This document discusses methods for combining cash flow factors to solve more complex engineering economy problems involving shifted uniform series and gradients. It explains how to combine factors to find the present, future, or annual worth for cash flows that do not begin immediately or that change over time. Examples are provided to illustrate combining factors for shifted uniform series, non-standard shifted series, shifted gradients, and decreasing gradients. The use of spreadsheet functions to simplify the calculations is also discussed.

1. 1
Combining Factors
Chapter 3

2. 2
Combining Factors
o 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.
08/07/14 2

3. 3
Shifted Uniform Series
If 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)
08/07/14 3
0 1 2
J
J+1
J+2
J+3 N-1 N
P= ?
A
P’A

4. 4
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.

5. 5
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.

6. 6
Shifted Uniform Series - Example
First: 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)

7. 7
Non-standard Shifted SeriesNon-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?
08/07/14 70 4 8 12 16 20
$1000
F= ?

8. 8
Non-standard Shifted SeriesNon-standard Shifted 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) = $7013
Method 3: convert the annuity to an equivalent yearly
annuity
A = 1000(A/F,4%,4) = $235.49
F = 235.49 (F/A,4%,20) = $7012
08/07/14 8

9. 9
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.
08/07/14 9

10. 10
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)
08/07/14 10

11. 11
Shifted Uniform Series by Excel
 NPV=PV of benefits - PV of costs
 When 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)

12. 12
Shifted Uniform Series by Excel
 NPV=PV of benefits - PV of costs
 When 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)

13. 13
Shifted Uniform Series by Excel

14. 14
Shifted Gradient
The present worth of an arithmetic
gradient will always be located two
periods before the gradient starts.

15. 15
Shifted Gradient

16. 16
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)

17. Shifted Decreasing Gradient
 The 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.

18. 18
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)

19. 19
Shifted Decreasing Gradient

20. 20
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

21. 21
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)=$12,023.75
For 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).

22. 22
Shifted Decreasing Gradient
 Pw=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.

23. Shifted Gradient by Excel
 Determine 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.

24. Shifted Gradient by Excel
 The 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

25. Shifted Gradient by Hand
The 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) = $1726
PA2= P’A2(P/F,15%,8) = A2(P/A,15%,5)(P/F,15%,8)
= 1500(3.3522)(0.3269)= $1644
PT = 1726 + 1644 = $3370

26. 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) = $10,622
 FA2 = A2(F/A,15%,5) = 1500(6.7424) = $10,113
 PT = (FA1 + FA2)(P/F,15%,13) = 20,735(0.1625) = $3369

27. Shifted Gradient – using Intermediate
Year Method
Find 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