Detailed Solution: Computer Organization and Architecture Homework

This document presents a solved homework assignment focusing on Computer Organization and Architecture. The solution covers various topics including number system conversions, base conversions (decimal, binary, octal, hexadecimal), and Boolean algebra simplifications. Specifically, it addresses problems related to determining the base of a number system given an equation, converting numbers between different bases (base-16 to base-3, base-7 to base-2, base-10 to octal, and base-8 to decimal), and determining the range of representable numbers in one's complement, two's complement, and signed magnitude representations. The solution also includes simplification and analysis of digital circuits using Boolean algebra and De-Morgan's laws. Desklib provides this resource to aid students in understanding key concepts and improving their performance in computer architecture courses.

Running head: COMPUTER ORGANIZATION AND ARCHITECTURE
Computer Organization and Architecture
Name of Student-
Name of University-
Author’s Note-

1COMPUTER ORGANIZATION AND ARCHITECTURE
Answers for Question 1:
a) For calculating the X value:
(152) x= (6A) 16
 X2 + (5 * X1) + (2 * X0) = (6 * 161) + (10 * 160)
 X2 + 5X + 2 = 106
 X2 + 5X - 104 = 0
 X2 + 13X - 8X – 104 = 0
 X(X + 13) – 8(X + 13) = 0
 (X - 8) (X + 13) = 0
 X = 8 and X = -13
Therefore, the value that can be determined for X is 8.
(152)8= (6A) 16
b) Conversions:
i) BED16 converting to base-3
The value of BED = (B * 162) + (E * 161) + (D * 160)
= 2816 + 224 + 13
= (3053)10
(3053)10 =

2COMPUTER ORGANIZATION AND ARCHITECTURE
So, (BED)16 = (11012002)3
ii) 3217 conversion to base-2
(321)7 = (3 * 72) + (2 * 71) + (1 * 70)
= (162)10
Again, (162)10 =
After converting the value of (162)10 in binary is (10100010)2

3COMPUTER ORGANIZATION AND ARCHITECTURE
iii) (1235)10 conversion to octal representation
Therefore, after converting the value of (1235)10 = (2323)8
iv) 21.218 conversion to decimal representation
21.218 = (2 * 81) + (1 * 80). (2 * 8-1) + (1 * 7=8-2)
= 17 + 0.25 + 0.015625
= 17.265625
c) i) In a computer of 3 bit size, lowest negative number that is possible for representing one’s
complement number = 100
In a computer of 3 bit size, highest positive number that is possible for representing one’s
complement number = 011
ii) In a computer of 3 bit size, lowest negative number that is possible for representing two’s
complement number = 101

4COMPUTER ORGANIZATION AND ARCHITECTURE
In a computer of 3 bit size, highest positive number that is possible for representing two’s
complement number = 011
iii) In a computer of 3 bit size, lowest negative number that is possible for representing signed
magnitude number = 111
In a computer of 3 bit size, highest positive number that is possible for representing signed
magnitude number = 011
Answers of Question 2:
a) From the given question, output of L.H.S is
a b c d
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0
From the given R.H.S., the output of the circuit is

5COMPUTER ORGANIZATION AND ARCHITECTURE
a b c d e
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
Therefore, the output of given L.H.S. is same as the output of the given R.H.S.
b) The given circuit that can be minimized from given circuit is:
c) X’ + Y’ + XYZ’
= X’ + Y’ + (X’ + Y’ + Z)’ [De-Morgan’s Law]

6COMPUTER ORGANIZATION AND ARCHITECTURE
= (XY (X’ + Y’ + Z))’ [De-Morgan’s Law]
= (XX’Y + XYY’ + XYZ)
= (0 + 0 + XYZ)
= (XYZ)’
= X’ + Y’ + Z’ [De-Morgan’s Law]
Hence, X’ + Y’ + XYZ’ = X’ + Y’ + Z’ [PROVED]

7COMPUTER ORGANIZATION AND ARCHITECTURE
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