Computer Organization and Architecture Assignment Solution Analysis

This document provides a solution to a computer organization and architecture assignment. The solution addresses several key topics, including number representation in different formats (one's complement, two's complement, signed magnitude, and unsigned magnitude) and conversions between number systems (binary, octal, hexadecimal, and decimal). The assignment also includes a boolean algebra problem, demonstrating simplification using algebraic manipulation. Furthermore, the document presents a truth table and a minimized boolean expression derived from it, along with a corresponding circuit diagram, likely representing a digital logic circuit. This comprehensive solution provides a detailed analysis of fundamental concepts in computer organization and architecture.

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

1COMPUTER ORGANIZATION AND ARCHITECTURE
Question 1.
(a)
I. In a computer of 8 bit, smallest negative of one’s compliment is -127 and largest number
is 127.
II. In a computer of 8 bit, smallest negative of two’s compliment is -128 and largest number
is 127.
III. In a computer of 8 bit, smallest negative of signed magnitude is -127 and largest number
is 127.
IV. In a computer of 8 bit, smallest negative of unsigned magnitude is 0 and largest number
is 255.
(b)
I. (5AB)16 = (2653)8
II. (101101.101)2 = (45.625)10
III. (12348)10 = (1100000111100)2
IV. (679810)10 = (133223220)5
V. (976.6310)10 = (1111010000.10100)2
VI. (1001001011)2 = (24B)16
VII. (10011110) = -(98)10
Question 2:
(a) X’(X+Y)+(XX+Y)(Y’+X)=Y+X
L.H.S. – X’(X+Y) + XXY’+ XXX+ YY’+ XY [Distributive law]
= X’X + X’Y + XXY’+ XXX + YY’+ XY [Distributive law]

2COMPUTER ORGANIZATION AND ARCHITECTURE
= 0 + X’Y + XY’ + X + 0 + XY [Since, X’X = 0 AND XX = X]
= X’Y + XY’ + X + XY
= X’Y + XY’ + X (1 + Y)
= X’Y + XY’ + X [Since, 1 + Y = 1]
= X’Y + X (Y’ + 1)
= (X + X’) (X + Y)
= (X + Y) [Since, X + X’ = 1] = R.H.S.
(b) Let, the Assignment, Blog, Discussion Forum and Quiz be A, B, C, and D. P be the
possibility of students who can pass the exam.
So, the truth table will be:
A B C D P
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1

3COMPUTER ORGANIZATION AND ARCHITECTURE
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
The expression for students passing in the following table will be
AB’CD + ABC’D + ABCD’ + ABCD
Minimizing the equation we get,
A [D (B’C + BC) + BCD’]
The circuit diagram for the equation is