Depreciation of assets: Flat rate, unit cost and reducing balance for VCE General Mathematics Unit 3

What this dot point is asking

VCAA wants you to model the falling value of an asset (a car, a machine, a computer, a coffee machine) with a first-order linear recurrence relation, and to do it three ways: flat rate, unit cost and reducing balance depreciation. You must be able to write the recurrence relation and the rule for each method, use them to find the value after $n$ periods, read a given recurrence and say which method and rate it describes, compare the methods in a table or on a graph, and work backwards to find an unknown rate, cost per unit or time. Every one of these skills appears regularly in Examination 1 and Examination 2.

The study design groups depreciation with the rest of Recursion and financial modelling because it uses exactly the same recurrence you meet for interest, loans and annuities: $u_0 = a$, $u_{n+1} = Ru_n + d$, which is printed on the formula sheet. Depreciation is simply that recurrence with $R \le 1$ and $d \le 0$.

The answer

Three methods, one recurrence

The value of an asset after $n$ periods is written $V_n$, with $V_0$ the purchase price. The three methods differ only in what happens each period.

Three pieces of vocabulary appear in questions:

• Book value (or future value) is the value recorded for the asset at a given time, $V_n$.
• Scrap value (or salvage value) is the value at which the owner stops depreciating or disposes of the asset.
• Write-off value is the value at which the asset is removed from the books, often $0 or a nominated scrap value.

Flat rate depreciation

Flat rate depreciation removes the same dollar amount every period. That amount is usually quoted as a percentage of the purchase price, so a $48 000 van depreciated at a flat rate of $12.5\%$ per annum loses $0.125 \times 48\,000 = 6000$ dollars every year, forever the same amount:

Applying the step $n$ times removes $6000$ a total of $n$ times, which gives the rule

Because a constant is subtracted, the values $48\,000, 42\,000, 36\,000, \ldots$ drop by equal steps and lie on a straight line. The van's value would reach zero when $6000n = 48000$, that is after $n = 8$ years.

The trap with flat rate is the base of the percentage. "$12.5\%$ flat rate" is $12.5\%$ of the original price every year, not $12.5\%$ of this year's value. If you take the percentage of the current value you have switched to reducing balance.

Unit cost depreciation

Unit cost depreciation ties the loss in value to use, not time. The asset loses a fixed amount for every unit of use: every kilometre driven, every hour operated, every cup of coffee brewed, every page printed.

So a $52 000 delivery van that depreciates by $0.20 per kilometre is worth $52\,000 - 0.20 \times 83\,500 = 35\,300$ dollars after it has travelled $83\,500$ km, however long that took.

When the question also tells you the usage per period, unit cost becomes a recurrence exactly like flat rate. If the van covers $25\,000$ km a year it loses $0.20 \times 25\,000 = 5000$ dollars a year:

This is why a recurrence such as $V_{n+1} = V_n - 1314$ could be either flat rate or unit cost: the recurrence alone cannot tell them apart. The context (a rate per kilometre, per hour or per cup) tells you it is unit cost, and the per-period amount is the product of the rate and the usage. The 2023 coffee machine question above is built on exactly this idea: $1314 per year at $0.04 per cup means $1314 \div 0.04 = 32\,850$ cups a year.

Reducing balance depreciation

Reducing balance depreciation removes a fixed percentage of the current value each period. After losing $r\%$, the asset keeps $(100 - r)\%$, so its value is multiplied by

A tractor bought for $120 000 and depreciated at $15\%$ per annum by reducing balance has $R = 0.85$:

The first few values are $120 000, $102 000, $86 700, $73 695. The dollar loss shrinks each year ($18 000, then $15 300, then $13 005) because $15\%$ of a smaller number is smaller. That is the signature of geometric decay: big early losses, then a long tail that flattens out. Since multiplying a positive value by $0.85$ always leaves a positive value, the value never reaches zero. This is why reducing balance questions ask when the value first falls below a threshold, not when it reaches zero.

Reducing balance is the depreciation twin of compound interest. Compound interest multiplies by $R = 1 + \frac{r}{100}$; reducing balance depreciation multiplies by $R = 1 - \frac{r}{100}$. That is why the finance solver handles it: enter the rate as a negative interest rate.

Reading a recurrence relation

Exam 1 often gives you a recurrence relation and asks what it models. Read it in two steps.

1. Is the previous term multiplied by something other than 1? If yes ($V_{n+1} = 0.92V_n$), it is reducing balance and the rate is $(1 - R) \times 100\%$, here $8\%$. If the multiplier is greater than 1 it is growth (compound interest or appreciation), not depreciation.
2. Is a constant subtracted? If yes ($V_{n+1} = V_n - 750$), it is flat rate or unit cost, losing $750 per period. Use the context to decide which, and divide by the purchase price if a flat rate percentage is asked for.

The explicit rule follows immediately: $V_n = V_0 - nd$ for a constant subtraction and $V_n = R^nV_0$ for a constant multiplier. VCAA writes the explicit rule with $V_0$ (or $u_0$), not with a "first term" $t_1$, so the power is $n$, not $n - 1$. Check any rule by substituting $n = 0$: it must return the purchase price.

Comparing the methods

Because flat rate is linear and reducing balance is geometric, their graphs always have the same relationship. Reducing balance loses more in the early years, so for a while its value is lower. Flat rate keeps subtracting the same amount while the reducing balance losses shrink, so eventually the straight line crosses below the curve.

The figure shows a $40 000 machine under a flat rate of $5000 per year and a reducing balance rate of $20\%$ per year. Reducing balance is lower at every year up to year 5 ($\$13\,107.20$ against $\$15\,000$), but at year 6 the flat rate value of $10 000 is below the reducing balance value of $10 485.76. The question "after how many years is one method first lower than the other" is answered by tabulating both rules side by side until the order flips.

Finding an unknown

Most multi-mark questions give you enough information to find something other than the final value.

• Flat rate or unit cost, unknown depreciation per period: $d = \dfrac{V_0 - V_n}{n}$. Divide by $V_0$ and multiply by 100 for a flat rate percentage; divide by usage per period for a cost per unit.
• Reducing balance, unknown rate: solve $V_n = R^nV_0$ for $R$, so $R = \left(\dfrac{V_n}{V_0}\right)^{1/n}$ and $r = (1 - R) \times 100$. On the CAS use solve, or the finance solver with $PMT = 0$.
• Unknown time: for flat rate and unit cost, solve the linear equation and round up to the next whole period if the question asks when the value first drops below a threshold. For reducing balance, use a table, solve, or the finance solver for $N$, then round up.

Using the CAS

Two techniques cover almost everything.

Sequence or table. Enter the rule ($48000 - 6000n$ or $120000 \times 0.85^n$) as a function of $n$ and view the table. This is the fastest way to find when a value crosses a threshold or when one method overtakes another.

Finance solver (reducing balance only). Set $N$ to the number of years, $I(\%)$ to the depreciation rate with a negative sign, $PV$ to the purchase price as a negative (money paid out), $PMT = 0$, and $PpY = CpY = 1$. Solving for $FV$ gives the depreciated value. For the tractor: $N = 3$, $I(\%) = -15$, $PV = -120\,000$, $PMT = 0$ gives $FV = 73\,695$. You can equally solve for $I(\%)$ or $N$ when those are unknown. Flat rate and unit cost are not compound processes, so do them with the rule, not the finance solver.

How exam questions ask about depreciation

• "Write a recurrence relation for $V_n$ in terms of $V_0$, $V_{n+1}$ and $V_n$." Give both parts: the starting value $V_0 = \ldots$ and the step $V_{n+1} = \ldots$. Missing $V_0$ loses the mark.
• "Write a rule for $V_n$ in terms of $n$." $V_n = V_0 - nd$ or $V_n = R^nV_0$, with the numbers substituted.
• "What is the annual flat rate depreciation percentage?" Annual dollar loss divided by the purchase price, times 100.
• "By what amount does it depreciate per hour/kilometre/unit?" Total loss divided by total units used.
• "After how many years will the value first be less than ...?" Table or solve, then round up to a whole number of years.
• "Show that the value after 3 years is ..." Show the recurrence steps or the substitution into the rule; the answer is given, so the working earns the mark.

Why the methods behave differently, in one line each

• Flat rate: the loss is a fixed share of the purchase price, so it never changes and the value drops in a straight line to zero.
• Unit cost: the loss depends on use; with steady use it behaves exactly like flat rate, and with uneven use the value can fall faster in busy years and slower in quiet ones.
• Reducing balance: the loss is a share of what is left, so it shrinks every period and the value approaches zero without reaching it.

Real businesses choose reducing balance for assets that lose most of their value early (cars, computers) and unit cost for assets that wear out with use (vehicles by kilometres, machines by hours). Tax rules sometimes specify the method, which is why VCAA contexts often begin "for taxation purposes".