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Lecture Notes for Computer Architecture II - St. Cloud State University

Lecture Notes for Computer Architecture II - St. Cloud State University

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CPU clock speed<br />

1 cycle<br />

1 Hertz = <br />

Second<br />

1<br />

Cycle time = <br />

Clock rate<br />

Page | 120<br />

Several different cases<br />

- One instruction per cycle<br />

- Many instructions per cycle<br />

- Several cycles per instruction<br />

Example:<br />

2GHz machine<br />

If 1 instruction per cycle 2 GIPS<br />

If 2 instructions per cycle 4 GIPS<br />

Per<strong>for</strong>mance can be average or peak<br />

GIPS - Peak Per<strong>for</strong>mance<br />

GFLOPS - Peak Per<strong>for</strong>mance (GFLOPS million floating point operations)<br />

CPU time = Number of instructions * cycle time<br />

To increase per<strong>for</strong>mance decrease the CPU time per program<br />

- reduce cycle time<br />

- reduce number of instruction per program<br />

Example:<br />

Machine M1 10 seconds to run Program A<br />

Machine M2 15 seconds to run Program A<br />

How much faster M1 compared to M2?<br />

Solution<br />

Per<strong>for</strong>mance M1 CPU time M2 15<br />

= = = 1.5<br />

Per<strong>for</strong>mance M2 CPU time M1 10<br />

CPU time = CPU clock cycles per program * clock cycle time<br />

Ef<strong>for</strong>ts to decrease CPU clock cycles per program may increase clock cycle time<br />

Per<strong>for</strong>mance<br />

- means different things to different people<br />

- two important measures of per<strong>for</strong>mance<br />

- throughput

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