ch3 combinig factors_rev2

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Combining FactorsCombining Factors

Chapter 3

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Combining Factorso Most estimated cash flow series do not fit

exactly the series for which the factors and equations were developed. Therefore, it is necessary to combine the equations.

o For a given sequence of cash flows, there are usually several correct ways to determine the equivalent present worth P, future worth F, or annual worth A.

o This chapter explains how to combine the engineering economy factors to address more complex situations involving shifted uniform series and gradient series. Spreadsheet functions are used to speedup the computations.

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Shifted Uniform SeriesIf the cash flow does not begin

until some later date, the annuity is known as Shifted Uniform Series.

Annuity is shifted for J periods.PA = A (P/A, i%, N-J) ( P/F, i%, J)

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0 1 2J

J+1J+2

J+3 N-1 N

P? =

AP’A

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Shifted Uniform Series As stated above, there are several methods that

can be used to solve problems containing a uniform series that is shifted. However, it is generally more convenient to use the uniform-series factors than the single-amount factors. There are specific steps that should be followed in order to avoid errors: ◦ 1. Draw a diagram of the positive and negative cash

flows. ◦ 2. Locate the present worth or future worth of each

series on the cash flow diagram. ◦ 3. Determine n for each series by renumbering the cash

flow diagram. ◦ 4. Draw another cash flow diagram representing the

desired equivalent cash flow. ◦ 5. Set up and solve the equations.

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Shifted Uniform Series - Example

PA present worth of annual series A.

P’A present worth at time other than 0.

PT is total present worth at time 0.

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Shifted Uniform Series - ExampleFirst: find P’A for shifted series

◦P’A = 500(P/A, 8%,6) , P’A is located in year 2

Second: Find PA in year 0, using “future” PA.◦ PA= P’A(P/F,8%,2)

Third: Find total present worth PT=P0+PA = 5000+ 500(P/A,8%,6)

(P/F,8%, 2) =5000+500(4.6229)(0.8573)

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Non-standard Shifted Series Non-standard Shifted Series Solution Methods:

◦ 1) Treat each cash flow individually◦ 2) Convert the non-standard annuity or gradient to

standard form by changing the compounding period◦ 3) Convert the non-standard annuity to standard by

finding an equal standard annuity for the compounding period

Example:◦ How much is accumulated over 20 years in a fund that

pays 4% interest, compounded yearly, if $1,000 is deposited at the end of every fourth year?

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$1000

F? =

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Non-standard Shifted Non-standard Shifted Series Series

Method 1: consider each cash flows separately F = 1000 (F/P,4%,16) + 1000 (F/P,4%,12) +

1000 (F/P,4%,8) + 1000 (F/P,4%,4) + 1000 = $7013

Method 2: convert the compounding period from annual to every four years

ie = (1+0.04)4 -1 = 16.99%

F = 1000 (F/A, 16.99%, 5) = $7013Method 3: convert the annuity to an equivalent

yearly annuity A = 1000(A/F,4%,4) = $235.49 F = 235.49 (F/A,4%,20) = $701204/09/23 8

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Computer Solutions (Excel)Computer Solutions (Excel)

• Use of a spreadsheet similar to Microsoft’s Excel is fundamental to the analysis of engineering economy problems.

• Appendix A of the text presents a primer on spreadsheet use.

• All engineers are expected by training to know how to manipulate data, macros, and the various built-in functions common to spreadsheets.

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Excel’s Financial FunctionsExcel’s Financial Functions To find the present value P: PV(i%, n, A, F)

To find the present value P of any series: NPV(i%, second_cell, last_cell) + first_cell

To find the future value F: FV(i%, n, A, P)

To find the equal, periodic value A: PMT(i%, n, P, F)

Nesting of 1 function for the argument of another is allowed.• To find the number of periods n:

NPER(i%, A, P, F) • To find the compound interest rate i:

RATE(n, A, P, F) • To find the compound interest rate i:

IRR(first_ cell:last_ cell) 04/09/23 10

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Shifted Uniform Series by ExcelNPV=PV of benefits - PV of costsWhen the uniform series A is shifted, the

NPV function is used to determine P, and the PMT function finds the equivalent A value.

PMT(i%,n,NPV(i%,second_cell:lasCcell)+firsCcell,F)

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Shifted Uniform Series by Excel

NPV=PV of benefits - PV of costsWhen the uniform series A is shifted, the

NPV function is used to determine P, and the PMT function finds the equivalent A value.

PMT(i%,n,NPV(i%,second_cell:lastCell)+first_cell,F)

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Shifted Uniform Series by Excel

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Shifted Gradient The present worth of an

arithmetic gradient will always be located two periods before the gradient starts.

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Shifted Gradient

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Shifted GradientShifted Gradient

The period in which the gradient first appears is labeled period 2.

Gradient year 2 is placed in year 5 of the entire sequence in Figure 3-10c. It is clear that n = 5 for the P/G factor. The PG arrow is correctly placed in gradient year 0, which is year 3 in the cash flow series.

The general relation for PT is taken from Equation. The uniform series A =$100 occurs for all 8 years, and the G = $50 gradient present worth appears in year 3.

PT =PA + PG= 100(P/A,i,8) + 50(P/G,i,5)(P/F,i,3)

Shifted Decreasing GradientThe base amount is equal to the largest

amount in the gradient series, that is, the amount in period 1 of the series.

The gradient amount is subtracted from the base amount instead of added to it.

The term - G(P/G,i,n) or -G(A/G,i,n) is used in the computations and in Equations [2.18] and [2.19] for PT and AT, respectively.

The present worth of the gradient will still take place two periods before the gradient starts, and the equivalent A value will start at period I of the gradient series and continue through period n.

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Shifted Decreasing Gradient

PT = $800(P/F,i, 1) + 800(P/A,i,5)(P/F,i,1) - 100(P/G,i,5)(P/F,i,1)

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Shifted Decreasing Gradient

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Shifted Decreasing Gradient

Investment sequence G=$500,Base amount = $2000 , n =5

Withdrawal sequence up to year 10, G=$-1000, Base =$5000, n=5

Years 11 and 12: A=$1000 Solve each sequence separately

first

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Shifted Decreasing Gradient

For investment sequence (Pi) ◦Pi=present worth of deposits ◦=2000(P/A,7%,5) +500(P/G,7%,5)◦=

2000(4.1002)+500(7.6467)=$12,023.75For withdrawal series (Pw) represent

the present worth of the withdrawal base amount and gradient series in year 6-10 (P2), + present worth of withdrawal in year 11 and 12 (P3).

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Shifted Decreasing GradientPw=P2+P3= PG(P/F,7%,5)+P3==[5000(P/A,7%,5) – 1000(P/G,7%,5)](P/F,7%,5)]

+1000(P/A,7%,2)(P/F,7%,10)=[5000(4.1002)-1000(7.6467)]

(0.7130)+1000(1.8080)(0.5083)=$9165.12+919.0=$10,084.12

Net present worth = Pw-Pi= 10,084.12-12,023.75= $-1939.63

The A value may be computed using the (A/P,7%,12) factor

A=P (A/P,7%,12) = 1939.63(A/P,7%,12) = $-244.20

The interpretation of these results is as follows: In present-worth equivalence, you will invest $1939.63 more than you expect to withdraw. This is equivalent to an annual savings of $244.20 per year for the 12-year period.

Shifted Gradient by ExcelDetermine the total present worth PT in

period 0 at 15% per year for the two shifted uniform series in Figure 3-17. Use two approaches: by computer with different functions and by hand using three different factors.

Shifted Gradient by ExcelThe cash flows are entered into cells, and

the NPV function is developed using the format

NPV(i%,second_cell:last_celI) + first_ cell or NPV(Bl,B6:B18)+ B5

Solution by Hand: 1) using present worth

Shifted Gradient by HandThe use of P/A factors for the uniform

series, followed by the use of P/F factors to obtain the present worth in year 0, finds PT

PT=PA1 +PA2

PAl = P’A1(P/F,15%,2) = A1(P/A,15%,3)(P/F,15%,2)

= 1000(2.2832)(0.7561) = $1726PA2= P’A2(P/F,15%,8) =

A2(P/A,15%,5)(P/F,15%,8)= 1500(3.3522)(0.3269)= $1644PT = 1726 + 1644 = $3370

Shifted Gradient – using Future

Use the F/A, F/P, and P/F factors.PT = (FA1 + FA2)(P/F,15%,13)

FA1= F’A1(F/P,15%, 8) = A1(F/A,15%,3)(F/P,l5%,8)=1000(3.4725)(3.0590) = $10,622FA2 = A2(F/A,15%,5) = 1500(6.7424) = $10,113

PT = (FA1 + FA2)(P/F,15%,13) = 20,735(0.1625) = $3369

Shifted Gradient – using Intermediate Year MethodFind the equivalent worth of both series at year 7

and then use the P/F factor.PT=(FA1+PA2)(P/F,15%,7)The PA2 value is computed as a present worth; but

to find the total value PT at year 0, it must be treated as an F value. Thus,

FA1 = F’A1(F/P,15 %,2) = A1(F/A,l5%,3)(F/P,15%,2)= 1000(3.4725)(1.3225) = $4592

PA2 = P’A2(P/F, 15 %,l ) = A2(P/A, l5%,5)(P/F,l5%,1)= 1500(3.3522)(0.8696) = $4373

PT = (FA1 + PA2)(P/F,15%,7)= 8965(0.3759) = $3370