CS401 assignment solution 1

Question No. 1:
a)
Suppose your computer has a processor with 24-bit address lines. What is
maximum amount of memory that can be attached in your system? (Show the
step(s) for calculation of maximum addressable memory) (2.5 marks)
2^24 = 16777216
16777216/8 = 2097152bytes
2097152/1024=2048KB
2048/1024 = 2MB
b)
How many address bits are required for accessing 1GB RAM? (Show the step(s)
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for calculation of required address bits) (2.5 marks)
1GB*1024=1024MB
1024MB*1024=1048576KB
1048576KB*1024=1073741824bytes
1073741824Bytes*8=8589934592bits
2^33 = therefore 33 address lines exist
Question No. 2:
What are the contents of memory locations 0151, 0152, 0153, ………….,0158 if
0151 is starting address for Label1. (1 mark for each location)
Label1: dw 8494 = 0151
db 42 = 0153
dw 54 = 0154
dw 7500 = 0156
db 01 = 0158
Question No. 3:
a)
Calculate physical address using the following segment offset pairs. (1
mark each)
1. 00EA:02A4=01144=1144
2. 0100:AA23=0BA23=BA23
3. D3B8:F222=E2DA2=2DA2
4. 00A0:1234=01C34=1C34
5.8FEf:0FFF=90EEF=0EEF
b)
What is effective address generated by the following instructions? Every
instruction is independent of others. Initially bx = 0x0101, bp=0x0222,
si=0x1234, var1=0x1771
(1 mark each)
1. mov ax, [bx+si] =[01010+1234]=[02244]=[2244]
2. mov ax, [bx+100] (100 is in decimal)
First convert the 100 to hex 64 than add[01010+64]=[01074]=[1074]
3. mov ax, [bp+si] =[02220+1234]=[03454]=[3454]
4. mov ax, [var1+bp] =[01771+0222]=[01993]=[1993]
5. mov ax, [si+var1] =[12340+1771]=[13AB1]=[3AB1]