Enhancing Computer Performance: Volume Calculation and Instruction Optimization

Answers to Test 2, Question 1 • The functions receives the radius and height of a cylinder (גליל) and returns its volume (נפח). const float PI=3.14 float cyl_vol(float r, float h) { return(r*r*h*PI); }

(CarryIn (0 if addition 1 if subtraction CarryIn Operation Operation a a 0 0 Result Result b b 1 1 C C a a r r r r y y O O u u t t Question 2

Question 3 • ETA = 324M * 2.5 * 2ns = 1620M ns = 1.62 sec • ETB = 290M *3.2 *(1/550M)ns = 1687M ns = 1.687s • A is faster than B by 4% (1687/1620) • Computer A: FE=1.00 (all the instructions are effected), SE = 2/1.8 = 1.11.(1.00/1.11 + 1 -1)*1.62 = 1.458 sec • Computer B: FE=0.50 , SE=3.2/2.1 = 1.52(0.50/1.52 + 0.50)*1.687 = 1.398 secNow B is faster than A by 4%

Question 4 • mov $t0,$t1 ->add $t0,$t1,$zero • bgt $t0,$t1,L1 -> slt $t2,$t1,$t0 bne $t2,$zero,L1 • seq $t0,$s1,$s2 ->bne $s1,$s2,L1 addi $t0,$zero,1 j L2 L1: addi $t0,$zero,0 L2: • div $t0,$t1,$t2 -> div $t1,$t2 mflo $t0

Question 5 • Register $s0 contains0111 1111 1111 1111 1111 1111 1111 1111 Register $s1 contains1000 0000 0000 0000 0000 0000 0000 0001 slt $t0,$s0,$s1 compares 2 signed numbers. $s0 contains 231-1 and $s1 contains -231-1. 231-1 > -231-1 so $t0 equals 0. sltu $t1,$s0,$s1 compares 2 unsigned numbers. $s0 contains 231-1 and $s1 contains 231+1 . 232-1 < 231+1 so $t1 equals 1. • 0x80000000 = 2147483648 so both $t0 and $t1 will contain 0.

Question 6 • 16KB=214 , thus 14 address lines are needed. The width of the SRAM is 4 bits, so 4 input lines and 4 output lines are needed. • 2MB = 221, thus 21 address lines, 1 data input line, and 1 data output line are needed. • To read 4 words takes 1 + 4*12 + 4*2 = 57 cycles • If memory is interleaved in 2 banks we can read each 2 words in parallel, although we will have to still send them one at a time. So to read 4 words now takes 1 + 2*12 + 4*2 = 33 cycles.

Question 7 sum: subi $sp,$sp,8 # make room for 2 items sw $ra,4($sp)# push the return address sw $a0,0($sp)# push the argument n slt $t0,$a0,1 # test for n<1 beq $t0,$zero,L1 # if n>=1 goto L1 li $v0,0 # pseudoinstruction $v0=0 addi $sp,$sp,8 # pop 2 items off stack jr $ra The following is the recursive call to sum(n-1) L1: subi $a0,$a0,1 # n-- jal sum # call sum(n-1) lw $a0,0($sp) # return from fact(n-1) lw $ra,4($sp) # pop n and return address addi $sp,$sp,8 # pop 2 items off stack add $v0,$a0,$v0 # return n + sum(n-1) jr $ra

Question 8 • What is reduced is the number of bits for the opcode, this reduces the number of possible instructions. • RISC processors have instructions that are the same length (1 word) as opposed to GPRs that have variable length instructions.RISC processors have 3 operand instructions as opposed to 2 operand instructions.RISC processors access memory only through Load/Store instructions, as opposed to GPRs where all instructions (even ALU instructions) can have memory operands. • RISC programs might have more instructions.

Enhancing Computer Performance: Volume Calculation and Instruction Optimization