Ch2 (part2)arithmetic gradient

1. 1
Gradient Exampels

2. 2
Arithmetic Gradient Factors

3. 3

4. 4
Arithmetic Gradient Problem

5. 5
• Two computations must be made and added:
the first for the present worth of the base amount
PA and a second for the present worth of the
gradient PG. The total present worth PT occurs in
year 0. This is illustrated by the partitioned cash
flow diagram.

6. 6

7. 7
A)
PT = 500(P/A,5%, IO) + 100(P/G,5%,10)
= 500(7.7217) + 100(31.652)
= $7026.05 ($7,026,050)
B) Here, too, it is necessary to consider the
gradient and the base amount separately. The
total annual series AT is found by
AT = 500 + 100(A/G,5%,10) = 500 + 100(4.0991)
= $909.91 per year ($909,910)
• And AT occurs from year 1 through year 10.
• If PT is known AT can be calculated directly
AT = PT(A/P,5%,1O) = 7026.05(0.12950)
= $909.87 ($909,870)

8. 8
Geometric Gradient
• It is common for cash flow series, such as
operating costs, construction costs, and
revenues, to increase or decrease from period to
period by a constant percentage, for example,
5% per year. This uniform rate of change defines
a geometric gradient series of cash flows. In
addition to the symbols i and n used thus far, we
now need the term g
• g = constant rate of change, in decimal form, by
which amounts increase or decrease from one
period to the next

9. 9
Geometric Gradient

10. 10
Geometric Gradient
• In summary, the engineering economy relation
and factor formulas to calculate Pg in period t = 0
for a geometric gradient series starting in period
1 in the amount A I and increasing by a constant
rate of g each period are

11. 11
Geometric Gradient Problem

12. 12
Geometric Gradient Problem