Navigating between oscilloscope-land (Volts) and RF-land (dBm)

Posted on August 18, 2023 by admin

Have you ever been annonyed by the dBm and V units when you try to measure output from your benchtop signal generator (or arbitrary waveform generator, or ARB) with a spectrum analyzer, or measure a RF signal (sine) generator with an oscilloscope?

TLDR

For $50\Omega$ systems only. The fastest way is to work with squared voltages in linear scale with the provided ‘magic’ scaling factor (10 for amplitude and 20 for rms) and not mess with factorization in the log (dB) or anti-log ($10^{\frac{\mathrm{dB}}{10}}$) process:

P[\mathrm{mW}]=10V_{pp}^2=20V_{rms}^2

\mathrm{dBmW}=\mathrm{dB}\left(10V_{pp}^2\right)=\mathrm{dB}\left(20V_{rms}^2\right)

This happens a lot during calibration. An oscilloscope needs a 500Mhz signal at $600mV_{pp}$ to test if the front-end rolls off too early (i.e. have a lower bandwidth than advertised), yet my \$12k RF signal generator (20Ghz sweeper) only does dBm, and in coarse increments of 0.1dBm.

Decibel is supposed to be a dimensionless quantity, aka just a ratio.
The confusion came from people in each field abusing notations by skipping the reference quantity suffix after dB. e.g. RF people write dBW and dBmW (a dimensioned quantity with loaded assumptions of a $50\Omega$ scaling) as dB and dBm (which would naturally read as a dimensionless ratio)
When RF people talk about decibels, especially when they abuse notations by not mentioning the reference (basis) quantity, you can be pretty sure that they are talking about ratios between power quantites (dBW), not voltage quantities (such as dBV).
There’s no such thing as a $20log_{10}$ decibel unit. It’s always $10log_{10}$ dimensionless. The factor of 2 came from the square (power of 2) factoring out of logarithms as a multiple of 2 when ratios of power are expressed as ratios of voltages.
Always work in power ratios to avoid the confusion. dB is dimensionless but often the unspoken rule is that it’s a ratio of power quantities. Routines or formula like mag2db really mean magnitude_squared_in_dB because it’s doing $10log_{10}(A^2)$.
The other advantage of actually squaring the term first before taking the log is that you do not have to check the input for negative values as the squaring operation made it absolute value (square). If you are doing a logarithm, an extra multiplication (over an addition) is insignicant in terms of computational burden.
$V_{rms}$ maps directly to power quantities, scaled by the impedance (which is typically $50\Omega$). Everything else involves an extra crest factor (which is $\sqrt{2}$ for sinusoids)

Continue reading →

An easy way to remember Euclid GCD algorithm: laying tiles

Posted on November 21, 2021 by admin

I was a little embarrassed that I’ve never came across Euclid’s GCD algorithm until years after I graduated with Math and Engineering degrees. The descriptions I’ve found online (especially Wikipedia) are convoluted and confusing, and the mechanical description of the algorithm does not shine light to any intuition. Then there comes proofs and abstract algebra properties that even after I followed the reasoning, I’d soon forget after I stop looking at it in a few weeks.

Just by staring at one simple description of the algorithm:

The classic algorithm for computing the GCD, known as Euclid’s algorithm, goes as follows: Let $m$ and $n$ be variables containing the two numbers, divide $m$ by $n$ . Save the divisor in $m$ , and save the remainder in $n$ . If $n$ is $0$ , then stop: $m$ contains the GCD. Otherwise repeat the process, starting with division of $m$ by $n$ .
K.N. King’s C++ Programming (Chapter 6 Exercise 2)

I reversed engineered one line of reasoning how one could come up with the Euclid GCD on his own. Turns out there is one twist/angle that most traditional explanations glossed over: quotients do NOT matter and WHY it does NOT matter!

It goes back to what GCD $(a,b)$ stands for: greatest common divisor. You are looking for the largest factor that simultaneously divides $a$ and $b$ evenly. If you rethink in terms of multiplication, you are finding the biggest tile size $c$ that can simultaneously cover $a$ and $b$ without gaps:

Find $c$ s.t. $a = pc$ and $b= qc$ where $p,q$ are integers.

Imagine you are a lazy and cheap contractor who have unlimited tile samples of all sizes to try before making a large order:

[Goal] you have to cover with two floors with potentially different sizes $a$ and $b$ gaplessly
[Constraint: divisible] your customer won’t let you cut (fractional) tiles because it’s ugly
[Constraint: common denominator] you can only order ONE tile size to take advantage of bulk discounts
[Objective: greatest] more tiles means more work to lay it, so you want to shop for the biggest tile size that does the job.

For simplicity of the argument, we also assume the divisor $b$ is the smaller number. If $b$ is the bigger number, the roles will reverse in the next round because the remainder from dividing by a larger number is the dividend itself, which becomes the divisor in the next round while the old divisor (the larger number) becomes the dividend.

For relatively prime pairs, the algorithm will eventually stop with a GCD of $1$ because $1$ divides all integers.

Here’s the analog of Euclid algorithm:

start out with 1 big tile covering the entire smaller floor $b$
see if we can cover $a$ without gaps using big tiles of size $b$
if yes, we are done. $b$ , the macro-tile is your GCD (perfect-sized tile).
if there are gaps (non-zero remainder), pick a tile-size that covers the ugly gap (the remainder becomes the divisor), then repeat the same process above over the last tile size $b$ (the divisor becomes the dividend) until there are no gaps (remainder become zero).

The algorithm is taking advantage of the fact the remainder is smaller than the last tile size (divisor) so we can rewrite the last tile size in multiples of the remainder (old gap) plus the new gap. By the time the new gap evenly divides the last tile (the big tile right before it), all numbers before that can be written as multiples of the new gap (i.e. the last gap evenly divides all the numbers in the chain all the way up to $b$ and $a$ ).

Let’s start with a simple case GCD $(a=50, b=15)$ that terminates early:

$50 (dividend) = [15+15+15 (divisor)] + 5 (remainder)$

We focus on the last tile $15$ and write it in terms of the old remainder $5$ :

$15 = (5+5+5) + 0$

Since the new remainder is $0$ , the old remainder $5$ divides the last tile $15$ . Since we are dividing $a$ in terms of the remainder $5$ , we have

\begin{aligned} b &= 15 \\ &= (5+5+5) \\ &= 3r \\ \\ a &= 50 \\ & = (15 + 15 + 15) + 5\\ & = 3b + r\\ &= (5+5+5) + (5+5+5) + (5+5+5) + 5\\ &= 3(3r)+r \\ & = 10r \end{aligned}

So both $a$ and $b$ can be written in terms of integer multiples of $r$ right before the algorithm stops.

GCD $(50, 18)$ takes a few more steps:

\begin{aligned} b &= 18\\ a &= 50\\ &= (18 + 18) + 14\\ &=2b + r\\ r &= 14 \\ \\ b_1 &= r = 14\\ a_1 &= b = 18\\ &= (14) + 4\\ &= b_1 + r_1\\ r_1 &= 4\\ \\ b_2 &=r_1 = 4\\ a_2 &= b_1 = 14\\ &= (4+4+4)+2\\ &= 3b_2 + r_2\\ r_2 &= 2\\ \\ b_3 &= r_2 = 2\\ a_3 &= b_2 = 4\\ &=(2+2) + 0\\ r_3 &= 0\\ \end{aligned}

The algorithms stops at $r_3=0$ which means we successfully divided the last tile $a_3$ with $r_2$ (logistically stored in $b_3$ ) with no gaps left. $r_2$ (or $b_3$ is therefore the greatest common factor (divisor) that are shared by all the numbers involved all the way up to $a$ and $b$ .

This is the beauty of Euclid’s GCD algorithm: once the gap (remainder) is sliced small enough to evenly divide the tile (divisor) right before it, every term before it be can written in terms of integer multiples of the last number that divides the last tile without gap (no more remainder).

In our example, $4, 14, 18, 50$ can be written as a multiple of $2$ , aka $r_2$ because integer multiples of numbers that are already integer multiples of $r_2$ can be written in terms of a larger integer multiple of $r_2$ . This is why quotients do not matter in the Euclid GCD algorithm:

\begin{aligned} b_2 &= 2r_2\\ 4 &= (2+2) \\ \\ b_1 &= 3b_2 + r_2\\ &= 3(2r_2) + r_2\\ & = 7r_2\\ 14 &= [(4+4+4)+2] \\ &= [(2+2)+(2+2)+(2+2)+2] \\ \\ b &= b_1 + r_1\\ &=7r_2 + 2r_2\\ &=9r_2\\ 18 &= 14 + 4 \\ &= [(4+4+4)+2] + 4 \\ &= [(2+2)+(2+2)+(2+2)+2]+[(2+2)] \\ \\ a &= 2b + r\\ &=2(9r_2) + 7r_2\\ &= 25r_2\\ 50 & = 18 + 18 + 14 \\ &= [(2+2)+(2+2)+(2+2)+2]+[(2+2)] \cdots \\ &+ [(2+2)+(2+2)+(2+2)+2]+[(2+2)] \cdots \\ &+ [(2+2)+(2+2)+(2+2)+2] \end {aligned}

The spirit of the algorithm is that every time we see a gap, we fill it with a small tile that’s exactly the gap size, and attempt to replace the last tile by covering it in terms of the new smaller tile (that was used to fill the gap). We are recursively slicing the last tile with the gaps produced until there are no gaps left. Once you find the GCD tile, you can replace all the bigger tiles before in terms of it with no gaps.

There’s an interesting perspective of using the Euclid GCD algorithm to solve Diophantine Equations (integer written as a linear combination of integers with integer coefficients): turns if $x=1$ in $c=ax-by$ is the same as writing $a = by+r$ . We can flip the sign of $y$ by saying substituting $y'=-y$ .

In summary,

Euclid GCD’s algorithm is sub-dividing the last (or any one) large tile (last divisor) with the gap (last remainder) until there are no gaps left. Then any bigger tiles up the chain can be expressed as an integer multiple of the last piece (the smallest piece searched so far) that fills the last gap.

The first large tile is of course the smaller of the two numbers involved in the gcd calculation. The larger of the two number is the floor to be covered.

The other observation is that when you see the remainder being 1, you already know the two numbers are relatively prime. The last step to get the remainder to 0 is redundant (because it’s a guaranteed last step).

The other observation is that subdividing the last big tile (divisor) with the last gap (remainder) till it’s evenly divided chains up, because everything is an integer multiple of the smallest piece that evenly divides the gap (remainder) and the last big tile (divisor), therefore the algorithm is recursive.

\gcd(r_1,r_2) = \gcd(r_2, r_3) = ... = \gcd(r_{n-1}, r_n)

But this kind of recursion is not mandatory. It’s a tail recursion* that can be unrolled into iterations (looping), either by the compiler or the programmer, which is what we initially did.

Demystifying comparisons between eBay managed payments and the old way Hint: it's still the old final value fees (before cap) + payment processing fees (analogous to Paypal)

Posted on October 18, 2020 by admin

Most often when we sell on eBay, we receive Paypal payments, so we ended up paying eBay’s commissions (also called the final value fee, FVF for short) and Paypal processing fees as a percentage of the sale.

Now eBay pushes out Managed Payments (MP) which combines payment processing fees with the eBay commissions (FVF) because eBay now manages the payment gateway. The rest is every time you sell something, you get a payout (sales – fees) deposited directly to your bank (it used to be collecting paypal balances then withdraw it).

They have a different calculation formula which they claimed the sellers are better off in most cases, but should we take eBay’s word for it? Regardless of whether you are enrolled in managed payments, the fee percentage for each sale depends on:

eBay store subscription level
category

It’s impractical to do the side by side fee structure comparison to see when we are better off for each sale, plus we cannot easily switch between the two plans.

It’d be very helpful if we can put the managed payment fee structure on the same form as the conventional eBay+Paypal fee structure, and figure out under what conditions we are better off or worse off with managed payments.

Initially I was ready to do the derivations to put both plans on the same scale, but I spotted that managed payment (combined) percentage is simply the vanilla (non-managed) eBay category final value fee + 2.35% payment processing fees! That’s how they’ve calculated the combined managed payments percentage!

This makes life a lot easier. Since I can factor out the 2.35% that applies to the whole sum (which also include shipping and sales tax) regardless of the fee cap, this works exactly the same as Paypal (which charges 2.9%) and we are getting a 0.55% discount in payment processing fees.

For managed payments, since we’ve already separated out the payment processing fees, the fee cap applies to the vanilla final value fee portion which is equivalent to the old eBay final value fee. Keep that in mind.

The part that has changed is the fee cap. The old way caps the commissions/FVF directly regardless of product category, yet the new way (managed payments) caps the sale price that are charged commissions. This means the fee cap goes up or down a little depending on the final value fee percentage class applying to your sale.

eBay set the (commissioned) sale price cap to closely match the realized commission cap in the old way, for example non-store subscriber who will have their raw final value fees capped at $750 will see the same cap for the most common 10% categories (12.35%-2.35% = 10%) because the (commissioned) sale price cap is $7500, which $7500*10%=$750.

Big industrial equipment also have the same cap regardless because $15,000*2% = $300.

For non-store subscribers, 10% is the anchor (iso-fee-cap) category. So you are a little worse of with books (price cap +2%) and better off with musical instruments (price cap -6.5%).

For store subscribers, you get a bit more break over the fee-cap (lower) because your final value fee percentage is lower than the anchor (likely they chose a breakeven point at 10% when determining the sales price to cap final value fee over. Just easy numbers, not rocket science):

	Managed	Conventional	Discount*
Common (9.15%)	$2,500*9.15% = $228.75	$350 (Small stores) $250 (Big stores)	$121.25 (Small stores) $21.25 (Big stores)
Heavy gears (1.5%)	$15,000*1.5% = $225	$250	$25
Books (12%)	$2,500*12% = $300	$350 (Small stores) $250 (Big stores)	$50 (Small stores) -$50 (Big stores worse off)
Guitars (3.5%)	$2,500*3.5% = $87.5	$350 (Small stores) $250 (Big stores)	$262.5 (Small stores) $162.5 (Big stores)

I call Basic/Premium ‘small-stores’ and Anchor/Enterprise ‘big-stores’.

So here are the observations, which is all you need to know:

Under managed payments, small-stores gets a better deal than big-stores, because the advantage of the $100 lower fee cap with big stores are eliminated with variable fee caps.
The breakeven line for small stores ($350 cap) is 14% fees ($350/$2500), which the highest category is 12% so far. This means small-stores are ALWAYS better off switching to managed payments.
The breakeven line for big stores ($250 cap) is 10% fees ($250/$2500), which the only category above that right now is books (12%). Big-stores selling BOOKS are worse off with managed payments.

So in other words,

everybody is better off with managed payments (fee-wise) EXCEPT big-stores selling books
under managed payments, there’s no final value fee advantage for being a big-store

* Remember you got 0.55% discount over the payment processing portion of the fees too and is not shown here since we’re just talking about savings in vanilla final value fees.

As far as books (12%) is concerned, if you are a big-store owner, your raw commissions cap raised from $250 to $300 because $300 = $2500 * 12%. But if you factor the 0.55% discount, if the sale price is $x,

Since the raw commissions stops at $300 ($2500 * 12%), the additional $50 cap increase starts to get offset (and turn out positive as the processing fee discounts outweighs the commissions cap increase) when the payment processing discounts ABOVE $2500 covers all of it:

So the range of sale price x which the only setup (books for big-store sellers) can do worse is

There are some ambiguities (technically incorrect documentation) on eBay’s website which implied vanilla final value fees (portion) are charged for sales tax. I made a sale and checked the numbers and it’s not true. Only the 2.35% payment processing fee portion is charged against the sales tax (like paypal for handling extra money), the category final value fee (in my case 9.15%) is not applied to the sales tax.

They actually meant “the 2.35% payment processing fee portion” when they said “managed payment final value fees”. This is also part of the reason why I wrote this article, because they do not use the language that conceptually separate the two portion of the combined final value fees (vanilla final value fee + payment processing fee) in managed payments, thinking that they are simplifying the math for sellers, without realizing if the two concepts really fused into one, they’ll be shortchanging sellers over sales tax.

The $0.3 fixed per-transaction fee applies to both managed payments and the conventional way (Paypal also charge +$0.3 fixed per-transaction), so nothing has changed.

Mathematics of calculating overheads to charge

Posted on October 17, 2020 by admin

It’s very common to have a percentage fee (called tax rate below) slapped on our earnings:

Commissions (e.g. eBay final value fees)
Sales Tax
Incremental income taxable within the same bracket.

Many adults cannot directly figure out how much EXTRA do they need to earn to get the target amount of that are rightfully theirs (after fees).

They are missing out these handy intuitions:

If your EXTRA earnings are taxed at 40%, not wasting an extra $100 is almost as good as earning $166.67!
You always undercharge the fee overhead if you just multiply your target amount with it (intuitive but WRONG). The amount you get shortchanged goes up like a rocket for rates past 10%!

All you need is grade school math to figure it out, but the results are highly non-linear and are not easy to remember. This means the tutorial below makes an excellent discussion material to motivate mathematics education in grade school!

If you want to copy or use the materials, just remember to reference “Rambling Nerd with a Plan” blog page or “Humgar LLC”. You don’t have to ask and you are welcome. I certainly appreciate comments about this tutorial.

Before I start, hints for working with percentages:

divide percentages by 100. Think multipliers/factors in your head.
a factor of $1$ represents the full amount
division by $x$ is multiplication by reciprocal $1/x$
preferably work in terms of shrink/expand factors (multiplications only) over the FULL amount, rather than working out the excesses (additions/subtractions)

Assigning intuitive names are very important for interpreting algebraic expressions with familiar life concepts!

Let’s define these variables first:

$T$ is the target price you want to get AFTER the fees
$D$ is the damage you should charge
$r$ is the rate (fee/tax percentages)
$E$ is the excess that goes to the fee/tax collector

These derived variables will be discussed in the body of the tutorial

$s$ is the shrinkage (multiplier), where
$s\equiv(1-r)$
$m$ is the magnifier (multiplier),
$m\equiv\frac{1}{s}$
$c$ is the compensation (multiplier) for billing,
$c=m-1$

If somebody pays you the total damage $D$ before tax/fees (at rate $r$ ), you are home free with $T$ after subtracting the fees $D \times r$ from $D$

$D - rD = (1-r)D \equiv sD = T$

In other words, you the FULL damages gets shrunk by a factor $s \equiv (1-r)$ which becomes the target amount $T$ you get to keep for yourself.

Don’t be scared by the algebra! It’s just the other way of saying, if your tax rate is 30%, it means you get to keep 70% of your income, because 100%-30% = 70%, so the shrinkage $s$ is $0.7$ .

Since you have the target price $T$ in mind and want to undo the shrinkage (multiplication) to calculate the full amount you should charge $D$ , you divide the target price by the $T$ by the same factor to magnify it back:

$D = \frac{T}{s} = \frac{T}{1-r}$

It is NEVER helpful to directly multiply the fee/tax percentage with the target price $T$ like what most people would intuitively do. We can think of the division above as multiplying by the reciprocal of the loss factor $s$ or $(1-r)$ , which we will call $m$ the magnifier:

$m = \frac{1}{s} = \frac{1}{1-r}$

An alternative view is to break the damage $D$ into target price $T$ along with excess $E$ to bill the customers for the fees/taxes (which goes to the collector):

$D = T + E$

My preferred way to calculate the excess is simply compute $D$ first by finding calculating the magnifier $m$ , or simply dividing the target price $T$ by the shrinkage $s$

$D = mT = \frac{T}{s} = \frac{T}{1-r}$

then subtract the target price $T$ from the damage $D$ to get the excess:

$E \equiv D-T$

which you can write it in terms of target price $T$ which you know so you don’t have to work out the total damages $D$ first:

$E \equiv D - T = mT - T = T(m-1)$

We can define compensation (multiplier) $c$ as how much you should multiply the target price $T$ with:

$c \equiv m-1$

to recover your fee/tax costs $E$ :

$E = cT$

e.g. if you need to charge the customer 1.43 times to breakeven after fees (given a 30% tax/fees rate), the overhead ‘item’ on the bill should be 0.43 (43%) of the target price. For a $100 item, you’ll need to bill your customer $43 more just to cover a 30% fees.

In summary, we have

$P \equiv T + E = (1+c)T$

with $c$ that can be figured out quickly as

$c = \frac{r}{s}$

where $r$ and $s$ are both bounded between 0 and 1 which adds up to 1:

$r + s = 1$

Take an example of 30% tax rate ( $r=0.3$ ), the shrink is 70% ( $s=0.7$ ), because the factors are complements to each other which adds up to 1.

$0.3 + 0.7 = 1$

If you want to write it all in terms of shrinkage $s$ :

$c = \frac{1}{s}-1 = \frac{1-s}{s}$

Since $1-s = r$ , we can rewrite the compensation (factor) in these complement terms $(r,s)$ , which I think it’s the easiest way to remember:

$c = \frac{r}{s}$

or simply the rate $r$ divided by shrinkage $s$ ,

You can also write it in terms of fees/tax rate $r$ in full:

$c = \frac{r}{1-r}$

e.g. if you have a 30% tax rate , you can find the compensation (factor) $c$ by calculating $3/7$ , because it’s simply 30% over 70%.

To estimate the overhead charge to offset the fee percentage, or compensation $c$ by multiplying with the target price $T$ :

WRONG way	$r$
CORRECT way	$\frac{r}{1-r}$

The gap (amount you get shortchanged by computing compensation $c$ wrong) is:

$g \equiv c-r = \frac{r}{1-r}-r = \frac{r^2}{1-r}$

The numbers look close enough (for the intuitive but wrong way) when it’s like around 10%, but your loss (overheads not charged) shoots up (non-linearly) as the fee/tax percentage goes up beyond 10%!

Fee percentage $r$	Correct overhead % to charge $c = \frac{r}{1-r} = \frac{r}{s}$	Undercharged $\frac{r^2}{1-r}$
5%	5.26%	0.26%
10%	11.11%	1.1%
20%	25%	5%
30%	42.86%	12.86%
40%	66.67%	26.67%
50%	100%	50%

Intuitively, you need to charge DOUBLE the target price if half of it gets taxed to breakeven, NOT adding half of the target price as the overhead! The FULL target price itself is the right overhead to charge for a 50% tax rate!

Since the numerator $r$ is bounded because $0 \leq r < 1$ , the denominator $(1-r)$ can blow up (the overhead needed to charge your customer) as it gets small (i.e. high tax/fee percentage). Here’s a plot of

$\frac{1}{1-r}$

e.g. A 66.6% imports tax rate means giving two cars to the government when you import one!

To push it to the extreme, 90% tax rate would mean 900% overhead, which means if you buy one car, you give the government 9 cars, assuming they don’t tax themselves. Your estimates are off by 10 times if you use the wrong method to compute the excess $E$ making up for the tax that gets charged.

If people can think in terms of the excess you will need to earn EXTRA to offset the tax rate, you will think twice about letting the government dip in another 2% or so. That’s why welfare-state through taxes can be highly counter-productive when it get past certain point.

Advanced analysis for the intrigued

Just for fun, let’s investigate the impractical case (communist tyrants) when tax rates gets to 100% and above ( $r \geq 1$ ):

Fee percentage $r$	Correct overhead % to charge $c = \frac{r}{1-r} = \frac{r}{s}$	Undercharged $\frac{r^2}{1-r}$
$\vdots$	$\vdot$	$\vdots$
90%	900%	810%
100%	$\infty$	$\infty$
200%	-200%	-400%
300%	-150%	-450%
$\vdots$	$\vdot$	$\vdots$
$\infty$	-100%	– $\infty$

When charged 100% tax rate, the government seize any gains from your production. There’s never enough overhead you can charge your customer to breakeven.

The more perverted case is what happens when you go above 100%, say at 200% tax rate? When you sell $100 worth of stuff, you pay the government $200 to do so. You are paying a $100 penalty to work/produce $100 for the government. Twisted!

Then how do we explain the negative numbers in the compensation (factor) $c$ ? In reality, it means you ALWAYS make a loss no matter what. There’s no way you can bill your customer to make up for the loss.

In the unrealistic (unicorn) case, if the benevolent crazy dictator is willing and able to compensate your losses (at the sales tax rate), aka giving you money back every time you lose money,

$P = (1+c)T$

means the the magnifier $m \equiv (1+c) < 0$ for sure when $r \ge 1$ . It’s illustrated in the graph as a deadzone

$-1 < \frac{r}{1-r} < 0$

which you can verify the same by solving this inequality

$-1 < \frac{r}{1-r} \leq 0$

and realize you get $r-1 \ge r$ when $(1-r) \leq 0$ , which boils down to saying $-1 \ge 0$ which is a contradiction.

That means to achieve target earnings $T$ under tax rates above 100%, you should charge the customer negative amounts (i.e. giving away goods + money) in order to get the negative taxes (subsidy).

e.g. At 300% tax rate ( $r=3$ ), it your sticker price (also target price) is $100, you’ll need to GIVE the customer $50 so the government will triple match your loss as a subsidy (which is $150), which is enough to cover the $50 giveaway (to the customer) plus the $100 so you’ll breakeven.

Having negative tax (subsidy) is a perverse incentive. Forget about maintaining a target earning (same as the original sticker price for the goods you are selling), just lose (give away) money as fast as you can to suck up as much subsidy as you can before the system collapse.

I still fondly recall my economics 301 (intermediate microeconomics professors) James Montgomery saying that “if you can graph it, it can possibly happen” in the lecture.

Even more sophisticated math for the geeks

Remember from above, the magnification factor $m$ can be rewritten as the reciprocal of shrinkage factor $s$ ?

$m = \frac{1}{s} = \frac{1}{1-r}$

which reminds me of geometric series

$\frac{1}{1-r} = 1 + r + r^2 + r^3 + \cdots}$

given $|r| < 1$ ? For non-perverted cases, we are considering $0 \leq r < 1$ , which is well within the domain.

This gives an idea of approximating the compensation factor $c$ :

$m \equiv 1+c = 1 + r + r^2 + r^3 + \cdots$

which after taking $1$ off both sides:

$c = r + r^2 + r^3 + \cdots$

This explains exactly why people are wrong to just assume $c$ is just $r$ , because their false intuition kept ONLY the first order term in the power series approximation!

The gap (how much they are off with the WRONG way), is

$g \equiv c-r = \frac{r^2}{1-r} = r^2 + r^3 + \cdots$

So while $0.1^2$ dies fast enough (0.01) by 2nd order term, $0.9^2=0.81$ , $0.9^3 = 0.729$ … are still very significant up to say 6th order term which makes it off by a factor of $8.1$ .

Yes, you are off by 1% when the tax rate is 10%, but you can be wrong by a factor of $8.1$ times if you just multiplied the tax rate $r$ with the target price $T$ to estimate the amount of excess $E$ to collect to make up for the fees!

Summary

The total damage (bill) you should charge your customer can be calculated by dividing the target amount (of money you want to home free with) by the shrinkage, which is (1 – tax rate). e.g. 30% tax rate means 70% shrinkage. Divide the target by 0.7 to get the final bill
A shortcut to calculate how much you should put in (by multiplying with the target price) for an charge item on the bill to make up for the fees is (tax rate)/(shrinkage) e.g. 30% tax rate means 70% shrinkage. Multiply the target by 30/70 (or 3/7, which is 42%) to get the excess which covers the fee.
Do NOT attempt to use the rule of thumb to calculate the excess by just multiplying with the tax rate unless it’s 10% or less. The ‘rule of thumb’ assumed $r^2$ (and thus the higher powers) are insignificant, which is nowhere true for 0.1 and above based on this graph!

Rambling Nerd with a Plan

Hoi Wong's blog

Category Archives: Layman Mathematics

Navigating between oscilloscope-land (Volts) and RF-land (dBm)

TLDR

An easy way to remember Euclid GCD algorithm: laying tiles

Demystifying comparisons between eBay managed payments and the old way Hint: it's still the old final value fees (before cap) + payment processing fees (analogous to Paypal)

Mathematics of calculating overheads to charge