Solutions to HW #1 Problems 3 and 4

3.	Find the cost of memory in dollars/MByte:

	DRAM - answers ranging from $5-$10/MByte
               ex 4Meg 72 pin SIMM for $24 = $24/4MByte = $6/MByte

	Hard Disk Drive - range $.09 - $.40
               ex Segate IDE 2.5 GB for $279 = $279/2500MByte = $.11/MByte

	DAT/DLT tape drive plus tapes - range $.01 - $.12 perMByte
	       ex Conner 4mm DAT 2GBYTE SCSI2 for $535
                  each 2GByte tape = $18 - need 30 to reach half the budget
                  $535 + $18(30) = $1075    30 * 2GB = 60GB or 60,000MBytes
                  $1075/60000MBytes = $.02/MByte


4.	Find the number of bits that can dance on the head of a pin.
	Estimate the number of bits that would be contained in the area
	the size of a pinhead for Semiconductor (DRAM) and Magnetic
	Memory (Hard Disk) technologies:

	Formulas    (always start with the formula for what you ultimately 
        --------     want to find!)


	bits/pin = bits/mm^2 * mm^2/pinhead

        *** Now show how to derive things in main formula ***

	bits/mm^2 = bytes of media          bits
                    -------------------  * ------
                    storage area (mm^2)     byte
 	storage area:
         /DRAM = length*width (cm)^2     100mm^2
                 -----------------  *   ---------
                  chip                   1 cm^2

         /HD   = pi*radius(cm)^2/platter * # platters   100mm^2
                 ------------------------------------ * -------
                 HD                                      1cm^2

  
	Estimates - Provide values for variables (words) above
	_________
	
	mm^2/pinhead - 1mm^2 (by inspection)
        DRAM - length - 1 cm
               width  - 1 cm
               area = 1cm^2    by thumbnail estimate

	HD   - radius of 1 platter - 1.75  by inspection
               # platters  - 10   - average of current technology

   	Bytes/media DRAM - 4 MB from problem #3
              please note that there are 10^6 MBytes/Byte
                    HD   - 2 GB from problem #3
                          there are 10^9 Gbytes/Byte

	Calculations  - Now just put the values into the formulas and
	------------    do the math!

	reasonable answers for both were approx N*100,000 Bits/pinhead
                                 (where N is 1-10)

	common errors:
            1 MByte = 1024 bytes (2^10) No! 1 MByte = 2^20 or 10^6
            1mm = 10cm     No!   1cm = 10mm  and 1cm^2 = 100mm^2
            problems with common sense
                some calculated that there were 100-200 pins on a Hard Disk
                    does that really make sense?  picture it!