The Algebra of Units
Dimensional Analysis for Fun and Profit
The first few times I taught this class, I was surprised to find
students turning in homework solutions with no notion of what units they
were using --- seconds, cycles, instructions, programs, microseconds,
kilograms, furlongs per fortnight, etc. I realized
that there was no single course (except perhaps freshman Physics) in
which students were taught about units.
Students are often confused by all the "milli", "micro", "nano", "mega",
"µ" prefixes they see. Here's
table defining them. A more humorous (although still correct)
discussion is at the Free Online Dictionary of Computing.
In computer evaluation, the important units are typically
All other units will be based on these: for example, if the price
performance of a computer were measured in "MIPS per kilobuck", that
"million instructions per second per thousand dollars", i.e.
- machine cycles
106 * instruction/(second * 103 * dollar)
or equivalently 103 * instructions / (seconds * dollars).
One particular unit that often confuses people is the Hertz (abbreviated
"Hz"), a common unit of frequency. A Hertz is technically 1
divided by a second, but what people usually mean is 1 "cycle"
per second, whatever a "cycle" is. For example, a radio
station broadcasting at 93.9 MHz is producing a signal that oscillates
93,900,000 times per second, and a 300 MHz
processor has a clock that pulses 300,000,000 times per second.
The electrical side of the course uses different units:
Each of these, of course, can be modified by SI prefixes; the most
are "kilohms" (103 Ohms), "milliamps"
(10-3 Amperes), and "µsec" (10-6
- seconds, a unit of time
- Joules, a unit of energy (= 1 kg m2/sec2)
- Coulombs, a unit of charge (a certain, very large, number of
- Amperes, a unit of current (= 1 Coulomb/second)
- Volts, a unit of electrical potential (= 1 Joule/Coulomb)
- Ohms, a unit of resistance (= 1 Volt/Ampere)
- Watts, a unit of power (= 1 Joule/sec = 1 Volt * 1 Ampere)
Choosing appropriate units: what's "good"?
When I lived in Canada, somebody told me the joke
"Canada has the most efficient postal system in the world.
Where else can you send a letter for only a penny a day?"
[Canada Post at the time charged something like 40 cents to mail a
Whether we're trying to send a letter arriving as soon as possible for
as little money as possible, or buy a computer that gets the job done as
soon as possible for as little money as possible, we need a sensible way
to balance cost with time. Miles per gallon is a good way to measure a
car's fuel efficiency, because we want to travel a lot of miles and burn
as few gallons of gas as possible; doubling both means you've gone on a
longer trip, but your efficiency rating should be the same. By
contrast, cents per day is a bad way to measure a postal service's
efficiency, because you'd really like to decrease both the cost
and the delay, rather than being willing to increase one of
them as long as the other increases too.
Let's say we want large values of our measurement to indicate a "good"
computer, combining cost with speed. We choose a unit that increases as
the amount of work (measured in, say, instructions) increases, so we'd
better have instructions in the numerator; on the other hand, the more
seconds the computer takes, the worse it is on this measure, so
seconds had better be in the denominator. Dollars are likewise "bad",
so we put them in the denominator too. Thus we're forced to something
like "instructions / (dollars * seconds)". We might then try some
real-world examples -- say, a computer that executes 3000000
instructions in 0.1 seconds and costs 4000 dollars, coming to
7500 instructions per dollar-second -- and conclude that these numbers
were too large to handle conveniently, especially as computer
performance improves in the future. So we would perhaps decide to scale
the unit by a factor of a thousand, getting
7.5 * 103 instructions / (dollars * seconds)
which is in exactly the unit described above. (I called it "MIPS
per kilobuck" not only to illustrate the prefixes, but also because
people often use MIPS and kilobucks by themselves, so it's easy for
people to understand dividing one by the other.)
The Rules of Algebra
I did some algebraic manipulations in the previous paragraph that you
may not understand. The first rule of thumb is to
Treat each unit as though it were a different, unrelated
For example, 3x + 4x can be written more simply as 7x, but 3x + 4y can't
be simplified unless x and y are related somehow. Similarly, it makes
sense to add 3 instructions to 4 instructions, getting 7 instructions,
but adding 3 instructions to 4 dollars gives you no new insight (and in
fact is pretty meaningless).
This first rule of thumb gives rise to the more specific
Don't add, subtract, or compare two quantities unless they are
measured in exactly the same units
If they're in units that can be converted into one another,
like seconds and hours, then convert one into the other before adding or
2 hours + 15 seconds =
2 hours * (3600 seconds/hour) + 15 seconds =
7200 seconds + 15 seconds = 7215 seconds
Why am I justified in multiplying part of the equation by 3600
seconds/hour? Because 3600 seconds is one hour, so 3600
seconds divided by one hour equals 1, and I can multiply anything I want
by 1. (I could legally have multiplied by 1 hour/3600 seconds, but then
I would wind up with 1/1800 hour2/second, which
while technically correct is still not addable to 15 seconds.)
If, on the other hand, they're in units that cannot be
converted into one another, like seconds and instructions, then their
sum and difference are meaningless and you're probably doing something
wrong by even wanting to add or subtract them.
In a well-known Far Side cartoon, Albert Einstein is staring
puzzledly at a blackboard on which are the equations
"E=mc3", "E=mc4", etc.
successively crossed off, while a cleaning lady clears his desk and says
something like "That's better, got that all squared away.
Yessirree, all squared away."
It's a good cartoon, but anybody who knows his units (as Einstein
certainly did) couldn't possibly have believed
"E=mc3" or "E=mc4" for a
moment: the units on the two sides of the equals sign don't match.
Mass (the m in the equation) is measured in grams or
kilograms. Speed (e.g. the speed of light, c) is
measured in centimeters per second or meters per second. Energy is
measured in joules or ergs, which are in turn defined in terms of mass,
distance, and time units: in particular,
1 erg = 1 g * cm2 / (second2)
1 joule = 1 kg * m2 / (second2)
In other words, energy is measured in
g*cm2/sec2, exactly the same units as
mc2. But mc3 would have units of
g*cm3/sec3, which isn't equal to any
amount of energy for the same reason that a mile is not equal to
any amount of time, money, or mass.
Which is not to say that the equation "E=mc2"
is trivial. Saying that there is a fixed ratio between matter
and energy was not obvious before Einstein, nor was saying that the
ratio was the square of the speed of light. However, if there was to be
such a ratio, it would have to be the square of some
speed, just to make the units come out right.
You can multiply or divide any units
If you drive 35 miles in half an hour, your speed can be described as
(35 miles) / (0.5 hours) = 70 miles/hour.
We all know that miles/hour is a reasonable unit; so is
miles/gallon. What if you multiplied your speed by your gas mileage?
The result would be measured in miles2 / (hour * gallon), and
might be a very reasonable measure of the fuel efficiency of your car,
taking into account your desire to get somewhere quickly. Similarly,
the above example of MIPS per kilobuck is quite a plausible unit if you
want to measure how much computing speed you're getting for your money.
Decide what are reasonable units for the answer before you work it
If a question in the textbook asks "How long does Machine X take to run
Program Y?", the answer had better come out in seconds (or hours or
milliseconds or something like that). If you work through all the
arithmetic and algebra and come out with an answer in seconds per cycle,
or dollars per instruction, you've done something wrong.
Let units guide you in your algebra
If you've decided that the answer to your problem should come out in
seconds, and you have at your disposal a number of instructions, a
number of cycles per instruction, and a number of instructions per
second, the only way it makes sense to combine them is to
divide the number of instructions by the number of instructions per
(3.5 million instructions) / (5 million instructions/second)
= 0.7 seconds
after cancelling the instructions in numerator and denominator.
Again, we're treating units essentially like variables. (Note that the
number of cycles per instruction wasn't actually used in the calculation
If you guessed wrong, and multiplied the number of
instructions by the number of instructions per second, you'd wind up
with an answer measured in instructions2/second, not in
seconds, so you'd know you had done something wrong.
Many problems in the textbook ask you "How much faster is Machine X than
Machine Y?" or questions like that. By the above rules, you can't
answer this question until you've come up with some measure of "speed"
for both machines, measured in the same units. But once you've got that
(let's say, in MIPS, although that's a questionable way to measure
speed), how do you compare the two? My students in the past have done
one of two things:
- Subtract SpeedY from SpeedX, producing an
answer like "Machine X is 4.5 MIPS faster than Machine Y". This is
technically correct, but not particularly useful, especially when you
say it to your boss who doesn't know a MIPS from his elbow.
(Actually, all too many students have started with the wrong units and
wound up with answers like "Machine X is 8 seconds faster than Machine
Y", which is utterly meaningless! This sort of comparison
makes sense only if you're talking about a single task which one machine
performs in 8 seconds less than the other one does. For a different
task, or a larger or smaller version of the same task, it provides no
useful information at all.)
- Divide SpeedX by SpeedY. Since both speeds
are in exactly the same units, the units cancel and we're left with just
a number like 1.3, interpreted as "Machine X is 1.3 times faster than
Machine Y" or "Machine X is 30% faster than Machine Y". This sort of
answer is really much more helpful: if your speed figures are truly
representative of machine performance, it tells you that any
task you choose to run will finish 30% faster on Machine X than on