CS2100
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CS2100 – COMPUTER ORGANISATION FINAL EXAMS
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CS2100 Computer Organization Final Exam: Questions and Solutions, Exams of Computer Programming
A final exam for a computer organization course (cs2100), covering topics such as boolean functions, mips code, sequential and combinational circuits, and cache memory. It includes multiple-choice questions, multiple-response questions, and problem-solving tasks. The exam assesses understanding of digital logic design, assembly language programming, and memory hierarchy concepts. It is designed for university students studying computer engineering or computer science, providing a comprehensive review of key concepts in computer organization. The exam questions require students to apply theoretical knowledge to practical problems, enhancing their analytical and problem-solving skills. The document serves as a valuable resource for exam preparation and self-assessment in computer architecture.
Typology: Exams
2024/2025
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CS2100 – COMPUTER ORGANISATION FINAL EXAMS
(b)
(c)
A : 000
B : 001 00
C : 001 01 xx, 001 10 xx, 001 11 xx 3×4= {010 – 111} xxxx 6×16 = 96 Total = 1 + 1 + (12 + 96) = 110.
1 1 X 1
X X 1 0
0 X X 0
MUX
S 0
S 1
MUX
S 0
S 1
S 2
E^7
S 2
S 1
S 0
3x
DEC
(a)
(b)
(c)
(d)
A B
HA
C
S
A B C in
FA
C out S
A
HA B
C S
.data A: .word 10,21,12,17,9,1,20,33 # after: 10,18,12,2,7,-1,20, B: .word 100,3,20,15,2,2,65, size: .word 8
.text main: la $s0, A la $s1, B
**# $s0 is the base address of array A
$s1 is the base address of array B
Box 1**
syscall li $v0, 10 syscall
# system call to print
# system call to exit
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(a)
(b)
(c)
(d)
(e)
(f)
(g)
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**# $s0 = array A; $s1 = arrayB; $s2 = size; $s3 = A[i]; $s4 = B[i]
$s5 = count
for (i = 0 to size-1) {
if (A[i] is odd) {
A[i] = A[i] - B[i];
count = count + 1;
}
}
.data A: .word 10,21,12,17,9,1,20,33 # after: 10,18,12,2,7,-1,20, B: .word 100,3,20,15,2,2,65, size: .word 8**
end: li $v0, 1 # system call code for print_int add $a0, $s5, $0 # transfer $s5 to $a0 for printing syscall # print $s li $v0, 10 # system call code for exit syscall
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0
4 1
3
2
Present state Input Next State
F G x DF = F +^ DG = G + 0 0 0 1 1 0 0 1 0 0 0 1 0 1 0 0 1 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 1 1 0
(a)
(b)
(c)
(d)
(e)
(f)
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