Faster cpu wastes more time as compared to slower cpu

Question

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.

Answer 1

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).

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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

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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.

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1. 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.

2. 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.

3. 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.

Answer 2

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.

Answer 3

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".

Answer 4

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.

Answer 5

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..).

Answer 6

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.