Find Mean Formula With Penalty as Parameter

[susanto3311]

hi all...

i have payment schedule with parameter like this :
e.g. original price $\$$5000

- if i pay in <=6 months i have penalty (add penalty) 1% from original price
- if i pay in >6 months i have penalty (add penalty) 10% from original price

e.g.

my problem is :

i have $\$$5000 and payment in 8 months, how find mean for each month that must i pay?i have $\$$5000 and payment in 12 months, how find mean for each month that must i pay?

illustration :$\$$5000 + penalty = ...? (in 8 months), mean= .../month$\$$5000 + penalty = ...? (in 12 months), mean= .../monthhow to make math formula?any assistance, thanks in advance.

susanto

 

[MarkFL]

If we let $P$ be the monthly payment, and we let $n$ be the number of months within which we pay off the debt, then we could use a piecewise defined function to get the monthly payment:

\(\displaystyle P(n)=\begin{cases}\dfrac{1.01\cdot5000}{n}, & 0<n\le 6 \\[3pt] \dfrac{1.1\cdot5000}{n}, & 6<n \\ \end{cases}\)

 

[susanto3311]

If we let $P$ be the monthly payment, and we let $n$ be the number of months within which we pay off the debt, then we could use a piecewise defined function to get the monthly payment:

\(\displaystyle P(n)=\begin{cases}\dfrac{1.01\cdot5000}{n}, & 0<n\le 6 \\[3pt] \dfrac{1.1\cdot5000}{n}, & 6<n \\ \end{cases}\)

ihi Mark, I'm sorry i mean with 2 condition above, how to make formula (join them) in a SINGLE formula.

it's possible? would you like help me?

 

[MarkFL]

You could use:

\(\displaystyle P(n)=\frac{50}{n}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

 

[susanto3311]

You could use:

\(\displaystyle P(n)=\frac{50}{n}\left(109\left\lfloor\arctan\left(\frac{\pi}{12}\left(x-\frac{1}{2}\right)\right)\right\rfloor+100\right)\)

hi Mark, i confuse use your formula, can you make explain more simple, & where come from :

number 50 and (x-1/2),

i'm not good about math and for just testing the final result how to convert your formula in MS Excel formula?

 

[MarkFL]

hi Mark, i confuse use your formula, can you make explain more simple, & where come from :

number 50 and (x-1/2),

i'm not good about math and for just testing the final result how to convert your formula in MS Excel formula?

The \(\displaystyle x-\frac{1}{2}\) should be \(\displaystyle n-\frac{1}{2}\), and I've edited my previous post to correct this typo.

I've never had any use for Excel, so I can't tell you what to do with it. I would think though that some kind of conditional statements are allowed in its cells, and that would be the simpler way to go.

 

[susanto3311]

The \(\displaystyle x-\frac{1}{2}\) should be \(\displaystyle n-\frac{1}{2}\), and I've edited my previous post to correct this typo.

I've never had any use for Excel, so I can't tell you what to do with it. I would think though that some kind of conditional statements are allowed in its cells, and that would be the simpler way to go.

hi Mark, one question..where come from 50/n? how about 10000? 2000?

 

[MarkFL]

hi Mark, one question..where come from 50/n? how about 10000? 2000?

If we let $A$ be the loan amount, then we could write (notice I have made another correction ):

\(\displaystyle P(n)=\frac{A}{100n}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

The term:

\(\displaystyle \left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor\)

is something I contrived to be $0$ for $0<n\le6$ and $1$ for $6<n$. It has as a factor the interest for the longer term loan, minus the interest factor for the short term loan (9%):

\(\displaystyle \frac{9A}{100}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor\)

Now the interest factor for the short term loan is always there (1% plus the actual principal of 100%):

\(\displaystyle \frac{101A}{100}\)

So, adding the two, we obtain the total amount that must be paid back:

\(\displaystyle \frac{9A}{100}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+\frac{101A}{100}\)

If we factor out \(\displaystyle \frac{A}{100}\) we have:

\(\displaystyle \frac{A}{100}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

Now, to determine the amount of the $n$ payments, we divide this total by $n$ to finally obtain:

\(\displaystyle P(n)=\frac{A}{100n}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

 

[susanto3311]

If we let $A$ be the loan amount, then we could write (notice I have made another correction ):

\(\displaystyle P(n)=\frac{A}{100n}\left(109\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

The term:

\(\displaystyle \left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor\)

is something I contrived to be $0$ for $0<n\le6$ and $1$ for $6<n$. It has as a factor the interest for the longer term loan, minus the interest factor for the short term loan (9%):

\(\displaystyle \frac{109A}{100}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor\)

Now the interest factor for the short term loan is always there (1%):

\(\displaystyle \frac{101A}{100}\)

So, adding the two, we obtain the total amount that must be paid back:

\(\displaystyle \frac{109A}{100}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+\frac{101A}{100}\)

If we factor out \(\displaystyle \frac{A}{100}\) we have:

\(\displaystyle \frac{A}{100}\left(109\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

Now, to determine the amount of the $n$ payments, we divide this total by $n$ to finally obtain:

\(\displaystyle P(n)=\frac{A}{100n}\left(109\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

i have tring your formula but stuck,

could you help how to easy using your formula,,,

if amount = 2000000 in 9 month the result is...?
if amount = 2000000 in 6 month the result is...?

thanks

 

[susanto3311]

hi Mark, how to prove it if your formula correct.

that's my problem

amount 10000000
n= 7 month
determine total?

would you like help me..

 

[MarkFL]

hi Mark, how to prove it if your formula correct.

that's my problem

amount 10000000
n= 7 month
determine total?

would you like help me..

I have (hopefully) one final correction to the "single formula":

\(\displaystyle P(n)=\frac{A}{100n}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

According to W|A, this formula results in a payment of \(\displaystyle P(7)=\frac{11000000}{7}\)

Wolfram|Alpha - (10000000/700)(9*floor(arctan((pi/12)(7-1/2)))+101)

If we compute it manually, we obtain:

\(\displaystyle P(7)=\frac{1.1\cdot10000000}{7}=\frac{11000000}{7}\)

 

[susanto3311]

I have (hopefully) one final correction to the "single formula":

\(\displaystyle P(n)=\frac{A}{100n}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

According to W|A, this formula results in a payment of \(\displaystyle P(7)=\frac{11000000}{7}\)

Wolfram|Alpha - (10000000/700)(9*floor(arctan((pi/12)(7-1/2)))+101)

If we compute it manually, we obtain:

\(\displaystyle P(7)=\frac{1.1\cdot10000000}{7}=\frac{11000000}{7}\)

hi Mark, this last question there is something missing from me..., i mean if payment in between 1-6 months you get penalty 1% and otherwise if payment ONLY in 7th month (not cumulative month) get penalty 10%

example :
for number 10,000,000, month 7 :
so the final result for 10,000,000 month 7 is 10,460,000 not 11,000,000.

before 10,000,000/6 month = 1,000,000 x 6 x 1% = 60,000 then 10,000,000 - 6,000,000 = 4,000,000
then 4,000,000 x 10% (passed 6 month) = 400,000 so the total is 10,000,000+60,000+400,000 = 10,460,000

i hope you would like help me for the last question?
what's is formula?

 

[MarkFL]

It seems to me that in this new system, the total amount that will be repaid is:

\(\displaystyle A+0.01A+0.09A\left(\frac{n-6}{n}\right)\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor\)

Or:

\(\displaystyle \frac{A}{100}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

And so the amount of $n$ equal payments would be:

\(\displaystyle \frac{A}{100n}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

 

[susanto3311]

It seems to me that in this new system, the total amount that will be repaid is:

\(\displaystyle A+0.01A+0.09A\left(\frac{n-6}{n}\right)\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor\)

Or:

\(\displaystyle \frac{A}{100}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

And so the amount of $n$ equal payments would be:

\(\displaystyle \frac{A}{100n}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

hi Mark thank for your time, i have using Wolfram but i can't see the result cause to buy pro version
would you like to show me how to compute it manually...

i am sure this last question...

 

[MarkFL]

hi Mark thank for your time, i have using Wolfram but i can't see the result cause to buy pro version
would you like to show me how to compute it manually...

i am sure this last question...

What are you wanting to compute?

 

[susanto3311]

It seems to me that in this new system, the total amount that will be repaid is:

\(\displaystyle A+0.01A+0.09A\left(\frac{n-6}{n}\right)\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor\)

Or:

\(\displaystyle \frac{A}{100}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

And so the amount of $n$ equal payments would be:

\(\displaystyle \frac{A}{100n}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

what's total amount of :

10,000,000 for 6 month = ...?
10,000,000 for 7 month = ...?

how to compute manually it?

 

[MarkFL]

what's total amount of :

10,000,000 for 6 month = ...?
10,000,000 for 7 month = ...?

how to compute manually it?

The monthly payment amount is given by:

\(\displaystyle P(A,n)=\frac{A}{100n}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)\)

10,000,000 for 6 month = ...?

Here, $A=10000000,\,n=6$ hence:

\(\displaystyle P(10000000,6)=\frac{10000000}{100\cdot6}\left(\frac{9(6-6)}{6}\left\lfloor\arctan\left(\frac{\pi}{12}\left(6-\frac{1}{2}\right)\right)\right\rfloor+101\right)=\frac{100000}{6}(0+101)=\frac{10100000}{6}\)

10,000,000 for 7 month = ...?

Here, $A=10000000,\,n=7$ hence:

\(\displaystyle P(10000000,7)=\frac{10000000}{100\cdot7}\left(\frac{9(7-6)}{7}\left\lfloor\arctan\left(\frac{\pi}{12}\left(7-\frac{1}{2}\right)\right)\right\rfloor+101\right)=\frac{100000}{7}\left(\frac{9}{7}+101\right)=\frac{100000}{49}\left(9+707\right)=\frac{71600000}{49}\)