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Scheduling decisions to be made

blitz6804 — Tue, 20 Oct 2009 00:19:17 GMT

Opinion question for you all:

Would you rather have one class a day on Monday, Wednesday, Thursday, and Friday, or have one each on Wednesday and Friday and two on Thursday?

The latter is nice for four-day weekends, the former is nice in that its only one class a day.

Please comment your opinion here.

Of Cool 'n' Quiet and Its Effects

blitz6804 — Thu, 16 Oct 2008 02:04:42 GMT

People often ask why I seem to be on a personal crusade about Cool 'n' Quiet ["CnQ"]. The conventional modus operandi for overclockers is to turn off CnQ, and then start cranking up the bus speed. I wonder why people take this first step. The engineers at AMD, similar to their counterparts at Intel, designed CnQ to make a processor use less electricity when a user does not require the entirety of the processor's potential. I will discuss Enhanced Intel SpeedStep Technology ["EIST"] later on here.

To me, the benefits of this technology are self-evident. Most people, the exception here being folders, will not be using the entirety of their processing capacity all the time. The rest of us however, since we do not game with our computers our every waking hour, are likely using less than half the processing power. I am sure there are those on OCN who do use their computers solely for gaming: when it is not gaming, the computer is not on. The question then becomes, how do these users access OCN? Do they use a lesser computer?

I, personally, would rather not be bothered with two different computers if I did not need to. I find myself hard-pressed to believe that Internet Explorer, Firefox, Chrome, Safari, Navigator, or any other browser for that matter, requires the use of a two to three gigahertz processor. I find it hard to believe that any of these browsers requires much more than a one-gigahertz processor. This is why I do not use my three-and-a-quarter gigahertz processor. Instead, I use a CnQ-like program � more on that later � to reduce the processor's multiplier and voltage when the load goes below a certain boundary. This is dissimilar from Intel's EIST. EIST will lower the voltage and then lowers the multiplier or Front-Side Bus ["FSB"] or both.

Both systems have their positives and negatives. The advantage of CnQ is that it is predictable. There are three power states on most chipsets: low, middle, and high. The disadvantage of course is that it is a disjoint, non-linear application. Once you cross the bright-line threshold, you are increasing your tier. The difference between a 1799 MHz load and a 1801 MHz is insignificant to say the least, but the latter will set you back another 100 mV. EIST replaces the non-linear application with a continuum. This is similar to the PowerNow technology of older AMD laptops. Rather than adjust only the multiplier, the operating system can also vary the FSB. Using varying combinations of FSB and multiplier permit literally almost any speed the computer would need on the fly. The disadvantage here of course is that it is hard to predict where the processor will be at any given time. Since the FSB is changing real time, your RAM speed is also changing, which can lead to unusual behavior at times. Voltage is also on a similar continuum. The advantage here is that you are sure you are only giving the processor the voltage and speed it needs to do its work without a large excess.

CnQ is not immune from the RAM speed variable since the AM2 era. Unlike socket 939 processors, which adjust only the multiplier and voltage, socket AM2 K8's � that is, dual-core processors � also adjust the RAM divider. I cannot speak for socket AM2 K10's � that is, the tri- and quad-core processors � as I have not yet had the chance to play with one. More on the divider effect later.

As to why overclockers usually turn off CnQ: not all motherboard chipsets can handle it properly. While many chipsets will permit you to overclock the HyperTransport Bus ["HTT"] with CnQ enabled, many will not permit you to change the multiplier nor the voltage. In my personal experience, my LANParty DK 790FX-M2RS will permit me to change voltage while I have CnQ enabled, the multiplier must remain fourteen. Despite the setting I give it in the BIOS, Windows reads the multiplier as fourteen. My older nForce3 chipset, on the other hand, rather than ignoring the BIOS settings ignored CnQ. Setting the multiplier or voltage explicitly disabled CnQ for that function. While this is not a deal-breaker for all people, as many prefer to keep the stock multiplier, it will interfere with those who want to decrease their multiplier for a higher HTT or those with Black Edition processors.

What is a user to do? Enter CrystalCPUID. CrystalCPUID has what its programmer calls Crystal-n-Quiet ["C-n-Q"]. C-n-Q permits you to use Windows code to emulate CnQ. The advantage to this is that I can use C-n-Q regardless of the setting of CnQ in the BIOS. The power of this is obvious from the previous paragraph: I can use C-n-Q to reduce power consumption, but still run a non-stock multiplier or voltage. In my particular instance, that is thirteen and 1.360 V respectively. When I tried running with CnQ enabled, the computer would not boot. As previously discussed, it was trying to run the computer at three-and-a-half gigahertz rather than the three-and-a-quarter I told it to run. The other advantage of C-n-Q to CnQ is that it is customizable. Rather than setting the vague percentage limits in Windows Vista or the Boolean switch of Windows XP, I can set each of the power states exactly how I want them.
For those of you who have XP, your power choices are essentially "Always On," which locks the multiplier and RAM to their BIOS-set values, and "Minimal Power Management," which enables CnQ. The other options are some blends of the first two containing different elements not pertinent to CnQ, such as monitor timeouts et cetera. In Vista, you have a "Processor Power Management" setting in your Power Plans. There are options for minimum state and maximum state available for both wall-power and batter-power states. Obviously, if you are not using a laptop or a computer without a battery backup of some flavor you will not have the battery option. You can set the minimum state to any number between 1 and 100%, which determines the slowest speed you permit your processor to clock. The maximum state may be any number between 1 and 100% that is greater than or equal to the minimum state. When both numbers match, it disables CnQ and locks the multiplier and voltage in the position that best matches that percentage. (Recall here that EIST will change the FSB as well to get closer to the percentage than CnQ can do.) Setting both numbers to 100% results in the "Always On" of XP, that is, locking the multiplier and voltage exactly what they are stated in the BIOS.

With my Athlon64 X2 4400+ Toledo, my power states were 5x / 1.10 V, 9x / 1.25 V, and 11x / 1.35 V. Similarly, my Athlon64 X2 5400+ Black Edition Brisbane has 5x / 1.10 V, 9x / 1.25 V, and 14x / 1.315 V at stock. With CnQ, the book is closed. With C-n-Q however, I have been able to change the last one to 13x / 1.360 V. Unlike with CnQ, which uses some strange voodoo to determine what state to be in, I can set the percentage changeovers myself. When I ran stock, the computer seemed to favor the middle option, that is, 1800 MHz at 1.25 V. After my configuration of C-n-Q, the computer favors the lowest option, 1250 MHz at 1.10 V. Further, the advantage of C-n-Q over CnQ is that you can explicitly lock the processor to any state you want, be it low, middle, or high. You do not fiddle with the Vista percentages at all either, in fact, since CnQ is disabled in the BIOS, the percentages do not even appear.

Here is the breakdown:
Low: 5x250 = 1250 @ 1.136 V; used from 0-1125 MHz, where it kicks to medium
Med: 9x250 = 2250 @ 1.280 V; used from 1125-2138 MHz, where it kicks to high or drops to high
High: 13x250 = 3250 @ 1.360 V; used from 2138-3250 MHz, where it drops to medium

Why bother? Simple, I am saving money. As I have said, I am often in the lowest power state. Why should I use more wattage than is needed? According to American Power Conversion PowerChute, I use 311 W during normal load with C-n-Q turned on. If, however, I set the processor to the maximum setting, akin to disabling C-n-Q and thus acting like the average overclocker, I use 346 W under the same load. This is a minute difference to be sure. For sake of argument, let us investigate the matter further. Unfortunately, the computer is not always at idle, and to consider only the 35 W difference as shown would be dishonest.

My computer is on, according to Everest Ultimate, approximately forty-seven percent of the time. From my rudimentary calculations of my average usage patterns, my computer is in the lowest power state about seventy percent of the time, the middle about twelve percent of the time, and the maximum state about eighteen percent of the time. To find out how much money I am saving, we need a weighted average to determine my average power draw. A weighted average is of course the sum of possible values multiplied by their probability of occurrence.

I used various programs to chew up processing cycles to give me the power states I wanted and typical loads for the same. Here is my data:

Low power state, 375 MHz usage: 311 W
Middle power state, 1575 MHz usage: 346 W
High power state, 3150 MHz usage: 380 W

Thus, we can use a weighted average to determine my average power consumption with C-n-Q enabled:

(70/100) * 311 W + (12/100) * 346 W + (18/100) * 380 W = 327.62 W

Thus, the average power consumption while using C-n-Q is 327.62 W. Then I recorded the same speeds but with C-n-Q disabled. My data is as follows:

375 MHz usage: 346 W
1575 MHz usage: 354 W
3150 MHz usage: 380 W

Again, we use a weighted average to compute power consumption:

(70/100) * 346 W + (12/100) * 354 W + (18/100) * 380 W = 353.08 W

With C-n-Q turned off, I bring in 353.08 W. The difference between C-n-Q disabled and C-n-Q enabled is 25.46 W, proving my earlier comment about the dishonesty of assuming that what is true for idle is true for load.
(Continued in the comments.)

Athlon64 RAM Speed Computations

blitz6804 — Fri, 03 Oct 2008 04:50:31 GMT

I am not a blogger by nature; my personal feelings tend to stay just that. However, there are things from time to time that annoy me enough to justify a rant.

This is one of them.

As we all know: Athlon64 processors, whether they are Socket 754 or AM2+, use integrated memory controllers. While this permits a lower latency and a higher bandwidth than the competition, computing the RAM's actual speed might leave some Dazed and Confused. As we all know or can figure out, with an Intel, RAM is a simple matter. If the FSB is 333 MHz by default, and you are running it at 400 MHz, RAM set to DDR2 667 would be running at 400 MHz, which is DDR2 800 speeds. What about when you have that same scenario but with your RAM set to a DDR2 533 divider? Simple: 266*(400/333) = 319.5 = DDR2 639. Why am I talking about Intel when you came here for AMD? I am building, I'll get there.

Athlon64 processors, as previously mentioned, have an integrated memory controller. The side effect of course is that the RAM is no longer dependent solely on the RAM divider and the HTT (HTT:AMD::FSB:Intel), but also, the total speed of the Athlon64 in question. After real-world observations of three Skt 939 systems, my Skt AM2 system, and my tenure in the Socket 939 Appreciation Club and Knowledgebase, I have discovered the formula. Looks like my degree came in handy for something! Now, this is not a know-all, catch-all formula. I have seen it mess up on at least one occasion. (Exempli gratia infra.) Sure, everyone else probably knows this by now, but after retyping and re-posting it about five times in the last week, I have decided to just write it here so I have a simple link in future.

The general formulae are:

Eq0: HTT_OC*Multi = CPU

Eq1: (HTT_OC/HTT_Stock)*Divider = RAM'

Eq2: CPU/RAM' = Divider'

Eq3: Ceiling(Divider') = Divider

Eq4: CPU/Divider = RAM

The astute will notice that Eq1 is identical as to Intel. The astute will notice further that when dealing with Athlon64s, the HTT_Stock is ALWAYS 200 MHz. I do not care if you have a Barton or an Toliman, the stock reference clock is always 200 MHz. What do we do with them? I guess it might be better to give you some numbers. Consider the following hypothetical system:

In a Newcastle 2800+, the CPU multiplier is 9x, the RAM is on a DDR 333 divider (200:166), and the HTT is 250. We see then:

HTT_OC*Multi = 250*9 = 2250 = CPU

(HTT_OC/HTT_Stock)*Divider = (250/200)*166 = 207.5 = RAM'

CPU/RAM' = 2250/207.5 = 10.84 Divider'

Ceiling(Divider') = Ceiling(10.84) = 11 = Divider

CPU/Divider = 2250/11 = 204.54 = RAM

That is, the RAM in this system is running at DDR 409, since DDR stands for Double-Data Rate, and thus, is twice the speed of the RAM's clock. (Commonly referred to "dual pumped" since the RAM is doing one action on the "up" and one action on the "down." AMD CPUs are also dual pumped. Your HyperTransport is, in reality, twice of what the numbers indicate. (If you are using an Intel and are still reading this, your RAM is dual pumped but your processor is "quad pumped": it does two actions on the up and two actions on the down. This is why your "rated FSB" is four times the FSB in the BIOS.)

Confused? Repetition should cure this. Consider a phenomenal Opteron. (I HAVE to cut back on the puns.) You are running an Opteron 185 Denmark at 325 HTT with a 10x multiplier. Your RAM is running on a DDR 266 divider. What is the RAM actually at? Let us do the math:

HTT_OC*Multi = 325*10 = 3250 = CPU

(HTT_OC/HTT_Stock)*Divider = (325/200)*133 = 216.1 = RAM'

CPU/RAM' = 3250/216.1 = 15.04 = Divider'

Ceiling(Divider') = Ceiling(15.04) = 16 = Divider

CPU/Divider = 3250/16 = 203.13 = RAM

Thus, the RAM here would be running at DDR 406.

If you are younger than I, you are likely wondering why I am still talking about Socket 939. Other than it being the best Socket I know of, it is the one I learned how to overclock on. However, let us consider now Socket AM2 for the younger readers.

Consider a 6400+ BE Windsor with an HTT of 217 and a 17x multiplier. Your RAM is running on a DDR2 533 divider because after buying the CPU and the motherboard, you were broke. Anyway, here is how that breaks down:

HTT_OC*Multi = 217*17 = 3689 = CPU

(HTT_OC/HTT_Stock)*Divider = (217/200)*266 = 288.6 = RAM'

CPU/RAM' = 3689/288.6 = 12.78 = Divider'

Ceiling(Divider') = Ceiling(12.78) = 13 = Divider

CPU/Divider = 3689/13 = 283.8 = RAM

Thus, your system would be running DDR2 568. Really a disservice to that awesome, likely underwater, processor.

Sometimes, you will have different RAM dividers giving the same results. To borrow GuardianOdin's system, consider this:

Opteron 165 Denmark with an HTT = 354 MHz, Multi = 8.5, and then RAM either equaling DDR 286 or DDR 300. (DFI boards have more dividers than most.) Since we have been at it for a while, I will do the short-form math and leave you to fill in the blanks:

DDR 286: Ceil(3009/253.1)=12; 3009/12~DDR 502
DDR 300: Ceil(3009/265.5)=12; 3009/12~DDR 502

Interesting!

And as promised, I close with Pioneerisloud's computer.

He has an Opteron 165 Denmark. It has an HTT of 333, a multiplier of 9, and a RAM divider of DDR 200. We expect:

Ceil(2997/166.5)=18; 2997/18~DDR 333

This is, however, not the case. For some weird reason, his motherboard says that 2997/18 = 167.8 MHz. This is the first, and thus far, only time I have found my formula to fail.

I hope this adventure in mathematics has been as fun for you as it has for me. Stick around until next we meet.