OPC Calculations 3

We are now considering converting our OPC car to normal car. The purpose for this is to facilitate bringing Baby1 to her grandparents’ place during the daytime when Dear2’s maternity leave is over and needs to return to work. It would be rather inconvenient if we have to bring Baby1 out before 7am each day and only bring her home after 7pm each day. So now, we need to consider if it is worth forking out around $15000 in cash to convert our OPC to normal.

There are many factors that will affect our decision to convert or not. Some of the more intangible factors include the comfort and flexibility of driving, the availability of car seat in our car (versus taxi), and the need to transport baby equipment frequently. However, for today, I’ll just give an analysis on one of the more tangible factor – the value of the converted car. In other words, how much of the cash top up will depreciate and how much can be recovered eventually.

Let’s ask our buddy \LaTeX to help us out. ;)

(This analysis builds upon some results from earlier in this series. See also OPC Calculations & OPC Calculations 2.)

Let C be the COE of the car, O be the OMV of the car, and P be the PARF of the car. Let t be the number of years passed before scrapping the car, and n be the number of years passed before converting from OPC to normal plate. Hence, let P_{n,t} denote the PARF of the car, which was converted after n years, after t years.

The amount of top up required to be paid to LTA can be express as \frac{10-n}{10} \times 17000. This will be apportioned to C and O in the same ratio that C and O was deducted initially. For example, assuming C<17000 and C+O>17000, then the amount apportioned to C will be \frac{C}{17000} \times \frac{10-n}{10}\times 17000 and the amount apportioned to O will be \frac{17000-C}{17000} \times \frac{10-n}{10} \times 17000.

Further, let m=\left\{\begin{array}{c  l} \frac{75}{100}&\quad\mbox{if }0<t<5\\ \frac{100-(t\times5)}{100}&\quad\mbox{if  }5<t<10\end{array}\right.

Now, we can formally define P_{n,t}.

\begin{array}{rcl}P_{n,t}&=&(\frac{C}{17000}\times\frac{10-n}{10} \times 17000)\times \frac{10-t}{10}\\&&+m\times (O-(17000-C)+\frac{17000-C}{17000} \times \frac{10-n}{10} \times 17000)\\&=&C\frac{10-n}{10}\frac{10-t}{10}+m(O+C-17000+(17000-C)\frac{10-n}{10})\\&=&C\frac{10-n}{10}\frac{10-t}{10}+mO(1+\frac{C-17000}{O}+\frac{17000-C}{O}\frac{10-n}{10})\\&=&C\frac{10-n}{10}\frac{10-t}{10}+mO(1+(-1+\frac{10-n}{10})(\frac{17000-C}{O}))\\&=&C\frac{10-n}{10}\frac{10-t}{10}+mO(1-(1-\frac{10-n}{10})(\frac{17000-C}{O}))\end{array}

Next, let us define the P value of a normal car (i.e. n=0).


We substitute P_{0,t} in P_{n,t}.


Here, note that C\frac{10-t}{10}+m(17000-C) is the residue value of the initial OPC rebate after t years. Remember the assumption that C<17000, therefore 17000-C is the amount of O deducted as part of the initial OPC rebate.

Notice that (1-\frac{10-n}{10})(C\frac{10-t}{10}+m(17000-C))>0, therefore P_{n,t} < P_{0,t}. That is, the PARF of an OPC car that is converted to normal is always less than that of an identical car that is normal from the beginning. Additionally, the PARF of the converted car at any time t will be further reduced by a factor of 1-\frac{10-n}{10}. That is, a factor of 1-\frac{10-n}{10} will permanently be lost from the PARF of the car, and this factor increases with n.

To visualize this, I have plotted the PARF of an OPC converted to normal with respect to n and t.

Graph of PARF against t at varying n.

I’ve used our car as an example in this diagram. “Converted at 0 year” refers to a normal car from the start. “Converted at 10 year” refers to an OPC that did not convert to normal. Notice the gap between the normal car and the OPC which converted to normal. This is what I refer to as unrecoverable loss. The pro-rated cash top up amount paid to LTA to convert an OPC to a normal car does not restore the car to its original value. So for people who are considering the convert, do keep this unrecoverable loss in mind.

Where does this bring us? Well, from the percentage of loss point of view, it makes sense to convert as early as possible. But from the absolute amount of loss point of view, later is better. So, to convert or not to convert? There is really no straightforward answer. But now, at least, we know that we must be prepared to bear some immediate losses from the minute of the conversion.



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