CS2100 Exam Questions Analysis and Cache Mapping Scenarios
Explore past year exam questions from CS2100, including code analysis, instruction execution, data forwarding scenarios, and cache mapping examples. Understand concepts like rotating array elements, instruction sequences, data caching, and more.
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CS2100 Past Year Exam Questions (Prepared by Aaron Tan)
AY2015/16 Semester 1 Q11 addi $t1, $a0, 12 # I1 lw $t0, 12($a0) # I2 Loop: lw $t2, -4($t1) # I3 sw $t2, 0($t1) # I4 addi $t1, $t1, -4 # I5 bne $t1, $a0, Loop # I6 sw $t0, 0($t1) # I7 (a) $a0 stores the base addr of integer array A. What does the code do? Rotates A[0], A[1], A[2], A[3] one element to the right. A[0] A[1] A[2] A[3] (b) If first instruction executed is I1, which is the seventh instruction executed? Instruction I3. 2
AY2015/16 Semester 1 Q11 addi $t1, $a0, 12 # I1 lw $t0, 12($a0) # I2 Loop: lw $t2, -4($t1) # I3 sw $t2, 0($t1) # I4 addi $t1, $t1, -4 # I5 bne $t1, $a0, Loop # I6 sw $t0, 0($t1) # I7 (c) No data forwarding. Branch resolved at stage 4. Total number of cycles = 41 1 2 D F 3 E D 4 M W E F 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 I1 F I2 I3 I4 I5 I6 I3 M W D E M W F D F E D M W E M W F D E M W F D E M W 3
AY2015/16 Semester 1 Q11 addi $t1, $a0, 12 # I1 lw $t0, 12($a0) # I2 Loop: lw $t2, -4($t1) # I3 sw $t2, 0($t1) # I4 addi $t1, $t1, -4 # I5 bne $t1, $a0, Loop # I6 sw $t0, 0($t1) # I7 (d) With data forwarding. Branch resolved at stage 2, predicted taken. Total number of cycles = 23 1 2 D F 3 E D F 4 M W E D F 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 I1 F I2 I3 I4 I5 I6 I3 M W E D F M W E D M W E F M W D F E D M W E M W 4
AY2015/16 Semester 1 Q12 Base addresses: A = 0x00000080 B = 0xFFFF0040 C = 0x12345688 int sum = 0; for (int i=0; i<32; i++) { sum = sum + (A[i]*B[i]) C[i]; } Direct-mapped cache, 8 blocks, 4 words per block. First four iterations: A[0]..A[3] mapped to block 0. B[0]..B[3] mapped to block 4. C[0], C[1] mapped to block 0, C[2], C[3] mapped to block 1. 0x00000080 = 0000 00001 000 0000 0xFFFF0040 = 0000 00000 100 0000 0x12345688 = 0101 01101 000 1000 A[0] (miss), A[1] (miss), A[2] (miss), A[3] (hit). B[0] (miss), B[1]..B[3] (hits) C[0] (miss), C[1] (miss), C[2] (miss), C[3] (hit). Hit rate = 5/12 5
AY2015/16 Semester 1 Q12 int sum = 0; for (int i=0; i<32; i++) { sum = sum + (A[i]*B[i]) C[i]; } A: 0000 00001 000 0000 B: 0000 00000 100 0000 C: 0101 01101 000 1000 Block 0 C[30] C[31] unknown unknown Block 1 B[20] B[21] B[22] B[23] Block 2 B[24] B[25] B[26] B[27] Block 3 B[28] B[29] B[30] B[31] Block 4 A[16] A[17] A[18] A[19] Block 5 A[20] A[21] A[22] A[23] Block 6 A[24] A[25] A[26] A[27] Block 7 A[28] A[29] A[30] A[31] 6
