Ch3 Lecture Notes

8/2/2019 Ch3 Lecture Notes

1/21

Combining Factors

4/14/2012


2/21

Learning Objectives

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3/21

Uniform Series that are SHIFTED

A shifted series where present worth (PW) point in

time is NOT t= 0.

Shifted either to the left of 0 or to the right of t= 0.

Dealing with a uniform series:

The PW point is always one period to the left of thefirst series value

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4/21

Shifted Series

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0 1 2 3 4 5 6 7 8

A = $500/year

Consider:

P of this series is at t = 2 (P2 or F2)P2 = $500 (P/A,i%,4) or could refer to as F2

P0 = P2 (P/F,i%,2) or F2 (P/F,i%,2)

P2P0


5/21

Series with other single cash flows

It is common to find cash flows that are

combinations of series and other single cash

flows.

Solve for the series present worth values thenmove to t = 0.

Solve for the PW at t = 0 for the single cash flows.

Add the equivalent PWs at t = 0.

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6/21

Series with other cash flows

Consider:

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0 1 2 3 4 5 6 7 8

A = $500 F5 = $400

F4 = $300

Find the PW at t = 0 cash flow.

i = 10%

P =$1583.56

*


7/21

The Linear Gradient - revisited

The Present Worth of an arithmetic gradient (linear

gradient) is always located:

One period to the left of the first cash flow in the

series ( 0 gradient cash flow) or,

Two periods to the left of the 1G cash flow.

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8/21

Shifted Gradient: Example

Given:

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Base Annuity = $100

G = +$100

0 1 2 3 4 .. .. 9 10

Let Cash Flow (CF) start at t = 3:

Base annuity is $100/yr; increasing by$100/year through year 10; i = 10%; Find thePW at t = 0.


9/21

Shifted Gradient: Example

PW of the Base Annuity

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Base Annuity = $100

0 1 2 3 4 .. .. 9 10

P2= $100 (P/A,10%,8) = $100(5.3349) = $533.49

nannuity = 8 time periods

P0= $533.49 (P/F,10%,2) = $533.49(0.8264) =$440.88

P2 = $100(P/A,10%,8)


10/21

Gradient Present Worth

For the gradient component

10

0 1 2 3 4 .. .. 9 10

G = +$100

PW of gradient is at t = 2:

P2= $100(P/G,10%,8) = $100(16.0287) = $1,602.87

P0 = $1,602.87 (P/F,10%,2)=$1,324.61

Total Present Worth: $440.88 + $1,324.61=

$1,765.49

P2 = $100(P/G,10%,8)


11/21

Shifted Geometric Gradient

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0 1 2 3 n

A1

Present worth point is at t = 2 for this example


12/21

Geometric Gradient Example

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0 1 2 3 4 5 6 7 8

A = $700/yr

12% Increase/yr

i = 10%/year

A1 = $400 @ t = 5


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Geometric Gradient Example

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0 1 2 3 4 5 6 7 8

A = $700/yr

12% Increase/yr

i = 10%/year

PW point for thegradient

PW point for theannuity

A1 = $400 @ t = 5


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The Gradient Amounts

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Present Worth of the Gradient at t = 4*not in table

P4 = $400{ P/A1,12%,10%,4 } = 1,494.70$

3.73674

P0 = $1,494.70(P/F,10%,4) = $1,494.70(0.6830)

P0 = $1,020,88

PW of the Annuity

P0 = $700(P/A,10%,4)

= $700(3.1699) = $2,218.94

Total Present Worth: $1,020.88 + $2,218.94 = $3,239.82


15/21

Geometric gradient

P4 = $400{ P/A1,12%,10%,4} ; gi

P4 = $400 [1 (1.12/1.10)4 ] / (.10 -.12)

P4 = $400 [1 - 1.0181824 ] / -0.02

P4 = [1 - 1.074735] / -0.02 = 3.736745

P0 = $400 * 3.736745 = $1494.698

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16

Shifted Decreasing Linear Gradients

Given the following shifted, decreasing gradient:

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

Find the Present Worth @ t = 0


17/21

17


Given the following shifted, decreasing gradient:

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

PW point @ t = 2


18/21

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0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

P2 or, F2: Take back to t = 0

P0 here


19/21

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0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year


P0 here

Base Annuity = $1,000


20/21

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Time Periods Involved

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year


P0 here

1 2 3 4 5

Dealing with n = 5.


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Time Periods Involved

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

1 2 3 4 5

P2 = $1,000( P/A,10%,5 ) 100( P/G,10%.5 )

$1,000 G = -$100/yr

P2= $1,000( 3.7908 ) - $100( 6.8618 ) = $3,104.62

P0 = $3,104.62( P/F,10%,2 ) = $3104.62( 0 .8264 ) = $2,565.65

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