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Math Quiz Thread

 
 
 
lpetrich said:
A triangle with geodesics as its sides.
Click to expand...

OK,, but it confused me, it would seem that by definition the sides of a triangle are geodesics. planar or spherical.

By bivector do you mean bisecting lines?

Can you give a link to R or is that your hypothesis?
 
Not a quiz ... Just a question. How do we incorporate the element of time in the return on investment formula, and can it have a beneficial application? For instance, if I buy2 cars (two separate investments) on jan 1 for $3,000 each and one is sold for $4,000 on March 31st and the other is sold for $4,200 on oct 31st, then how do i compare investments that take into account the time it took to sell the investment with the higher ROI? Is it as simple as converting it into an APR, and if so, is this nothing but an after-the-fact calculation, or can it have presale value?
 
fast said:
Not a quiz ... Just a question. How do we incorporate the element of time in the return on investment formula, and can it have a beneficial application? For instance, if I buy2 cars (two separate investments) on jan 1 for $3,000 each and one is sold for $4,000 on March 31st and the other is sold for $4,200 on oct 31st, then how do i compare investments that take into account the time it took to sell the investment with the higher ROI? Is it as simple as converting it into an APR, and if so, is this nothing but an after-the-fact calculation, or can it have presale value?
Click to expand...

I think you'd probably want to compare their  Internal rate of return (or  Modified internal rate of return).
 
beero1000 said:
fast said:
Not a quiz ... Just a question. How do we incorporate the element of time in the return on investment formula, and can it have a beneficial application? For instance, if I buy2 cars (two separate investments) on jan 1 for $3,000 each and one is sold for $4,000 on March 31st and the other is sold for $4,200 on oct 31st, then how do i compare investments that take into account the time it took to sell the investment with the higher ROI? Is it as simple as converting it into an APR, and if so, is this nothing but an after-the-fact calculation, or can it have presale value?
Click to expand...

I think you'd probably want to compare their  Internal rate of return (or  Modified internal rate of return).
Click to expand...
I checked it out, but the section on problems with the internal rate of return seems to indicate that it's not the right tool for the job.
 
1000 dollars profit in 3 months, or 1200 dollars profit in 10 months? Both are higher than many funds (or I'm looking at it wrong).
 
beero1000 said:
fast said:
I checked it out, but the section on problems with the internal rate of return seems to indicate that it's not the right tool for the job.
Click to expand...

Did you look at the second link?
Click to expand...
There's still the problem of there being irrelevant variables in the equation.

The second investment yielded $200 more profit than the first investment, but the second investment took seven months longer than the first. Which was the better investment? If we do not consider time, the second investment is better. If we consider time, the first investment is better. If we consider both profit and time, then the first investment is better because even though it was less profitable (dollar-wise), it took considerable more time to make the additional profit. My quick and crude calculations tell me that the second car would have been sold about no more than 18 days later than the first for both investments to be equally attractive when considering both time and money. The per day average profit gradually declines with each passing day.
 
Kharakov said:
1000 dollars profit in 3 months, or 1200 dollars profit in 10 months? Both are higher than many funds (or I'm looking at it wrong).
Click to expand...
I'm just using the sale of a couple vehicles with somewhat arbitrary numbers to serve as an example to discuss how to find a ROI formula that incorporates time into the formula.
 
fast said:
beero1000 said:
Did you look at the second link?
Click to expand...
There's still the problem of there being irrelevant variables in the equation.

The second investment yielded $200 more profit than the first investment, but the second investment took seven months longer than the first. Which was the better investment? If we do not consider time, the second investment is better. If we consider time, the first investment is better. If we consider both profit and time, then the first investment is better because even though it was less profitable (dollar-wise), it took considerable more time to make the additional profit. My quick and crude calculations tell me that the second car would have been sold about no more than 18 days later than the first for both investments to be equally attractive when considering both time and money. The per day average profit gradually declines with each passing day.
Click to expand...

What irrelevant variables? You certainly need the number of periods and the cash flows. If you just want to use those, use IRR. If you want to include the reinvestment rate and finance rate, use MIRR. If instead, you want to use the discount rate, use NPV. If you know the discount rate, the best option is NPV. Most individuals don't, but can get better estimates of their reinvestment/finance rates (your savings account yield, and/or loan interest rates), so MIRR is a better option for them.

For example -

Car 1:
n = 3 months
Cash flows, -3000 at start of interval 0, +4000 at end of interval 2.

Car 2:
n = 10
Cast flows, -3000 at start of interval 0, +4200 at end of interval 9.

IRR
Car 1: (monthly) IRR solves to IRR = 10.06%
Car 2: (monthly) IRR solves to IRR = 3.42%

If you know your monthly reinvestment and finance rates (say, a completely made up .08%, and .4%, respectively)

MIRR
Car 1: (monthly) MIRR solves to MIRR = 3.27%
Car 2: (monthly) MIRR solves to MIRR = 3.76%

If you know your monthly discount rate (say, .16%)

NPV
Car 1: NPV solves to $980.86
Car 2: NPV solves to $1133.39
 
beero1000 said:
fast said:
There's still the problem of there being irrelevant variables in the equation.

The second investment yielded $200 more profit than the first investment, but the second investment took seven months longer than the first. Which was the better investment? If we do not consider time, the second investment is better. If we consider time, the first investment is better. If we consider both profit and time, then the first investment is better because even though it was less profitable (dollar-wise), it took considerable more time to make the additional profit. My quick and crude calculations tell me that the second car would have been sold about no more than 18 days later than the first for both investments to be equally attractive when considering both time and money. The per day average profit gradually declines with each passing day.
Click to expand...

What irrelevant variables? You certainly need the number of periods and the cash flows. If you just want to use those, use IRR. If you want to include the reinvestment rate and finance rate, use MIRR. If instead, you want to use the discount rate, use NPV. If you know the discount rate, the best option is NPV. Most individuals don't, but can get better estimates of their reinvestment/finance rates (your savings account yield, and/or loan interest rates), so MIRR is a better option for them.

For example -

Car 1:
n = 3 months
Cash flows, -3000 at start of interval 0, +4000 at end of interval 2.

Car 2:
n = 10
Cast flows, -3000 at start of interval 0, +4200 at end of interval 9.

IRR
Car 1: (monthly) IRR solves to IRR = 10.06%
Car 2: (monthly) IRR solves to IRR = 3.42%

If you know your monthly reinvestment and finance rates (say, a completely made up .08%, and .4%, respectively)

MIRR
Car 1: (monthly) MIRR solves to MIRR = 3.27%
Car 2: (monthly) MIRR solves to MIRR = 3.76%

If you know your monthly discount rate (say, .16%)

NPV
Car 1: NPV solves to $980.86
Car 2: NPV solves to $1133.39
Click to expand...
My short-lived belief that prompted me to comment about irrelevant variables was born of ignorance. Your explanation and example was very helpful. Thank you.
 
Fast- either way, the second investment still generates more returns than many top tier mutual funds. So... unless you are paying to store the car, or you need the 3 grand for another equally awesome investment, you might as well take the loot for both of the cars.

sheesh, noticed your post at the top of this page a bit late. NM.... I still think the returns are good on either one. :D
 
fast said:
Not a quiz ... Just a question. How do we incorporate the element of time in the return on investment formula, and can it have a beneficial application? For instance, if I buy2 cars (two separate investments) on jan 1 for $3,000 each and one is sold for $4,000 on March 31st and the other is sold for $4,200 on oct 31st, then how do i compare investments that take into account the time it took to sell the investment with the higher ROI? Is it as simple as converting it into an APR, and if so, is this nothing but an after-the-fact calculation, or can it have presale value?
Click to expand...



If you had more data points you would have a 'time series' which you could turn into an equation or a model correlating the appreoation with other variables.

Jan 1 $3000
Feb 1 ?
March 1 $4000
April 1 ?
....
Oct 1 $4200.....

The best you could do is try to fit a curve to the data points ad extrapolate to the ROI point. It is basically how a company determines ROI for investing in a piece of machinery that reduces cost.

When dealing with time series the first thing I do is simply plot the data and see if tier is an observable trend.


That is what economists and stock market analysts do.

If you waited another month it may depreciate.

There is no formula that answers your question. Each time series is a problem that has to be analyzed.

http://en.wikipedia.org/wiki/Time_series
'
'...Time series analysis comprises methods for analyzing time series data in order to extract meaningful statistics and other characteristics of the data. Time series forecasting is the use of a model to predict future values based on previously observed values...'

Market research companies charge upwards o fhundreds of thousands of dollars for large scale corporate time series analysis.

http://www.freedownloadmanager.org/download/financial-time-series-with-open-office-4276224.html

Time series analysis is a major area of applied math.





.
 
Last edited:
\(k1 -[ k2*y(t)] - [k3 * \frac{dy}{dt} ] - [ \frac{1}{k4} \int_0^t y dt] = 0 \)

k1 = 8
k2 = 1
k3 = 100e-6
k4 = 0.1e-6


Initial condition y(0) = 0


What is y at t = 50e-6?
 
Last edited:
I found the solution.


Let y = dY/dt, with Y = integral from 0 to t of y

Thus, Y(0) = Y'(0) = 0, and we must find Y'(50e-6).

We must solve
\(k1 - k2 \frac{dY}{dt} - k3 \frac{d^2 Y}{dt^2} - \frac{1}{k4}Y = 0\)

It has the form
\(Y = k1 k4 + c1 e^{q1 t} + c2 e^{q2 t}\)

where q1 and q2 are q values that satisfy
\(k2 q + k3 q^2 + \frac{1}{k4} = 0\)

With the boundary conditions, we find
\(Y = k1 k4 \left( 1 - \frac{q2 e^{q1 t} - q1 e^{q2 t}}{q2 - q1} \right) \)
\(\frac{dY}{dt} = - k1 k4 q1 q2 \frac{e^{q1 t} - e^{q2 t}}{q2 - q1} \)

Working out the numbers with Mathematica, I find
\( q1,q2 = -5000. \pm 316188. i \)

and y(50e-6) = - 0.0199559


 
steve_bnk said:
\(k1 -[ k2*y(t)] - [k3 * \frac{dy}{dt} ] - [ \frac{1}{k4} \int_0^t y dt] = 0 \)

k1 = 8
k2 = 1
k3 = 100e-6
k4 = 0.1e-6


Initial condition y(0) = 0


What is y at t = 50e-6?
Click to expand...




The only twist was showing the last term as the integral instead of the derivative. One you see it is a second order differential equation then you can google a textbook solution. See Ipetrich's solution. Second order system solutions are complex exponentials.

Another solution is La Place Transforms.

In the complex frequency domain division is 1/S where S is the commonly used variable. The time domain equation transforms to

k2 + S*k3 + 1/[S*k4]
S*k2*k4 + S^2*k3*k4 + 1

Transform to time domain by integration or by using transform pair tables.

The natural frequency of the system is set by k2K3. K2 is the damping factor and determines the exponential envelope decay. K2 = 0 and you get a lossless oscillator.

The third method is a numerical solution such as Euler's Method and differentials.

http://en.wikipedia.org/wiki/Euler_method


Second order systems appear everywhere in the real world. Shock absorbers, electric circuits.

I used Euler's Method giving y(50e-6) = -.021

Two other points

y(5e-6) = +0.247

y(15e-06) = -0.234

Time step or resolution 0.1e-6 seconds.


 
Comparison to steve_bnk's more recent results:


My results:
5e-6 -> 0.246754
15e-6 -> -0.234623
50e-6 -> -0.0199559

In good agreement with his results.

 
steve_bnk said:
steve_bnk said:
\(k1 -[ k2*y(t)] - [k3 * \frac{dy}{dt} ] - [ \frac{1}{k4} \int_0^t y dt] = 0 \)

k1 = 8
k2 = 1
k3 = 100e-6
k4 = 0.1e-6


Initial condition y(0) = 0


What is y at t = 50e-6?
Click to expand...




The only twist was showing the last term as the integral instead of the derivative. One you see it is a second order differential equation then you can google a textbook solution. See Ipetrich's solution. Second order system solutions are complex exponentials.

Another solution is La Place Transforms.

In the complex frequency domain division is 1/S where S is the commonly used variable. The time domain equation transforms to

k2 + S*k3 + 1/[S*k4]
S*k2*k4 + S^2*k3*k4 + 1

Transform to time domain by integration or by using transform pair tables.

The natural frequency of the system is set by k2K3. K2 is the damping factor and determines the exponential envelope decay. K2 = 0 and you get a lossless oscillator.

The third method is a numerical solution such as Euler's Method and differentials.

http://en.wikipedia.org/wiki/Euler_method


Second order systems appear everywhere in the real world. Shock absorbers, electric circuits.

I used Euler's Method giving y(50e-6) = -.021

Two other points

y(5e-6) = +0.247

y(15e-06) = -0.234

Time step or resolution 0.1e-6 seconds.


Click to expand...




// Euler's Method
clear;
clf;

n_points = 1000;

k1 = 8
k2 = 1;
k3 =100e-6;
k4 = .1e-6;

dt = 0.1e-6; // integration time step

y = 0; // initial condition
f3_integral = 0; // initial condition

for n = 1:n_points;

f1 = y * k2;
f2 = k1 - f3_integral - f1;
dy = (f2/k3)*dt;
y = y + dy;
d_f3 = (y * dt)/k4;
_y(n) = y;
f3_integral = f3_integral + d_f3;

end;

// indexes into array for 3 poinTS
n_50 = 50e-6/dt;
n_5 = 5e-6/dt;
n_15 = 15e-6/dt;

_y(n_50)
_y(n_5)
_y(n_15)

plot2d(_y);

results

-->_y(n_50)
ans =

- 0.0208652
-->_y(n_5)
ans =

0.2468425
-->_y(n_15)
ans =

- 0.2346924 -->_y(n_50)
ans =

- 0.0208652
-->_y(n_5)
ans =

0.2468425
-->_y(n_15)
ans =

- 0.2346924



//

 
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