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Additional Solved Problems Lump Sum Future Value

Additional Solved Problems

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Additional Solved Problems. Lump Sum Future Value. The Problem. You've received a $40,000 legal settlement. Your great-uncle recommends investing it for retirement in 27-years by “rolling over” one-year certificates of deposit (CDs) Your local bank has 3% 1-year CDs - PowerPoint PPT Presentation

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Page 1: Additional Solved Problems

Additional Solved Problems

Lump Sum

Future Value

Page 2: Additional Solved Problems

The Problem

– You've received a $40,000 legal settlement. Your great-uncle recommends investing it for retirement in 27-years by “rolling over” one-year certificates of deposit (CDs)

– Your local bank has 3% 1-year CDs

– How much will your investment be worth?

– Comment.

Page 3: Additional Solved Problems

Categorization– Your capital gains will be reinvested. There is

no cash-flow from the settlement for 27 years, so this is a lump sum problem.

– There is some uncertainty in the cash flows because interest rate are static for just the first year, but we assume that it will be 3% until you retire

– If you are unable to shelter your earnings, the IRS will want their cut

Page 4: Additional Solved Problems

Data Extraction

• PV = $40,000

• i = 3% (or 3% * (1- marginal tax rate)?)

• n = 27-years

• FV = ?

Page 5: Additional Solved Problems

Solution by Equation

56.851,88$

)03.01(000,40

)1(27

niPF

Page 6: Additional Solved Problems

Calculator Solution

N I PV PMT FV

27 3% 40,000 0 ?$88,851.56

Page 7: Additional Solved Problems

Comments

• Your great uncle's a financial idiot• Given a 27-year investment, you should

either – Invest the money more aggressively to

accumulate the money you need to survive, or

– Live! Blow the money on that red convertible!

Page 8: Additional Solved Problems

3 Additional Solved Problems

Lump Sum

Interest Rate

Page 9: Additional Solved Problems

Problem 1

• If you have five years to increase your money from $3,287 to $4,583, at what interest rate should you invest?

Page 10: Additional Solved Problems

Algebraic Solution

%87.60687.0132874583

1)1(

5

1

1

i

PF

iiPFnn

Page 11: Additional Solved Problems

Problem 2

• An investment you made 12-years ago is today worth its purchase price. It has never paid a dividend.

• Closer inspection reveals that the share price has been highly periodic, moving from $150 when purchased, to $300 in the next year, to $75 in the next, back to $150, before repeating

Page 12: Additional Solved Problems

Cyclical Price Movement

0

50

100

150

200

250

300

0 2 4 6 8 10 12

Year

Pri

ce (

$)

Page 13: Additional Solved Problems

12-Year and Average Returns

0150

0150150

iodReturnHoldingPer

StartCF

DividendsStartCFEndCF

%67.41%100%100%75%100121

7575150

7575150

30030075

150150300

121

Compare with Average HPR

Page 14: Additional Solved Problems

Comments

– Here we have the average holding period return being 41.67% per year, while the security has returned you nothing over the whole period!

– Averages seduce us with their intuitiveness– The correct average to have used was the

geometric average of return factors, not the arithmetic average of return rates

Page 15: Additional Solved Problems

Averages Must be Meaningful 1

– You walk 1 mile at 2 mph and another at 3 mph. What was your average speed? (2+3)/2 = 2.5 mph.

– NO!– The first leg lasts 1/2 hour, and the second leg

lasts 1/3 hours, total 5/6 hours.– So average speed is 2/(5/6) = 2.4 mph.

Page 16: Additional Solved Problems

Averages Must be Meaningful 2

– A little analysis shows that the correct mean for the walker is the harmonic mean

– The correct mean for the return problem may be shown to be the geometric mean of the (1+return)’s

– The appropriate mean requires thought

Page 17: Additional Solved Problems

Problem 3– In 1066 the First Duke of Oxbridge was

awarded a square mile of London for his services in assisting the conquest the England. The 30th Duke wished to live a faster paced life, and sold his holding in 1966 for £5,000,000,000. Examination of original project’s cost showed only the entry “1066 a.d.: to repair armor, £5”

– What was rate of capital appreciation ?

Page 18: Additional Solved Problems

Categorization

– We may assume that the Dukes lived quite well from leasing land to their tenants, but we are not interested in the revenue cash flows here, just the capital cash flows

– There is a present cash flow, a future cash flow, and no annuity payments, so the problem is the return on a lump-sum invested for a number of periods

Page 19: Additional Solved Problems

Data Extraction

• PV = 10

• FV = 5,000,000,000

• n = (1966 - 1066) = 1900

• i = ?

Page 20: Additional Solved Problems

Solution by Equation

%10.1%096667999.1

15

5000000000

11

1900

1

1

n

PF

niPFnn

Page 21: Additional Solved Problems

Solution by Calculator

n i PV Pmt FV

1900 ?1.09666%

-5 0 5,000,000,000

Page 22: Additional Solved Problems

Comments

• Note that a capital gain of only 1.1% per year results in a huge value over time

• Time plus return is very potent

• The real issue here is what is missing, namely the revenue streams

Page 23: Additional Solved Problems

Additional Solved Problems

Lump Sum

Number of periods

Page 24: Additional Solved Problems

The Problem

• How many years would it take for an investment of $9,284 to grow to $22,450 if the interest rate is 7% p.a. ?

• p.a. = per annum = per year

Page 25: Additional Solved Problems

Categorization– This is a lump sum problem asking for a

solution in terms of time. Most of these problems are useful models of reality if expressed in real terms, not nominal terms

– In any nominal situation, the terminal $22,450 will not be a constant, but will depend on the unknown time

– We will assume that the numbers and rates are in real terms

Page 26: Additional Solved Problems

Data Extraction

• PV = $9,284• FV = $22,450

• i = 7% p.a.

• n = ?

Page 27: Additional Solved Problems

Solution by Equation

yearsn

iPF

niPF n

05.1307.01ln

284,9450,22

ln

1ln

ln1

Page 28: Additional Solved Problems

Additional Solved Problems

Lump Sum

Present Value

Page 29: Additional Solved Problems

The Problem

– If investment rates are 1% per month, and you have an investment that will produce $6,000 one hundred months from now, how much is your investment worth today?

Page 30: Additional Solved Problems

Categorization

– This is the most basic of financial situations, and involves finding the present value of a future payment given no periodic payments

– The issue of risk is a little fuzzy. It is assumed that the rate given is for the project’s risk category

Page 31: Additional Solved Problems

Data Extraction

• FV = $6000

• PV = ?

• n = 100 months

• i = 1%

Page 32: Additional Solved Problems

Solution by Equation

27.218,2$)01.01(

000,6

)1(*)1(

)1(

100

P

iFi

FPiPF n

nn

Page 33: Additional Solved Problems

Calculator Solution

N I PV PMT FV

100 1% ?-2,218.27

0 6000

Page 34: Additional Solved Problems

Additional Solved Problems

Lump Sum

Special Case: Doubling

Rule of 72

Page 35: Additional Solved Problems

The Problem

• Consider the following simple example:– Sol Cooper Investments have offered you a

deal. Invest with them and they will double your investment in 10 years. What interest rate are they offering you?

– We could solve this using

• but this is over-kill

1 nPFi

Page 36: Additional Solved Problems

Data Extraction

• Doubling

• n = 10

• i = ?

Page 37: Additional Solved Problems

Some Algebra

72.012

ln2

08.02* :8%i double, To

ln2

2

2*2

ln

251

231

22

ln

1ln

ln1

53

in

PF

ii

ii

PF

n

ii

ii

ii

PF

iPF

niPF n

Page 38: Additional Solved Problems

Solution by Equation

%16.71212

answerAccurate

%20.7%1072

72.0*

10

11

n

xx

i

i

ni

Page 39: Additional Solved Problems

The Secret Reveled– Now you have seen the derivation of the rule

of 72, you are now able to produce your own personal rules. Example:

• “The Rule of a Magnitude”To increase your wealth by 10 times, the product of

interest and time is 240, that is about (2.08/2)*ln(10)

Example, how long will it take to increase your money ten times, given interest rates of 10%?

N = 240/10 = 24 years, real answer is 24.16 years

Page 40: Additional Solved Problems

How good is the Rule of 72?

– We have derived a rule using approximation methods, but have no idea how accurate it is

– There are two tests we could apply• we could take some range, and determine the

absolute maximum error of the rule in that range

• we could simply graph the error

– Graphs are fun:

Page 41: Additional Solved Problems

Doubling your Money

0.00%

10.00%

20.00%

30.00%

40.00%

50.00%

60.00%

70.00%

80.00%

90.00%

100.00%

0 5 10 15 20 25 30

Years

Inte

res

t

rule

algebra

Page 42: Additional Solved Problems

Absolute Error

-30.00%

-25.00%

-20.00%

-15.00%

-10.00%

-5.00%

0.00%

5.00%

0 5 10 15 20 25 30

Years

% E

rro

r

error

Page 43: Additional Solved Problems

Graph of Rule of 72 Error

– The high error in a part of the graph that does not interest us is hiding the error in the part that does. We have two choices

• plot absolute error on a log scale

• truncate the graph and re-scale

– Truncation is fun

Page 44: Additional Solved Problems

Absolute Error

-0.40%

-0.35%

-0.30%

-0.25%

-0.20%

-0.15%

-0.10%

-0.05%

0.00%

0.05%

0.10%

5 10 15 20 25 30

Years

% E

rro

r

Page 45: Additional Solved Problems

Another Example

– You are a stockbroker wishing to persuade a young client to reconsider her $50,000 invested in 3%-CDs.

– Your client believes that stock mutual funds will return about 12% for the foreseeable future, but is averse to the volatility risks. Her money will remain fully invested for the next 48 years.

Page 46: Additional Solved Problems

Step 1

– The first step requires the calculation of how long is required to obtain a single doubling

• CDs: 72/3 = 24 years to double

• Mutual fund: 72/12 = 6 years to double

Page 47: Additional Solved Problems

Step 2

– The second step requires the calculation of how many doublings will occur during the lives of the investments

• CDs: 48/24 = 2 doublings

• Mutual fund: 48/6 = 8 doublings

Page 48: Additional Solved Problems

Step 3

– The third step calculates the value of the investment in 48 years

– CDs: 2 doublings of $50,000 • = $200,000

– Mutual fund: 8 doubling of $50,000• =256 * $50,000

• =$12,800,000 in 48 years

Page 49: Additional Solved Problems

Conclusion– We shall discover that her risk is smaller than

she imagines, but she will be about 64 times more wealthy if she accepts that risk

– Using the accurate method, her respective wealths are $206,613 and $11,519,539,

– The lesson is to start to invest early, and accept some risk

Page 50: Additional Solved Problems

Growth at 3 and 12 %

– The following graph shows her wealth increases over 10 years at a 3% and 12%

• The graph was cut at 10 years because the 12% rate of growth is so large that it dwarfs the 3% growth, making the graph meaningless

Page 51: Additional Solved Problems

Growth of $50,000 for 10 Years @ 3% and 12%

Ten Years Growth @ 3% & 12%

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

0 1 2 3 4 5 6 7 8 9 10Holding Period in Years

Val

ue

at

en

d o

f H

old

ing

Pe

rio

d

CD Stock

Page 52: Additional Solved Problems

Log Transformation of Y-Axis

– A common way to plot two such cash flows on the same graph is to use a semi-log graph. This prevents scale problems from hiding one of the graphs

– Note that the two graphs appear to be straight lines, and this is in fact the case

Page 53: Additional Solved Problems

Growth of $50,000 at 3% and 12% for 48 Years (Log Scale)

48 Years Log scale

10000

100000

1000000

10000000

100000000

0 5 10 15 20 25 30 35 40 45 50

Holding Period in Years

Va

lue

at

en

d o

f H

old

ing

Pe

rio

d (

Lo

g S

cale

)

CD Stock

Page 54: Additional Solved Problems

What is the use of the Rule?

– A significant source of avoidable error in financial calculations results from blindly “running the numbers” without reviewing them for empirical reasonableness

– It is a good practice to estimate values before computing them

– The rule of 72 is one tool that sometimes gives you “numerical feel” of a problem

Page 55: Additional Solved Problems

• Your reaction to learning the rule of 72 is– “Why bother, I’ve got the latest and best HP

financial calculator.”

• In a business meeting, the unilateral drawing of a financial calculator has a chilling effect on your opponents flexibility in a negotiation – It is amazing how many real problems you

can solve in your head using the rule of 72

Page 56: Additional Solved Problems

Additional Solved Problems

Irregular Cash Flows

Backwards Method

Page 57: Additional Solved Problems

The Problem

– You have been offered a video business, and estimate that video rental technology will be obsolete in 8 years when cable bandwidths and video compression will permit “movie-on-demand.” You require a 20% return on this class of risk. The cash flows, starting 1-year from now,are: 90, 110, 140, 140, 130, 90, 70, 30 (thousands of $s)

Page 58: Additional Solved Problems

A Faster Method of Discounting

– This is basically a present value of a lump sum repeated 8-times

– The most straightforward method would be to crunch the answer or use an Excel worksheet

– A good method to use on a calculator is the following algorithm:

Page 59: Additional Solved Problems

A Faster Method of Discounting (Continued)

– Input the last cash flow, and divide by the interest factor to “bring it to” a year earlier

– Iterate:• Add this discounted cash flow to the cash flow that

is already there, and discount the total for another period by dividing by the interest factor. Stop when you reach the current time

• Doing this is a lot simpler than it sounds

Page 60: Additional Solved Problems

Data and Computation: Backwards

year Flow Accumulation8 $30.00 $30.007 $70.00 $97.276 $90.00 $178.435 $130.00 $292.214 $140.00 $405.643 $140.00 $508.772 $110.00 $572.521 $90.00 $610.470 $0.00 $554.97

Page 61: Additional Solved Problems

Equations

rate 0.1RateFactor =1+rate

year Flow Accumulation8 30 =0 + B87 70 =C8/RateFactor + B96 90 =C9/RateFactor + B105 130 =C10/RateFactor + B114 140 =C11/RateFactor + B123 140 =C12/RateFactor + B132 110 =C13/RateFactor + B141 90 =C14/RateFactor + B150 0 =C15/RateFactor + B16

Page 62: Additional Solved Problems

Data and Computation:Traditional

year_ Flow_ Discounted0 $0.00 $0.001 $90.00 $81.822 $110.00 $90.913 $140.00 $105.184 $140.00 $95.625 $130.00 $80.726 $90.00 $50.807 $70.00 $35.928 $30.00 $14.00

Sum $554.97

Page 63: Additional Solved Problems

Equations

year_ Flow_ Discounted0 0 =Flow_*(1+rate)^-year1 90 =Flow_*(1+rate)^-year2 110 =Flow_*(1+rate)^-year3 140 =Flow_*(1+rate)^-year4 140 =Flow_*(1+rate)^-year5 130 =Flow_*(1+rate)^-year6 90 =Flow_*(1+rate)^-year7 70 =Flow_*(1+rate)^-year8 30 =Flow_*(1+rate)^-year

Sum =SUM(C19:C27)

Page 64: Additional Solved Problems

Calculator SolutionThe computation a BAII+ calculator is

30/1.1+ 70/1.1+ 90/1.1+ 130/1.1+ 140/1.1+ 140/1.1+ 110/1.1+ 90/1.1=

The solution is $554.97

Calculators differ in the way they string computations, you may need to add “=“ after the dollar amounts

See the savings on computational time!

Page 65: Additional Solved Problems

Comments– It is particularly useful to know the backwards

method when the yield curve is not flat. (Use the forward rates). The level of computation savings are even greater in this case

Page 66: Additional Solved Problems

Additional Solved Problems

Deceptive Interest Rates

Page 67: Additional Solved Problems

The Problem

• Advertisement:

– American Classic Cars! Finance Special! Sprite Conversion! Now Only $15,000! Just $1,000 Down, and 3-years to pay! Only 3% per year! (Compounded monthly with your good credit.)

Page 68: Additional Solved Problems

The Problem (Continued)

• Classic Car News has an almost identical car advertised for $9,000, but it needs $3,000 of work to match the condition of the car offered by ACC.

• What implied rate of interest, (per year, compounded yearly) would you be paying if you purchased the car from ACC?

Page 69: Additional Solved Problems

Explanation• When purchasing from Smart, you are

buying a bundle of financing and car

• To un-bundle the package, you use the cost of acquiring the competing car– Cash value of car = $9,000 + $3,000 =

$12,000

• Next, determine the cash flows associated with the financed car

Page 70: Additional Solved Problems

Calculator

N I PV PMT FV

36 3%/12=

0.25%

($15,000-$1,000)

=$14,000

?-407.14

0

Page 71: Additional Solved Problems

Analysis Continued

• The equivalent value of each cash flow is• $(12,000-1,000)

• -407.14

• …

• -407.14 (36 equal payments in all)

Page 72: Additional Solved Problems

Calculator (Continued)

N I(monthly)

PV PMT FV

36 3%/12 =0.25%

($15,000-$1,000) =$14,000

?-407.14

0

36 ?1.6419%

($12,000-$1,000) =$11,000

-407.14 0

Page 73: Additional Solved Problems

True Interest Rates

p.a. compounded p.a. %58.211016419.1

monthly compounded p.a. %70.1912*6419.112

i

i

The true interest rate on this loan is much higher than that in the advertisement

An enterprising lady sold jewelry in a factory where she worked. The people she sold to were poor creditrisks. She gave them interest-free loans, one third down.She marked up her prices to cost + 200%. No Risk!

Page 74: Additional Solved Problems

Series of Annuities• The next problem evaluates a project that

has a sequence of annuities– The method of solution is to evaluate each

annuity to the date one year before its first cash flow, and then to discount these lump sum equivalent amounts to today’s date

– The cash flow feature of a financial calculator may also be used

Page 75: Additional Solved Problems

The Problem

• Expected cash flows from a project requiring a 20% return

• Years Cash Flow Each Year

• 0 $(20,000,000)• 1 to 5 $3,000,000• 6 to 30 $2,000,000• 31 to 49 $1,000,000• 50 $(2,000,000)

Page 76: Additional Solved Problems

From To Amount PV0 0 (20,000,000) (20,000,000)1 5 3,000,000 8,971,8366 30 2,000,000 3,976,649

31 49 1,000,000 20,40450 50 (2,000,000) (220)

Rate 0.2 Sum (7,031,331)

Present Values of Components

Page 77: Additional Solved Problems

From To Amount PV0 0 -20000000 =Amount/Rate/(1+Rate)^(From-1)*(1-(1/(1+Rate)^(To-From+1)))1 5 3000000 =Amount/Rate/(1+Rate)^(From-1)*(1-(1/(1+Rate)^(To-From+1)))6 30 2000000 =Amount/Rate/(1+Rate)^(From-1)*(1-(1/(1+Rate)^(To-From+1)))31 49 1000000 =Amount/Rate/(1+Rate)^(From-1)*(1-(1/(1+Rate)^(To-From+1)))50 50 -2000000 =Amount/Rate/(1+Rate)^(From-1)*(1-(1/(1+Rate)^(To-From+1)))

Rate 0.2 Sum =SUM(E5:E9)

Excel Equations

Page 78: Additional Solved Problems

Note

• A single lump sum is just a degenerate annuity. The above equations made use of this fact

• The project is not at all attractive at the given rate

• At what discount rate does the project become attractive?

Page 79: Additional Solved Problems

Present Value of Project v Rate

(10,000,000)

0

10,000,000

20,000,000

30,000,000

40,000,000

50,000,000

0.00% 5.00% 10.00% 15.00% 20.00%

Discount Rate

Pre

sen

t V

alu

e

Page 80: Additional Solved Problems

The Problem• What is the present value of the following

project? The cash flow starts in year 1

• $20,000, $20,000, $20,000, $20,000, $20,000, $20,000, $20,000, $15,000, $20,000, $20,000, $20,000, $20,000, $20,000, $20,000, $20,000.

• The discount rate is 12% p. a.

Page 81: Additional Solved Problems

Analysis

• This project is basically an annuity with a hiccup.– Add $5,000 to the hiccup, – Evaluate the annuity, and then – Subtract the PV of the $5,000

Page 82: Additional Solved Problems

Algebraic Solution

87.197,134$

42.019,212.1000,5

29.217,13612.11

112.0000,20

11

1

_

8

15

_

pertannadj

pert

n

annadj

PP

P

iipmt

P

This is the project’s value

Page 83: Additional Solved Problems

The Problem• Mary will retire in 12-years, has $100,000

saved, and will put $12,000 into an account (at the end of every year) until she retires.

• She will take a $20,000 cruse in year-5.

• She expects to live 8 years after she retires, and will leave $30,000 to “bury her.” What will be her retirement income?

• The bank pays 3%

Page 84: Additional Solved Problems

Key to Solution

– After Mary’s wake, there is no money left. The future value of all her cash flows is the zero. The present value of all cash flows must also be zero

– We will discount all flows to the current year– You may prefer to use Mary’s retirement or

death day as the reference

Page 85: Additional Solved Problems

Solution Outline

• 0 = 100000 +

• 12000*PVIFA(3%, 12-years) -

• 20000*PVIF(3%, 5-years) -

• X*PVIFA(3%, 8-years)*PVIF(3%, 12-years)

• - 30000*PVIF(3%, 20-years)

Page 86: Additional Solved Problems

Solution by Equation

20

1285

12

03130000

0310311030

03120000

0311030

120001000000

-

--

).*(

).(*..X

).*(

- ..

Page 87: Additional Solved Problems

Solution using Excel

This is set to zero using the Tools “Solve” function

This was set to an arbitraryvalue, and then solved for

int 0 ret -37694

year CF Pert bal0 1000001 12000 1150002 12000 1304503 12000 1463644 12000 1627545 12000 -20000 1596376 12000 1764267 12000 1937198 12000 2115319 12000 229876

10 12000 24877311 12000 26823612 12000 28828313 -37694 25923714 -37694 22932015 -37694 19850616 -37694 16676717 -37694 13407618 -37694 10040419 -37694 6572220 -37694 -30000 0