Example 3 - page 144 (contd.)

• Average inst time for machine B:

1) CPI has been extended.

= 40% of the times 1 more cycle

2) Clock rate is faster: 1.05 times: less than machine A. By how much?

Avg instruction time for machine B: (1 + 40/100*1) * (clock cycle time /1.05)

= 1.4 * x/1..05 = 1.3x

Proved that A is faster.

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