Build and compare linear financial models

A model starts at $6000 and loses $750 each year. Which number belongs in V₀?

Move through years 0 to 4. Each point shows the value after that many complete years. Compare the change from one selected year to the next

What stays the same at every one-year step?

Two rules, two different jobs

A recurrence tells you how to get the next value from the previous one. An explicit rule tells you the value directly from the year number

1. 1Recurrence: use the previous value State the starting value and how to get the next value. If a model starts at 6000 and loses 750 each year, subtract 750 from the previous value \(\displaystyle V_0=6000,\quad V_{n+1}=V_n-750\)
2. 2Explicit rule: use the index Start with the original amount and apply the change n times. This gives any term without finding all the earlier terms \(\displaystyle V_n=6000-750n\)

Which is an explicit rule that gives the value directly from n?

Read a model from a graph

First read the value at year zero. Then find how much the value changes over a known number of years. Divide by that number of years to get the yearly change

1. 1Do not mistake two years for one If the value falls by $1500 over two years, the yearly loss is $750 \(\displaystyle 1500\div2=750\)
2. 2Use the exact plotted values Read the axes and coordinate key. Do not estimate a different starting amount or confuse the year number with the value

Changing the index changes the rate

Suppose a vehicle loses 12 cents per kilometre and travels 6000 km each year. A year and a kilometre are different steps

1. 1Convert the currency 12 cents is $0.12 \(\displaystyle 12\div100=0.12\)
2. 2One yearly step 6000 kilometres produce a yearly loss of $720 \(\displaystyle 6000\times0.12=720\)
3. 3One kilometre step For a distance-indexed rule, use $0.12 per kilometre, not $720 per year

Loss per operating hour

\(\displaystyle \$0.25\)

Operating hours per year

\(\displaystyle 800\)

What positive dollar amount does the machine lose in one year?

Find a rate that is missing

A vehicle falls from $16000 to $13000 while travelling 15000 km. Find the total loss first; the original price itself is not the loss

1. 1Total loss Subtract remaining value from original value \(\displaystyle 16000-13000=3000\)
2. 2Loss per kilometre Divide the $3000 loss by 15000 km \(\displaystyle 3000\div15000=0.20\)
3. 3Several years of use If use is given per year, multiply by the number of years before dividing the total loss

Compare two balances

Let the balances follow these rules. To find B minus A, subtract the whole of A’s rule

1. 1Starting difference B starts $500 lower \(\displaystyle 4500-5000=-500\)
2. 2Yearly difference B gains $70 more each year \(\displaystyle 270-200=70\)
3. 3Difference rule Combine the starting gap and the extra yearly gain \(\displaystyle D_n=B_n-A_n=-500+70n\)

Difference rule

\(\displaystyle D_n=-500+70n\)

Year

\(\displaystyle n=4\)

Find D₄. Keep its sign: D is B minus A.

Six equal payments total exactly the target. When do accumulated payments first strictly exceed it?