Model the value of a depreciating asset with straight-line (prime-cost) depreciation S = V0 - Dn, find the salvage value after n years, the annual depreciation D and the time to reach a given value, work with a rate given as a percentage of the original cost, and contrast it with declining-balance depreciation

What this dot point is asking

NESA wants you to model how an asset loses value over time with the simplest method,
straight-line depreciation (also called the prime-cost method). An asset is something
the business owns, such as a work vehicle or a piece of equipment. The idea is that the asset
drops in value by the same dollar amount every year, so its value plots as a falling
straight line. The whole dot point is that line. You write its equation $S = V_0 - Dn$, find
the value after $n$ years, find the annual depreciation $D$, and find when the value
reaches a chosen amount. You also contrast it with the declining-balance method, where a
fixed percentage of the current value is removed each year instead.

This is the mirror image of simple interest,
where the same amount is added each period. There the line rises; here it falls. The
graph behaves exactly like a simple-interest graph
turned upside down, with a negative gradient. It belongs strictly to the Year 11 Money Matters
topic (the syllabus code MS-F1). The full treatment of declining balance as its own model is a
Year 12 topic, so here it is only a brief contrast.

The answer

Straight-line depreciation subtracts the same amount every year, so three facts answer
almost every question.

• Value against time is a falling straight line. Starting from the purchase price $V_0$,
  each year removes a fixed depreciation $D$, so after $n$ years the value (the salvage or
  book value $S$) is

At $n = 0$ no time has passed, so $S = V_0$: the line starts at the purchase price on the
vertical axis.

• The annual depreciation $D$ is the gradient (going down). The number multiplying $n$ is
  $-D$, so the line falls by $D$ dollars for every $1$ year across. If you are given the start
  value, the salvage value and the useful life instead of $D$, recover the annual amount from

• The model stops at the useful life. A straight line would keep falling forever and
  eventually go negative, but an asset cannot be worth less than nothing. So the equation only
  applies up to the useful life, at the end of which the asset is worth its salvage value
  (its trade-in or scrap value).

A landscaping ute bought for $48,000 that depreciates by $5500 a year has the equation
$S = 48\,000 - 5500n$. After $3$ years it is worth $48\,000 - 5500 \times 3 = 31\,500$, i.e.
$31,500.00, and over a $7$-year useful life it falls to a salvage value of
$48\,000 - 5500 \times 7 = 9500$, i.e. $9500.00.

Depreciation as a falling straight line

Because the same $5500 comes off every year, the value of the ute drops in equal steps, the
signature of a straight line. Building the table from $S = 48\,000 - 5500n$ gives

Each value is $5500 below the one before it. Plotting these points and ruling a line gives
the graph below, built up in stages: draw the axes, mark the purchase price, fall by the annual
depreciation to the salvage value, then read values off.

Stage 1, draw and label the axes. Put time $n$ across the bottom and value $S$ up
the side, the standard convention. Scale the time axis to the useful life ($0$ to $7$ years) and
the value axis high enough to fit the purchase price ($48,000).

Stage 2, mark the purchase price. At $n = 0$ the ute is worth its full purchase price, so
plot the starting point at $48,000 on the vertical axis. This is where the line begins, the
vertical intercept $V_0$. A depreciation line starts at the top, not the origin, because the
asset is most valuable when new.

Stage 3, fall by the annual depreciation to the salvage value. From the purchase price,
take off $5500 each year, plotting $(1, 42\,500)$, $(2, 37\,000)$ and so on, down to the
salvage value $(7, 9500)$. Rule a straight line through the points: it has a negative
gradient of $-5500$, falling $5500 for every $1$ year across. The line ends at the salvage
value, where the useful life is up.

Stage 4, read the graph. To read the value at a chosen time, go up from that time to the
line and across to the value axis. At $3$ years the line reads $31,500 and at $5$ years
it reads $20,500. To go the other way and find when the value hits a target, start at a
value on the vertical axis, read across to the line, then down to the time axis.

The read-offs match the formula to the cent: at $n = 3$, $S = 48\,000 - 5500 \times 3 = 31\,500$,
i.e. $31,500.00, and at $n = 5$, $S = 48\,000 - 5500 \times 5 = 20\,500$, i.e. $20,500.00.
Because the line is straight, a read-off between plotted points is exact, not an estimate.

Why the gradient is the annual depreciation

The gradient of the depreciation line is worth a closer look, because it is the annual
depreciation, just with a minus sign. Gradient is rise over run: pick any two points and divide
the change in value by the change in time. Going from $1$ year to $3$ years is a run of $2$
years, and the value falls from $42,500 to $31,500, a fall of $11,000.

The line drops $5500 per year, so its gradient is $-5500$, the negative of the annual
depreciation $D = 5500$. This is the reverse of a simple-interest graph,
whose gradient is the positive interest added each year. So if a graph gives you the line but
not $D$, find the gradient from two points and take its size: $D = |\text{gradient}|$.

A rate given as a percentage of the original cost

Sometimes the depreciation is quoted not as a dollar amount but as a percentage of the
original cost each year. This is still straight-line depreciation, because the percentage is
always taken from the same purchase price, so the dollar amount removed each year is
constant. Convert the rate to a dollar amount once with $D = \text{rate} \times V_0$, then use
$S = V_0 - Dn$ as normal. For office equipment bought at $15,000 depreciating at $20\%$ of
the original cost a year, the annual amount is $D = 0.20 \times 15\,000 = 3000$, so the equipment
loses the same $3000 every year and the value falls in a straight line.

This is the trap that catches students: $20\%$ of the original cost is straight-line and
gives a falling line, but $20\%$ of the current value is declining balance and gives a
curve. Watch the wording. Notice too that taking a fixed fraction of the original each year must
eventually reach zero, here at $15\,000 \div 3000 = 5$ years, after which the formula would give a
negative value. The model only makes sense up to that point: an asset is never worth less than
nothing.

Contrast: straight-line versus declining balance

The other depreciation method, declining balance, removes the same percentage of the
current value each year rather than a fixed dollar amount. Because the value it is taken from
shrinks every year, the dollar amount removed shrinks too, so declining balance falls fast at
first and then flattens, tracing a curve instead of a straight line. The graph below puts
both on the same $48,000 ute: straight-line at $5500 a year, and declining balance at
$20\%$ of the current value a year.

In the first year declining balance is much steeper, because $20\%$ of the still-high starting
value is a bigger drop than the straight line's constant $5500. As the years pass each
declining-balance step shrinks (it is always $20\%$ of a smaller value), so the curve flattens
and never quite reaches zero, whereas the straight line keeps its constant $5500 step all the
way to the fixed salvage value of $9500. The two graphs sit on the same ute but tell different
stories: a constant fall to a known salvage value, versus a front-loaded drop that tails off. The
full declining-balance method, where a fixed percentage of the current value is lost each year,
is covered in Year 12;
here you only need to recognise the difference in shape.

How exam questions ask about this

Each wording points at one move on the model. Learn the translation:

• "Write the depreciation equation" or "find a formula for the value after $n$ years."
  Substitute the purchase price $V_0$ and the annual depreciation $D$ into $S = V_0 - Dn$.
• "Find the value (or book value, or salvage value) after $n$ years." Substitute that $n$
  into $S = V_0 - Dn$.
• "What is the salvage value at the end of its useful life?" Substitute $n = \text{useful
  life}$ into the equation.
• "Find the annual depreciation / how much it depreciates each year." If you are given the
  start and salvage values and the years, use $D = \dfrac{V_0 - S}{n}$. If you are given a rate,
  use $D = \text{rate} \times V_0$.
• "How much did it originally cost?" You are given $S$, $D$ and $n$; rearrange $S = V_0 - Dn$
  to $V_0 = S + Dn$.
• "When will it be worth $X?" Substitute $S = X$ and solve $X = V_0 - Dn$ for $n$.
• "It depreciates at $r\%$ of its original cost each year." This is straight-line: find
  $D = r\% \times V_0$ once, then use the equation. ($r\%$ of the current value would be
  declining balance, a curve.)
• "Find the total depreciation over $n$ years." It is $Dn$, which also equals $V_0 - S$.
• "Compare the straight-line and declining-balance methods." Straight-line is a falling
  straight line (constant dollar drop to a fixed salvage value); declining balance is a curve
  (constant percentage drop, steep then flattening, never reaching zero).

Edge case: rearranging for the original cost, and where the line stops

Two subtleties round out the topic. First, the same equation answers a backwards question.
If you know the current value, the annual depreciation and the age, you can find what the asset
originally cost. Rearrange $S = V_0 - Dn$ into $V_0 = S + Dn$. For example, a van now worth
$36,000 after $3$ years at $4650 a year cost $V_0 = 36\,000 + 4650 \times 3 = 49\,950$,
i.e. $49,950.00. You just add back the depreciation that has already been taken off. Second,
a straight-line model is only valid up to the useful life, the number of years the asset is
kept in use. The mathematical line would keep falling and cross below zero, but an asset cannot
have a negative value. So the line stops at the salvage value, and the formula is not used beyond
it. If a rate is a percentage of the original cost, that zero point is exactly
$V_0 \div D = 1 / \text{rate}$ years away; for $20\%$ of the original, that is $1 / 0.20 = 5$
years. Both points follow from the value being a straight line, which is the defining feature of
straight-line depreciation.