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        riddles >> general problem-solving / chatting / whatever >> Financial dealings
        
(Message started by: BenVitale on Jul 10th, 2009, 10:22am)
Title: Financial dealings
Post by BenVitale on Jul 10th, 2009, 10:22am
Hi,

I have a series of financial problems, I hope you'll enjoy them as much as I have.

The first one is about choosing a raise:

Problem #1 :


Suppose that your starting salary is $1,500 and the question is, would you choose a $300 raise every year, or a $75 raise every 6 months?


P.S.

In these types of problems there are ways to pose the question so that the "paradoxical" answer is the correct one, but the whole thing may then become an exercise in lateral thinking which has little to do with mathematics.
Title: Re: Financial dealings
Post by Obob on Jul 10th, 2009, 11:22am
Can you clarify:

is $1500 the monthly salary or annual salary?

is the $300 raise to the monthly salary or to the annual salary?

is the $75 raise to the monthly salary, to the 6-montly salary, or to the annual salary?

At any rate, it should be clear that the $300 raise is better unless you do not expect to be employed there long-term or you are able to double investments over a period of 6 months.
Title: Re: Financial dealings
Post by BenVitale on Jul 10th, 2009, 12:19pm
Well, you could get a $75 raise on your annual salary of $1,500 or a $300 raise on your annual salary of $1,500.

If a potential employer offers you:

Option #1: a $75 raise every 6 months
Option #2 : a $300 raise every year

If it is less than one year, the solution is clear.

But, over a 3-year period, which option would you choose?
Compute the total wages that would be paid under option #1 and option #2 ... and compare the two

If a potential employer offers you these 2 options, you may think that it is a trick question, could you be convinced that option #1 is better? If not, then disprove.
Title: Re: Financial dealings
Post by rmsgrey on Jul 10th, 2009, 2:29pm

on 07/10/09 at 12:19:03, BenVitale wrote:
Well, you could get a $75 raise on your annual salary of $1,500 or a $300 raise on your annual salary of $1,500.

If a potential employer offers you:

Option #1: a $75 raise every 6 months
Option #2 : a $300 raise every year

If it is less than one year, the solution is clear.

But, over a 3-year period, which option would you choose?
Compute the total wages that would be paid under option #1 and option #2 ... and compare the two

If a potential employer offers you these 2 options, you may think that it is a trick question, could you be convinced that option #1 is better? If not, then disprove.


Assuming your base salary is 0, paid continuously, and defining a such that, under the first scheme, you get a over the second 6 months, 2a over the third, etc, then, under the second scheme, you get 4a in each of the third and fourth 6 month periods, 8a in the fifth and sixth, and so on...

So in each 6 month period, you get:

+option 1 gain (accumulated total) / +option 2 gain (accumulated total)

+0(0)/+0(0)
+a(a)/+0(0)
+2a(3a)/+4a(4a)
+3a(6a)/+4a(8a)
+4a(10a)/+8a(16a)
+5a(15a)/+8a(24a)
...

For the first 6 months, neither scheme offers any benefit. Between 6 months and a year, option #1 gets you money faster and gives you a higher accumulated total. Between a year and 15 months, option #2 gets you money faster but hasn't yet cancelled option #1's lead. After 15 months, option #2 is getting you money faster, and gives you a higher accumulated total.

If the first increment happens sooner, the overtake will happen sooner. If money isn't earned continuously over time, then you need to complete the first pay period to end at or after 15 months...

If there's some other trickiness going on, then things will be different (like the $75 being an increase to the 6-monthly salary, while the $300 is an increase to the annual salary - in which case both increase your salary rate by $300 over the course of a year, so option #1, which gets the increases sooner, will never be worse)
Title: Re: Financial dealings
Post by BenVitale on Jul 11th, 2009, 5:48pm
The trick is the first increase coming along sooner in the twice-a-year scheme.

'Game:

I take Strategy #1, the '$75-raise-twice-a-year'
You (rmsgrey) take Strategy #2, the '$300-raise-a-year'

By working out the total wages paid over a period of 3 years:

I get : $5,625

rmsgrey gets : $5,400

Imagine a situation where you would be offered  a certain amount $$$ raise a year vs. an amount $ raise per week.
Title: Re: Financial dealings
Post by pex on Jul 12th, 2009, 3:01am

on 07/11/09 at 17:48:38, BenVitale wrote:
I get : $5,625

I think you only get

6*(1500/12)
+ 6*((1500+75)/12)
+ 6*((1500+2*75)/12)
+ 6*((1500+3*75)/12)
+ 6*((1500+4*75)/12)
+ 6*((1500+5*75)/12)
= 5062.50.

By the same reasoning, for the other scheme,

12*(1500/12)
+ 12*((1500+300)/12)
+ 12*((1500+2*300)/12)
= 5400,

as you got.

To get to $5625, the $75 increase would have to be in the semiannual salary (corresponding to a $150 raise in annual salary).
Title: Re: Financial dealings
Post by ThudanBlunder on Jul 14th, 2009, 4:18am
There is actually a mistaken answer in Rouse Ball (http://www.amazon.com/Mathematical-Recreations-Essays-Rouse-Ball/dp/0025064304) :o, caused by the loose interpretation of a similar question. But I am not in UK at the moment, and so I don't have the book to hand in order to be more specific.
Title: Re: Financial dealings
Post by BenVitale on Jul 18th, 2009, 10:25pm

on 07/14/09 at 04:18:31, ThudanBlunder wrote:
There is actually a mistaken answer in Rouse Ball (http://www.amazon.com/Mathematical-Recreations-Essays-Rouse-Ball/dp/0025064304) :o, caused by the loose interpretation of a similar question. But I am not in UK at the moment, and so I don't have the book to hand in order to be more specific.


I didn't read that book.

I do think, now,  that the problem is an ill-posted one.

I write:
Option #1:  a $300 raise every year
Option #2:  a $75 raise every 6 months <=> $150 a year.

As we can see the first option is still much better.

Suppose the problem read: 'Current' instead of 'starting' -- it wouldn't make any difference ... unless it is different for the first option and for the second one.

The problem could have been more interesting if we added the following element :

- the starting salary for the 1-st option $1,000 and $1,500 for the 2-nd option

We write:
Option #1:  A starting salary of $1,000 with a $300 raise every year
Option #2:  A starting salary of $1,500 with a $75 raise every 6 months or $150 a year

Or we could add the question: what if the raises were proportional to salaries?

However, if and when the starting or current salary is the same for both options and the raises are constant and without any relation to the initial amount, then I deduce that option #1 is better.
Title: Re: Financial dealings
Post by BenVitale on Jul 24th, 2009, 11:14pm
We could also have this interpretation :


Option #1: a $75 raise every 6 months

Option #2: a $300 raise every year


And over a period of 3 years with ...

... option #2 I could get:

(1,500 + 300)*3 = $5,400

... Option #1 I could get:

Year 1:
--------
(1,500/2 + 75) + (1,500/2 + 75) + 75 = 1,500 + 3*75 = $1,725

Year 2:
-------
((1,500/2 + 75) + 75 + )+ 75 +
1500/2 + 3*75 + 1500/2 + 4*75 = 1500 + 7*75 = $2,025

Year 3:
--------
1500/2 + 5*75 + 1500/2 + 6*75 = 1500 + 11*75 = $2,325

Total wages that could be paid out = $1,725 + $2,025 + $2,325 = $6,075

Now, let's see if we can use these expressions to find a general pattern.

Once we do that, we can create modelling equations, and compare the growth patterns.

Let's key it to a variable. For instance, the $300-a-year raise pattern is:

year 1:
-------
pay: $1500
total: $1500

year 2:
-------
pay: $1500 + $300
total: $1500 + $1500 + $300 = 2*$1500 + $300

year 3:
-------
pay: $1500 + $300 + $300
total: 2*$1500 + $300 + $1500 + $300 + $300 = 3*$1500 + 3*$300

year 4:
--------
pay: $1500 + $300 + $300 + $300
total: 3*$1500 + 3*$300 + $1500 + 3*$300 = 4*$1500 + 6*$300

year 5:
-------
pay: $1500 + 4*$300
total: 5*$1500 + 10*$300

The multipliers are growing as:

year: 1, 2, 3, 4, 5,...

$1500: 1, 2, 3, 4, 5,...

$300: 0, 1, 3, 6, 10,...

For year "x", you have:

$1500: x*$1500

$300: (1/2)x(x - 1)

...where that last formula is the one you'll recognize as ''the sum of the first x whole numbers''.

Now, would anyone like to create a formula for the total earned in "x" years?

Then do something similar for the other option. Once you have the two formulas, graph them and see what you get.
Title: Re: Financial dealings
Post by BenVitale on Feb 5th, 2010, 5:32pm
Here's a new problem : Compound Interest

Say, you decide to invest 10,000 dollars into a 25 year interest savings account that pays 3% annually.

Assume the interest is paid continuously.
That is, there is a continuous amount of money being put into your account such that if you have $X at the beginning of the year, then you will have $1.03X at the end of the year.

Furthermore, suppose that at the beginning of each new year after you initially place the $10,000 you decide to deposit $500 into this savings account.

Thus, 24 payments of $500 will be deposited until all the money is withdrawn after 25 years.

Question: How much money will you receive after this 25 year period?
Title: Re: Financial dealings
Post by rmsgrey on Feb 6th, 2010, 1:53am

on 02/05/10 at 17:32:10, BenVitale wrote:
Here's a new problem : Compound Interest

Say, you decide to invest 10,000 dollars into a 25 year interest savings account that pays 3% annually.

Assume the interest is paid continuously.
That is, there is a continuous amount of money being put into your account such that if you have $X at the beginning of the year, then you will have $1.03X at the end of the year.

Furthermore, suppose that at the beginning of each new year after you initially place the $10,000 you decide to deposit $500 into this savings account.

Thus, 24 payments of $500 will be deposited until all the money is withdrawn after 25 years.

Question: How much money will you receive after this 25 year period?


f25(9,500) where f(x)= (x+500)*1.03
Title: Re: Financial dealings
Post by BenVitale on Feb 6th, 2010, 11:53am

on 02/06/10 at 01:53:44, rmsgrey wrote:
f25(9,500) where f(x)= (x+500)*1.03


I don't get your answer.

Please read this page for explanation:

Regular Deposits (http://mathforum.org/dr.math/faq/faq.interest.html)

the value P of the account at the end of n years is given by

P = M([1 + (i/q)]nq - 1)(q/i)
Title: Re: Financial dealings
Post by towr on Feb 6th, 2010, 2:07pm

on 02/06/10 at 11:53:35, BenVitale wrote:
I don't get your answer.
It's recursive.
fn(x) = fn-1(f(x))
Title: Re: Financial dealings
Post by BenVitale on Feb 6th, 2010, 3:59pm

on 02/06/10 at 14:07:48, towr wrote:
It's recursive.
fn(x) = fn-1(f(x))


Oh, I see. Thanks for the clarification.


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