CSE 140 Questions and Answers Updated 2026

CSE 140 Questions and
Answers Updated 2026
for(iB=B0;BiB<B100;Bi++)

{

sum+=A[i];

}



$s0B=B&A

$v0B=BsumB-BAnswerBaddB$t0,B$t0,B$zeroB#letBiB=B0

addiB$t1,B$zero,B100B#tempB=B100

LOOP:BlwB$t3,B0($s0)B#temp1B=BA[i]

addB$v0,B$v0,B$t3B#sumB+=Btemp1

addiB$s0,B$s0,B4B#addrBofBA[i+1]

addiB$t0,B$t0,B1B#iB=Bi+1

bneB$t1,B$t0,BLOOPB#ifBiB<B100



ToBimplementBtheBfollowingBCBcode,BwhatBareBthe

instructionsBthatBweBshouldBputBinBtheBbox?



for(iB=B0;BiB<B100;Bi++)

A[i]B=Bi;



addB$t0,B$zero,B$zero

addiB$t1,B$zero,B100

LOOP:B??

??

addiB$t0,B$t0,B1

,beqB$t0,B$t1,BLOOPB-BAnswerBswB$t0,B0($s2)

addiB$s2,B$s2,B4



WhichBofBtheBfollowingBisBNOTBcorrectBaboutBtheseBtwo

ISAs?

A.Bx86BprovidesBmoreBinstructionsBthanBMIPS

B.Bx86BusuallyBneedsBmoreBinstructionsBtoBexpressBaBprogram

C.BAnBx86BinstructionBmayBaccessBmemoryBforB3Btimes

D.BAnBx86BinstructionBmayBbeBshorterBthanBaBMIPSBinstruction

E.BAnBx86BinstructionBmayBbeBlongerBthanBaBMIPSBinstructionB-
BAnswerBB.Bx86BusuallyBneedsBmoreBinstructionsBtoBexpressBaBprogram



AssumeBthatBweBhaveBanBapplicationBcomposedBwithBaBtotalBofB500000

instructions,BinBwhichB20%BofBthemBareBtheBload/storeBinstructionsBwithBan

averageBCPIBofB6Bcycles,BandBtheBrestBinstructionsBareBintegerBinstructions

withBaverageBCPIBofB1Bcycle.BIfBtheBprocessorBrunsBatB1GHz,BhowBlongBis

theBexecutionBtime?

A.B500000Bns

B.B1000000Bns

C.B1750000Bns

D.B3500000Bns

E.BNoneBofBtheBabove

500000B-BAnswerBB.B1000000Bns



500000B*B(0.8*1+0.2*6)B*B1BnsB=B1000000Bns



http://cseweb.ucsd.edu/classes/su12/cse141-a/slides/Performance_20120809.pdf

, •BAssumeBthatBweBhaveBanBapplicationBcomposedBwithBaBtotalBofB500000

instructions,BinBwhichB20%BofBthemBareBtheBload/storeBinstructionsBwith

anBaverageBCPIBofB6Bcycles,BandBtheBrestBinstructionsBareBinteger

instructionsBwithBaverageBCPIBofB1Bcycle.

•BIfBweBdoubleBtheBclockBrateBtoBbeB2GHzBwithoutBimproveBthe

memoryBlatency,BtheBaverageBCPIBforBload/storeBinstructionBwillBalso

beBdoubledBtoB12Bcycles.BWhat'sBtheBperformanceBimprovementBafter

thisBchange?

A.BNoBchange

B.B1.25

C.B1.5

D.B2

E.BNoneBofBtheBaboveB-BAnswerBB.B1.25



ETnewB=B500000B*B(0.8*1+0.2*12)B*B0.5BnsB=B800000Bns

SpeedupB=BETold/ETnew=1000000/800000B=B1.25



http://cseweb.ucsd.edu/classes/su12/cse141-a/slides/Performance_20120809.pdf



WhyBdoesBanBIntelBCoreBi7B@B2.8BGHzBusually

performBbetterBthanBanBIntelBCoreB2BExtremeB@B3.2

GHzBorBAMDBPhenomBIIBX4@3.4GHz?

A.BBecauseBtheBinstructionBcountBofBtheBprogramBareBdifferent

B.BBecauseBtheBclockBrateBofBPhenomBIIBisBhigher

C.BBecauseBtheBCPIBofBCoreBi7BisBbetter

D.BBecauseBtheBclockBrateBofBPhenomBIIBisBhigherBandBCPIBofBCoreBi7BisBbetter

E.BNoneBofBtheBaboveB-BAnswerBC.BBecauseBtheBCPIBofBCoreBi7BisBbetter