coxc0015@ucselx.sdsu.edu (Imagine) (05/01/91)
I'm new on the IBM scene and in my motherboard manual it says that when
running on less that 4 slots filled the cpu speed is reduced by 10%-20%. I
have a Dieco 286-16Mhz, I used the norton system info utility. The benchmark
was at 8.4(Mhz, I think). If my mathmatics is correct that a little less than
50% reduced speed. Can someone please explain why such a great lose in speed.
Thanks,
Trung
david@kessner.denver.co.us (David Kessner) (05/01/91)
In article <1991Apr30.181315.2889@ucselx.sdsu.edu> coxc0015@ucselx.sdsu.edu (Imagine) writes: > I'm new on the IBM scene and in my motherboard manual it says that when >running on less that 4 slots filled the cpu speed is reduced by 10%-20%. I >have a Dieco 286-16Mhz, I used the norton system info utility. The benchmark >was at 8.4(Mhz, I think). If my mathmatics is correct that a little less than >50% reduced speed. Can someone please explain why such a great lose in speed. > > Trung Well. First off. Nortons does not give it's results in Mhz. Rather, it gives an index that it only good for comparing with other Norton results. Further, Nortons Si 4.5 results are very different from Norton Si 5.0 results. By the number you gave, I assume you have version 5.0-- the one I am least familiar with. You say four slots, I assume you mean four SIMM sockets. Here's the poop on memory. On most 16mhz machines, the memory cannot keep up with the CPU so the CPU often was to wait for the memory to spit out the data. This is known as a wait-state. A one-wait state system will have the CPU wait one additional clock cycle for each memory access. Most 16mhz machines use one or two wait-states. There are many ways of getting rid of these wait-states. You motherboard probably uses an "Interleaved" scheme, while most 386/486's use a cache. In a normal system, memory can be thought of as one bank-- and all memory accesses are using that bank with the associated wait states. In an interleaved system, there is two banks where all odd bytes are in one bank, and all even bytes in the other bank-- hence the term 'interleaved'. When the CPU is accessing one bank, the system _starts_ a memory access on the other bank (for the next byte). It takes the same amount of time for each memory access, but since they are overlapping you get an effective zero-wait-state. It's not perfect, however. The motherboard assumes that the CPU will want the _next_ byte, and starts to access that byte. This is a good guess and works most of the time, but there are times when the next byte is not wanted and the system must start over and access the correct byte. This is why caches are better (but more expensive). The SIMM's in your computer come in two sizes-- 256K and 1meg-a-byte. Each one will provide 8-bits of data at a time (I'm ignoring parity), but the 286 CPU wants data 16-bits at a time. Therefore, you will (almost) always see them in groups (banks) of two. Since interleaving requires two banks of RAM, and each bank has two SIMMs, you need at least four SIMMs to have interleaved memory-- and that is why you need four 'sockets' filled to benifit from the speed that interleaved memory will provide. Common memory configurations are: Total Size of Number Interleaved? RAM SIMM's SIMM's ----- ------ ------ ------------ 512 256K 2 No 1meg 256K 4 Yes 2meg 1meg 2 No 4meg 1meg 4 Yes 8meg 1meg 8 Yes Actual configurations may vary-- consult your manual to be absolutely sure.. I hope that helped! -- David Kessner - david@kessner.denver.co.us | do { 1135 Fairfax, Denver CO 80220 (303) 377-1801 (p.m.) | . . . If you cant flame MS-DOS, who can you flame? | } while( jones);