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Topic: Financial dealings (Read 1632 times) |
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Benny
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Financial dealings
« on: Jul 10th, 2009, 10:22am » |
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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.
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Obob
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Re: Financial dealings
« Reply #1 on: Jul 10th, 2009, 11:22am » |
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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.
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Benny
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Re: Financial dealings
« Reply #2 on: Jul 10th, 2009, 12:19pm » |
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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.
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| « Last Edit: Jul 10th, 2009, 12:20pm by Benny » |
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rmsgrey
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Re: Financial dealings
« Reply #3 on: Jul 10th, 2009, 2:29pm » |
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on Jul 10th, 2009, 12:19pm, 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. |
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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)
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Benny
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Re: Financial dealings
« Reply #4 on: Jul 11th, 2009, 5:48pm » |
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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.
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| « Last Edit: Jul 11th, 2009, 7:07pm by Benny » |
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pex
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Re: Financial dealings
« Reply #5 on: Jul 12th, 2009, 3:01am » |
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on Jul 11th, 2009, 5:48pm, BenVitale wrote:
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).
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ThudnBlunder
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Re: Financial dealings
« Reply #6 on: Jul 14th, 2009, 4:18am » |
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There is actually a mistaken answer in Rouse Ball  , 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.
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| « Last Edit: Jul 14th, 2009, 4:38am by ThudnBlunder » |
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Benny
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Re: Financial dealings
« Reply #7 on: Jul 18th, 2009, 10:25pm » |
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on Jul 14th, 2009, 4:18am, ThudanBlunder wrote:There is actually a mistaken answer in Rouse Ball , 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.
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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.
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| « Last Edit: Jul 18th, 2009, 10:27pm by Benny » |
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Benny
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Re: Financial dealings
« Reply #8 on: Jul 24th, 2009, 11:14pm » |
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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:
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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.
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| « Last Edit: Jul 24th, 2009, 11:15pm by Benny » |
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Benny
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Re: Financial dealings
« Reply #9 on: Feb 5th, 2010, 5:32pm » |
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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?
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| « Last Edit: Feb 5th, 2010, 5:33pm by Benny » |
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rmsgrey
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Re: Financial dealings
« Reply #10 on: Feb 6th, 2010, 1:53am » |
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on Feb 5th, 2010, 5:32pm, 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? |
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f25(9,500) where f(x)= (x+500)*1.03
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Benny
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Re: Financial dealings
« Reply #11 on: Feb 6th, 2010, 11:53am » |
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on Feb 6th, 2010, 1:53am, rmsgrey wrote:
f25(9,500) where f(x)= (x+500)*1.03 |
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I don't get your answer.
Please read this page for explanation:
Regular Deposits
the value P of the account at the end of n years is given by
P = M([1 + (i/q)]nq - 1)(q/i)
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Benny
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Re: Financial dealings
« Reply #13 on: Feb 6th, 2010, 3:59pm » |
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on Feb 6th, 2010, 2:07pm, towr wrote:
It's recursive.
fn(x) = fn-1(f(x))
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Oh, I see. Thanks for the clarification.
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