when multipled by whatever number(10 in my example) to cover the initial cost of ownership of the asset plus the cost of Investment 'A' itself. Which means as I change A to solve for C, then C changes because A is used to determine C's total cost.

This is a circular infinite equation^ but in words only. Im not sure how to write it out. and I especially dont know how to do it so the equation only goes circular for a few times.

 

[JeffM]

I dont think I made the question clear enough. Im not trying to solve what you have in mind.

I want Investment 'A' when multipled by whatever number(10 in my example) to cover the initial cost of ownership of the asset plus the cost of Investment 'A' itself. Which means as I change A to solve for C, then C changes because A is used to determine C's total cost.

This is a circular infinite equation^ but in words only. Im not sure how to write it out. and I especially dont know how to do it so the equation only goes circular for a few times.

I'm afraid I still don't understand. I spent almost my entire professional career dealing with investments including mergers and acquisitions.

"The initial cost of the ownership of the asset" refers to what asset?

In your first post, you said "c asset is 100." Is that an initial cost or a final value?

In fact, it is unclear whether you are dealing with two assets, a and d, whose costs when added together equal c, or whether a and c are market values of the same asset at different times.

Perhaps if you explained in more concrete terms what you are dealing with. For example, a is the purchase price of a stock, and c is the sum of the purchase prices of the stock and a new boat.

I'm sure it's quite clear in your mind what a and c stand for, but it is not for us.

 

[KgBrother]

I'm afraid I still don't understand. I spent almost my entire professional career dealing with investments including mergers and acquisitions.

"The initial cost of the ownership of the asset" refers to what asset?

In your first post, you said "c asset is 100." Is that an initial cost or a final value?

In fact, it is unclear whether you are dealing with two assets, a and d, whose costs when added together equal c, or whether a and c are market values of the same asset at different times.

Perhaps if you explained in more concrete terms what you are dealing with. For example, a is the purchase price of a stock, and c is the sum of the purchase prices of the stock and a new boat.

I'm sure it's quite clear in your mind what a and c stand for, but it is not for us.

I understand this investment opportunity is especially atypical but Ill do my best to describe it in a way that doesnt come with as many assumptions as the investment world does.

Lets say A is a gamblers bet(amount Im willing to put up). B is how much ill multiply my bet money if my bet is correct. C is the sum of A(Gamblers bet) and D(all the other bets I made)

I want to put just enough into A that when multiplied by B itll equal C. You might notice however that its impossible to solve since A changes C's value.


Sample of bet:
All other prior bets are $100
$10(bet)x10= ($100+$10(bet))

Clearly thats wrong^ so then I contine to raise the bet:
$11(bet)x10= ($100 +$11(bet))

still incorrect^ so I continue to raise the bet:
$11.10(bet)x10=($100+$11.10(bet))

Again this is incorrect^. But it doesnt matter to me that its incorrect to me personally. Im just trying to do the equation 3-5 times until I get a bet number(A) that when multiplied will only have a slight discrepancy from (C). This way Im losing as little money as possible. I would do this normally by performing the equation 5 times in a row but thats inefficient. Thats truly what I need help with: avoiding doing the equation 5 or 10 or 20, etc times in a row. If there was a formula where I could choose how many times the equation plays out(in this case 5 times), how would I write that?



Before you ask, "why would you invest(/bet) in this way?", know that Im not trying to make money with this, but instead provide a counter investment/bet that will mitigate a loss of money from my bigger prior investments/bets if they fail.Those prior investments/bets i truly care about would represent D.

P.S. I want to thank you for your patience. I truly do appreciate any attempt to help me resolve this. Let me know if theres anyway I can help resolve this.

 

[JeffM]

As you have described the problem, a, b, c, and d are not independent.

[MATH]a + d = c.[/MATH]
You can choose two of the three numbers, but then the third is determined. If you tell me a is two and d is 7, then c must be 9. Or if you tell me c is 11 and d is 4, then a must be 7. You have two degrees of freedom.

[MATH]a * b = c.[/MATH]
Again you have two degrees of freedom: you can choose two of the three numbers, but then the third is determined.

In terms of your problem, the payoff on the bet is b to 1 and not yours to choose. That reduces your degrees of freedom to one. As a practical matter, how much you are able and willing to bet in total is subject to a ceiling. This reduces your degrees of freedom with respect to a and d to none. If b = 10 and c = 100, then

[MATH]a * 10 = 100 \implies a = 10 \text { and } 10 + d = 100 \implies d = 90.[/MATH]
And that is the correct answer to what I suspect is your ultimate question. If all your other bets lose while the 10 to 1 bet pays off, you would break even.

In other words, given b and c, a and d are determined, but they are calculable.

 

[KgBrother]

As you have described the problem, a, b, c, and d are not independent.

[MATH]a + d = c.[/MATH]
You can choose two of the three numbers, but then the third is determined. If you tell me a is two and d is 7, then c must be 9. Or if you tell me c is 11 and d is 4, then a must be 7. You have two degrees of freedom.

[MATH]a * b = c.[/MATH]
Again you have two degrees of freedom: you can choose two of the three numbers, but then the third is determined.

In terms of your problem, the payoff on the bet is b to 1 and not yours to choose. That reduces your degrees of freedom to one. As a practical matter, how much you are able and willing to bet in total is subject to a ceiling. This reduces your degrees of freedom with respect to a and d to none. If b = 10 and c = 100, then

[MATH]a * 10 = 100 \implies a = 10 \text { and } 10 + d = 100 \implies d = 90.[/MATH]
And that is the correct answer to what I suspect is your ultimate question. If all your other bets lose while the 10 to 1 bet pays off, you would break even.

In other words, given b and c, a and d are determined, but they are calculable.

This does not answer my question. C= A+D that means when Im trying to choose a bet amount(A) I want it to cover the cost of A itself and all prior bets(in this latest scenario) A*B=C

Lets say my all my prior bets are $100 total(D)

My latest bet(A) I want to put is is incase my other bets lose. I want my prior $100 bets and my current bet being used to protect those bets and itself.

Everytime I raise the bet to cover itself, I have to raise it infinitely but raise it less and less.


Here is SCENARIO:
"Darn, I spent $100 dollars on some prior bet. If Im wrong Ill lose $100. Let me place a counter bet elsewhere so that if Im wrong, then this counter bet will produce enough to atleast cover itself and the prior bet."

Now I start doing the math:
This counter bet with give me 10x returns. So How much do I need to bet to cover the $100 AND this very bet Im about to make?

You might think 10 is the answer, but its not. Because 100+10 is 110 which is what I have to protect now if I make that bet. 10*10 doesnt equal 110.

So then i raise my counter bet to 11. but then my total Im trying to protect is 100+11=111. Clearly 11*10 doesnt equal 111 either.

so on and so forth until infinity.

Now i can do math like that^ but it will take forever if I have many cycles to go through and possibly many other bets/investments doing the same thing.

 

[JeffM]

You lost $100 on prior bets.

You have bet that pays 10 for every 1 you wager if you win. There is no exact answer because money is not infinitely divisible, but there is an approximate answer.

[MATH](a + 100) = 10a \implies 9a = 100 \implies a \approx 11.11.[/MATH]
So bet $11.12. If you win you get $111.20, which more than covers your prior loss of $100 plus your bet of $11.12 by 8 cents. Alternatively, you can bet $11.11. Then if you win, you collect $111.10, reducing your loss from $100 to 2 cents.

There is no infinite number of calculations required. You are mixing this up with a famous paradox, the so-called St. Petersburg paradox.

You are eventually guaranteed to win if you keep placing bigger and bigger bets according to this formula, provided the bookie does not go bankrupt and will keep accepting your bets on credit. But the proviso will never happen. It is theoretical rather than practical.

 

[KgBrother]

You lost $100 on prior bets.

You have bet that pays 10 for every 1 you wager if you win. There is no exact answer because money is not infinitely divisible, but there is an approximate answer.

[MATH](a + 100) = 10a \implies 9a = 100 \implies a \approx 11.11.[/MATH]
So bet $11.12. If you win you get $111.20, which more than covers your prior loss of $100 plus your bet of $11.12 by 8 cents. Alternatively, you can bet $11.11. Then if you win, you collect $111.10, reducing your loss from $100 to 2 cents.

There is no infinite number of calculations required. You are mixing this up with a famous paradox, the so-called St. Petersburg paradox.

You are eventually guaranteed to win if you keep placing bigger and bigger bets according to this formula, provided the bookie does not go bankrupt and will keep accepting your bets on credit. But the proviso will never happen. It is theoretical rather than practical.

Thank you! I feel like such a dummy for not figuring that out sooner! This takes me 90% of the way there because there is 1 step I left out that I didnt want to mention for fear that this conversation would become too confusing from the start(it was already hard enough to explain from my limited background in math)

Now that you understand what I was trying to do youll be able to help me with my final question:

Same scenario BUT instead I have a spread of "bets" that all need to be factored in the total(C) that Im trying to solve.

A is still bet amount, B is multiplier, and C is Bet(A) + All other bets(D) + E(a big bet made last month Im trying to protect, or just think of this as money lost that I need to recoop)

A,X,Z= are the bets Ill place(i need to figure out how much to bet)
B= (different for each bet)
C= A+D+E (A can be substituted with X or Z for each respective Bet)
E= $2,550
D= Is the two other bet amounts.
(NOTE: A, X, Z are all different bets)

Formula: (A, or X, or Z) *B=C
Bet1:
A*20= C

Bet2:
X*10= C

Bet3:
Z*7= C

If I do the formula in the way you described I will still have to redo the equation multiple times in a row because every time I change Bet1 A that changes D for the other two equations. So then i do Bet2 X to match the updated C. Which again changes D(making bet1 incorrect now). Then I do the same with Z. And yet again, C changes for all the equations, requiring me to go back and keep changing them seemingly forever.

Is there a way I can write this formula so that I only have to do it once and not repeat work over and over? I want to only bet as much on A, X and Z as needed to cover the total C.

P.S. my actual situation would be the equivalent of 12 bets, but in this scenario Im keeping it to 3 for simplicity.