SME
Engineering Economics · Problem type
Uniform series: (P/A, i%, n) and (A/P, i%, n)
Use this for “A $30,000 loan at 7% is repaid in 8 equal yearly payments. How much is each payment?”, or what a series of payments is worth today.
Chart
P(8) = 30,000
This is an example. Type your own values to start a new problem.
$
P
=
$
A
(P/A,
i
%,
n
)
(P/A)
Factor (P/A, i%, n)
calculated
(A/P)
Factor (A/P, i%, n)
calculated
Formulas
(
P
A
)
=
(1
−
(1 +
i
100
)
−
n
)
÷
(
i
100
)
5.9713
=
(1
−
(1 +
7
100
)
-8
)
÷
(
7
100
)
(
A
P
)
=
1
(
P
A
)
0.16747
=
1
5.9713
P
=
A
×
((
P
A
))
30,000 = 5,024.03 × 5.9713
Assumptions
•
The first payment comes one period after P, at the end of year 1.
•
The same amount A is paid at the end of every period for n periods.
•
(A/P) is the capital-recovery factor: the loan payment per dollar borrowed.
Step-by-step
WE KNOW
P = $30,000, i = 7%, n = 8 yr
FIND
factor (P/A, i%, n) ((P/A)), factor (A/P, i%, n) ((A/P)), payment each year (A)
Step 1 · Find factor (P/A, i%, n): (P/A)
Use (P/A) = (1 − (1 + i ÷ 100)⁻ⁿ) ÷ (i ÷ 100)
Add the present worths of $1 paid at the end of each year: the sum of a geometric series.
(
P
A
)
=
(1
−
(1 +
7
100
)
-8
)
÷
(
7
100
)
(
P
A
)
=
(1
−
(1 + 0.07)
-8
)
÷
(
7
100
)
(
P
A
)
=
1 −
1.07
-8
7 ÷ 100
(
P
A
)
=
1 − 0.582
0.07
(
P
A
)
=
0.418
0.07
(
P
A
)
=
5.9713
Step 2 · Find factor (A/P, i%, n): (A/P)
Use (A/P) = 1 ÷ ((P/A))
Spreading
P
over
the
years
undoes
adding
the
payments
up.
(
A
P
)
=
1
5.9713
(
A
P
)
=
0.16747
Step 3 · Find payment each year (A)
Use P = A × ((P/A))
Spread
the
present
worth
over
the
years:
the
same
as
P
×
(
A
P
).
A
=
P
(
P
A
)
A
=
30,000
5.9713
A
=
about
$5,024.03
Check
Put the numbers back in. Both sides should match.
5.9713
=
(1
−
(1 +
7
100
)
-8
)
÷
(
7
100
)
✓
0.16747
=
1
5.9713
✓
30,000 = 5,024.03 × 5.9713 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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