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1、Chapter12Solutions12.1VonNeumannarchitecturesrelyononeaddressbusandonedatabus,soactivitiesrequiringbususagemustwaituntilthebusisfree.Harvardarchitectureshavemultipleaddressanddatabuses.Asaresult,lessbuscontentionoccurs,anditemscanbecollectedfrommemoryinparallel.Also,Harvarda
2、rchitectureshaveseveralmemoriesthatcancontainprograminstructions,data,orboth;vonNeumannarchitecturesfeatureonlyonememorythatmustcontainbothprograminstructionsanddata.12.2Pipeliningreferstotheoverlappingofprocessortasksmadepossiblebymultiplebusesandmultiplememories.Aprogramin
3、structioncanbefetchedfromonememorywhileapieceofdataiscollectedfromanother,allwhileapreviousinstructionisbeingexecuted.12.3(a)4480310=1010111100000011229910=00000001001010112(b)Theproductofthetwonumbersis1339609710=1100110001101000100000012.Atleasttwenty-fourbitsareneededtore
4、presenttheproduct.(c)Roundedtoitssixteenmostsignificantbits,theproductbecomes1100110001101001000000002,wherebit8changestoonebecausethelowereightbits,10000001aregreaterthan10000000.Theroundedproductequals1339622410,foranerrorof12710.(d)The32-bitproductis0000000011001100011010
5、00100000012.Roundedtothesixteenmostsignificantbits,theproductbecomes000000001100110000000000000000002=1336934410.Anerrorof–2675310iscommitted.12.4PartBinaryDecimal(a)0010100141(b)0101001082(c)0001010020(d)0000101010Notethatrightshiftshalvethenumberwhileleftshiftsdoubleit.Whe
6、narightshiftcausesa“one”bittobelostattheright,precisionislost.Forexample,halfof41is20.5,whichcannotberepresented.Instead,theresultisatruncated20.12.5(a)23745(b)–18468(c)–1(d)–32768(e)3276712.6(a)00001100(b)10110001(c)10000000(d)11111111(e)0111111112.7Beforeshifting,011011102
7、=11010.Aftershifting,110111002=22010.12.8Beforeshifting,011011102=11010.Aftershifting,110111002=–3610.Thesignofthenumberchangesasaresultofshifting.12.9(a)Anexponentof–4canbeusedtonormalizeallnumbersintheblock.Thischoiceofexponentensuresonesignbitremainsineachnumber.(b)Before
8、NormalizationAfterNormalization–335–5360x2–4–1786–28576–904–14464–525–84001
