IPT - A Virtual Approach IPT A Virtual Approach by Peter Whitehouse
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Computer Systems eXercises #3

Memory and Addressing - Solutions

1.
1. A227: EDE0 normalised is B1050 which is in VID RAM
2. 725072 bytes above the base of memory
3. 708 Kb

2.
1.48.5Kb = 49664 bytes = C200 bytes
  so the address of the last byte is 60FEC + C200 = 6D1E0
2. 6D1E0 = 446956 bytes = 436.4 Kb

3.
1. F233B
2. E998F
3. 010E8
4. CB06C

4.
1. 90EC1 - 0F44B = 81A76 =  531062 bytes =  518Kb
2. 62539 bytes = 61Kb

5.
the address bus limits the size of the address that can be transmitted.  
By dividing up the address into a segment and an offset, and sending them 
separately through the address bus, we effectively allow access to LARGE sections 
of memory whilst transporting relatively few bits to do it.

6.
1. B000:8400
2. B000:9111
 
  

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