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Politics : Formerly About Advanced Micro Devices
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To: Scumbria who wrote (102370)4/5/2000 6:13:00 PM
From: chic_hearne  Read Replies (2) of 1575762
 
1.2 GHz Thunderdbird beats a 1.5 GHz Willamette.

The Willamette will have to reach significantly higher frequencies if it wants to beat the Thunderbird.
Some of the basic numbers:

BRANCH MISS-PREDICTION PENALTY.

1.2 GHz Thunderbird: 8.33 nanoseconds (10 cycles)
1.5 GHz Willamette: 13.33 nanoseconds (20 cycles)

ALU TROUGHPUT: OPERATIONS PER SECOND

1.2 GHz Thunderbird: 1.2 ? 3.6 Billion sustained
1.5 GHz Willamette: 1.0 ?.4.5 Billion sustained

MAXIMUM LEVEL 1 DATA CACHE LOAD / STORE BANDWIDTH

1.2 GHz Thunderbird: 3.6 Billion operands /second
1.5 GHz Willamette: 1.5? 3.0 Billion operands /second

The branch miss prediction penalty is much higher on the Willamette. The Trace Cache doesn't help that much. The Instruction Cache on the Athlon contains partly pre-decoded instructions as well (like the K6 did) Each instruction byte has 3 additional bits to help accelerate branch prediction and instruction alignment of x86 instructions for the decoders.

The sustained ALU trough put of the Willamette is higher but can contain more depencies and may actually end up somewhat lower. The Willamette ALU's don't perform all integer instructions themselves. (Shifts are much slower on the Willamette than on the P6 according to Intel)

The databand width to the data cache is higher on the Thunderbird because it has 3 Address Generators which can do both loads and stores. The Willamette has a single load address generator and a single store address generator.

Intel says: The Willamette will be 30% faster then Coppermine. In my opinion this is a 1.5 Willamette compared with a 1 GHz Coppermine. Thus 30% faster with equal semiconductor process and not with equal frequencies.

COMMENT: I don't know this deep technical stuff. Is there any weight to this?

chic
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