Lo 2 b factors

Cost estimation

EGN 3203 Engineering EconomicsLO 2 b Capital Investment Problems &Cash Flow

Factors: How Time and Interest Affect Money.

PURPOSE

Account for the time value of money using tabulated factors, spreadsheet functions, and calculator functionsTOPICSSingle amount factors F/P, P/FUniform series factors P/A, A/P, F/A, A/FGradients arithmetic and geometricSpreadsheet/calculator use

Fundamental equation determines amount F accumulated after t = 1, 2, , n years for a single present amount PF1 = P + Pi = P(1 + i)F2 = F1 + F1i= P(1 + i) + P(1 + i)i = P(1 + i)2

Ft = P(1 + i)t Fn = P(1 + i)n

Factors: How Time and Interest Affect Money: Single Payment Factors


Find F, given PFormula: F = P(1 + i)nNotation: F=P(F/P,i%,n)

(1+i)n = single-payment compound amount factor

Spreadsheet: = FV(i%,n,,P)Calculator: FV(i,n,A,P)Find P, given FFormula: P = F[1/(1 + i)n]Notation: P=F(P/F,i%,n)

1/(1+i)n = single-payment present worth factor

Spreadsheet: = PV(i%,n,,F)Calculator: PV(i,n,A,F)

Find F, given PFormula: F = P(1 + i)nNotation: F=P(F/P,i%,n)(1+i)n = single-payment compound amount factor

Spreadsheet: = FV(i%,n,,P)Calculator: FV(i,n,A,P)

Find P, given FFormula: P = F[1/(1 + i)n]Notation: P=F(P/F,i%,n)1/(1+i)n = single-payment present worth factor

Spreadsheet: = PV(i%,n,,F)Calculator: PV(i,n,A,F)


Example: Ada, Inc. can pay $10,000 now or $15,000 five years from now for power back-up equipment. Are P and F values equivalent at 5% per year?


Example: Ada, Inc. can pay $10,000 now or $15,000 five years from now for power back-up equipment. Are P and F values equivalent at 5% per year?


Single Payment Factors F/P :Example

Sol.

Single Payment Factors P/F : Example

Sol.

Notes

UNIFORM SERIES FORMULAS (A/F, F/A) (P/A, A/P )There are four uniform series formulas that involve A, where A means that:The cash flow occurs in consecutive interest periods, andThe cash flow amount is the same in each period.The formulas relate a present worth P or a future worth F to a uniform series amount A. The two equations that relate P and A are as follows.

The uniform series formulas that relate A and F are as follow.

Uniform Series Involving P/A and A/P

012345A = ?

P = GivenThe cash flow diagrams are:Standard Factor Notation P = A(P/A,i,n)A = P(A/P,i,n)Note: P is one period Ahead of first A value (1) Cash flow occurs in consecutive interest periodsThe uniform series factors that involve P and A are derived as follows:(2) Cash flow amount is same in each interest period

012345A = GivenP = ?

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Example: Uniform Series Involving P/AA chemical engineer believes that by modifying the structure of a certain watertreatment polymer, his company would earn an extra $5000 per year. At an interestrate of 10% per year, how much could the company afford to spend now to justbreak even over a 5 year project period?

(A) $11,170 (B) 13,640 (C) $15,300 (D) $18,950The cash flow diagram is as follows:P = 5000(P/A,10%,5)= 5000(3.7908)= $18,954Answer is (D)

012345A = $5000P = ?i =10%Solution:

13

Uniform Series Involving F/A and A/F 2012 by McGraw-Hill, New York, N.Y All Rights Reserved2-14 (1) Cash flow occurs in consecutive interest periodsThe uniform series factors that involve F and A are derived as follows:(2) Last cash flow occurs in same period as F

012345F = ?A = Given

012345F = GivenA = ?Note: F takes place in the same period as last ACash flow diagrams are:Standard Factor Notation F = A(F/A,i,n)A = F(A/F,i,n)

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Example: Uniform Series Involving F/AAn industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much will the savings amount to in 7 years? (A) $45,300 (B) $68,500 (C) $89,228 (D) $151,500 The cash flow diagram is:

A = $10,000F = ?i = 8%

0 1 2 3 4 5 6 7 Solution: F = 10,000(F/A,8%,7) = 10,000(8.9228) = $89,228Answer is (C)

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Uniform Series Involving P/A and A/P

Uniform Series Involving F/A and A/F

Uniform Series Involving F/A and A/F

GRADIENT FORMULASThe previous four equations involved cash flows of the same magnitude A in each interest period. Sometimes the cash flows that occur in consecutive interest periods are not the same amount (not an A value), but they do change in a predictable way. These cash flows are known as gradients, and there are two general types: arithmetic and geometric.

2012 by McGraw-Hill, New York, N.Y All Rights Reserved2-20Arithmetic GradientsArithmetic gradients change by the same amount each period The cash flow diagram for the PG of an arithmetic gradient is:

0

123nG2G

43G(n-1)G

PG = ?

G starts between periods 1 and 2(not between 0 and 1)

This is because cash flow in year 1 is usually not equal to G and is handled separately as a base amount (shown on next slide)Note that PG is located Two Periods Ahead of the first change that is equal to GStandard factor notation is PG = G(P/G,i,n)

20

2012 by McGraw-Hill, New York, N.Y All Rights Reserved2-21Typical Arithmetic Gradient Cash Flow

PT = ?i = 10%

0 1 2 3 4 5 400450500550600

PA = ?i = 10%

0 1 2 3 4 5 400400400400400

PG = ?i = 10%

0 1 2 3 4 5 50100150200+This diagram = this base amount plus this gradient

PA = 400(P/A,10%,5)PG = 50(P/G,10%,5)PT = PA + PG = 400(P/A,10%,5) + 50(P/G,10%,5)Amount in year 1 is base amountAmount in year 1 is base amount

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Converting Arithmetic Gradient to A 2012 by McGraw-Hill, New York, N.Y All Rights Reserved2-22i = 10%

0 1 2 3 4 5 G2G3G4Gi = 10%

0 1 2 3 4 5 A = ?

Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n) General equation when base amount is involved is

A = base amount + G(A/G,i,n)

0 1 2 3 4 5 G2G3G4GFor decreasing gradients, change plus sign to minusA = base amount - G(A/G,i,n)

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Example: Arithmetic Gradient The present worth of $400 in year 1 and amounts increasing by $30 per year through year 5 at an interest rate of 12% per year is closest to:(A) $1532 (B) $1,634 (C) $1,744 (D) $1,829

0

123Year430460

4490520

PT = ?5

400i = 12%G = $30 = 400(3.6048) + 30(6.3970) = $1,633.83Answer is (B)PT = 400(P/A,12%,5) + 30(P/G,12%,5)The cash flow could also be converted into an A value as follows:A = 400 + 30(A/G,12%,5) = 400 + 30(1.7746)= $453.24Solution:

23

P/G factor formula for gradient only

A/G factor for annual equivalent, gradient only

P/G and A/G factors are tabulated. There are no direct spreadsheet or calculator functions

Gradient Formulas:

Arithmetic Gradients

Example: Estimated annual revenue of $5,000 increases by $500 per year starting next year. Find P and A over an 8-year period at i = 10%.


Solution:

P = A(P/A,i%,n) + G(P/G,i%.n) = 5000(P/A,10%,8) + 500(P/G,10%,8) = $34,689

A = 5000 + 500(A/G,10%,8) = 5000 + 500(A/G,10%,8) = $6,502 per yearArithmetic Gradients


Geometric GradientsGeometric gradients change by the same percentage each period

0

123nA1A 1(1+g)14A 1(1+g)2A 1(1+g)n-1

Pg = ?

There are no tables for geometric factors

Use following equation for g i:Pg = A1{1- [(1+g)/(1+i)]n}/(i-g)where: A1 = cash flow in period 1 g = rate of increase

If g = i, Pg = A1n/(1+i)Note: If g is negative, change signs in front of both g valuesCash flow diagram for present worth of geometric gradient

Note: g starts between periods 1 and 2

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Example: Geometric Gradient

Find the present worth of $1,000 in year 1 and amounts increasing by 7% per year through year 10. Use an interest rate of 12% per year.(a) $5,670 (b) $7,333 (c) $12,670 (d) $13,550

0

1231010001070411451838

Pg = ?

Solution:Pg = 1000[1-(1+0.07/1+0.12)10]/(0.12-0.07) = $7,333Answer is (b)g = 7%i = 12%To find A, multiply Pg by (A/P,12%,10)

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Geometric gradient series Starts from base amount A1 and increases by constant percentage g in years 2 through n.

Geometric Gradients

Geometric gradient series

Gradient Formulas:

Present worth No tabulated factors

This is P for all cash flows, including A1

Term in brackets is (P/A,g,i%,n) factor

When g = i, relation reduces toP = A1[n/(1+i)]

Example: Payroll totals $250,000 this year; expected to grow at 5% per year through year 5. Find P at i = 12%.

Year 1: $250,000Year 5: 250,000(1.05)4 = 303,876

= 250,000(3.94005) = $985,013

Geometric Gradients

Geometric Gradients : Example

Notes