The following conditions exist.
Consider three diff erent processors P1, P2, and P3 executing the same
instruction set. P1 has a 3 GHz clock rate and a CPI of 1.5. P2 has
a
2.5 GHz clock rate and a CPI of 1.0. P3 has a 4.0 GHz clock rate and has a CPI of 2.2.
And the question is
Which processor has the highest performance when executing the same program?
I have learned to compare cpu execution when comparing computer performance.
However, since cpu execution time = CPI * instruction set * 1/clock rate, the size of the instruction set cannot be known only with the conditions in the above problem, I thought that the performance between processors could not be compared.
I looked for other issues similar to this one, and the issue is Which processor has the highest performance expressed in instructions per second? We compared the performance between processors under the condition of instructions per second as shown.
So what I want to know is whether it is possible to compare performance between processors without any special conditions. (Isn't the given problem wrong?) If possible, I wonder how they can be compared.
However, since cpu execution time = CPI * instruction set * 1/clock rate, the size of the instruction set cannot be known only with the conditions in the above problem, I thought that the performance between processors could not be compared.
For that formula; I'd assume that instruction set reflects how many instructions are needed to do the same work. E.g. if one instruction set can do a multiply in one instruction it might have instruction set = 1, and if another instruction set needs 20 instructions to do one multiply then it might have instruction set = 20 because you need 20 times as many instructions to get the same work done.
For your homework (3 processors that all execute the same instruction set), instruction set is irrelevant - they all take the same number of instructions to get any amount of work done. With this in mind you can just do performance = clock rate / CPI. More specifically:
P1 = 3 GHz / 1.5 = 2,000,000,000
P2 = 2.5 GHz / 1.0 = 2,500,000,000
P3 = 4 GHz / 2.2 = 1,818,181,818
Of course the homework is extremely over-simplified - stalls (time CPU spends doing nothing while waiting for things like cache misses and instruction fetch after a branch misprediction, etc) tend to have a larger impact on performance than clock speed or "theoretical maximum CPI"; and "measured CPI in practice" depends which specific instructions are used (and can never be a single number like "1.5 CPI" for all programs). In other words, in practice, it's easy to have a situation where a CPU is fastest for one program but slowest for another program.
This question is related Computer Organization and Architecture. i would truly appreciate some assistance
Processor A has a clock speed of 1 GHz, and takes 1 cycle for integer operations, 2 cycles for memory operations, and 4 cycles for floating point operations. Empirical data shows that programs run on Processor A typically are composed of 35% floating point operations, 30% memory operations; and 35% integer operations.
However you wish to design a new processor, Processor B, which is an improvement on Processor A. Processor B will run the same programs as Processor A. To complete your design, you are faced with two options for improving performance:
Increase the clock speed to 1.2 GHz, but memory operations take 3 cycles
Decrease the clock speed to 900 MHz, but floating point operations only take 3 cycles
Compute the speedup for both options and decide the option Processor B should take.
i have tried finding the Execution time for Processors A and B by using ICxCPIxCT. I used MIPS (Clock rate/CPIx10^6) as the IC in all cases. When i was finished with everything, the Processor B(1) was deemed the most efficient since it had the lowest CPU time(940.3x10^-9) but i'm not sure if my method was correct.
From what I understand, to calculate CPI, it's the percentage of the type of instruction multiplied by the number of cycles right? Does the type of machine have any part of this calculation whatsoever?
I have a problem that asks me if a change should be recommended.
Machine 1: 40% R - 5 Cycles, 30% lw - 6 Cycles, 15% sw - 6 Cycles, 15% beq 3 - Cycles, on a 2.5 GHz machine
Machine 2: 40% R - 5 Cycles, 30% lw - 6 Cycles, 15% sw - 6 Cycles, 15% beq 4 - Cycles, on a 2.7 GHz machine
By my calculations, machine 1 has 5.15 CPI while machine 2 has 5.3 CPI. Is it okay to ignore the GHz of the machine and say that the change would not be a good idea or do I have to factor the machine in?
I think the point is to evaluate a design change that makes an instruction take more clocks, but allows you to raise the clock frequency. (i.e. leaning towards a speed-demon design like Pentium 4, instead of brainiac like Apple's A7/A8 ARM cores. http://www.lighterra.com/papers/modernmicroprocessors/)
So you need to calculate instructions per second to see which one will get more work done in the same amount of real time. i.e. (clock/sec) / (clocks/insn) = insn/sec, cancelling out the clocks from the units.
Your CPI calculation looks ok; I didn't check it, but yes a weighted average of the cycles according to the instruction mix.
These numbers are obviously super simplified; any CPU worth building at 2.5GHz would have some kind of branch prediction so the cost of a branch isn't just a 3 or 4 instruction bubble. And taking ~5 cycles per instruction on average is pathetic. (Most pipelined designs aim for at least 1 instruction per clock.)
Caches and superscalar CPUs also lead to complex interactions between instructions depending on whether they depend on earlier results or not.
But this is sort of like what you might do if considering increasing the L1d cache load-use latency by 1 cycle (for example), if that took it off the critical path and let you raise the clock frequency. Or vice versa, tightening up the latency or reducing the number of pipeline stages on something at the cost of reducing frequency.
Cycles per instruction a count of cycles. ghz doesnt matter as far as that average goes. But saying that we can see from your numbers that one instruction is more clocks but the processors are a different speed.
So while it takes more cycles to do the same job on the faster processor the speed of the processor DOES compensate for that so it seems clear this is a question about does the processor speed account for the extra clock?
5.15 cycles/instruction / 2.5 (giga) cycles/second, cycles cancels out you get
2.06 seconds/(giga) instruction or (nano) seconds/ instruction
5.30 / 2.7 = 1.96296 (nano) seconds / instruction
The faster one takes a slightly less amount of time so it will run the program faster.
Another way to see this to check the math.
For 100 clock cycles on the slower machine 15% of those are beq. So 15 of the 100 clocks, which is 5 beq instructions. The same 5 beq instructions take 20 clocks on the faster machine so 105 clocks total for the same instructions on the faster machine.
100 cycles at 2.5ghz vs 105 at 2.7ghz
we want the amount of time
hz is cycles / second we want seconds on the top
so we want
cycles / (cycles/second) to have cycles cancel out and have seconds on the top
1/2.5 = 0.400 (400 picoseconds)
1/2.7 = 0.370
0.400 * 100 = 40.00 units of time
0.370 * 105 = 38.85 units of time
So despite taking 5 more cycles the processor speed differences is fast enough to compensate.
2.7/2.5 = 1.08
105/100 = 1.05
so 2.5 * 1.05 = 2.625 so a processor 2.625ghz or faster would run that program faster.
Now what were the rules for changing computers, is less time defined as a reason to change computers? What is the definition of better? How much more power does the faster one consume it might take less time but the power consumption might not be linear so it may take more watts despite taking less time. I assume the question is not that detailed, meaning it is vague meaning it is a poorly written question on its own, so it goes to what the textbook or lecture defined as the threshold for change to the other processor.
Disclaimer, dont blame me if you miss this question on your homework/test.
Outside an academic exercise like this, the real world is full of pipelined processors (not all but most of the folks writing programs are writing programs for) and basically you cant put a number on clock cycles per instruction type in a way that you can do this calculation because of a laundry list of factors. Make sore you understand that, nice exercise, but that specific exercise is difficult and dangerous to attempt on real world processors. Dangerous in that as hard as you work you may be incorrectly measuring something and jumping to the wrong conclusions and as a result making bad recommendations. At the same time there is very much the reality that faster ghz does improve some percentage of the execution, but another percentage suffers, and is there a net gain or loss. Or a new processor design faster or slower may have features that perform better than an older processor, but not all feature will be better, there is a tradeoff and then we get into what "better" means.
Suppose I have a program that has an instruction to add two numbers and that operation takes 10 nanoseconds(constant, as enforced by the gate manufactures).
Now I have 3 different processors A, B and C(where A< B < C in terms of clock cycles). A's one clock cycle has 15 nanosec, B has 10 nanosec and C has 7 nanosec.
Firstly am I correct on my following assumptions-
1. Add operation takes 1 complete cycle of processor A(slow processor) and wastes rest of 5 ns of the cycle.
2. Add operation takes 1 complete cycle of processor B wasting no time.
3. Add operation takes 2 complete cycles(20 ns) of processor C(fast processor) wasting rest of the 20-14=7 ns.
If the above assumptions are correct then isn't this a contradiction to the regular assumption that processors with high clock cycles are faster. Here processor C which is the fastest actually takes 2 cycles and wastes 7ns whereas, the slower processor A takes just 1 cycle.
Processor C is fastest, no matter what. It takes 7 ns per cycle and therefore performs more cycles than A and B. It's not C's fault that the circuit is not fast enough. If you would implement the addition circuit in a way that it gives result in 1 ns, all processors will give the answer in 1 clock cycle (i.e. C will give you the answer in 7ns, B in 10ns and A in 15ns).
Firstly am I correct on my following assumptions-
1. Add operation takes 1 complete cycle of processor A(slow processor) and wastes rest of 5 ns of the cycle.
2. Add operation takes 1 complete cycle of processor B wasting no time.
3. Add operation takes 2 complete cycles(20 ns) of processor C(fast processor) wasting rest of the 20-7=13 ns.
No. It is because you are using incomplete data to express the time for an operation. Measure the time taken to finish an operation on a particular processor in clock cycles instead of nanoseconds as you are doing here. When you say ADD op takes 10 ns and you do not mention the processor on which you measured the time for the ADD op, the time measurement in ns is meaningless.
So when you say that ADD op takes 2 clock cycles on all three processors, then you have standardized the measurement. A standardized measurement can then be translated as:
Time taken by A for addition = 2 clock cycles * 15 ns per cycle = 30 ns
Time taken by B for addition = 2 clock cycles * 10 ns per cycle = 20 ns
Time taken by C for addition = 2 clock cycles * 07 ns per cycle = 14 ns
In case you haven't noticed, when you say:
A's one clock cycle has 15 nanosec, B has 10 nanosec and C has 7 nanosec.
which of the three processors is fastest?
Answer: C is fastest. It's one cycle is finished in 7ns. It implies that it finishes 109/7 (~= 1.4 * 108) cycles in one second, compared to B which finishes 109/10 (= 108) cycles in one second, compared to A which finishes only 109/15 (~= 0.6 * 108) cycles in one second.
What does a ADD instruction mean, does it purely mean only and only ADD(with operands available at the registers) or does it mean getting
the operands, decoding the instruction and then actually adding the
numbers.
Getting the operands is done by MOV op. If you are trying to compare how fast ADD op is being done, it should be compared by time to perform ADD op only. If you, on the other hand want to find out how fast addition of two numbers is being done, then it will involve more operations than simple ADD. However, if it's helpful, the list of all Original 8086/8088 instructions is available on Wikipedia too.
Based on the above context to what add actually means, how many cycles does add take, one or more than one.
It will depend on the processor because each processor may have the adder differently implemented. There are many ways to generate addition of two numbers. Quoting Wikipedia again - A full adder can be implemented in many different ways such as with a custom transistor-level circuit or composed of other gates.
Also, there may be pipelining in the instructions which can result in parallelizing of the addition of the numbers resulting in huge time savings.
Why is clock cycle a standard since it can vary with processor to processor. Shouldn't nanosec be the standard. Atleast its fixed.
Clock cycle along with the processor speed can be the standard if you want to tell the time taken by a processor to execute an instruction. Pick any two from:
Time to execute an instruction,
Processor Speed, and
Clock cycles needed for an instruction.
The third can be derived from it.
When you say the clock cycles taken by ADD is x and you know the processor speed is y MHz, you can calculate that the time to ADD is x / y. Also, you can mention the time to perform ADD as z ns and you know the processor speed is same y MHz as earlier, you can calculate the cycles needed to execute ADD as y * z.
I'm no expert BUT I'd say ...
the regular assumption that processors with high clock cycles are faster FOR THE VAST MAJORITY OF OPERATIONS
For example, a more intelligent processor might perform an "overhead task" that takes X ns. The "overhead task" might make it faster for repetitive operations but might actually cause it to take longer for a one-off operation such as adding 2 numbers.
Now, if the same processor performed that same operation 1 million times, it should be massively faster than the slower less intelligent processor.
Hope my thinking helps. Your feedback on my thoughts welcome.
Why would a faster processor take more cycles to do the same operation than a slower one?
Even more important: modern processors use Instruction pipelining, thus executing multiple operations in one clock cycle.
Also, I don't understand what you mean by 'wasting 5ns', the frequency determines the clock speed, thus the time it takes to execute 1 clock. Of course, cpu's can have to wait on I/O for example, but that holds for all cpu's.
Another important aspect of modern cpu's are the L1, L2 and L3 caches and the architecture of those caches in multicore systems. For example: if a register access takes 1 time unit, a L1 cache access will take around 2 while a normal memory access will take between 50 and 100 (and a harddisk access would take thousands..).
This is actually almost correct, except that on processor B taking 2 cycles means 14ns, so with 10ns being enough the next cycle starts 4ns after the result was already "stable" (though it is likely that you need some extra time if you chop it up, to latch the partial result). It's not that much of a contradiction, setting your frequency "too high" can require trade-offs like that. An other thing you might do it use more a different circuit or domino logic to get the actual latency of addition down to one cycle again. More likely, you wouldn't set addition at 2 cycles to begin with. It doesn't work out so well in this case, at least not for addition. You could do it, and yes, basically you will have to "round up" the time a circuit takes to an integer number of cycles. You can also see this in bitwise operations, which take less time than addition but nevertheless take a whole cycle. On machine C you could probably still fit bitwise operations in a single cycle, for some workloads it might even be worth splitting addition like that.
FWIW, Netburst (Pentium 4) had staggered adders, which computed the lower half in one "half-cycle" and the upper half in the next (and the flags in the third half cycle, in some sense giving the whole addition a latency of 1.5). It's not completely out of this world, though Netburst was over all, fairly mad - it had to do a lot of weird things to get the frequency up that high. But those half-cycles aren't very half (it wasn't, AFAIK, logic that advanced on every flank, it just used a clock multiplier), you could also see them as the real cycles that are just very fast, with most of the rest of the logic (except that crazy ALU) running at half speed.
Your broad point that 'a CPU will occasionally waste clock cycles' is valid. But overall in the real world, part of what makes a good CPU a good CPU is how it alleviates this problem.
Modern CPUs consist of a number of different components, none of whose operations will end up taking a constant time in practice. For example, an ADD instruction might 'burst' at 1 instruction per clock cycle if the data is immediately available to it... which in turn means something like 'if the CPU subcomponents required to fetch that data were immediately available prior to the instruction'. So depending on if e.g. another subcomponent had to wait for a cache fetch, the ADD may in practice take 2 or 3 cycles, say. A good CPU will attempt to re-order the incoming stream of instructions to maximise the availability of subcomponents at the right time.
So you could well have the situation where a particular series of instructions is 'suboptimal' on one processor compared to another. And the overall performance of a processor is certainly not just about raw clock speed: it is as much about the clever logic that goes around taking a stream of incoming instructions and working out which parts of which instructions to fire off to which subcomponents of the chip when.
But... I would posit that any modern chip contains such logic. Both a 2GHz and a 3GHz processor will regularly "waste" clock cycles because (to put it simply) a "fast" instruction executed on one subcomponent of the CPU has to wait for the result of the output from another "slower" subcomponent. But overall, you will still expect the 3GHz processor to "execute real code faster".
First, if the 10ns time to perform the addition does not include the pipeline overhead (clock skew and latch delay), then Processor B cannot complete an addition (with these overheads) in one 10ns clock cycle, but Processor A can and Processor C can still probably do it in two cycles.
Second, if the addition itself is pipelined (or other functional units are available), then a subsequent non-dependent operation can begin executing in the next cycle. (If the addition was width-pipelined/staggered (as mentioned in harold's answer) then even dependent additions, logical operations and left shifts could be started after only one cycle. However, if the exercise is constraining addition timing, it presumably also prohibits other optimizations to simplify the exercise.) If dependent operations are not especially common, then the faster clock of Processor C will result in higher performance. (E.g., if a dependence stall occurred every fourth cycle, then, ignoring other effects, Processor C can complete four instructions every five 7ns cycles (35 ns; the first three instruction overlap in execution) compared to 40ns for Processor B (assuming the add timing included pipelining overhead).) (Note: Your assumption 3 is incorrect, two cycles for Processor C would be 14ns.)
Third, the extra time in a clock cycle can be used to support more complex operations (e.g., preshifting one operand by a small immediate value and even adding three numbers — a carry-save adder has relatively little delay), to steal work from other pipeline stages (potentially reducing the number of pipeline stages, which generally reduces branch misprediction penalties), or to reduce area or power by using simpler logic. In addition, the extra time might be used to support a larger (or more associative) cache with fixed latency in cycles, reducing miss rates. Such factors can compensate for the "waste" of 5ns in Processor A.
Even for scalar (single issue per cycle) pipelines clock speed is not the single determinant of performance. Design choices become even more complex when power, manufacturing cost (related to yield, adjusted according to sellable bins, and area), time-to-market (and its variability/predictability), workload diversity, and more advanced architectural and microarchitectural techniques are considered.
The incorrect assumption that clock frequency determines performance even has a name: the Megahertz myth.
I have to compare the speed of execution of the following code (see picture) using DLX-pipeline and single-cycle processor.
Given:
an instruction in the single-cycle model takes 800 ps
a stage in the pipeline model takes 200 ps (based on MA)
My approach was as follows.
CPU time = CPI * CC * IC
Single-cycle:
CPU time = 1 * 800 ps * 10 instr. = 8000 ps.
Pipeline:
CPI = 21 cycles / 10 instr. = 2.1 cycles per instruction
CPU time = 2.1 * 200 ps * 10 = 4200 ps.
CPU time single-cycle / CPU time pipeline = 8000/4200 = 1.9, so the pipeline code runs 1.9 faster.
But I was said, I have to work with clock cycles and not with the time -- "It doesn't matter how much time a CC takes".
I don't see how to make a comparison otherwise. Could you please help me?
Your analysis is indeed correct, but I guess your professor is looking for an explanation like this:
Suppose the single cycle processor also has the stages that you have mentioned, namely IF, ID, EX, MA and WB and that the instruction spends roughly the same time in each stage as compared to the pipelined processor version. Now you can draw a pipeline diagram for this single cycle processor, and see that it would take 50 cycles on a single cycle processor (which can work on 1 instruction at a time) compared to the 19 cycles on a pipelined processor.
Again, I prefer the way you have analyzed it (as the single cycle processor wouldn't really have each of those stages in a different clock cycle, it would just have a very long clock cycle to cover all the stages). Also, you've not mentioned whether this is a stalling-only MIPS pipeline (for which your answer is correct) or if this is a bypassed-MIPS pipeline. If this is the latter, you can shave off a few more cycles and get it down to 15 cycles.