CS401 Assignment Idea Solution

Question No. 1:
a. supposes architecture A has 14-bit address bus. What is maximum size of
memory that can be accessed in this architecture? (Show the steps for calculating
maximum accessible memory) (5 marks)
Sol:
Accessible memory addresses = 2number of address bits
214 = 16384 bytes
16384 / 8 = 2048 bytes
2048 / 1024 = 2 KB
b. An architecture B has a maximum limit of 2GB memory. How many address
bits are required for this architecture? (Show the steps for calculating required
number of address bits) (5 marks)
Sol:
2GB * 1024 = 2048 MB
2048MB * 1024 = 2097152KB
2097152KB * 1024 = 2147483648bytes
2147483648bytes * 8 = 17179869184 bits
Question No. 2:
What are the first and the last physical memory addresses accessible using the
following segment values? (2 mark each)
a. 0000
b. FFFF
Question No. 3:
Calculate physical address using the following segment offset pairs.
(1 mark each)
a. ABCD:1234
Solution:
1)
A B C D 0
0 1 2 3 4 +
A C F 0 4 = physical address
b. 1234:ABCD
Solution:
2)
1 2 3 4 0
0 A BC D +
1 C F 0 D = physical address
Question No. 4:
What is effective address generated by the following instructions? Every
instruction is independent of others. Initially BX = 0xFF00, SI=0x00FF
(1 mark each)
a. mov ax, [BX+SI]
Sol :
Effective address = [bx+si]
= [FF00 + 00FF]
= [FFFF]
b. mov ax, [BX+1024] (1024 is in decimal)
Sol :
Effective address = [bx + 1024]
After converting 1024 into hexadecimal
Effectice address = [FF00 +400 ]
= [FF00 + 400]
= [10300]