Computer Architecture & Organization Assignment

This computer architecture and organization assignment tests understanding of number representation in different systems (ones complement, two's complement, signed magnitude, unsigned magnitude). Students must convert hexadecimal to octal, binary fractions to decimal, and integers to binary. Additionally, the assignment involves proving a Boolean algebra identity and designing a circuit for a grading system based on passing criteria for multiple assessments.

UNIVERSITY AFFILIATION
DEPARTMENT OR FACULTY
COURSE CODE
COMPUTER ARCHITECTURE AND ORGANIZATION
STUDENT NAME
STUDENT REGISTRATION NUMBER
DATE OF SUBMISSION
1

QUESTION 1
(a) Given a tiny computer that has a word size of 8 bits, what are the smallest negative numbers
and the largest positive numbers that this computer can represent in each of the following
representations? (2+2+2+2=8 marks)
i. One's complement
For the one’s complement,
01111111=27 −1=127 … for largest positive number
10000000=−(01111111)=−(27−1)=−127 … for largest negative number
ii. Two's complement
For the two’s complement,
01111111=27 −1=127 … for largest positive number
10000000=−(10000000)=−(27)=−128 … for largest negative number
iii. Signed magnitude
−12710 ¿+12710
The leftmost bit is the sign bit. It represents the sign of the magnitude,
Binary equivalent of 12710=111 11112
Therefore,
−12710=1 111 11112
to
+12710 =0 11111112
iv. Unsigned magnitude
smallest unsigned=010 , largest unsigned=25510
000000002=010 ¿ 111111112=25510
(b) Convert the followings: (7 marks)
i. 0x5AB into Octal
0 x 5 AB∈hexadecimal=5 AB16
Conversion: Hexadecimal to Binary to Octal
2

To binary,
0 x 5=0101 b
0 xA=1010 b
0 xB=1011b
Therefore,
0 x 5 AB=010110101011 b
To octal,
¿ 010 110 101011 b → octal
¿ 26538
ii. 101101.1012 into Decimal
Factor 25 24 23 22 21 20 2-1 2-2 2-3
Binary 1 0 1 1 0 1 . 1 0 1
Decimal 32 0 8 4 0 1 . 0.5 0 0.125
¿ 32+8+ 4+1+0.5+ 0.125
¿ 45.62510
iii. 12348 into Binary
¿ octal ¿ binary
1 2 3 4
001 010 011 100
¿ 001 010 011100 b
¿ 10100111002
iv. 679810 into Base 5 number
¿ 679810
¿ 6798
5 =1359+ 3
5 →3
¿ 1359
5 =271+ 4
5 → 4
¿ 271
5 =54+ 1
5 → 1
¿ 54
5 =10+ 4
5 → 4
¿ 10
5 =2+0
5 → 0
3

¿ 2
5 → 2
¿ 2041435
v. 976.6310 into Binary
Dividing the section into two,
The whole part,
¿ 976
2 =488 R → 0
¿ 488
2 =244 R → 0
¿ 244
2 =122 R → 0
¿ 122
2 =61 R → 0
¿ 61
2 =30 R → 1
¿ 30
2 =15 R → 0
¿ 15
2 =7 R → 1
¿ 7
2 =3 R→ 1
¿ 3
2 =1 R → 1
¿ 1
2 =1 R →1
¿ 11110100002=97610
The fractional part,
¿ .63∗2=1 .26
¿ .26∗2=0 .52
¿ .52∗2=1.04
¿ .04∗2=0.08
4

¿ .08∗2=0 .16
¿ .16∗2=0 .32
¿ .32∗2=0 .64 …. starts recurring
¿ 0.10100002=0.6310
Therefore,
¿ 1111010000.10100002
vi. 1001001011 into Hexadecimal
Dividing into four-bit words,
¿ 0010 0100 1011
¿ 0 x 24 B
vii. 10011110 (8-bit 2’s complement representation) to Decimal
1 0011110=signed value ( negative )
Changing the sign to get the magnitude of the number,
1 0011110→ 011000012 +12=011000102
Converting to decimal,
011000102 →9810
For a negative number,
¿−9810
QUESTION 2
(a) Using basic Boolean algebra identities for Boolean variables X and Y, prove that
X ’ (X +Y )+( XX+Y )(Y ’+ X )=Y +X . Please show all steps and mention the identities
used. (6 marks)
Simplifying the Boolean algebra,
X ’ (X +Y )+( XX+Y )(Y ’+ X )=Y +X
( x' x +x' y ) + ( xx + y ) ( y' + x )
But , x' x → 0
Therefore,
¿ x' y + ( xx + y ) ( y'+ x )
5

¿ x' y + ( xx y'+ xxx + y y' + xy )
But , xxx → x
¿ yy ' →0
while , xx y' → x y'
Hence,
¿ x' y +x y' + x + xy
¿ x' y + x ( y' +1 ) +xy
¿ ( y' +1 ) =1
¿ x' y + x + xy
¿ y ( x'+ x ) + x
¿ ( x'+ 1 )=1
¿ y +x
(b) A subject has 4 assessments, such as Assignment, Blog, Discussion Forum and Quiz. In
order to pass in the subject, a student must get as below:
 Assignment component must pass (50 or above)
 Any 2 more assessments must pass
 Fail otherwise
Construct a truth table and find the minimized Boolean function. Draw a circuit diagram
for the Boolean function. (9 marks)
Let,
A=assignment component
B=Blog
C=discussion Forum
D=Quiz
A B C D Output
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
1 1 0 0 0
6

1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
¿ ABCD+ ABC D' + ABC' D+ A B' CD
Using the Karnaugh map to simplify the output,
y= ACD+ ABC + ABD
The circuit design,
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