Performance – Last Lecture Bottom line performance measure is time Performance A = 1/Execution Time A Comparing Performance N = Performance A / Performance.

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Presentation transcript:

Performance – Last Lecture Bottom line performance measure is time Performance A = 1/Execution Time A Comparing Performance N = Performance A / Performance B

Example If a machine A runs a program 25 seconds and machine B runs the same program in 20 seconds, how much faster is machine B that machine A?

Metrics of performance Compiler Programming Language Application Datapath Control TransistorsWiresPins ISA Function Units (millions) of Instructions per second – MIPS (millions) of (F.P.) operations per second – MFLOP/s Cycles per second (clock rate) Megabytes per second Answers per month Useful Operations per second Each metric has a place and a purpose, and each can be misused

Relating Metrics Instead of reporting execution time in seconds, we often use cycles So, to improve performance (everything else being equal) you can either ________ the # of required cycles for a program, or ________ the clock cycle time or, said another way, ________ the clock rate.

Our favourite program runs in 10 seconds on computer A, which has a 400 Mhz. clock. We are trying to help a computer designer build a new machine B, that will run this program in 6 seconds. The designer can use new (or perhaps more expensive) technology to substantially increase the clock rate, but has informed us that this increase will affect the rest of the CPU design, causing machine B to require 1.2 times as many clock cycles as machine A for the same program. What clock rate should we tell the designer to target?" Example

Following formula relates most basic metrics to CPU time: CPU time= Seconds= Instructions x Cycles x Seconds Program Program Instruction Cycle CPU time= Seconds= Instructions x Cycles x Seconds Program Program Instruction Cycle

Execution Time After Improvement = Execution Time Unaffected +( Execution Time Affected / Amount of Improvement ) Example: "Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster?" How about making it 5 times faster? Principle: Make the common case fast Amdahl's Law

Suppose we enhance a machine making all floating-point instructions run five times faster. If the execution time of some benchmark before the floating-point enhancement is 10 seconds, what will the speedup be if half of the 10 seconds is spent executing floating-point instructions? We are looking for a benchmark to show off the new floating- point unit described above, and want the overall benchmark to show a speedup of 3. One benchmark we are considering runs for 100 seconds with the old floating-point hardware. How much of the execution time would floating-point instructions have to account for in this program in order to yield our desired speedup on this benchmark? Example

Performance is specific to a particular program/s Total execution time is a consistent summary of performance For a given architecture performance increases come from: increases in clock rate (without adverse CPI affects) improvements in processor organization that lower CPI compiler enhancements that lower CPI and/or instruction count Pitfall: expecting improvement in one aspect of a machine’s performance to affect the total performance Remember