1.1-1.26qufac dikjo wpnry slgxz tbhme v1.27-1.44jbfji edgcj dcjbf haa3.1、1: t0=02: t1=13: if (a0<t1) t2=1; else t2=04: if (t2= =0) goto finish5: t0=t0+t16: t1=t1+27: jump to loop8: v0=t03.2、Find the element which appears most times, $v0 stores the the maxium number of theidentical integer in the array, $v1 stores its value.3.4、addi $t0,$t1,1003.5、lui $t1,61addi $t1,$t1,2304lw $t2,11($t1)add $t2,$t2,$t0sw $t2,10($t1)3.6、Loop: lw $v1, 0($a0)sw $v1, 0($a1)addi $a0, $a0, 4addi $a1, $a1, 4beq $v1, $zero, Endaddi $v0, $v0, 1j LoopEnd:3.7、1: I format,3: I format,4: I format,5: I format,6: I format,3.8、v0=0;v1=a0;v0++;a1=v1;a0++;a1++;if(v1= =0) goto loop;v0--;3.9、bne $t0,$s5,exitloop: add $t1,$s3,$s3add $t1,$t1,$t1add $s3,$s3,$s6lw $t0,0(t1)beq $t0,$s5,loop exit:3.11、slti $t5,$t0,1bne $t0,1,exitloop: addi $t0,$zero,0add $t4,$t0,$t0add $t4,$t4,$t4lw $t1,0($a0)add $t2,$t4,$t1add $t3,$a1,$t1sw $t2,0($t3)addi $t0,$t0,1slti $t5,$t0,101beq $t5,1,loopexit:3.23、loop: addi $t0,$zero,98lb $t1,0($a0)addi $a0,$a0,1beq $t1,$zero,exitbne $t1,$t0,loopbeq $t1,$t0,nextsub $v0,$a0,1exit: add $v0,$zero,$zeronext:3.24、bcount: addi $sp,$sp,-12sw $ra,8($sp)sw $a0,4($sp)sw $s0,0($sp)add $s0,$zero,$zeroNext: jal bfindadd $a0,$v0,$zerolb $t0,0($a0)beq $t0,$zero,End #encounter nulladdi $s0,$s0,1addi $v0,$v0,1j NextEnd: add $v0,$s0,$zerolw $s0,0($sp)lw $a0,4($sp)lw $ra,8($sp)addi $sp,$sp,12jr $ra4.1、000000000000000000000010000000004.2、111111111111111111111100000000014.4、-5004.5、-14.6、21474836474.10、loop: sub $t3,$zero,$t3slt $t0,$t3,$zerobeq $t0,$zero,loopadd $t2,$t3,$zero how to use 3 instructions?4.13、If A_lower is positive, A_upper_adjusted equals to A_upper.If A_lower is negative, A_upper_adjusted equals to A_upper+1.4.17、addu $t0, $t3,$t4sltu $t2,$t0,$t34.22、srl $t0,$s0,8andi $s1,$t0,40954.23、when it overflew , set should be converted .4.40、xor $s0,$s0,$s1xor $s1,$s1,$s0xor $s0,$s0,$s14.46、ripple carry c1=a0*c0+b0*c0+a0*b0 2Tc2=a1*c1+b1*c1+a1*b1 4Tc3=a2*c2+b2*c2+a2*b2 6TThe worst case : c4=a3*c3+b3*c3+a3*b3 8TCarry look ahead : gi=ai*bi T pi=ai+bi Tc1=c2=c3=c4=3T4.53、4.54、step1 6.42*101=0.642*102step2 9.51*102+0.642*102=10.152*102step3 10.152*102=1.0152*102step4 1.0152*103 is rounded to three significant digits 1.015*1034.55、6.42X101=0.642X1020.642+9.51=10.1526.42X101+9.51X102=10.152X102=1.0152X103so the answer is 1.02X1035.1、RegDst=0 lw sw and branchAluSrc=0 R-format beqMemtoReg=0 R-format sw and beqZero=0 R-format sw and lw5.2、RegDst=1 R-format sw and beqAluSrc=1 lw and swMemtoReg=1 lw sw and beqZero=1 R-format lw and sw5.6、修改两个信号：MemtoReg: 00 :Alu result01 :From Mem data10 :PC+4RegDst: 00 : Instruction[20-16]01 :Instruction[15-11]10 :$ra寄存器5.8、lw $t0 ,$t1,$t25.14、P374页图Instruction a b cR-formatlwswbeqjlongest8 10 10time5.165.17、5.22、jr rsIt is an R-type instruction.So the changes can be made like following .5.24、M1=5*22%+4*11%+4*49%+3*16%+3*2%=4.04M2=4*22%+4*11%+3*49%+3*16%+3*2%=3.33M3=3*22%+3*11%+3*49%+3*16%+3*2%=3So the CPU excution time isT1=CPI*Clockcycletime=4.04/(500*106 )=8.08nsT2=3.33/(400*106 )=8.325nsT3=3/(250*106 )=12nsHence M1 is the fastest machine.5.26、beq $t3,$zero,Exitadd $t3,$t3,$t3add $t3,$t3,$t3 #get 4*$t3add $t3,$t3,$t2 #caculate the address of destination[$t3] loop: lw $t4,0($t1)sw $t4,0($t2)addi $t1,$t1,4addi $t2,$t2,4bne $t2,$t3,loopExit:branch 3 lw 5 and the rest are 4The number of executed instructions is 4+5*100=504The nmber of cycles is3+3*4+(5+4*3+3)*100=20155.27、1. According to the figure above there will be two changes.1>Extend the multiplexor for ALUSrcA by adding ALUout to it .We mark it with 102> We add a multiplexor between B and Memory write data ,connect memory dataregister with the multiplexor and mark it with 1.The multi is controled by the ]signal bcp.2. By making these changes ,we can directly increase the address $t1,$t2,in ALUout by 4When the memory data loading from $t1is read into memory data register, we canquickly write it into memory by setting bcp to 0.3. By estination ,we can cut down clock cycles distinctively by implementing theinstruction in hardware above .The performance increase comes from adding internal registers.7.7、7.8、7.9、212*(32*4+16+1)=580Kb7.11、a. 1+16*10+16*1=177 clock cyclesb. 1+4*10+4*1=45 clock cyclesc. 1+4*10+16*1=57 clock cycles7.12、The CPI of a is 1.2+0.5%*177=2.085The CPI of b is 1.2+0.5%*45=1.425The CPI of c is 1.2+0.5%*57=1.485The processor using wide memory is 2.085/1,425=1.46 times faster than using nerrowand 1.04 times faster than the one using interleaved memories.7.15、AMAT=(1+0.05*20)*2ns=4ns7.20、7.22、7.24、The number of sets in the cache: S/ABThe number of index bits in the address: log2(S/AB)The number of bits needed to implement the cache : Tag =k- log2(S/AB)-log2B7.27、Cache1: X+(4%+0.5*8%)*7=2So,X=1.44Cache2: 1.44+(2%+0.5*5%)*10=1.85Cache3: 1.44+(2%+0.5*4%)*10=1.84Thus,machine1spemds the most cycles on cache mises.7.32、Page size is 16-KB=214B240/214=226Total size of bits 226*(36-14+4)=208MB7.33、Virtual addressCache hit8.1 A 200 1000/5B 150 750/58.2 (1)n*5*(1/1000)+n*(1/10)*(20/1000)=1 n=142(2)5*n*(1/750)+500/5*(1/10)*18*(1/1000)+(5n-500)/5/5/10*25/1000 =1 n=1168.3 7和8距离近的那张图，原因是路程短8.10 a=1+19.7ms/5ns+2+2=3.9*10^6 cyclesb=3.94*10^6*128/4=1.26*10^8cycles) 32/b=50.798/a=12.750.79 12.78.18 17.84。
