CPU Performance - Computer Organization and Design - Lecture Slides, Slides of Computer Aided Design (CAD)

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CPU Performance

How to Best Measure Performance?

• Time!
• What kind of “time”?
  • Response time: time between start and finish of a program, aka wall clock time. Good measure from the “user” perspective
  • Throughput: total amount of work done in a fixed amount of time. Good from the “computer center” perspective
• Note: a improvement in one measure usually results in an improvement in the other

Example

• Machine A: program takes 10 sec
• Machine B: same program takes 15 sec
• Hence: perf(A) / perf(B) = exec(B) / exec(A) = 15 / 10 = 1.
• Machine A is 1.5 times faster than machine B

How to Measure Time

• CPU time via the “CPU performance equation”
• CPU time = CPU clock cycles * clock cycle time
• Abbrev: exec = cycles * tclk
• Equivalently: exec = cycles / frequency

Example (2)

• cycles(B) = 1.2 * cycles(A) = 1.2 * 2e10 = 2.4e
• Also: exec(B) = cycles(B) / freq(B)
• So: freq(B) = cycles(B) / exec(B) = 2.4e10 / 6 sec = 4 GHz
• Need to double clock rate to meet the 6 sec requirement!

CPU Performance Equation (1)

• We rarely have access to the total clock cycles for a program P.
• Usually can get total number of instructions executed (by profiling the code)
• Usually can get the number of cycles per instruction (look it up in a datasheet for that processor)
• Hence, a better CPU performance equation is:

exec = IC * CPI * tclk

Ex: CPU Performance Equation (1)

• Suppose machine A has a tclk of 1 ns (1e-9 sec) and a CPI of 2.0 for a program P
• Suppose machine B has the same ISA as machine A, but has a tclk of 2 ns and a CPI of 1.2 for program P
• Which machine is “faster”?
• exec = IC * CPI * tclk
  • exec(A) = IC(A) * 2.0 * 1 ns = 2.0 * IC(A)
  • exec(B) = IC(B) * 1.2 * 2 ns = 2.4 * IC(B)
• Note that P is fixed, and the ISA is the same on both machines

Ex: CPU Performance Equation (2)

• Perf(A) / Perf(B) = exec(B) / exec(A)

= (2.4 * IC(B)) / (2.0 * IC(A)) = 1.2 IC(B) / IC(A)

• How to get IC(B) / IC(A)?
• IC(B) = IC(A) since P is fixed and the ISA is the same
• Hence, machine A is 1.2 times faster than machine B for P

Measuring Performance

Reconsidering the CPI

• Computing the CPI requires careful analysis of how each instruction behaves on the hardware
• Must understand how each assembly language instruction is implemented to know how many cycles it takes
• Note: many times you don’t have total IC and average CPI
• Instead you might have the IC for each separate instruction, or for each of a family of instructions
• And you might have the CPI for each of those separate types of instructions

Example (1)

• Suppose instruction A has a CPI = 1 (e.g. jmp, bne)
• Suppose instruction B has a CPI = 2 (e.g. add/sub)
• Suppose instruction C has a CPI = 3 (e.g. mult)
• Suppose a compiler writer has a choice between 2 approaches to implementing a single HLL instruction:
• Sequence 1 requires 2 A, 1 B, 2 C
• Sequence 2 requires 4 A, 1 B, 1 C

Example (2)

• Question1: which code sequence is longer?
• Question2: which code sequence is faster?
• Question3: what is the CPI of each sequence?
• Sequence 1 executes 2 + 1 + 2 instructions = 5
• Sequence 2 executes 4 + 1 + 1 instructions = 6
• Question1 answer: Sequence 2 is longer. If code size is our performance measure, then Sequence 1 is best.

Example (4)

• Average CPI(seq1) = (1 * 2 + 2 * 1 + 3 * 2) / 5 = 10/5 = 2
• Average CPI(seq2) = (1 * 4 + 2 * 1 + 3 * 1) / 6 = 9/6 = 1.
• Question3 answer: The average CPI of sequence 2 is lower than the average CPI of sequence 1

Another Example (1)

• Suppose an computer designer has two alternatives to implementing branches:
• CPU A: set a condition code based on the result of a “compare” instruction and then follow that compare immediately by a “branch” instruction that uses that code
• CPU B: the compare is “folded” into the branch in a single all-purpose branch instruction that does both in one step