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Simple math formula for selecting your CPU speed!

L

Leo V

Diamond Member
Hi,

After spending much free time studying the plummeting AMD Thunderbird prices 🙂, I thought: Which speed (MHz) offers the best bang for the buck? Now I have what I think is a rational formula for ranking the offers. Remember, this is only applicable to the same model of CPU, so don't compare apples to oranges (no pun intended 😉):

Desirability = (MHz squared) / price

This means, Tbird 800MHz rates (800*800)/($130) = 4923 Tbird-marks, which makes it the most cost-effective Thunderbird to buy right now amongst all others.

How did I come up with this formula? I'm assuming that Moore's law (CPU speed doubles every 18 months) holds true, meaning that CPU speed increases exponentially; history shows this assumption to be accurate. For my purposes, let's define your CPU's usefulness as the ratio of your CPU's power to the fastest available. For a Tbird 800MHz, this would equal 800/1200 = 2/3 (since 1.2GHz Tbird has just been released).

I'm assuming that you'll upgrade your CPU once its usefulness (as I defined it) becomes too low. (ie, it becomes obsolete). Your CPU's usefulness decreases over time, in an exponential fashion (it'll halve every 18 months by Moore's law). My goal is to minimize your average yearly cost of CPU upgrades.

As it turns out, the integral of your CPU's usefulness over its lifespan (translation: the total amount of benefit you get out of it) is directly proportional to the MHz. (Also representable as a sum of an exponentially decreasing series).

A separate factor here is bang-for-the-buck, which equals simply (MHz)/price. Bang-for-the-buck always favors slower CPU's (for instance, Tbird 700 gives you 7MHz per dollar, while Tbird 900 only gives 5MHz per same.

I think that the ultimate measure thus equals:
(lifetime_usefulness) * (MHz_per_dollar)
which gives (MHz squared)/price.

Any remarks/arguments/comments?
 
Interesting concept, and may be usefull information, but its no decision maker.

There are still far to many factors that arent' taken into affect. Cache, platform, chipset, ram, etc.

bart
 
Obviously, other things like chipsets/cache/etc are a factor. However, once you've settled on a specific model, MHz is the only variable left. This is when my formula could help--you compare identical CPU's at different speeds.
 
Leo V, interesting thinking! Mayby you could walk me through the math a little more...

"(lifetime_usefulness) * (MHz_per_dollar)
which gives (MHz squared)/price."

from the T-bird example and using your formula above:

.66 * 6.15 = 4.06

how are you relating this to 4923 T-bird marks?
 
Moore's law states that cpu complexity doubles every 18 months, not speed.
Which makes your lil formula completely useless!@#$% 😉
 
let's not forget, moore's law is B$ imho, and all my upgrade decisions are solely based on how much money I have when I'm upgrading (usually once per year or 2). That's some good thinking though Leo, but I think you have too much time on your hands (prolly in math class, because it's usally boring hehe 😉)
 
What I do is put all the prices and speeds in a spreadsheet, make a graphic, and look where the line goes up the sharpest. The point right before the big increase should be the optimun price x performance.

Didn't really read the formula so I can't say anything about it.
 
shabby, my assumption is quite right, because exponential complexity growth implies exponential performance growth. All evidence so far shows exponential growth in CPU performance, for the time being, so the formula is firmly grounded.

Pretender: no, because doubling everything doesn't affect relative comparisons. (It just cancels out).

WoundedWallet, I believe this formula (MHz*MHz/price) pretty much gives you the sweet spot you described--not the top MHz/dollar, but rather the elbow on the curve.
 
Danlz, while I measured the usefulness at any given moment as a changing ratio, I concluded that the total usefulness over time is proportional to your original clock frequency. This is why you saw 800*800/130, instead of (800/1200)*800/130.

However, it makes no difference: dividing each CPU's rating by the same constant (1200 in this case) has no effect on the actual rankings.
 
Leo, looks super to me. I think Shabby is saying that (proprietary)complexity such as Intel's MMX, SSE, Streaming Media and Internet enhancement, and computer ID bs, does not add to useable CPU performance for the majority of us. Maybe an additional multiplying factor such as 0.9 * bird marks for Intel, or 0.5 * bird marks for vaporware?😀
 
I dont get why you need to squre the clock speed, can you go over that a little bit more in detail?
 
LXi, good question, as this is the heart of the formula. I encourage everyone to read this reply:

Conceivably, Joe could claim that MHz/price (ie bang-for-the-buck) is the ultimate factor--and buy a real slow CPU (like Duron 600 for only $44). The obvious problem is that he'll never enjoy a decent "usefulness" percentage--his CPU will almost be obsolete from the start. Thus Joe would find himself constantly buying more cheap CPU's (to avoid complete obsolesence), which would cost a LOT per year.

On the other extreme, Jack could claim that the most expensive CPU will give him the greatest edge (% usefulness) and last him the longest--and blow $500 on a 1.2GHz TBird which won't outlast the $270 1GHz Tbird by much. Jack wouldn't upgrade frequently, but each upgrade would cost him a ridiculous amount--again amounting to an unreasonably high (average) yearly cost.

The formula I present accounts for both of these problems: it determines the balance between both usefulness and bang-for-the-buck. Moreover, one which (all non-MHz specs being equal) promises you the best performance on a fixed average yearly CPU-upgrade budget.


Danlz: LOL, that's a very good point! Regarding CuMines vs. Tbirds, if .9 is the "average" performance factor for Intel/Tbird performance (sounds realistic to me), you could compare Intel vs. Tbird. (Guess who'd win hands-down 😉).
 
Ok a few more questions, does higher mark represent better performance/clock speed ratio, and lower mark represents higher %usefulness? I did some quick calculations, the Duron 600 will get a 8181 marks, while the 1GHz will get 3703. The lower clock speed will always get higher marks regardless if you square it or not.
 
LXi, I warned not to compare different models. Duron offers much less performance per-MHz. If you use a realistic .7-.75 Duron->Tbird performance conversion factor, you'd only get ~4000-4600 TBird-marks, which actually makes sense. Remember, apples and oranges.
 
Oh... so there is this convertion factor thing.

How about a Tbird 700? Which cost $102. Getting a score of 4803, that's still a lot higher than 1GHz's 3703. The 1.2GHz gets 2909 marks. What does this tell us? Lower mark=more %usefulness, and higher mark=better price/MHz ratio?
 
No, higher mark = best ultimate cost-effectiveness, in either case. Cost effeciveness means that if you spend K dollars (on average) each year on CPU upgrades, you'll get the most performance out of it.

Also, CPU's that are considered already on the edge of obsolescence should be naturally disregarded (obviously a Celeron 300MHz on sale for $2 would get a great score--but were you considering it in the first place?)
 
Not squaring the MHz would indeed favour slower chips (so long as their MHz/price ratio is high). This is Joe's example. He buys the Duron 600 as it wins (MHz/price) hands-down. In 1-2 months, the Duron is obsoleted (suppose), and Joe must buy another $45 CPU. By the year's end, Joe has blown some $450 on ten CPU upgrades, and he never enjoyed a half-decent usefulness percentage (like getting 60FPS in a recent game). This is an extreme, but you get my idea.

(contd in a minute...)
 
In Jack's example, his CPU may take 18 months (suppose) to be obsoleted, then he buys another $500 upgrade. This hobby comes out to $333/year, rather unpleasant despite the high usefulness % he initially enjoys.

(contd...)
 
Now consider Jim, who buys a Tbird/800 for $130, as recommended by my formula. Suppose it lasts (while at a high % usefulness, unattainable by the Duron 600) for the better part of 1 year before being upgraded. The hobby amounts to only $150/year, a fraction of Joe's or Jack's costs, and the performance is highly satisfactory at all times. 🙂

Do you now see how this works?
 
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