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Politics : Formerly About Advanced Micro Devices -- Ignore unavailable to you. Want to Upgrade?

To: Elmer who wrote (32701)	5/16/1998 12:52:00 AM
From: Ali Chen	Read Replies (2) \| Respond to of 1571408

Elmer, <Sorry John on both points. PII is probably cheaper to produce,> You must be very technically challenged, Elmer-Boy, to believe in this statement. The cost of wafer must be about the same. Same chemicals, labor, substrate, ~$2500 - $3000 a piece, right? With die size 2X of those for K6, each P-II die must be twice as expensive. Even if you assume the P-II yield is two times higher, the cost of dies should be the same at most. If you think more realistically, say, AMD has >50%, you cannot get >100% for P-II, and the 2X-yield assumption looks as an apparent overestimation. Therefore, even at this point the P-II is not cheaper than K6. Now go to packaging. Again, the cost must be comparable: substrate, wire bonds or flip-chip, no big difference. Now make a 12-layers PCB. Solder 140 add-on components, on both sides, 5 pieces of cache, thermal plate, precision pins, spring holders, four plastic pieces... If you think that all this reduces the P-II cost, I rest my arguments.

To: Elmer who wrote (32701)	5/16/1998 1:15:00 PM
From: Kenith Lee	Read Replies (1) \| Respond to of 1571408

PII is probably cheaper to produce, and unless you are blasting mutant aliens, Elmer, May had been cheaper to produce when AMD was having yield problems. Given with its much larger die size, it wouldn't take much for K6 to catch up with PII in number of dice yielded per wafer. the PII easily out performs the K6-x clock for clock Are you sure about this? Are you hyping again? Celeron 266 (a PII) equals to P5MMX 233 and K6-200. Maybe PII (with cache) will beat vanilla K6 because of Slot 1 or PII (with large L1) beats vanilla K6-3D.