Lecture 5Geometric Gradient Series,

Finishing Chapter 2

Read 84-102Read 84-102

Problems 2.30, 32, 35, 38, 39, 47, 52Problems 2.30, 32, 35, 38, 39, 47, 52

  

Do the self-test in studying for Exam.Do the self-test in studying for Exam.

Geometric Gradient

Increasing/decreasing at a constant Increasing/decreasing at a constant percentage, not a constant amountpercentage, not a constant amount

g > 0, series will increase, g < 0, series will g > 0, series will increase, g < 0, series will decreasedecrease

A

A(1+g)

A(1+g)N-1

P

Or

A

A(1-g)

A(1-g)N-1

Present Worth Pn, of any cash

flow An, at i is:

) x... x a(xxP

xxxxaP

i

g

g

AP

igAP

igAiAP

N

N

nN

n

N

n

nn

nnnnn

132

32

1

1

1

1

11

x,by multiply

N termsfirst for the series geometrica of form closed theis which),....(

i1

g1 x and

g1

A a Letting

)1(

)1(

)1(

series thefrom removed be can that ermconstant ta is,g)A(1

)1()1(

n),i,(P/F, termeach torthpresent wopayment single of principle apply the thenWe

)1()1()1(

Then subtract the two equations from one another as we did in our earlier derivations.

g i if,

gi if),1/(

)1()1(1

get, we termsoriginal for the x and a replacing

1 x ,1

)(

)()1(

)(

1

1

1

iNAgi

igA

P

wherex

xxaP

xxaxP

xxaxPP

NN

N

N

N

Geometric gradient series present worth factor (P/A,g,i,N)

Unlike the linear gradient the annual amount Unlike the linear gradient the annual amount is imbedded in the equation.is imbedded in the equation.

Example:

Airplane ticket price will increase 8% in Airplane ticket price will increase 8% in each of the next four years. The cost at each of the next four years. The cost at the end of the first year will be $180. the end of the first year will be $180. How much should be put away now to How much should be put away now to cover a students travel home at the end of cover a students travel home at the end of each year for the next four years? each year for the next four years? Assume 5%.Assume 5%.

67.715$03.0

11928.0180

08.05.

)05.1()08.1(1180

)1()1(1

44

gi

igAP

nn

As a check we can also solve this problem without using the geometric As a check we can also solve this problem without using the geometric gradientgradient

YearYear TicketTicket 11 AA11 = = = 180= 180 22 AA22 = 180 + 8%(180) = 180 + 8%(180) = 194.40= 194.40 33 AA33 = 194.40 + 8% (194.50) = 194.40 + 8% (194.50) = 209.95= 209.95 44 AA44 = 209.95 + 8% (209.95) = 209.95 + 8% (209.95) = 226.75= 226.75

P = 180(P/F,5%,1) + 194.40(P/F,5%,2) + 209.95(P/F,5%,3) + P = 180(P/F,5%,1) + 194.40(P/F,5%,2) + 209.95(P/F,5%,3) + 226.75(P/F,5%,4)226.75(P/F,5%,4)

=$715.66=$715.66

There are no tables for the geometric gradient.There are no tables for the geometric gradient.

Future worth FactorSince F =P(1+i)Multiplying (P/A,g,i,n) by (1+i) will give F

 

gi

giAF

igi

igAiPF

nn

nnn

n

)1()1(

)1()1()1(1

)1(

Example

A graduating CE is going to make $35,000/yr A graduating CE is going to make $35,000/yr with Granite Construction. A total of 10% of with Granite Construction. A total of 10% of the CE salary will be placed in the mutual fund the CE salary will be placed in the mutual fund of their choice. The CE can count on a 3% of their choice. The CE can count on a 3% salary increase with the standard of living salary increase with the standard of living increases for the next 30 years of employment. increases for the next 30 years of employment. If the CE is aggressive and places their If the CE is aggressive and places their retirement in a stock index fund that will retirement in a stock index fund that will average 12% over the course of their career, average 12% over the course of their career, what can the CE expect at retirement?what can the CE expect at retirement?

A = 35,000 x 0.1 = 3,500i = 12%g = 3%n = 30

714,070,1$92.305350009.0

)03.1()12.1(3500

)1()1(

3030

gi

giAF

nn

Recall that all of the interest equations Recall that all of the interest equations can only be used when interest period is can only be used when interest period is the same as the compounding period.the same as the compounding period.

Problem 2.15 revisited

Many of you solved this problem using brute force, P = 1,000,000 + 800,000(P/F,8%,1) +….+ P = 1,000,000 + 800,000(P/F,8%,1) +….+

1,000,000(P/F,8%,10) = $6,911,5391,000,000(P/F,8%,10) = $6,911,539

You should just recognize that you could also solve it by You should just recognize that you could also solve it by

P = 1,000,000 + 800,000(P/A,8%,5) +1,000,000(P/A,8%,5)P = 1,000,000 + 800,000(P/A,8%,5) +1,000,000(P/A,8%,5)(P/F,8%,5)(P/F,8%,5)

Or

P = 1,000,000 + 100,000(P/A,8%,10) – P = 1,000,000 + 100,000(P/A,8%,10) – 200,000(P/A,8%,5)200,000(P/A,8%,5)

Recognizing multiple ways to solve a Recognizing multiple ways to solve a problem will be crucial on the exam!problem will be crucial on the exam!

More Complicated Example,More Complicated Example, Solve the following Cash Flow diagram Solve the following Cash Flow diagram

for Present Worth,for Present Worth,

Chapter 2 is now complete. All of the Chapter 2 is now complete. All of the basic equations have been presented.basic equations have been presented.

   Most of the basic equations are functions Most of the basic equations are functions

on the spread sheet programs like excel, on the spread sheet programs like excel, lotus, and there is a downloadable lotus, and there is a downloadable program made by the author of the program made by the author of the textbooktextbook