🚀 go-pugleaf

Michael Jones of SGI was generous enough to take the time.  He also
comments that the mainframe as a whole is a more usable configuration
than the sum of the PC's.

Michael Jones:

I bet we could estimate it closely this way: Amdahl's powerful
mainframe must use advanced technology parts in order to be
so powerful. Parts of 2x the performance usually cost > 2x the
price. Harwdare system vendors get 4x to 6x parts cost as sale
price, while PC parts at Frye's are close to wholesale (2x). 

Ergo:

   Amdahl mainframe at $a. (say $3M)
   Amdahl cost at a/4 (at the most)
   this figure (a/4) represents "potential energy" of a sort.

   Typical system at Frye's is $p (say $2000).
   Frye's price is 2x cost, say, so the question is:
      How many cheap PC's for 2*(a/4), == a/(2p).

   Answer: 3,000,000/(2*2,000) = 750 pc's == 1 Amdahl

How reasonable is this, I wonder?

       750*30 MIPS = 22.5 BIPS
         750*200MB = 150 GBytes of DIsk
         750*1Mb/s = 750 MBytes/sec Disk transfer
 750 floppy drives = 1.08 GBytes removable storage ;-)
       750*4Mb RAM = 3 Gbytes of RAM
   750*Lotus 1-2-3 = 750 concurrent users
      750*1 RS-232 = 750 network attachments

Sounds ok. Now how big would the room capable of housing
750 pc's need to be? Well, if they were stacked in a cube
and each was 24"x24"x8", we have 3x^3 = 750, x^3 = 250, 
and x = 6.3. Rounding to x = 7, we have a stack that's
7x7x(3x7) for a size of 7*24 = 14 x 14 x 14 feet. Well,
that's too big, lets stack them half as high. Now the
room is 2^(1/2) bigger on each side, so it must be 19.7
feet on a side. Rounding, we get:

  A room 20' x 20' stacked solid with PC's 7' high.

Well, we need some room to get to the machines, so we can
change the 750 floppies during backups. Let's make the 
room 3x the length, so we have long rows of stacked PC's
with 4' aisles. Now the room is:

 20' wide, 60' long, and 7' high.

The question is, are there any rooms this large in the
world?