Brad M. answered • 01/20/14

Investment, Valuation, Business Strategy Game, WACC, NPV, DCF, YTM

**Mary gets ~15k more**:)

Amy G.

asked • 01/19/14bob places $1000 a year in his IRA for ten years and then invests $2000 a year for the next ten years. Mary places $2000 a year in her IRA for ten years and then invests $1000 a year for the next ten years. They both have invested $30000. If they earn 8 percent annually, how much more will Mary have earned than Bob at the end of 20 years?

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Brad M. answered • 01/20/14

Tutor

4.9
(505)
Investment, Valuation, Business Strategy Game, WACC, NPV, DCF, YTM

Hi Amy -- using rough "mid-point" estimates, Bob cooks 10k for 15yrs and 20k for 5yrs ...

Mary cooks 20k for 15yrs and 10k for 5yrs ... 8% for 15yrs triples the mid-point outlays ...

BOB: 30k plus 21.6 ... 23.4 ... 25.4 ... 27.5 ... 30k ~ Bob 60k

MARY: 60k [2x Bob's 1st leg] plus 15k [half Bob's last leg] => **Mary gets ~15k more** :)

Steve S. answered • 01/19/14

Tutor

5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus

Bob, B = 1000( (1.08)^20 + (1.08)^19 + (1.08)^18 + (1.08)^17 + (1.08)^16 + (1.08)^15 + (1.08)^14 + (1.08)^13 + (1.08)^12 + (1.08)^11 )

+ 2000( (1.08)^10 + (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + (1.08) )

B = 1000 ((1.08)^11) ( (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08 + 1 )

+ 2000 (1.08) ( (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08 + 1 )

+ 2000 (1.08) ( (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08 + 1 )

Let A = (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08 + 1

(1.08) A = (1.08)^10 + (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08

(1.08) A - A = (1.08)^10 - 1

A = ( (1.08)^10 - 1 ) / 0.08

[N.B.: A is a geometric series. The formula for the first n terms is sum{1,n}(a*(r^n - 1)/(r - 1)). For A, a =1, r = 1.08, and n = 10. But every student should know how to derive that formula; which is essentially what I did.]

B = 1000 ((1.08)^11) A + 2000 (1.08) A

B = 1000 (1.08) A ( 1*(1.08)^10 + 2 )

Mary, M = 1000 (1.08) A ( 2*(1.08)^10 + 1 )

M - B = 1000 (1.08) A ( 2*(1.08)^10 + 1 ) - 1000 (1.08) A ( 1*(1.08)^10 + 2 )

M - B = 1000 (1.08) A ( 2*(1.08)^10 + 1 - 1*(1.08)^10 - 2 )

M - B = 1000 (1.08) A ( (1.08)^10 - 1 )

M - B = 1000 (1.08) ( (1.08)^10 - 1 ) ( (1.08)^10 - 1 ) / 0.08

M - B = 1000 (1.08) ( (1.08)^10) - 1 ) ^2 / 0.08

M - B ≈ $18,131.95

[N.B.: Every student on a precalculus - calculus track should make sure they understand this solution completely; they should be able to replicate it for similar problems without notes.]

Steve S.

Tom, It's ALWAYS good to have confirmation. Thanks!

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01/20/14

William S. answered • 01/19/14

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4.4
(10)
Experienced scientist, mathematician and instructor - William

Amy,

Since you did not specify how often the interest is compounded, I might not be able to give you an exact number, but let's try:

I'll assume (for simplicity's sake) that the interest is compounded annually.

Then A = P(1 + r/n)^{nt }= ($30,000)(1 + r/n)^{nt}

Since I have assumed the number of times the interest is compounded is only once per year, n = 1 and our equation becomes:

A = ($30,000)(1 + 0.08)^{t}

At the end of the first year

A_{1} = ($30,000)(1.08) = 32,400 for both Bob and Mary

Here's where things get tricky. At the beginning of year 2 Bob adds another $2,000 and Mary adds $1000 to the principal.

Bob's numbers in green, Mary's numbers in orange:

And for Mary

At the end of 10 years, Bob will have $91,740.87 and Mary will have $78,25431.

Beginning at year 11, the situation reverses itself and Bob starts adding $1,000 per year while Mary increases her contributions to $2,000.

Here's where things get a bit embarrassing. At the end of 20 years, I have Bob with $211,375.52 and Mary with only $200,236.17, which I don't think is the way things were supposed to work out. I'll continue working on this to see if I can find a flaw in my logic.

Tom D. answered • 01/19/14

Tutor

New to Wyzant
Very patient Math Expert who likes to teach

1)Bob puts in $1000 for 10 years and THEN transitions to $2000/yr later

Bob_Total = $1000*(1.08)^20 + $1000*(1.08)^19 + ... $2000*(1.08)^10 + $2000*(1.08)^9 + ...

2)Mary puts in $2000 for 10 years and THEN transitions to $1000/yr later

Mary_Total = $2000*(1.08)^20 + $2000*(1.08)^19 + ... $1000*(1.08)^10 + $1000*(1.08)^9 + ...

The difference (Diff) between the two investment strategies can be computed (Diff=Mary_Total-Bob_Total)

Diff= $1000[ (1.08)^20)+(1.08)^19+...(1.08)^11] - $1000[ (1.08)^10)+(1.08)^9+...(1.08)^1]

Note that the 2nd 10 terms can be subtracted from the 1st 10 terms of each 20 term total

Diff1= $1000[ 1.08^20 + 1.08^19 +... 1.08^11 ]=$1000*(33.777) (note: used geometric series for 33.777 factor)

Diff2= $1000[ 1.08^10 + 1.08^9 + ... 1.08^1 ] =$1000*(15.6455) (note: used geometric series for 15.6455 factor)

Diff= $1000*(33.777-15.6455)= $18,132 difference despite investing the same total amount.

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Tom D.

01/20/14