CS401 Assignment No. 1 Fall 2011 solution
Saturday, October 29, 2011 Posted In CS and IT Edit ThisQuestion 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)
Solution: -
Accessible memory addresses = 2number of address bits
224 = 16777216 bytes
16777216 / 1024 = 16384 KB
16384 / 1024 = 16 MB
b)
How many address bits are required for accessing 1GB RAM? (Show the step(s) for calculation of required address bits) (2.5 marks)
Solution: -
As you know 1GB means 1024MB so we can write it as
1GB = 1024 x 1MB
As you know that 1MB is 220
So we can write the expression as
1024 x 220
This can be further simplified as
210 x 220
=230
Hence, 30 address lines are required to access 1GB RAM.
Question No. 2:
What are the contents of memory locations 0151, 0152, 0153, ………….,0158 if 0151 is starting address for vusolutions Label1. (1 mark for each location)
Label1: dw 8494
db 42
dw 54
dw 7500
db 01
Solution: -
Memory location Contents
0151 94
0152 84
0153 42
0154 54
0155 00
0156 00
0157 75
0158 01
Question No. 3:
a)
Calculate physical address using the following segment offset pairs. (1 mark each)
1. 00EA:02A4
2. 0100:AA23
3. D3B8:F222
4. 00A0:1234
5. 8FEf:0FFF
Solution: -
Memory Location | Contents |
0151 | 94 |
0152 | 84 |
0153 | 42 |
0154 | 54 |
0155 | 00 |
0156 | 00 |
0157 | 75 |
0158 | 01 |
Question No. 3:
a)
Calculate physical address using the following segment offset pairs. (1 mark each)
1. 00EA:02A4
2. 0100:AA23
3. D3B8:F222
4. 00A0:1234
5. 8FEf:0FFF
Solution: -
1)
00EA0
002A4 +
_______
01144 = physical address
2)
01000
0AA23 +
_______
0BA23 = physical address
3)
D3B80
0F222 +
______
E2DA2 = physical address
4)
00A00
01234 +
________
01C34 = physical address
5)
8FEF0
00FFF +
_________
90EEF = physical address
b)
What is effective address generated by the following instructions? Every instruction is independent of others. vusolutions Initially bx = 0x0101, bp=0x0222, si=0x1234, var1=0x1771 (1 mark each)
1. mov ax, [bx+si]
2. mov ax, [bx+100] (100 is in decimal)
3. mov ax, [bp+si]
4. mov ax, [var1+bp]
5. mov ax, [si+var1]
Solution: -
1)
Effectice address = [bx+si]
= [0101 + 1234]
= [1335]
2)
Effectice address = [bx + 100]
After converting 100 into hexadecimal
Effectice address = [bx + 64]
= [0101 + 64]
= [0165]
3)
Effective address = [bp+si]
= [0222 + 1234]
= [1456]
4)
Effective address = [var1+bp]
= [1771 + 0222]
= [1993]
5)
Effective address = [si + var1]
= [1234 + 1771]
= [29A5]
Q.2:
Answer:
A) Segment: 0000: Range: 00000-0FFFF
B) Segment: FFFF: Ranges: FFFF0-FFFFF and 00000-0FFEF
