MHB Find Mean Formula With Penalty as Parameter

AI Thread Summary
The discussion revolves around calculating monthly payments for a loan with penalties based on the payment duration. A piecewise function is proposed to determine the monthly payment, incorporating penalties of 1% for payments made within 6 months and 10% for payments beyond that. Users seek clarification on how to consolidate the penalty conditions into a single formula and how to compute payments manually. The final formula provided accounts for the penalties and allows for calculations based on different loan amounts and payment durations. The conversation emphasizes the need for clear explanations and practical computation methods for users unfamiliar with complex mathematics.
susanto3311
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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
 
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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:

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

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

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?
 
You could use:

$$P(n)=\frac{50}{n}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)$$
 
MarkFL said:
You could use:

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

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?
 
susanto3311 said:
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 $$x-\frac{1}{2}$$ should be $$n-\frac{1}{2}$$, 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.
 
MarkFL said:
The $$x-\frac{1}{2}$$ should be $$n-\frac{1}{2}$$, 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?
 
susanto3311 said:
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 :o):

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

The term:

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

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%):

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

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

$$\frac{101A}{100}$$

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

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

If we factor out $$\frac{A}{100}$$ we have:

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

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

$$P(n)=\frac{A}{100n}\left(9\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)$$
 
MarkFL said:
If we let $A$ be the loan amount, then we could write (notice I have made another correction :o):

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

The term:

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

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%):

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

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

$$\frac{101A}{100}$$

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

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

If we factor out $$\frac{A}{100}$$ we have:

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

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

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

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
 
  • #10
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..
 
  • #11
susanto3311 said:
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":

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

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

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

If we compute it manually, we obtain:

$$P(7)=\frac{1.1\cdot10000000}{7}=\frac{11000000}{7}$$
 
  • #12
MarkFL said:
I have (hopefully) one final correction to the "single formula":

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

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

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

If we compute it manually, we obtain:

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

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?
 
  • #13
It seems to me that in this new system, the total amount that will be repaid is:

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

Or:

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

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

$$\frac{A}{100n}\left(\frac{9(n-6)}{n}\left\lfloor\arctan\left(\frac{\pi}{12}\left(n-\frac{1}{2}\right)\right)\right\rfloor+101\right)$$
 
  • #14
MarkFL said:
It seems to me that in this new system, the total amount that will be repaid is:

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

Or:

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

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

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

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...
 
  • #15
susanto3311 said:
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?
 
  • #16
MarkFL said:
It seems to me that in this new system, the total amount that will be repaid is:

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

Or:

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

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

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

what's total amount of :

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

how to compute manually it?
 
  • #17
susanto3311 said:
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:

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

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

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

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

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

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

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