1
The loan amount is the present value of the payments.
Whatever you will pay, discounted back to today, must equal what you borrowed.
the bank
pays ₱100,000 once
today
you
pay R every month
for 24 months
fair deal
the two are equal in value
at the agreed rate
This is the annuity present value from two lessons ago, rearranged. Nothing new is
being introduced — only the direction the money flows.
2
The amortisation formula gives the payment.
i = 0.1212 = 0.01, n = 24
per month, and how many
R = P × i ÷ [1 − (1 + i)⁻ⁿ]
the formula
1.01⁻²⁴ ≈ 0.787566
the discount factor
R = 10000.212434 ≈ ₱4,707.35
every month for two years
The exponent is negative
Using 1.01²⁴ instead of 1.01⁻²⁴ turns the bottom into a large positive number and
the payment collapses to a few hundred pesos. The minus sign is what discounts the
future back to today.
3
Every payment is split, and the split shifts.
payment 1
₱1,000 interest, ₱3,707 principal
1% of the full ₱100,000
payment 12
about ₱591 interest
the balance has shrunk
payment 24
about ₱47 interest
almost all principal now
The payment never changes, but what it is doing does. Early payments are mostly
interest; later ones mostly repay the debt. That table is called an amortisation
schedule.
4
Total paid minus borrowed is the cost.
24 × 4,707.35 = ₱112,976.40
everything handed over
− 100,000
what was received
= ₱12,976.40 of interest
about 13% of the loan
Note that a 12% yearly rate cost about 13% of the loan over two years, not 24%. You
only owe the full amount at the start; by the end you owe very little.
5
Variations: the loan you can afford, the balance part-way, a down payment first.
The loan is the unknown: ₱5,000 a month is affordable for 12 months at 12% compounded monthly.
i = 0.01, n = 12
per month
P = 5000 × (1 − 1.01⁻¹²) ÷ 0.01
the loan is the present value
= 5000 × 11.255077 ≈ ₱56,275.39
the most that can be borrowed
Less than 12 × 5,000 = ₱60,000: part of every payment is interest.
The balance part-way: what is still owed on the ₱100,000 loan after 12 of its 24 payments?
12 payments of ₱4,707.35 remain
the balance is their present value
balance = 4707.35 × 11.255077
the same 12-month factor
≈ ₱52,981.59
more than half the loan
Halfway through the term, more than half the loan is still owed, because the early payments were mostly interest.
A down payment first: a ₱600,000 car, 20% down, the rest over 3 years at 10% compounded annually.
loan = 600,000 − 120,000 = ₱480,000
only the balance is borrowed
R = 480000 × 0.10 ÷ (1 − 1.1⁻³)
yearly payments, n = 3
= 48000 ÷ 0.248685 ≈ ₱193,015.11 a year
the payment
Applying the formula to the full ₱600,000 is the slip; the down payment is cash, not borrowed money.
6
A longer term lowers the payment and raises the cost.
2 years
₱4,707 a month
₱12,976 total interest
3 years
₱3,321 a month
₱19,571 total interest
5 years
₱2,224 a month
₱33,467 total interest
₱4,707.35 a month on the opening problem
Stretching the same loan from two years to five cuts the monthly payment by more
than half, and nearly triples what the borrowing costs. Both facts are true at once,
and an advertisement will usually only show you the first one.
On the exam
Periodical exams ask for i and n, the regular payment, the total interest, or what the first payment is mostly doing; entrance tests prefer the reasoning — why a longer term costs more, or why the exponent is negative. Find i and n first, then use R = P × i ÷ [1 − (1 + i)⁻ⁿ], and total interest is R × n − P. For a 10-second check, the payment must be more than the loan divided by n, but not by much.
Remember the flow
Rate per period and number of periods · R = P × i ÷ [1 − (1 + i)⁻ⁿ] · Total paid minus borrowed is the interest
The exponent is negative. A positive one collapses the payment to nonsense.
