TI Nspire CX II: Distributing Money Among Friends
The Problem
This post is inspired from a Instagram post by @gercekboss at www.eylemmath.weebly.com. The problem states that:
$100 is to be distributed among 100 friends:
1% is given to the 1st friend
2% is given to the 2nd friend
3% is given to the 3rd friend
and so on.
I have a TI Nspire CX II Document that covers this problem, which can be downloaded here: https://drive.google.com/file/d/1aewZ-8wU2JjZxgGyBpNHivEYLU5QAt31/view?usp=sharing
Name of the document: distribute percent.tns
Using the TI Nspire CX II to set up two sequences to model this problem as two sequences:
Balance Sequence (Problem 1.3, plot is in green)
u1(n) = u1(n-1) - u1(n-1) * n%
Intial Terms: 100
0 ≤ n ≤ 99 step 1
Amount Sequence (Problem Page 1.7, plot is in red)
u2(n) = u1(n-1) *n%
Initial Terms: 0
0 ≤ n ≤ 99 step 1
By analysis, the 10th friend will get the biggest distribution under this plan, $6.28. By the time we get to the 38th friends, the distribution becomes meaningless because the calculated amount will be less than one penny.
Regression Analysis
I performed curve fitting analysis for the balance and amount distributed on problems page 1.5 and 1.9, respectively.
Balance can be estimated by the logistic equation:
balance(n) ≈ 112.84/(1 + 0.1*e^(0.23n)) - 0.54 for 1 ≤ n ≤ 40
The maximum absolute error is 0.67.
Amount can be estimated by the quartic polynomial:
amount(n) ≈ -4.01E-5*n^4 + 4.31E-3 * n^3 - 0.15 * n^2 + 1.89 * n - 1.26
for 1 ≤ n ≤ 40
I did slightly better, the maximum absolute error is 0.52.
Generalizing the Problem
Now, let's take any money of money and any amount of friends for whom to distribute that money. The rate can increase per friend by at different rates:
a*n% + b%
If you want 2%, 4%, 6%, etc, let a = 2, b = 0
If you want each friend to get 1/2% more than the last, let a = 1/2, b = 0.
If you want the first friend to get 2%, second to get 3%, etc, let a = 1, b = 1.
The sequences are set up as:
Balance Sequence
u1(n) = u1(n-1) - u1(n-1) * (a*n% + b%)
Intial Terms: c
0 ≤ n ≤ l step 1
Amount Sequence
u2(n) = u1(n-1) *(a*n% + b%)
Initial Terms: 0
0 ≤ n ≤ l step 1
Variables:
a = rate parameter
b = rate parameter
c = initial amount of money
l = number of friends
You can customize the problem in Problem 2 of the document. Change the variables on Problem Page 2.2 and see the results on Problem Pages 2.3 and 2.4.
Source:
Question 251. eleymmath. https://eylemmath.weebly.com/algebra/100-question Retrieved July 21, 2019.
Eddie
All original content copyright, © 2011-2019. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
The Problem
This post is inspired from a Instagram post by @gercekboss at www.eylemmath.weebly.com. The problem states that:
$100 is to be distributed among 100 friends:
1% is given to the 1st friend
2% is given to the 2nd friend
3% is given to the 3rd friend
and so on.
I have a TI Nspire CX II Document that covers this problem, which can be downloaded here: https://drive.google.com/file/d/1aewZ-8wU2JjZxgGyBpNHivEYLU5QAt31/view?usp=sharing
Name of the document: distribute percent.tns
Using the TI Nspire CX II to set up two sequences to model this problem as two sequences:
Balance Sequence (Problem 1.3, plot is in green)
u1(n) = u1(n-1) - u1(n-1) * n%
Intial Terms: 100
0 ≤ n ≤ 99 step 1
Amount Sequence (Problem Page 1.7, plot is in red)
u2(n) = u1(n-1) *n%
Initial Terms: 0
0 ≤ n ≤ 99 step 1
By analysis, the 10th friend will get the biggest distribution under this plan, $6.28. By the time we get to the 38th friends, the distribution becomes meaningless because the calculated amount will be less than one penny.
Regression Analysis
I performed curve fitting analysis for the balance and amount distributed on problems page 1.5 and 1.9, respectively.
Balance can be estimated by the logistic equation:
balance(n) ≈ 112.84/(1 + 0.1*e^(0.23n)) - 0.54 for 1 ≤ n ≤ 40
The maximum absolute error is 0.67.
Amount can be estimated by the quartic polynomial:
amount(n) ≈ -4.01E-5*n^4 + 4.31E-3 * n^3 - 0.15 * n^2 + 1.89 * n - 1.26
for 1 ≤ n ≤ 40
I did slightly better, the maximum absolute error is 0.52.
Generalizing the Problem
Now, let's take any money of money and any amount of friends for whom to distribute that money. The rate can increase per friend by at different rates:
a*n% + b%
If you want 2%, 4%, 6%, etc, let a = 2, b = 0
If you want each friend to get 1/2% more than the last, let a = 1/2, b = 0.
If you want the first friend to get 2%, second to get 3%, etc, let a = 1, b = 1.
The sequences are set up as:
Balance Sequence
u1(n) = u1(n-1) - u1(n-1) * (a*n% + b%)
Intial Terms: c
0 ≤ n ≤ l step 1
Amount Sequence
u2(n) = u1(n-1) *(a*n% + b%)
Initial Terms: 0
0 ≤ n ≤ l step 1
Variables:
a = rate parameter
b = rate parameter
c = initial amount of money
l = number of friends
You can customize the problem in Problem 2 of the document. Change the variables on Problem Page 2.2 and see the results on Problem Pages 2.3 and 2.4.
Source:
Question 251. eleymmath. https://eylemmath.weebly.com/algebra/100-question Retrieved July 21, 2019.
Eddie
All original content copyright, © 2011-2019. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
