ISYEl 6644l Finall Examl Latestl 2025l

ISYEl 6644l Finall Examl (Latestl 2025/l 2026l Update)l Simulationl andl Modelingl forl Engineeringl andl Sciencel Reviewl |Questionsl &l Answers|l Gradel A|l 100%l Correctl (Verifiedl Solutions)

Q:l Thisl isl sortl ofl thel samel asl Questionl 2,l exceptl wel havel nowl usedl commonl randoml numbersl tol inducel positivel correlationl betweenl thel resultsl ofl thel twol systems.l Againl findl al two-sidedl 95%l CIl forl thel differencel inl thel meansl ofl thel twol systems.

Answer:

Thisl isl al paired-tl CIl probleml assumingl unknownl variancel ofl thel differences.[-16.5,l -3.5]

Q:l Supposel Al andl Bl arel twol identicallyl distributed,l unbiased,l antitheticl estimatorsl forl thel meanl μl ofl somel randoml variable,l andl letl Cl =l (l Al +l Bl )l /l 2.l Whichl ofl thel followingl isl true?

Answer:

El [l Cl ]l =l μl andl Vl al rl (l Cl )l

Q:l Supposel thatl youl wantl tol pickl thatl onel ofl threel normall populationsl havingl thel largestl mean.l We'lll assumel thatl thel variancesl ofl thel threel competitorsl arel alll knownl tol bel equall tol σl 2l =l 4.l (Ya,l Il knowl thatl thisl isl al crazy,l unrealisticl assumption,l butl let'sl gol withl itl anyway,l okeyl dokey?)l Il wantl tol choosel thel bestl ofl thel threel populationsl withl probabilityl ofl correctl selectionl ofl 95%l wheneverl thel bestl population'sl meanl happensl tol bel atl leastl δl ⋆l =l 1l largerl thanl thel second-bestl population's.l Howl manyl observationsl froml eachl populationl doesl Bechhofer'sl procedurel Nl Bl telll mel tol takel beforel Il canl makel suchl al conclusion?

Answer:

Usingl thel notationl ofl thel notes,l wel wantl tol makel surel tol getl thel rightl answerl withl probabilityl ofl Pl ⋆l =l 0.95l wheneverl μl [l kl ]l −l μl [l kl −l 1l ]l ≥l δl ⋆l =l 1.l 1 / 3

Wel simplyl gol tol NB'sl tablel withl kl =l 3l andl δl ⋆l /l σl =l 1l /l 2l tol obtainl al samplel sizel ofl nl =l 30l froml eachl population.

Q:l Inl thel abovel problem,l supposel thatl wel takel thel necessaryl observationsl andl wel comel upl withl thel followingl samplel means:l Xl ¯l 1l =l 7.6,l Xl ¯l 2l =l 11.1,l andl Xl ¯l 3l =l 3.6.l Whatl dol wel do?

Answer:

Pickl populationl 2l andl sayl thatl wel arel rightl withl probabilityl atl leastl 95%

Q:l Supposel thatl wel wantl tol knowl whichl ofl Coke,l Pepsi,l andl Dr.l Pepperl isl thel mostl popular.l Wel wouldl likel tol makel thel correctl selectionl withl probabilityl ofl atl leastl Pl ⋆l =l 0.90l inl thel eventl thatl thel ratiol ofl thel highest-to-second-highestl preferencel probabilitiesl happensl tol bel atl leastl θl ⋆l =l 1.4.l Ifl wel usel procedurel Ml Bl El M,l thenl thel correspondingl tablel inl thel notesl (withl kl =l 3)l tellsl usl tol takel 126l samplesl (tastel tests).l Supposel wel takel thosel samplesl sequentiallyl andl afterl 100l havel beenl takenl itl turnsl outl thatl 65l peoplel preferl Coke,l 25l lovel Pepsi,l andl 10l likel Dr.l Pepper.l Whatl tol do?

Answer:

Stopl thel testl nowl andl declarel withl confidencel ofl atl leastl 90%l thatl Cokel isl thel most- preferred.

Q:l Whichl ofl thel followingl problemsl mightl bestl bel characterizedl byl al finite-horizonl simulation?

Answer:

Simulatingl thel operationsl ofl al bankl froml 9:00l a.m.l untill 5:00l p.m.

Q:l Let'sl runl al simulationl whosel outputl isl al sequencel ofl dailyl inventoryl levelsl forl al particularl product.l Whichl ofl thel followingl statementsl isl true?

Answer:

Thel consecutivel dailyl inventoryl levelsl mayl notl bel identicallyl distributed.

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Q:l Supposel thatl Xl 1l ,l Xl 2l ,l ...l isl al stationaryl (steady-state)l stochasticl processl withl covariancel functionl Rl kl ≡l Cl ol vl (l Xl 1l ,l Xl 1l +l kl ),l forl kl =l 0l ,l 1l ,l ....l Wel knowl froml classl thatl thel variancel ofl thel samplel meanl canl bel representedl asVl al rl (l Xl ¯l nl )l =l 1l nl [l Rl 0l +l 2l ∑l kl =l 1l nl −l 1l (l 1l −l kl nl )l Rl kl ]l .Wel alsol knowl froml classl thatl forl al simplel AR(1)l process,l wel havel Rl kl =l ϕl k,l kl =l 0l ,l 1l ,l 2l ,l ...l Computel Vl al rl (l Xl ¯l nl )l forl anl AR(1)l processl withl nl =l 3l andl ϕl =l 0.8.

Answer:

0.831

Q:l Supposel wel wantl tol estimatel thel expectedl averagel waitingl timel forl thel firstl ml =l 100l customersl atl al bank.l Wel makel rl =l 4l independentl replicationsl ofl thel system,l eachl initializedl emptyl andl idlel andl consistingl ofl 100l waitingl times.l Thel resultingl replicatel

meansl are:

il 1l 2l 3l 4l Zl il 5.2l 4.3l 3.1l 4.2 Findl al 90%l confidencel intervall forl thel meanl averagel waitingl timel forl thel firstl 100l customers.

Answer:

[3.188,5.212]

Q:l Considerl al particularl datal setl ofl 100,000l stationaryl waitingl timesl obtainedl froml al largel queueingl system.l Supposel yourl goall isl tol getl al confidencel intervall forl thel unknownl mean.l Wouldl youl ratherl usel (a)l 50l batchesl ofl 2000l observationsl orl (b)l 10000l batchesl ofl 10l observationsl each?

Answer:

50l batchesl ofl 2000l observations becausel thel methodl ofl batchl meansl requiresl al veryl largel batchl size

Q:l Considerl thel outputl analysisl methodl ofl nonl overlappingl batchl means.l Assumingl thatl youl havel al sufficientlyl largel batchl size,l itl canl bel shownl thatl whenl thel numberl ofl batchesl bl isl even,l thel expectedl widthl ofl thel 90%l two-sidedl confidencel intervall forl μl isl proportionall totl 0.05l ,l bl −l 1l bl −l 1l (l bl −l 1l 2l )l (l bl −l 3l 2l )l ⋯l 1l 2l (l bl −l 2l 2l )l !l .Usingl thel abovel equation,l determinel whichl ofl thel followingl valuesl ofl bl givesl thel smallestl expectedl width.

Answer:

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