Smart People: Help me do math homework

[Canofbier]

Originally Posted by La literatura

So now I just get to plug in numbers for A, B, C, and D?

Yep. If you want a discrete solution, fill in 3 of them and leave the fourth as a variable, and Wolfram will calculate it.

[Jenson71]

Originally Posted by Canofbier

Yep. If you want a discrete solution, fill in 3 of them and leave the fourth as a variable, and Wolfram will calculate it.

How does it feel to be smart? Is it a burden at times?

[Frosty]

Cheater!

[Ace Gunner]

Originally Posted by Frosty

Cheater!

this!

[DaKCMan AP]

There are multiple solutions, but a mix of:

1 - 0.575262544
2 - 0
3 - 0.252042008
4 - 0.172695448

Gives an ER of 0.065.


Solver is your friend. You have 4 unknowns, 2 equations, and conditions on the 4 unknowns.

[Jenson71]

Originally Posted by DaKCMan AP

I'm fairly certain there are multiple solutions, but a mix of:

1 - 0.575262544
2 - 0
3 - 0.252042008
4 - 0.172695448

Gives an ER of 0.065.

Did you figure that out doing the A(1.085)+B(1.092)+C(1.031)+D(1.048)=1.065 or did you use another method?

[DaKCMan AP]

Originally Posted by La literatura

Did you figure that out doing the A(1.085)+B(1.092)+C(1.031)+D(1.048)=1.065 or did you use another method?

That's the formula for expected return (you can also use the same formulat without the leading 1's, i.e. A(0.085) + B(0.092)...) and 1 of your two equations.

Your 2nd equation is A + B + C + D = 1.

[Mike in SW-MO]

You can use a mix of Investment 1 & I3 or I4 then a mix of I2 plus I3 or I4. That gives you 4 mixes. You can combine any of those solutions to make a fifth.
Use xI1 + (1-x)I3=.65 solve for x.

Substitute any 2 investment options above and below target.

[prhom]

It's just a question of weighted averages. You must have at least one of the investments that are higher than 6.5 and at least one of the investments that are less than 6.5. To actually solve this you need more information about total dollars invested, the maximum number of individual investments that can be made or a ratio of dollars invested between funds.

[Coogs]

If you just average your 4 percentages, you are close to begin with at 6.4%.

So let's say you invest $4000.

25%, or $1000, at each of the 4 rates....

1000 at 8.5% = 85.00
1000 at 9.2% = 92.00
1000 at 3.1% = 31.00
1000 at 4.8% = 48.00

for a total of 256.00.

256.00/4000 = .064 = 6.4%

260.00/4000 = .065 = 6.5%

So therefore you need to find a way to get $4.00 more back on you investment.

I'm going to use the 9.2% and the 4.8% and the 1000.00 at each rate

We made 140 dollars at those two rates combined. We need 144 dollars to get to 260 dollars.

X = amount invested at 9.2%
2000 - X = amount invested at 4.8%

.092X + .048(2000 - X) = 144.00
.092X + 96.00 - .048X = 144.00
.044X = 144.00 - 96.00
.044X = 48.00
X = 48.00/.044
X = 1090.91


2000 - 1090.91 = 909.09

What percent of 1000 is 1090.91?

109.1%

109.1% of your original investment, which was 25%, is

25 times 1.091 = 27.275

What percent of 1000 is 909.09?

90.9%

90.9% of you original investment, which was 25%, is

25 times .909 = 22.725

So, your your percentages to be invested at each rate to give you a 6.5% return...

25% of your money at 8.5%
25% of your money at 3.1%
27.275% of your money at 9.2%
22.725% of your money at 4.8%

This is just one solution, but it will work for any amount that you choose to invest.

[Dayze]

reading through this thread makes me realize how stupid I am.

[Jenson71]

Originally Posted by Dayze

reading through this thread makes me realize how stupid I am.

Especially Coog and Saul Good. Give me a history book, and I'll have that subject mastered. Give me a few numbers, and I'll start to sweat.

[Saul Good]

You're making this too hard. You don't have to blend all four. It's a lot easier to blend 2 and put it into a simple algebraic equation.

For example, you can use holdings 1 (x) and 3 (y).

x(8.5) + y(3.1) = 6.5% is your equation.

Now, you have to express x in terms of y. Because x+y= 100% (1.00) of your investment, x=1.00-y

Now, you re-write the equation with only the y variable:

(1-y)(8.5)+y(3.1)=6.5

Do the multiplication:

8.5-8.5y+3.1y=6.5

Simplify further

8.5-5.4y=6.5

Subtract 6.5 from both sides of the equation

2-5.4y=0

Add 5.4y to both sides

2=5.4y

Divide both sides by 5.4 to solve for y

.37=y

37% of your investment should be in y. The rest (1-y) should be in x.

Check your work: .37(.031)+.63(.085)=.065

[Coogs]

Originally Posted by Saul Good

You're making this too hard. You don't have to blend all four. It's a lot easier to blend 2 and put it into a simple algebraic equation.

For example, you can use holdings 1 (x) and 3 (y).

x(8.5) + y(3.1) = 6.5% is your equation.

Now, you have to express x in terms of y. Because x+y= 100% (1.00) of your investment, x=1.00-y

Now, you re-write the equation with only the y variable:

(1-y)(8.5)+y(3.1)=6.5

Do the multiplication:

8.5-8.5y+3.1y=6.5

Simplify further

8.5-5.4y=6.5

Subtract 6.5 from both sides of the equation

2-5.4y=0

Add 5.4y to both sides

2=5.4y

Divide both sides by 5.4 to solve for y

.37=y

37% of your investment should be in y. The rest (1-y) should be in x.

Check your work: .37(.031)+.63(.085)=.065

That's all good if he is allowed to only use 2 of the 4 options. But if he is required to use all 4 of the percentages listed, he better go with something like along the lines I provided for him.

EDIT: We are essentially doing the same thing... I just eliminated 2 of the options by giving each a 25% value. Didn't mean to come across as my way was better Saul Good! Hope you didn't take it that way.

[Saul Good]

Originally Posted by Coogs

That's all good if he is allowed to only use 2 of the 4 options. But if he is required to use all 4 of the percentages listed, he better go with something like along the lines I provided for him.

EDIT: We are essentially doing the same thing... I just eliminated 2 of the options by giving each a 25% value. Didn't mean to come across as my way was better Saul Good! Hope you didn't take it that way.

If you have to use all four, simply divide the percentages in half, do the same thing for options 1 and 2, and add them together.

Standard algebra is really all that is needed here, and it's about a 30 second solution. I don't feel like math instructors do a good enough job of translating their teachings into practical applications. There really isn't any point in learning algebra if you don't know how to apply it. If you do, it's actually extremely useful and comes in handy a lot more often than most people probably realize.

Then, when you use it, people look at you like you just explained string theory even though you're just really busting out something you learned in 7th grade.