1999_ExamPM With My Answers

8/6/2019 1999_ExamPM With My Answers

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Fall, 1999

CLASS Information Technology Engineer Examination - Afternoon

Select questions to answer as described in the following chart:

Question Nos. Q1,Q2 Q3 to Q6 Q7 to Q10 Q11 to Q14 Q15 to Q17

Question Selection Compulsory Select 2 of4 Select 1 of4 Select 1 of4 Select 1 of3

Examination Time 13:00 ~ 15:30 150 minutes

Notes:

1. Use an HBpencil.Ifyouneed tochange an answer, erase yourprevious answercompletely andneatly. Wipe

away anyeraserfragments.2. Markyouranswersin accordancewiththeinstructions below. Yourtestwill not begradedifyoufail tocomplywiththeinstructions. Donotmarkorwriteonthe answersheetoutsideofthe

prescribedplaces.(1)Examinee Number

Writeyourexamineenumber inthespaceprovided, andmark the appropriatespace beloweachdigit.

(2)Date of Birth

Writeyourdateofbirth (innumbers)exactly asitisprintedonyourexamticket, andmarktheappropriatespace beloweachdigit.

(3)Question Selection

Markthe questionsthatyouselectto answerasfollows: [SYMBOL](4)Answers

Markyouranswerasshowninthefollowing Howto MarkYourAnswerssample question.

[SampleQuestion] Selectthecorrect answertoinsertinthe fromthechoicesprovided.

The Fall Information Technology EngineerExaminationisheldin

Answergroup: A. October B. November C. December

Since the correct answer is "A" (October), mark your answer sheet as

follows:

a [B.] [C.]

3. Assembler specifications are provided as a reference at the end of this booklet. Theinformationprovidedonpage 69 and beyond, however, isonlyintendedforusewithQ17.

a

Do not open the exam book until instructed to do so.

Inquiries about the exam questions will not be

Faint

Right Wrong

C

COBOL

Fortran

Assembler


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Q1. Examine the flow chart that follows and read the explanation below before answering the

sub-question.

[ExplanationofFlow Chart]

(1) Thisflowchartinsertsthecharacterstringthatisstoredin Array B in a specifiedpositionin

thecharacterstringstoredin Array A.

(2) ThesizeofArray A isstoredin AMAX, the lengthofthecharacterstringisstoredin AX, and

theindividual characters arestoredin A[1], A[2], ..., A[AX].

(3) The lengthofthecharacterstringto beinsertedisstoredin BX, andtheindividual characters

arestoredin B[1], B[2], ..., B[BX].(4) Thepositionwherethecharacterstringisto beinsertedisstoredin PX (1PXAMAX).

(5) IfPX isgreaterthan AX+1, a spaceisinserted at A[AX+1] ~ A[PX1].

(6) Anytextthatoverflowsfrom Array A as a resultoftheinsertionisdiscarded.

(7) Theoperationofthe MIN functionthatisusedinthisflowchartis asfollows:

MIN(X,Y): Returns X when X


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[Flow Chart]

St

art

Moveprocessing

Moveprocessing

Spaceinsertionprocessing

Space

End

Characterstringinsertionprocessing

Spaceinsertionprocessing

Note:The specification of the loop

conditionsis asfollows:variable name: initial value,step, endvalue

Note

(Note)

Characterstringinsertionprocessing


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Sub-question: Selectthecorrect answersto

insertinthe intheflowchartfromthechoicesprovided.

Answergroupfora:

A. AMAXBX B. AMAXBX+1 C. AMAXBX1

D. AMAXPX E. AMAXPX+1 F. AMAXPX1

Answergroupforb:

A. AX+BX B. AX+BX+1 C. AX+BX1

D. PX+BX E. PX+BX+1 F. PX+BX1

Answergroupforc:

A. B[X+PX] B. B[X+PX+1] C. B[X+PX1]D. B[XPX] E. B[XPX+1] F. B[XPX1]


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Ans: p n a- A, b-A, c-E. u bi yu cu chn xu B vo trong xu A vi cc thng s sau:

+ di mng A l AX, di ln nht mng A l AMAX, di mng B l BX, v tr cn

t vo l PX (1 cc phn t mng A khng b nh hng

+ Trng hp PX > AX + 1 -> Thm vo cc phn t trng t AX + 1 n PX 1

+ Trng hp PX < AX + 1 -> Cc phn t ca mng A t PX n AX cn dch chuyn i BX v

tr, Do , ta xt theo chiu ngc li, vi AX + BX > AMAX th dch chuyn cc phn t t

PX n AMAX BX ln BX n v, cn khng th dch chuyn cc phn t t PX n AX ->

xt phn t Y = MIN(AMAX-BX, AX). (p n a-A)

\ Chn cc phn t mng B vo

+ Trng hp AX + BX AMAX, cc phn t ca B ch c chn cho ht AMAX -> xt phn

t Y = MIN(AX + BX, AMAX) (p n b-A)

Phn t 1 ca mng B c chn v tr A[PX], phn t 2 ca mng B c chn v tr A[PX + 1]

-> mi quan h ca phn t B[j] vi A[X] l X = j + PX -1 hay j = X PX + 1 (p n c-E)

Q2. Examine the flow chart that follows and read the explanation below before answering thesub-questions.

[ExplanationofFlow Chart]

Thisflowchartusesthe arraysshownin Fig. 1 and Fig. 2 tocalculatepassengerfaresfor

travel overa givendistance. Fractionsofa kilometertraveled areroundeduptothenextinteger.

The arraysuffixesstartfrom "1." Theelementsin Array D givedistances, theelementsin Array

P areusedinthefarecalculation.


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[Flow Chart]

Fig.1 Array D Fig.2 Array P

L: Upperlimitonsuffixd: Inputdistance (unit: km)f: Calculatedfare (unit: yen)

Start

Errorprocessing

End


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Sub-question 1. Whatisthefarefora passengerwhotravels 35km? Selectthecorrect answer

fromthechoicesprovided.

Answergroup:

A. 650 B. 700 C. 750

D. 850 E. 1,450

Sub-question 2. Change Array P (Fig. 3) and the flow chart so that the fare is calculated by

adding a fixed additional fareforeachincrementofadditional distancetraveled, asshowninthe

table below. (Donotchange Array D.)

Table. Relationship between Increment of

Distance and Additional Fare

Incrementofdistance(km) Additional fare (yen)

Fig. 3 Array P


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Selectthecorrect answerstoinsertinthe intheflowchartfromthechoicesprovided.

[Flow Chart after Changes]

Answergroupfora:

A. 0 p L

2 pi

0 pf

B. 0 p L

2 pi

P(i)pf

C. 5 p L

1 pi

0 pf

D. 5 p L

2 pi

0 pf

E. 5 p L

1 pi

P(i)pf

Answergroupforb:

A. D(i) f B. P(i) f

C. f+ D(i) f D. f+ P(i) f

E. f+ D(i 1)P(i) f

Ans: Sub1 C, Sub2- a-C, b-D . Thut ton u tin tnh s tin tng ng vi khong cch c

th cho vi D(i) l qung ng tng ng, v P(i) l s tin cn tr cho mi km bn D(i).

Vi Sub1 th ta c d = 35km ng vi khong t D(2) -> D(3). S tin phi tr = P[1] + P[2]*(D[2] D[1]) + P[3] * (d D[2]) = 150 + 30 * 10 + 20 * 15 = 150 + 300 + 300 = 750.

Thut ton th 2 th s tin tng ng c ng theo cc phn khong cch khc nhau. N

s so snh tng ng xem d nm trong khong no (D[i -1]->D[i]) v f=tng P[j] vi j = 1 ti

i. Khi to ban u, L = 5, i = 1 v f = 0 (p n a-C). Sau f = P[i] + f v vi mi ln lp th

Start

End

Errorprocessing


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f s tng ln P[i]. (p n b-D)

Q3. Carefullyreadtheexplanationofthefollowingprogram, theexplanationofthe

pseudo-languagedescriptionformat, andtheprogram, andthen answerSub-questions 1 and 2.

[ExplanationofProgram]

Thisprogram merges twoproductfiles intooneproductfile. Theprogram merges file A

(Fig. 1)withfile B (Fig. 2)tocreate file C (Fig. 3). Assume thatthere arenorecordsinfile Aandfile B thatsharethesameproductcode.

Select two questions fromQuestion 3 toQuestion 6, and mark which questions you have

selectedonyour answersheet. Ifyouselectthreeormore questions, onlythefirsttwo questions

Productcode ProductnameRecord 1 100 ToothbrushRecord 2 200 Rinse

Fig.1 file A

Productcode ProductnameRecord 1 150 TowelRecord 2 250 CupRecord 3 350 Pen

Fig.2 file B

Productcode ProductnameRecord 1 100 ToothbrushRecord 2 150 TowelRecord 3 200 RinseRecord 4 250 CupRecord 5 350 Pen

Fig. 3 file C


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[ExplanationofArtificial Language Description Format]

Format Explanation

Declaration block

This area iswhereprocedures, variables, names, types, etc.,

aredeclared.

Declaresnames, types, etc., ofprocedures, variables, etc.

Processing block

This area iswhereprocessingisperformed.

variablenexpression Assignsthevalueofanexpressionto a variable.

procedurename (argument, ) Calls a procedure.

conditional expression

Yes processing 1

No processing 2

Executesprocessing 1 whentheconditionistrue.

Executesprocessing 2 whentheconditionisfalse.

loopconditional expression

processing

Repeatedly executes theprocessing while the condition is

true.

Indicates a blockofprocessing.

=, Comparators that are used in conditional expressions and

loopconditional expressions.

{ } Comments.


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[Program] Program name: File Merge Integers:stsA, stsB, endf Files:fileA, fileB, fileC {sequential accessfiles}

Records:rcdA, rcdB Procedures: Recordinput (file, record, status )

{Thisprocedurereadsonerecordfromthefilespecified by"file", andstoresthedata inthe area specified by"record" . When a recordisinput, a "0" isstoredinthevariable

specified by"status"; whennorecordisinput, a "1"isstoredinthevariablespecifiedby "status".}

Procedure: Recordoutput (file, record){Thisprocedurestoresthecontentsofthe area specified by"record"inthefilespecified

by"file".}

Recordinput (fileA, rcdA, stsA) Recordinput (fileB, rcdB, stsB )

endf 0 endf = 0 {Repeatwhileendf = 0.}

stsA = 1 Yes stsB = 1

Yes endf 1No Recordoutput (fileC, rcdB)

Recordinput (fileB, rcdB, stsB )

No stsB =1 Yes Recordoutput (fileC, rcdA)

Recordinput (fileA, rcdA, stsA)No rcdAproductcode < rcdBproductcode

Yes Recordoutput (fileC, rcdA) Recordinput (fileA, rcdA, stsA)

No Recordoutput (fileC, rcdB)

Recordinput (fileB, rcdB, stsB )

Sub-question 1. Howmanytimeswill the "Yes"portionoftheprogram (indicated by " ") beexecuted? Selectthecorrect answerfromthechoicesprovided.

Answergroup

A. 1 B. 2 C. 3 D. 4 E. 5

Declarationblock

Proce

ssingblock

E

F


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Sub-question 2. Selectthecorrect answertoinsertinthe inthefollowingdescription


(1) Assumethattheconditional expressionindicated by "F" intheprogramwasincorrectly

asfollows.

rcdAproductcode > rcdBproductcode

Thecontentsoffile C wouldthen be a .

(2) Assumethat, intheoriginal program, thefileshownin Fig. 4 wasinput asfile A insteadof

thefileshownin Fig. 1. Thecontentsoffile C wouldthen be b .

Answergroup

A.Product

codeProduct

name100 Toothbrush150 Towel200 Rinse250 Cup350 Pen

Productcode Productname

200 Rinse100 Toothbrush

Fig. 4 file A

B.Product

codeProduct

name100 Toothbrush150 Towel200 Rinse350 Pen250 Cup

C.Product

codeProduct

name100 Toothbrush200 Rinse150 Towel350 Pen250 Cup

D.Product

codeProduct

name150 Towel100 Toothbrush200 Rinse350 Pen250 Cup

E.Product

codeProduct

name150 Towel200 Rinse100 Toothbrush250 Cup350 Pen

F.Product

codeProduct

name150 Towel250 Cup350 Pen100 Toothbrush200 Rinse

G.Product

codeProduct

name200 Rinse150 Towel250 Cup350 Pen100 Toothbrush

H.Product

codeProduct

name350 Pen250 Cup200 Rinse150 Towel100 Toothbrush


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Ans: Sub1 C, Sub2 a F, b E. bi yu cu vit gi m cho chng trnh ghp 2 file A,B

thnh 1 file C vi gi s khng c 2 record ging nhau v trng product code trn 2 file.

Theo thut ton trn, phn iu kin yes s c thc hin 3 ln

rcdA stsA rcdB stsB C

1 100 0 150 0 (100 < 150 -> doc A)

2 200 0 150 0 (200>150->doc B)

3 200 0 250 0 (200doc A)

4 xxx 1 250 0 (doc B)

5 xxx 1 350 0 (docB)

6 xxx 1 xxx 1 -> stop; endf = 1

Vy iu kin stsA = 1 s c thc hin 3 ln.(p n Sub1 C)

Nu iu kin l rcdA product code > rcdB product code th ta c file C s c ghi nh sau:

rcdA stsA rcdB stsB C

1 100 0 150 0 (100 < 150 -> doc B) (150)

2 100 0 250 0 (100 < 250->doc B) (250)

3 100 0 350 0 (100doc B) (350)

4 100 0 xxx 1 (doc A ) (100)

5 200 0 xxx 1 (doc A) (200)


p n Sub2 a- F.

Vi iu kin thay i u vo, ta c file C s c ghi nh sau:rcdA stsA rcdB stsB C

1 200 0 150 0 (200 > 150 -> doc B) (150)

2 200 0 250 0 (200 < 250->doc A) (200)

3 100 0 250 0 (100doc A) ( 100)

4 xxx 1 250 0 (doc B ) (250)

5 xxx 1 350 1 (doc B) (350)


p n Sub2-b-E

Q4. Carefullyreadthefollowingexplanationofdatabasecommitmentcontrol, andthen answerthesub-question.

In a centralizeddata base, consistencyofthedata ismaintainedthroughcommitmentcontrol,

whichdetermineswhethertoperform all ornoneoftheprocessing associatedwith a transaction.


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Thispreventspartial execution of theprocessing (Fig. 1). In the following diagram, m and n

represent data, m' and n' represent the respective values after being updated, andparentheses

indicatethattheupdatingofthedata valueshasnotyet beenfinalized.

In a distributeddata base, theprocessingfor one transactionfrom the mainsite (therequest

originator)updatesthedata bases atmultiplesites. Thisgivesriseto thefollowingproblem: If

commitmentprocessingisperformedforeachofthedata bases atsite A andsite B inresponseto a

commitmentrequestformthemainsite (in Fig. 2), buttheprocessingendsnormally atsite A and

ends abnormally atsite B, thedata atthetwositeswill no longerbein agreement as a resultofthe

updateprocessing.

TimeTransaction 1 Transaction 2

Data base

Fault

Update Commitment Update Rollback

Fig.1 Centralized Data Base Update

Mainsite

Update

Time

Site A

Fault

Commitment

Site B

Fig.2 Distributed Data Base Update 1

Rollbackafterrecoverin


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To addressthisproblem, consideran arrangementinwhicheachsitedoesnotfinalizetransaction

processingimmediately. Instead, eachsiteestablishes anintermediatestateinwhichcommitment

androllbackarestill bothpossible, andthenexecutesfinalizationprocessing later. (Fig. 3) Thisis

called "two-phasecommitmentcontrol."

Sub-question. Selectthecorrect answerstoinsertinthe in Figs. 2 and 3 fromthechoices

provided.

Answergroupfora:

A. n B. (n) C. n D. (n) E.n

Answergroupforb andc:

A. 200 B. 300 C. 400 D. 500 E. 600

F. 700 G. 800 H. 900

Mainsite

Update

Intermediate stateconfirmation

Commitment

Site BSite ATime

Rollback

Intermediatestateconfirmation

Update

Fault

"ok" and "no" in thisdiagram indicate thesuccess and failure,respectively, of anupdate.

Fig. 3 Distributed Data Base Update 2


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Ans: p n a-A, b-C, c-E. V phn a, ta thy khi xy ra li th DB s c rollback li gi tr

c trc khi xy ra li, a y l n (p n a-A). Cn phn b, th gi tr s l gi tr s ca

transaction trc = 400 (p n b-C). Tng t, gi tr khi rollback s l gi tr t1, v t1

y = t + 200 -> t = 600( p n c-E).

Q5. Carefullyreadthefollowingexplanationofparallel processingofaprogram, andthen answer

thesub-question.

Consideraprogramthatprocessesimagedata consistingof600 lines, with 800 dotsperline. This

programprocessestheimageinunitsoflines, withtheresultofprocessingforone linehavingno

effect on theprocessing of any of the other lines. The overall time needed to execute can be

reduced bydividingthetaskofprocessingeachofthese 600 mutuallyindependent lines among

multiplecomputers andprocessing the lines inparallel. System X is a singlecomputerwith aprocessorwith 200MFLOPS processingcapability, memory, anddiskdrive. System Y consistsof

fourcomputerswiththesameprocessingcapability assystem X, all runninginparallel. Asshown

in Fig. 1, eachofthefourcomputershasitsownprocessorandmemory, but all fourshare a single

diskdrive.

Sub-question.Selectthecorrect answertoinsertinthe inthefollowingdescriptionfrom

thechoicesprovided.

Theprogramconsistsofthreeparts: animagedata readingsection, theone-linecomputation

Disk Processor

Bus

Shareddisk

Processor

Parallel ComputerSystem Y

Processor

Processor

Processor

Mem r

Fig.1 System Configuration

System X

Memor

Memor

Mem r


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section, a resultoutputsection (Fig. 2).requires 40 mstoexecute.requires 120mstoexecute.

requires execution time equivalent to 400,000 floatingpoint operations. The computation

sectionneedsto beexecutedfor600 linesinordertoprocesstheimage. Runningtheprogramon

system X will require a mstoexecute.

Ifweexecutethesameprogramonsystem Y, sections andcannot beperformedinparallel,

sotheywill beperformed bycomputerNo. 1. Section, whichneedsto beexecuted 600 times,

can beprocessedinparallel sincetheprocessingforeachofthe 600 linesisindependentofall theother lines. Therefore, thisprocessingcan bedividedequally among thefourcomputers. The

timeneeded bycomputer1 toexchangedata withtheotherthreecomputers (thecommunications

time)is 150mstodistributethedata to all ofthecomputers, and 150mstocollecttheresultsfrom

all ofthecomputers. Runningtheprogramonsystem Y will require b mstoexecute.

It is possible that if the amount of arithmetic operations that need to be performed in

sectionissmall, thecommunicationstime becomes longon a comparative basis, sothatparallel

processing actually requires more time to execute. Converting the processing in

sectiontofloatingpointoperations, ifthenumberoffloatingpointoperationsisfewerthan

c operations, runningtheprogramonsystem Y will actuallytake longerthanrunningthe

programonsystem X.Answergroupfora and b:

A. 300 B. 340 C. 460 D. 610 E. 760

F.1,200 G. 1,320 H. 1,360

Answergroupforc:

Start

Imagedatareadingsection

Imageprocessing

One-linecomputationsection

Imageprocessing

Fig.2 Processing Procedure

End

Resultoutputsection

Repeat 600 times

(can beexecutedinparallel)


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A. 5 B. 10 C. 13 D. 30 E. 6,000

F.8,000

Ans: p n a-H, b-E, c-C. 1 giy thc hin c 200 triu lnh, thc hin 400000 lnh trongthi gian l 400000/ 200000000 = 4/2000 = 2ms. Thi gian thc hin h thng X s l 40 +

120 + 2*600 = 1200 + 160 = 1360 ms (p n a-H). Do h thng Y chy song song v cc dng

c phn b u trn cc my nn s lng dng mi my thc hin l 600/4 = 150 dng.

Nh vy thi gian thc hin in cc dng trn mi my l 150 * 2 = 300 ms. Thi gian thchin trn h thng X = thi gian thc hin cng vic 1 + thi gian thc hin cng vic 3 +

thi gian cng vic 2 trn mi my + thi gian phn phi v tp hp d liu = 40 + 120 + 300+ 150 + 150 = 760 ms (p n b-E). Thi gian tng ng vi s cu lnh ca phn 2 ca h

thng X l y * 600 (ms)+ 160 (ms), cn vi h thng Y l y * 150 + 460 (ms). Thi gian trn h

thng X nh hn thi gian trn h thng Y khi y * 600 + 460 > y * 150 + 160 -> 450 * y y < 2/3 (ms) -> s lnh thc hin = 400000 /2 * 2/3 = 400000/3 ~ 13 vn lnh (pn c-C)

Q6. Carefully read the following explanation of inventory management, and then answerSub-questions 1 and 2.

Theplant where Mr. Y works recently decided to revamp its inventory management systems,whichpreviouslydependedontheexperience andhunchesofthemanagerinchargeofinventory.Mr. Y has been askedforhisrecommendationsconcerningtherawmaterialsorderingsystem.

Sub-question 1. There are 100 differentrawmaterialsused attheplantwhere Mr. Y works. Mr.Y decided to select the most vital items and propose a review of inventorymanagement for those items. Afterstudying various documents, it became clearthatthe best approachwould betocloselymanage (asvital items)thosematerialsforwhich theplant's annual expenditures (unitcostof theraw material x amountconsumedyearly)werehigh, whileexpending littleoninventory managementforthose materials for which annual expenditures were low. After researching theannual expendituresforeachrawmaterial, Mr. Yproducedthefollowingtable.

No. Itemcode Unitcost Annual consumption Annual expenditure1 AA01 30 56,380 1,691,4002 AA07 200 1,500 300,0003 AC01 1,500 23,400 35,100,000

100 ZQ80 10 2,875 28,750Total annual expenditure 231,730,960

Next, inordertoselectfromthetable asvital itemsthosematerialsthat accountedforthe largestpercentageofannual expenditures, Mr. Y analyzedthedata accordingtothefollowingprocedure:


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Hesorted all ofthematerials by annual expenditure, startingfromthehighest. He determined the cumulative annual expenditure for the list, position byposition, starting

fromthetop. He drew a bar graph with the position of each material on the x axis, and the annual

expenditureonthey axis. Heplotted a graph, indicatingthecumulative annual expenditurefrompositiontoposition.

Fromthechoicesprovided, selectthecorrect answerforthenameofthistypeofchartorgraph.

Answergroup

A. Arrowdiagram B. Ganttchart C. Control chart

D. Paretodiagram E. Portfoliograph


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Sub-question 2. Typical inventory management systems include the "periodic reordering

system" andthe "fixed-orderquantitysystem." Inordertoconsiderwhichofthesesystemsmight

be best suited for the list of vital items he created in Sub-question 1, Mr. Y summarized the

characteristics of each system. From the choicesprovided, select the correct answers to be

insertedinthe inthedescription below.

Inordertoselect anorderingmethod, itisnecessarytoconsidernumerousfactors, including

theunitcostoftherawmaterials, the amountofconsumptionofrawmaterialsin a giventime, the

sizeoffluctuationsinconsumption andtheeasewithwhichsuchfluctuationscan beforecast, the

lengthoftheprocurementinterval (thetimefromwhen a material isordereduntil itis accepted

intoinventory), costs associatedwithordering (such asthetransportcosts and administrativecosts

incurredforoneorder), andthecostofmaintaininginventory.

Becausethe "periodicreorderingsystem" requiresdetermining a regularinterval fororderingbasedontherawmaterialprocurementtime, forecastingtheconsumptionoftherawmaterial over

theperiod following each regular interval for ordering, and then determining a on

the basisofthatinformation, it becomespossibletomanageinventoryveryclosely. Thismethod

carries littleriskofrunningoutofmaterial evenwhenthere are largefluctuationsinconsumption.

Thismethodiseffectiveifthere are largechangesindemandinthemarketforaproduct, resulting

infrequentchangestotheproductionplan. Thismethodis alsoeffectiveformanaginginventories

ofrawmaterialsthatcarry a highunitcost.

The "fixed-order quantity system" orders quantities (predetermined on the basis of the

procurement interval) of materials that have fallen to b . Compared to the "periodic

reorderingsystem," thismethodissimpler. However, thereis a dangerofrunningoutofa materialthatundergoes a largefluctuation inconsumption. This method issuitableformanagementof

materialswhenthereisproductdemand andtherawmaterialprocurementinterval arestable, and

whenitispossibleto accuratelyknowthecurrentinventory level at anytime.

Answergroup

A. thesafeinventory level B. 50% ofthesafeinventory level C. theorderingtime

D. theorderingpoint E. theordering quantity F. the averageinventory level

Ans: p n Sub1 D, Sub 2 a E, b D. Vic s dng biu dng tng tch ly ca cc

item c gi tr t cao n thp l s dng lc Parero (p n Sub1 D). Phn 2 hi v

h thng t hng nh k v h thng t hng s lng c nh. Cu a l E, s lng thng. Cn cu b l D, im t hng.

Selectone questionfromQuestion 7toQuestion 10, andmarkwhich questionyouhave


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selectedonyour answersheet.

Ifyouselecttwoormore questions, onlythefirst questionwill begraded.

Q7. Carefullyreadtheexplanationofthefollowing C program, andthen answerthesub-question.


This program orders and outputs in ascending order the array "Stu" in the structure

"STUDENT", using "Height" asthekey.

Theprogramcreates a binarytreethatindicatestheorderwithoutsortingthe arraydata. The

following arrayvariables aredefinedtorepresentthe binarytree.

Fig. 1 illustrates thecasewhere theelements are arranged in ascendingorder, withtheelement

numbers in thesequence 3, 1, 0, 5, 2, 4. Fig. 2 shows thecontentsof the arrays "Lower" and

"Upper". Thevalue "1" indicatesthatthereisnoelementconnectednext.

Initially, theelements areconnectedonthe "Upper" sideintheorderinwhichthey appeared

inthedefinitionofthestructure array "Stu", creatingthe binarytreeshownin Fig. 3.

Root

Fig.1 Example of the

Element

Element No. Lower Upper

0 1 21 3 1

2 5 4

3 1 1

4 1 1

5 1 1

Fig.2 Contents of Arrays

Representing

C

Name

Age

Height

Weight


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Next, theprogramcallstherecursivefunction "BinTreeSort " andsortstheelementsinthe

binarytree accordingtoheightin ascendingorder. Theparameteristheelementnumberofthe

root. Theprototypeforthefunction "BinTreeSort " is asfollows:

Lastly, theprogramcallstherecursivefunction "DisplayData" andoutputsthecontentsof

the structure array in ascending order according to height, following the binary tree. The

parameteristheelementnumberoftheroot. Theprototypeforthefunction "DisplayData" is

asfollows:

[Program]

Root

Fig. 3 Connection Status Prior to Sorting

Root

C

Name

Age

Height

Weight

AikawaTaro

Ito Shiro

Kato Goro

TanakaSaburo

Yamanaka Jiro


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Name Age Height Weight


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C


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Sub-question: Select thecorrect answer to insert inthe in theprogram from thechoices

provided.

Answergroup

A. 1 B. Data C.Lower[Data]

D. Lower[Root] E.Next F. Root

G. Upper[Data] H. Upper[Root]

Ans: p n a G; b H; c B; d D. T tng ca thut ton ny l xt tt c cc nt bn

phi nt ROOT, nu n ln hn nt ROOT th chn vo nt con bn phi nt ROOT, cnkhng th chn vo nt con bn tri nt ROOT. Sau khi sp xp ht, ta s xt quy cc nt

con ca n.

Q8. Carefully read the explanation of the following COBOL program, and then answer the

sub-question.


Thisprogramcreates a salesperformancefilefromsalesfilesinwhichthedepartmental sales

amountsgenerated byeachstore andeachdepartment arerecordedseparately. Theprogram

thenrankseachdepartment accordingtothesizeofitsdepartmental sales amounts.

(1) Theformatofeachrecordinthesalesfileisshown below.

Storecode Departmental sales amounts

Department A Department B Department C Department D Department E

5 digits 9 digits 9 digits 9 digits 9 digits 9 digits

Eachstorehasfivedepartments, A through E.

Assumethatthere arenorecordsthathaveduplicatestorecodeswithinthesalesfile.

Assumethatthedepartmental sales amountswill notexceedninedigits.

Assumethatthesalesfiledoesnotcontainmorethan 9,999 records.

(2) Theformatofeachrecordinthesalesperformancefileisshown below. Storecode

Department A Department B Department E

Departmentrank

Departmentalsales amount

Departmentrank


Departmentrank


5 digits 4 digits 9 digits 4 digits 9 digits 4 digits 9 digits

Thedepartmentrank (with "1" as thehighest) is assigned according to thesales amountof the

C

COBOL


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departmentcomparedtothesamedepartmentsintheotherstores.

Ifthere areseveral storeswiththesamedepartmental sales amount, theyeachgetthesamerank.

Afterthesamedepartmentrankhas been assigned, thenextrankisonegreaterthanthenumberof

storesthathave beenrankeduptothatpoint.

(3) A workfileisusedtodeterminethedepartmental sales amountrankofeachdepartment.

[Program]


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COBOL


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COBOL


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Sub-question: Selectthecorrect answertoinsertinthe intheprogramfromthechoices

provided.

Answergroupfora andd:A. BANGO-BAN B. BANGO-SOR C. BUMON-BAND. BUMON-SOR E. GAKU-BAN F. GAKU-SORG.TENPO-BAN H.TENPO-SOR

Answergroupforb:A.MOVE BANGO-REC TO SORT-RECB.MOVE GAKU-WRKTO GAKU-SORC.READSORT-FILED.RELEASESORT-RECE.RETURN SORT-FILE

F. WRITESORT-REC

Answergroupforc:A.MOVE BUMON-SORTO LCN B.MOVE LCN TO BUMON-SORC.MOVE X TO BANGO-BAN D.MOVE X TO YE.MOVE Y TO BANGO-BAN F. MOVE Y TO X

Answergroupfore:A.TENPO-SEI = TENPO-BAN B.TENPO-SEI NOT = TENPO-BANC.TENPO-SEI = TENPO-SOR D.TENPO-SEI NOT = TENPO-SORE.TENPO-SOR = TENPO-BAN F. TENPO-SOR NOT = TENPO-BAN

COBOL


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Q9. Carefully read the explanation of the following Fortran program, and then answer the

sub-question.


A problem with the handling of matrices in the field of computing is that the volume of

calculations andthe amountofmainmemoryusedincreaseswiththesizeofthematrix. Herewe

will considera methodforcompressing a bandmatrix, a typeofsparsematrix thatcontai ns a large

numberofzeroelements, forstorage byomittingzeroelements.

Thisprogramdeterminesthe bandwidthofthe bandmatrix, storesthe bandwidthdata foreach

line in an array, and thenoutputs the contentsof that array. Here, a "band matrix" is a square

matrix withnon-zeroelementsdistributedinthevicinityoftheoppositecorners, asshownin Fig.

1.

Matrix dimension

Fig.1 Sample Input Data

(1) A bandmatrix withnon-zeroelementsdistributedinthevicinityoftheoppositecornersisinputin

the array "wa". Theprogramdeems anyelementwith an absolutevaluegreaterthan 10 -6 as being

a non-zeroelement.

(2) Theprogramdeterminesthe bandwidthofthematrix.

Fortherownumbered "row":

Theprogram searches from the left edge to the "row1" column, and assigns the column

numberofthefirstnon-zeroelementthatitencountersto "kl".

Theprogram searches from theright edge to the "row+1" column, and assigns the column

numberofthefirstnon-zeroelementthatitencountersto "kr".

Theprogram assignsthe largerof"rowk1" and "krrow" as "length".

The bandwidthisderived bydoublingthe largest "length" value and addingonetotheresult,

usingthe "length" valuesderivedforeachrow asdescribed above. Forexample, inthethird

lineofthe bandmatrix shownin Fig. 1, row = 3, kl = 1, andkr= 5, sorowkl = 2 andkrrow

= 2. Inthecaseofthis bandmatrix, the largest "length" valueforall oftherowsis "2," sothe

bandwidthis "5."

(3) Theprogram then allocates an area for the array wb(number of rows, bandwidth) to store a

numberofelementsequal tothe bandwidth.

(4) Theprogramstores a numberofelementsequal tothe bandwidthin "wb", orderingthemsothat

thediagonal elementof"wa" will becomethecentercolumnof"wb". A "wb" elementissetto 0

(nought), ifnowa elementcorresponds.

Matrix

Fortran


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Sub-question: Selectthecorrect answertoinsertinthe in theprogram from the choicesprovided.

Answergroupfora and b:A. krow .or. right==0

Answergroupforc:A. row B. row+maxC. row+wmax-1 D. row-wmaxE. row-wmax+1 F. row-wmax-1

Answergroupford:A. indexna B. indexnaC. index>1 .and. index1 .or. index1 .and. k1 .or. k


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Q10. Carefullyreadtheexplanationofthefollowing assembly lineprogram andtheprogramitself,

andthen answerthesub-question.


Thisprogram treats data stored in sixteen consecutive words as a 16 x 16-bit graphic. The

subprogramROTATErotatesthegraphic 90 degrees.

(1) Thestarting addressofthegraphicisstoredin GR1, whichispassedfromthemainprogram.

(2) Either"1" or"1" isstoredin GR3, whichispassedfromthemainprogram. When GR3 is "1," the

graphicisrotated 90 degreestotheright; when GR3 is "1," thegraphicisrotated 90 degreesto

the left.

(3) Theprogramstorestherotatedgraphicinthe area specified by GR2.

(4) Assumethatthe areasspecified by GR1 and GR2 donotoverlap.

(5) Fig. 1 shows anexampleoftheexecutionofthesubprogram ROTATE.

Fig.1 Execution Example of the Subprogram ROTATE

Original graphicOriginal graphic

Bit

When GR3 = 1

When GR3 = 1

Bit

Bit

Resultinggraphic

Resultinggraphic

Assembler


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(6) Whenrotatingthegraphictotheright by 90 degrees, asshownin Fig. 2, the bitstrin ginthefirst

wordintheoriginal graphicisshiftedinthedirectionindicated bythe arrowone bit at a time

tothe bitcolumn "a" (attherightedgeoftheresultinggraphic). Similarly, isthenshiftedto

"b," isshiftedto "c," etc. Whenrotatingthegraphictothe left by 90 degrees, asshownin Fig.

3, the bits areshiftedstartingfromthe lastword, whichisshiftedto bitcolumn "a," thenis

thenshiftedto "b," isshiftedto "c," etc.

Fig.2 Sequence of Processing WhenRotating to the Right

(7) When control returns to the mainprogram, the contents of the general-purpose registers GR1

through GR3 arereturnedtotheiroriginal states.

Assembler

Fig. 3 Sequence of Processing WhenRotating to the Left


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[Program]

Register save

Bit position initialization

Wordcounter

Word1 in original graphicp GR1

Word1 in resulting graphicp GR0

Sets bit in word1 of result

Bit counter

Bit check in original graphic

Clears bit that was set

Stores in resulting graphic

Next wo rd in resulting graphic p

GR0

Sets bit in word1 of result

Bit check in original graphic

Next word address in original

graphicp GR1

Bit position update

Address of resulting graphic p GR2

Registerrestore

Increment (+1 or 1)

Assembler


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Sub-question: Selectthecorrect answertoinsertinthe in theprogram from the choices

provided.

Answergroupfora andc:A.AND GR0,0,GR2 B.AND GR0,BITPOSC. EOR GR0,0,GR2 D. EOR GR0,BITPOSE. LEA GR0,0 F. LEA GR0,32768G. SLL GR0,1 H. SRL GR0,1

Answergroupforb andd:A. JMI LOOP1 B. JMI LOOP2C. JMP LOOP1 D. JMP LOOP2E. JNZ LOOP1 F. JPZ LOOP2G. JZE LOOP1 H. JZE LOOP2

Assembler


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Select one question fromQuestion 11 toQuestion 14, and mark which question you have

selectedonyouranswersheet. Ifyouselecttwoormore questions, onlythefirst q uestionwill be

graded.

Q11. Carefullyreadtheexplanationofthefollowing C program, andthen answerSub-questions 1

and 2.


Thefunction "wordwrap" outputs Englishtextsothat a wordisnotdividedinthemiddle attheend

ofa line.

(1) Theparametersforthefunction "wordwrap" are: "str[]", an arrayfortheinputstringinwhich

the English text isstored; and "max", the maximum number ofcharactersper output line (notincludingthe linefeedcharacter).

(2) Wordsintheinputstring aredelimited by a singlespace. There arenospaces atthe beginning and

endofthestring.

(3) Ifthemaximumnumberofcharactersperlineisgoingto beexceededinthemiddleofa word, that

wordisoutput atthe beginningofthenext line.

(4) Assuming "wx" asthemaximumword length and "sx" asthe lengthoftheinputstring, assume

thatthefollowingrelationshipistrue:

0 < wx < 16 < max < sx < 1024

Sampleinputstring:

Maximumnumberofcharactersper line: 30

Output:

((: spacecharacter)

C


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[Program](Linenumber)

/***Divides the character string into words ***/

output

C


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Sub-question 1. Select thecorrect answer to insert in the in the following description


Giventheinputcharacterstring andthemaximumnumberofcharactersper lineshown below,

line 21 isexecuted a times. Thevalueofthevariable "pos" whentheprogramendsis

b .

Inputcharacterstring:

Maximumnumberofcharactersper line:18

Answergroup

A. 5 B. 7 C. 8 D. 12 E. 13

F. 15 G. 16 H. 22 I. 27

Sub-question 2. Selectthecorrect answertoinsertinthe inthefollowingdescriptionfrom

thechoicesprovided.

Wewill nowchangetheprogramsothatitwill include any leadingspacesinthecharacter

string aspartofthefirstword (sothatspaces atthestartofthecharacterstring areoutput asis),

andsothatitwill outputonlyonespace betweenwordsevenifthere aremultiplespaces.

Thischangeis accomplished byinserting "E" immediately before line 12, and byinserting

"F" immediately before line 18.

Answergroup

A. Cnt++ B.cnt-- C.i D.i++ E.i--

F. i+1 H.i-1 I. idx++ J.idx--

Ans: p n Sub1 a H, b E Sub 2 c A, d-D, e-F. cnt ng vai tr m s k t trong 1 t

-> trong ton b chng trnh dng 21 c thc hin = s k t khc trng trong xu input

= 22 (p n Sub1- a-H). Bin pos y ng vai tr l ch s ca k t ang c c vo

C


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buffer output. D thy output ra s l:

Dng 1: AAAA_BBBBB_CCC

Dng 2: DDDD_EEE_FFF\0 (pos 1 = \0 -> pos l s k t trong dng cui cng khi kt thc

chng trnh = 13) (p n Sub1 b E)

Vi phn 2 th ta s cho cc du trng u vo xu u tin, y c-A, d-D. Khi kim tra ta s

loi b cc k t trng tip theo c th do e- F.

Q12. Carefully read the explanation of the following COBOL program, and then answer

Sub-questions 1 and 2.


Thisprogramoutputs a bargraphthatindicatesthenumberofstudents at a givenschool thathave

selectedeachoftheforeign languagesoffered.

(1) Thenumberofstudentsthathaveselectedeach languagecan beobtained bycalling a programnamed "KEISAN"; theresults arethenstoredinSENTAKU-T inthefollowingorder:

French (F), German (G), Chinese (C), Russian (R), Spanish (S)

(2) Theprintingformatforthe bargraphis asfollows:

The languages are listedonthe Y axis, andthenumberofstudentsisplottedonthe X axis.

Thenumberofstudentscanrangefrom 0 to 30; one asterisk"*" representsonestudent.

Thefollowingoutputsampleshowsthat 30 studentsselected French, 15 selected German, 20

selected Chinese, 0 selected Russian, and 5 selected Spanish.

Fig. Output Sample forNumber of Students Selecting Each Language

[Program]

C

OBOL


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Now the school modifies theprogramso that the bar graph is oriented vertically instead of

horizontally. Thedata sectionischanged asshown below, andthethree alternatives a, b, andcshowninthetable areconsideredforlines 27 through 35 intheproceduresection.

COBO

L


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Table Changes to the Procedure Section (Substitutions for Lines 27 through 35)


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Sub-question 1. Howshouldthe "IF" statementenclosedinthe box inthetable bechangedso

thattheconditionwould be "whenthevalueis a multipleo f10"? Selectthecorrect answerfrom

thechoices below.

C

OBOL

Statements


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AnswergroupA. COMPUTE K = S/10

IF K = 0 THEN MOVESTO KOMOKU END-IF

B. COMPUTE K = S/10 IF S = K*10THEN MOVESTO KOMOKU END-IF

C. COMPUTE K = S/10 COMPUTE K = S/ K

IF K = 10 THEN MOVESTO KOMOKU END-IF

D. COMPUTE K = S/10 COMPUTE K = K*10

MOVE KTO KOMOKU

Sub-question 2. Whichgraphwill beprintedwhenthechangesmadein a, b, andc, respectively,

are incorporated into theprogram? Selectonegraphfor a, b, andcfrom thechoicesprovided.

Assumethatthestudentdata fromthesampleoutput (item (2)intheexplanationoftheprogram)

isused.

Answergroup

A A A. B.

C. D.

COBO

L


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Q13. Carefully read the explanation of the following Fortran program, and then answer

Sub-questions 1 through 3.


A survey on food preferences was conducted. This program determines the correlation

coefficientforpreferences amongdifferentfoods andprintsouttheresults.

(1) Thenumberofrespondentsis between 2 and 100.

(2) Thesurvey asked therespondents toratehowmuch they likedninedifferentfoodson a

scalefrom 1 to 10.

(3) Table 1 showssamplesurveyresults. Thesurveyresultsfilecontainseachrespondents

numberandthescoretheygavetoeachoftheninefooditems.

Table 1 Survey Results

Respondent No. Food 1 Food 2 Food 3 Food 4 Food 5 Food 6 Food 7 Food 8 Food 9

001 5 3 7 6 8 2 7 2 5

002 10 3 7 10 10 3 6 4 6

003 7 3 1 7 4 2 7 9 2

(4) Theprogramcalculatesthecorrelationcoefficientforpreferences amongfooditems asdescribed

below.

Assumingthe averagescoreforfoods as Hs, thescoresthatrespondentn assignedtofood j and

foodkas Pnj and Pnk, andthenumberofrespondents as M, then Cjkiscalculated asfollows:

The correlation coefficient Rjk for food j and food k is calculated according to the following

formula. Assumethatthevalues Cjj and Ckk arenotzero.

(5) Sampleoutputresults areshown below.

Fortran


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[Program]

(Linenumber)

Fortran


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Sub-question 1. Whatwouldhappenifdata from 101 peoplewasgiventothisprogram? Select

thecorrect answerfromthechoices below.

Answergroup

A. Whendata for the 101st andsubsequentpeople is input, thedata thatwill bediscarded is

undefined. Therefore, itwill beunclearwhichrespondents'data wasusedtocalculatethe

correlationcoefficient.

B. The calculation of the correlation coefficient will include the data from the 101st and

subsequentrespondents.

C. Becausethedata fromthe 101st andsubsequentrespondentswill bestoredoutsidethe area

thatwas allocatedforthe array, theresultscannot beforecasted.

D. The correlation coefficient will be calculated on the basis the data from the last 100respondents.

E. The correlation coefficient will be calculated on the basis the data from the first 100

respondents.Fortran


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Sub-question 2. Selectthe answerthatwill produceresultsequivalentto lines 27 through 32 in

theprogramfromthechoicesprovided.

Answergroup

Fortran


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Sub-question 3. Assumethatthetwochangesdescribed below aremadetotheprogram. Select

B.

A.

C.

D.


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thecorrect answertoinsertin in Table 2 fromthechoicesprovided.

(1) Assumes the correlation coefficient between the same food items is 1.0 instead of

calculatingit.

(2) Usesthesymmetryoftheresultmatrix toskipcalculatingthecorrelationcoefficientfora

combinationoffoodsinwhichtheitems aresimplyreversedfrom anearlierpair, substituting

the appropriateresultinstead.

Table 2 Program Changes

Position Afterchange

Line 28 do k=l, j

Lines 32 to 36

Answergroup

FortranB.

A.

C.

D.


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Q14. Carefullyreadtheexplanationofthefollowing assembly languageprogram andtheprogram

itself, andthen answerSub-questions 1 and 2.


Thesubprogram ADDP acceptstwointegers a and b (Eu, bu0)in BCD notationfromthemain

program, andthencalculatesthesumofa and b in BCD notation.

(1) BCD notation expresses each digit of a decimal number as a four-digit binary number. For

example, thedecimal number5973 would beexpressedin BCD notation asfollows:

Thousandsdigit Hundredsdigit Tensdigit Onesdigit

5 9 7 3

q q q q

0101 1001 0111 0011

(2) a and b are bothfour-digit (16-bit)numbers. They arestoredin GR1 and GR2 respectively, andarepassedfromthemainprogram.

(3) AnexampleoftheexecutionofthesubprogramADDPisshown below.

Res ult

Fig. Sup-program ADDP Execution Example

(4) Whencontrol returnstothemainprogram, thecontentsofthegeneral-purposeregisters GR2 and

GR3 arereturnedtotheiroriginal states.

Assembler


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[Sub-program ADDP]

Extractfourbitsformintegera

Initialize results area

Extract four bits form integer a

Extract four bits from integer b

Four-bit addition

Resultu10?

Subtract 10 fromresult

Set carry

Store result of four-bit addition in bit position

Merge intermediate results

All digits completed?

Store intermediate result

Addcarry

Shift next four bits of integer to right

Shift next four bits of integer to right

Assembler


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Sub-question 1. Selectthecorrect answertoinsertinthe inthefollowingtextfromthe

choicesprovided.

IfthecontentsofGR1 and GR2passedfromthemain program are asfollows, then afterADDP

isexecutedthe bitstring a will besetin GR0 andthe bitstring b will besetin GR1 and

passed backtothemainprogram.

GR1 = 0101 0010 1001 1000

GR2 = 0100 1000 0101 0011

Answergroup

A. 0000 0000 0000 0000 B. 0000 0000 0000 0001

C. 0000 0001 0101 0001 D. 0100 1000 0101 0011

D. 0101 0010 1001 1000 E. 1001 1001 1001 1001

F. 1010 0001 0101 0001 G. 1111 1111 1111 1111

Sub-question 2. Using thesub-programADDP, a sub-programCALCwascreated toperform a

certainoperationontwointegersm andnexpressedin BCD notation (andwheremu0, nu 0).

(1)m andn arestoredin GR1 and GR2, respectively, and arepassedfromthemainprogram.

(2)Theoperationresultisstoredin GR1.

(3) Whencontrol returnstothemainprogram, thecontentsofthegeneral-purposeregisters GR2

and GR3 arereturnedtotheiroriginal states.

Thefollowingtableshowsthevaluesofm andnpassedfromthemainprogram, thevaluein GR1

thatisreturnedtothemain program, andthenumberoftimesthesub-programADDPwascalled.Notethat all valuesshowninthetable andinthe answergroup aregivenindecimal notation.

Selectthecorrect answertoinsertin inthetablefromthechoicesprovided.

Table Correspondence of Each Value

Valuepassedfromthemainprogram ValueofGR1 returned

fromthemainprogram

NumberoftimesADDP

iscalledm n

123 45 c d

51 3579 e f

Assem

bler


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[Sub-programCALC]

Answergroupforc ande

A. 0 B. 70 C. 78 D. 168

F. 737 G. 2529 H. 3528 I. 3579

J. 5535 K. 7158

Answergroupford andfA. 0 B. 1 C. 2 D. 3

F. 6 G. 9 H. 45 I. 51

J. 123 K. 3579

Register save

Initialize results

Loopcount

Overflow

Overflow

Restore register

Asse

mbl

er


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Selectone questionfromQuestion 15toQuestion 17, andmarkwhich questionyouhaveselected

onyour answersheet. Ifyouselecttwoormore questions, onlythefirst questionwill begraded.

Q15. Read thefollowingdescriptionconcerning internal design and then answer Sub-questions 1

through 3.

Intherelational data base logicdesignstage, a normalizeddata structureisthedata structure

thatdata analysis leadsto logically, butdoesnottakeintoconsiderationhowthedata basewill be

used bytheuserorapplicationprogram. Therefore, itisnecessarytore-examinenormalizeddata

structures in the construction of data bases that arepractical and efficient from a processing

standpoint byconsideringhowthedata basewill actually beused.

Forexample, ifa tableisintegrated, integrationprocessingisno longerneededwhensearching

for data, reducing the amount ofprocessing that must beperformed, but this is offset by an

increase in the processing that must be performed in order to maintain the consistency ofduplicateddata inthedata basewhenthedata isupdated. Whenre-examining a data structure, itis

sometimesnecessarytoconfirmwhetherthetableisoneinwhichsearchprocessingisfrequently

conductedornot andwhetherthetableisonethatisfrequentlyupdatedornot, andthentofocus

oneliminatingnormalizationinthosetablesinwhichsearch processingisfrequentlyconducted.

Company M isdeveloping anorderinputsystem andis attemptingtoimprovetheefficiencyof

orderinputwork. Company M hasotherworksystems, andhasdecidedtoutilizedata basetables

thatcan beshared. Thefollowingtables aretheresultofdata base logicdesignrelatingtoorder

inputwork.

[Resultofdata base logicdesign]

(1)Customer tableCustomercode Customername Address Telephone Statecode

(2)Order table

Ordernumber Customercode Orderdate Delivery Departmentcode

(3)Order details table

Ordernumber Productcode Orderquantity

(4)Product table

Productcode Productname Unitprice Closingdate

(5)Department table

Departmentcode Departmentname Numberofstaff

(6)Region code table

Regioncode Regionname

(7)State code table

Statecode Statename Regioncode

(8)Inventory table


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Productcode Inventory quantity

The following tables are used only by the order taking system: customer table, order table,

orderdetailstable, regioncodetable, andstatecodetable. Inordertakingwork, formanagement

purposesorders aredividedintonineregions, andtheseregions arefurtherdivided accordingto

theirstates.

Theproducttableisupdatedonceperweekbythesalessystem.

Thedepartmenttableisupdatedonceperyearbythepersonnel system.

Theinventorytableisupdatedfrequently bytheinventorymanagementsystem.

Inorderto betterunderstandtheprocessinginvolvedinordertaking andhoweachtableisused,

thecompanycreatedthefollowingtableshowingtherelationships betweenprocessing andeach

table.

Relationships between Processing and Each Table

Processingname

Processingtype

Frequencyof

processing

Tablename

Customer

Order

Orderdetails

Product

Department

Regioncode

Statecode

Inventory

Ordertaking Online 100times/

day

Ref App App Ref Ref Ref Ref Updt

Ordersearch Online 200times/

day

Ref Ref Ref Ref Ref Ref Ref _

Customersearch

Online 130times/

day

Ref Ref _ _ Ref Ref Ref _

Inventorysearch

Online 150times/

day

_ _ _ Ref _ _ _ Ref

Customerlistoutput

Batch 2times/

week

Ref _ _ _ _ Ref Ref _

Customertablemaintenance

Online 20times/

day

App,Updt,Del

_ _ _ _ Ref Ref _

Ref: Reference; App: Append; Updt: Update; Del: Delete

Sub-question 1. If the company wants to intentionally reduce normalization and integrate two

tablesintoonetableinordertoreduceintegrationprocessing, while alsokeepingmindtheneedto

minimizeincreasesinthe amountofdata, whichtwotables arethe bestcandidatesforintegration?

Select TWO answersfromthechoicesprovided.


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Productname QuantityU

Answergroup

A. Customertable B. Inventorytable C. Ordertable

D. Orderdetailstable E. Producttable F. Departmenttable

G. Regioncodetable H. Statecodetable

Sub-question 2. Order searchprocessing generates the mostprocessing requests in a day, and

demands thefastest response time. Fig. 1 shows theordersearch resultsdisplay screen. This

screenisdisplayedwhen anordernumberisinput.

Whichdata structureisthe bestcandidateforre-examinationinordertospeeduptheresponse

time? Selectthecorrect answerfromthechoicesprovided.

OrderSearch Results ScreenXx/xx/xxxx

OrderNo. Departmentname

Customername

Regionname Statename

Orderdate Delivery

Fig.1 Order Search Results Screen


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Ordernumber Dateoforder Delivery DepartmentnameU

Answergroup

A. Addthestatenametothedata itemsinthecustomertable.

B. Addthestatename andtheregionnametothedata itemsinthecustomertable.

C. Addthecustomernametothedata itemsintheordertable.

D. Addthecustomername andthedepartmentnametothedata itemsintheordertable.

E. Addtheproductnametothedata itemsintheorderdetail table.

F. Addtheproductname, unitprice, andclosingdatetothedata itemsintheorderdetail tabl e.

Sub-question 3. Selectthecorrect answertoinsertinthe inthefollowingtextfromthe

choicesprovided.

Fig. 2 shows the customer search results screen that was designed for customer search

processing. Thisscreenisdisplayedwhen a customercodeisinput.

Thisscreendisplaysthenumberofordersthatwerereceivedfromthatcustomer'sstateinthe

previousmonthinthefield "Previousmonth'sorders" andthecurrentnumberofcustomersinthecustomer'sstateinthefield "Currentnumberofcustomers."

CustomerSearch Results Screen

xx/xx/xxxx

Customercode Telephone

Customername

Address

Regionname Statename

Numberoforders

Previousmonth'sorders Currentnumberofcustomers

Fig.2 Customer Search Results Screen

The company seeks to speed up the response to online queries by studying methods for

calculatingdata itemsthatcan bederivedfromotherdata items.


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Thecompanystudiedthefollowingthreemethodsforcalculatingderivativedata items:

(1) Calculation atthetimeofreference

Derivativedata items arecalculated atthetimeofreference. Theresultsproduced bythese

calculations arenotstoredinthedata base.

(2) Calculation atthetimeofupdating

Derivativedata items arecalculatedwhenthedata itemsfromwhichthey arederived areupdated.

Theresultsproduced bythesecalculations arestoredinthedata base.

(3) Asynchronouscalculation

Derivativedata items arecalculated atsomepredeterminedtime, and arestoredin thedata base.

Evenifthedata itemsfromwhichthey arederived areupdated, thederivativedata items arenot

recalculated.

Thederivativedata item "Previousmonth'sorders" is a .Thederivativedata item "Currentnumberofcustomers" is b .

Answergroup

A. calculated atthetimeofupdating

B. calculated atthetimeofreference

C. calculated asynchronously


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Ans: p n Sub1 G, H, Sub2 B, Sub 3 a C, b - A . D thy cc bng Region code v State code

ch c dng tham chiu trong cc nghip v c s lng bn ghi c nh c lin quan

n nhau bi trng Region code. Chnh v vy, khi kt hp 2 bng , chng s khng lm

tng nhiu s lng data theo thi gian.(p n Sub 1 G,H). Phn 2, d thy phn ny n

tham chiu 6 bng theo th t sau:

+ Order (1)

+ Order details (1)

+ Customer (2)

+ Department (2)

+ Product (2)

+ State code (3)

+ Region code (4)

Nh vy, ta thy 2 bng state v region c tham chiu cui cng, lm tn thi gian, do vy ta

s a cc trng state v region vo bng customer. (p n Sub2 B)

Phn 3 ta thy gi tr ca Previous month's orders c xc nh theo 1 thi im xc nh,

do n l tnh ton khng ng b (p n Sub3 a C) cn Current number of

customers c tnh ton khi bng Customer c update, do n l tnh ton khi

update( p n Sub3 b A).

Q16. Read thefollowing description ofprogram design and then answer Sub-questions 1 and 2.

Company N isdeveloping a programtoperform job X thathandlesonlinetransactionprocessing

inreal time. Fig. 1 showsthesystemconfigurationinwhichthe job Xprogramruns.

11) Fig. 1 System ConfigurationforJob X Program

Fig.1 System Configuration for Job X Program

[ExplanationofFlowofTransaction Processing]

(1) The online control program receives a message from a terminal through the communications

control program.

Job XprogramTerminal

Serverusedforjobs

Terminal

Onlinecontrolprogram

Databasemanagement

program

Job Yprogram

Logfiles

Data base

Communications

controlprogram


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(2) Theonlinecontrol program analyzesthemessagethatitreceived andstartsupthe job program

thatisneededinordertoperformtherequestedprocessing.

(3) Ifthe jobprogramin questionrequires any I/O transactionswiththedata base, itsearchesforthe

locationwheretherequireddata isstored andperformsread/writeprocessingthroughthedatabase

managementprogram.

(4) Whenthe jobprogramin questioncompletesitsprocessing, theonlinecontrol programsendsthe

response message from the job program to the terminal through the communications control

program.


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Thefollowingdata bases andfile areused bythe job X program:

(1) Data bases

Customerattributedata base

Customeraccountdata base

Special customerdata base

(2) Logfile

Thedata directoryforthecustomerattributedata baseismaderesidentinmainmemory bythe

databasemanagementprogram. Beforereadingrecordsfromthecustomer accountdata base and

the special customer data base, input is required once from the disk in order to load the data

directory. The logfileis a sequential file, andmust beoutputtothediskoncepertransaction.

[Overviewofthe Job X Program](1) Theprogram reads the customer attribute data base, using as a key the customer code that is

includedinthemessage.

(2) Ifthecorrespondingcustomerrecordisnotfoundinthecustomerattributedata base, theprogram

generates anerrormessage andterminates.

(3) Ifthecorrespondingcustomerrecordisfound, theprogrameditsthemessageusingtheitemsfrom

thecustomer attributedata base.

(4) Usingthemessagethatit justedited, theprogramupdatesthecustomer'srecordinthecustomer

accountdata base.

(5) Iftheitemsinthecustomer attributedata baseinclude a flagthatindicatesthatthecustomeris a

special customer, theprogramuses themessage thatit justedited to alsoupdate thecustomer'srecordinthespecial customerdata base.

(6) Theprogramwritesthe loginformationinthe logfile.

(7) Theprogramgenerates a responsemessage andterminates.


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Sub-question 1. Fig. 2 shows a flowchartforthe job Xprogram. Selectthecorrect answerto be

insertedinthe intheflowchartwhenitiscompleted, selectingyour answerfromthe

choicesprovided.

Start

Customer'srecordfound?

Writetocustomer

accountdata base

Specialcustomer?

Createresponsemessage

End

End

Special

customerdatabase


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Fig.2Flow Chart for Job X Program

Answergroupfora andc

A. Createerrormessage.

B. Readcustomerattributedata base, usingcustomercode askey.

C. Readcustomeraccountdata base, usingcustomercode askey.

D. Editmessageusingitemsfromcustomerattributedata base.

E. Writetospecial customerdata base.

F. Readspecial customerdata base.

G. Usemessagetoeditcustomer accountrecord.

H. Usemessagetoeditspecial customerrecord.

I. Create loginformation.

J. Write loginformationin logfile.

Answergroupforb

A. Customerattributedata base B. Customeraccountdata base

C. Special customerdata base D. Logfile

Sub-question 2. Assume that thefollowing tableconcerningdisk I/O counts involving thefile

anddata bases listedhas beenfilledinforthecasewhere a messageconcerns a special customer

andnoerroroccurs. Selectthecorrect answersto beinsertedin fromthechoicesprovided.

Table Table of Disk I/O Counts for Each Data Base and file

Data base/filename Data directoryinput

Data base/fileinput

Data base/fileoutput

Total

Customer attribute data

base0

Customer account data

base1

Special customer data

based

Logfile

Total e

Answergroup

A. 1 B. 2 C. 3 D. 4 E. 5

F. 6 G. 7 H. 8 I. 9


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Ans: p n Sub1 a D, c J, b-B Sub 2 d- C, e - H. Phn 1 t s lung cng vic ta d thy

a-D, c J, b B. Phn 2, ta thy data directory input u cn phi c trong trng hp s

dng cc database, ta c bng nh sau:

Data base/file name Data

directory

input

Data base/file

input

Data base/file

output

Total

Customer attribute data

base

1 10

2

Customer account data

base

11

0 2

Special customer data

base

1 1 1 3

Log file 0 0 1 1

Total 3 3 2 8

Nh vy tng cng trong s lung ca chng trnh ta thy s ln cp nht I/O s l 8 ln.

p n Sub2 d C, e-H.

Q17. Read the following description of a microcontroller-controlled fan and then answer

sub-questions 1 through 3.

Fig. 1 showsthesystemconfigurationofthefan, and Table 1 liststhekeyfunctionsofthefan.

Table 2 describestheoperatingstatetransitionsofthefan, indicatingwhichoperatingstatethefan

will shifttofromitscurrentstatewhencertainconditions aremet. Forexample, ifthe ON/OFF

buttonispressedwhilethefanisinthe "continuousoperation" state, thefanwill shifttothe "off"

state. Table 3 liststhe LED lightingstatesforeachoperatingstate.


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COMET extensioninterruptrequestinput Timer

"On/Off"button

"Rest Timer"button

Inputport

Input/outputport Motorcontrol circuit

Bus

1Table 1.Button Functions

Buttonname Function

On/Off Switches betweencontinuousoperation andoff.

Rest Timer Sets/cancels the rest timer 30-minute mode and

60-minutemodesettings.

1Table 2.Fan Operating State Transition Table

OperatingstateCondition Off

Continuousoperation

30-minutemode

60-minutemode

"On/Off" buttonpressed Continuous

operationOff Off Off

"Rest Timer" buttonpressed 30-minutemode

60-minutemode

Continuousoperation

Rest timer remaining time = 0minutes Off

Rest timer remaining time = 30minutes

30-minutemode

Fig.1Fan System Configuration


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Note: A blankspaceindicates a conditionthatisnotpossibleorwhichhasnoeffect.

Table 3. LED Lighting States for Each Operating State

OperatingstateOff

Continuousoperation 30-minutemode 60-minutemodeLED

ON-LED On On On

30-LED On

60-LED On

A blankspaceindicatesthatthe LED isoff.

Sub-question 1. Table 4 explainstheinput/outputport. Thissystem allowsthemicrocontroller

toreaddata thatiswrittentotheinput/outputport. Selectthecorrect answertoinsertin inthe

text belowfromthechoicesprovided.

When answering, note that in the case of the COMET extension, bit 0 is the MSB (most

significant bit).

1Table 4.Explanation of the Input/Output Port

Bit No. Connectedto Remarks

0 ON - LED 0: Off

1: On1 30 - LED

2 60 - LED

3 Motorcontrol

circuit

0: Off 1: On

4~15 Unused Always "0"

Inorder tomanipulate the bits in the input/outputport, a programwaswrittenthatreads the

data in the input/output port, changes the target bit, and writes the new data back to the

input/outputport.

Forexample, inordertoturnoffthe 30-LED andturnonthe 60-LED, thisprogramwill AND

thedata fromtheinput/outputportwiththehexadecimal value a andthen ORthedata withthe

hexadecimal value b , andthenwritetheresult backtotheinput/outputport.

Answergroupfora

A. 7000 B. B000 C. D000 D. E000 E. F000

Answergroupforb

A. 1000 B. 2000 C. 4000 D. 6000 E. 8000


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Sub-question 2. Fig. 2 shows a flow chartfor the mainprogram. From thechoicesprovided,

insertthecorrect answersinthe in Fig. 2, assumingthatinterruptprocessinghasnoeffecton

theprogram.

Mainprogram

0 p TCStopp MODEOtherinitializationprocessing

MODE is a variablethatexpressesthecurrentoperatingstate. TCindicatesthenumberofsecondsuntil thefanshutsoff; TC is avariablethatiscounteddown bythetimerinterruptprocessing.

Wasthe"Rest Timer" button

pressed?

MODE= off

Continuousoperationp MODE Offp MODE

30-minutemodepMODE

60-minutepMODE

ON-LED onMotoron

All LEDsoffMotoroff

0 p TC

30-LED on60-LED off

1800 p TC

30-LED on60-LED off

3600 p TC

30-LED on60-LED off

0 p TC

Offp MODE

Fig.2 Main Program Flow Chart

Answergroupforc

A. Eitherbuttonwaspressed.

B. Neitherbuttonwaspressed.

C. Both buttonswerepressed.

D. "On/Off" buttonwaspressed.

E. "On/Off" buttonwasnotpressed.

AnswergroupfordthroughfA. 30-minutemode B. 60-minutemode

C. Off D. Continuousoperation


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Sub-question 3. Fig. 3 shows the flow chart for the timer interrupt processing. The timer

interruptisgeneratedevery 100ms. Selectthecorrect answerstoinsertinthe in Fig. 3 from

thechoicesprovided.

TimerinterruptCNT is a varia

lethatisinitialized

y themainprogram andisused as an interruptcounter. TCindicatesthenumberofsecondsuntil thefanshutsoff, andisdecrementedby oneeverysecond.

Offp

ODE

ff

M t roff

E

30-mi te modepMODE

30-

ED on

0-

ED off

i

Fig. 3 Timer Interrupt Processing Flow Chart

Answergroupforg

A. 6 B. 10 C. 60 D. 100 E. 1000

Answergroupforh

A.

30 B.

60 C.

180 D.

1800 E.

3600AnswergroupforI

A. 30 - LED B. 30 - LED and 60 - LED

C. 30 - LED and ON - LED D. 60 - LED

E. 60 - LED and ON - LED


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Ans: p n Sub 1 a- B, b B; Sub 2 c D, d D, e A, f B Sub 3 g B, h D, I - cS. Vi

phn 1, gi s ta c trng thi bit i l x -> mun bt bit th ta s OR vi 1, tt bit th ta

AND vi 0, gi nguyn bit th ta AND vi 1 hoc OR vi 0. Nh vy, y ta cn thay i

trng thi input x1x2...x16 , output l x101x40000...0. S a y s l 1011000000000000 hay

B000 cn b y l 0010000000000000 hay 2000. Vi phn 2, ta c c l p n D, n l

iu kin kim tra xem qut c bt khng? d l p n D, t ch chy lin tc chuyn

sang ch 30 minutes, e l p n A t ch 30 minutes sang ch 60 minutes, f l p

n B t ch 60 minutes chuyn sang ch continues. Vi phn 3, ta c CNT y

c m n g th s khi to li v TC s gim i 1 giy, vy g y tng ng vi 1

giy/100ms = 10. p n B. TC y c so snh vi h trc khi chuyn sang ch 30

minutes, vy h y chnh l chuyn t ch 60 minutes sang ch 30 minutes, hay h =

30 pht = 1800. p n D. Cn I y l chuyn t ch 30 minutes sang ch OFF, c

n ON v 30-LED u tt. p n C.

Assembly Language Specifications

1. COMET Hardware Specifications1.1 Processor

(1) COMET is a one-word/16-bitcomputer, ableto access an addressspacefrom 0 to 65535.

(2) The bitformatofa wordinthe COMET computeris asfollows:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(3) Numericvalues areexpressed as 16-bit binarynumbers. Negativenumbers areexpressedin

two'scomplement.

(4) Control issequential. COMET utilizes a two-wordinstructionword.

(5) The COMET hasthreetypesofregisters: GR (16 bits), PC (16 bits), and FR (2 bits).

There arefive GR (General Register)registers, GR0 through GR4. Thesefiveregisters are

usedfor arithmetic, logical, compare andshiftoperations. Of these, GR1 through GR4 are

alsoused asindex registers. GR4 isfurtherused asthestackpointer.

Thestackpointeris a registerthatisusedtostorethe addresscurrently atthetopofthestack.

The PC (Program Counter) stores thefirst address of the instruction word that iscurrently

beingexecuted. Oncethecurrentinstructionfinishesexecuting, thefirst addressofthenextinstructionwordissetinthe PC. Generally, when aninstructionsfinis hesexecuting, the PC is

incremented by two by meansof "address addition" (note); in thecaseof a branch, call, or

returninstruction, thenew addresstowhichcontrol branchesissetinthe PC.

The FR (Flag Register) stores information on whether data stored in a GRregister (as a

Bit No.Sign (Negative: 1, Positive: 0)


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resultofexecuting a load address instructionor anyof various arithmetic, logical, andshift

operationinstructions)isnegative, zero, orpositive, and alsostoresinformationresultingfrom

theexecutionofa compareoperationonthecomparativesizeoftwonumbers.

In other words, the bits in the FR register can be set as shown in the following chart,

depending on the result generated by certain

operation instructions. (In the case of compare

operations, however, refer to section 1.2(4),

"Compare Operation Instructions.")

Inthechartshown atright, whenthesign bitinthe GRregisteris "1," thevaluein questionis

negative. When all ofthe bitsinthe GRregisterare "0," thevalueiszero. Whenthesign bitin

the GRregisteris "0," and at leastoneofthedata bitsis "1," thevaluein questionispositive.

Thefirst (leftmost) bitinthe FRregistershowsthevalueinthe GRsign bit. Thesecond bit

indicateswhetherthevalueinthe GRregisteriszeroornot.Thevaluein FRisreferenced byconditional branchinstructions.

Theexecutionofinstructionsotherthanthosementioned abovedoesnotchangethevaluein

the FRregister.

(6) Instructionwordformat

The COMET utilizes a two-wordinstructionword. Thedetailedformatofinstructionsisnot

defined.

(7) Instructionnotation

Thefollowingnotationisusedinthedescriptionoftheinstructions.

GR General-purposeregisterwiththevalueofGR asitsnumber(where 0 eGRe4).

XR Index registerwiththevalueofXR asitsnumber(where 1 e XRe 4).SP Stackpointer(general-purposeregister4)

adr Label name (representing thatcorresponds to the label)or a decimal constant

(where

32768e adre65535 ; adrcanhave a valuefrom 0 to 65535, butthevaluesfrom 32768 to

65535 can also beusedtoexpressnegativedecimal numbers).

Effective address A valueproduced by adding, through "address addition," adrandthe

contentsofXR, orthe addresspointed at bythatvalue.

(X) Thecontentsofaddress X; ifX is a register, thenthecontentsofthatregister.

[ ] Square brackets ([ ])indicatethattheinformationcontainedinthe bracketsmay

beomitted.IfXRisomitted, modification bytheindex registerisnotperformed.

(Note) "Address addition": Treats the data to be added as unsigned data; the result is the

Data setinthe GRregister

Negativ

Zero Positi


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remainderthatisobtained by addingthetwovalues andthendividingthesum by 65536. (In

otherwords, theresultisthevalueofthe 16 lower bitsofthesum.) "Addresssubtraction" is

definedin a similarfashion.


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1.2 Instructions

Instructions andtheirfunctions aredescribedinthefollowingchart. Instructions aredescribedin

assemblernotation.

InstructionFormat

DescriptionofinstructionsOpcode

Operand

(1) Load, storeinstructionLoaD LD GR,adr

[,XR]

Sets (effective address)in GR.

STore ST GR,adr

[,XR]

Stores (GR)inthe addressindicated bytheeffective address.

(2) Load addressinstructionsLoad Effective Address LEA GR,adr

[,XR]

Setstheeffective addressin GR.

Sets FRin accordancewiththevalueofGR.

(3) Arithmetic and logical operationinstructions

ADD arithmeticADD GR,adr

[,XR]

Performsthespecifiedoperationon (GR) and(effective address), andsetstheresultin GR. Notethatinthesubtractionoperation, (effectiveaddress)issubtractedfrom (GR). Sets FRin accordancewiththevalueoftheoperationresult.

SUBtract arithmeticSUB GR,adr

[,XR]

ANDAND GR,adr

[,XR]

OROR GR,adr

[,XR]

Exclusive OREOR GR,adr

[,XR]

(4) CompareoperationinstructionsComPare Arithmetic CPA GR,adr

[,XR]

Performs an arithmeticcompareorlogicalcompareoperationon (GR) and (effectiveaddress), andsets FR asfollows, accordingtotheresultofthecompareoperation.

ComPare Logical CPL GR,adr

[,XR]

(5) ShiftoperationinstructionsShift Left Arithmetic SLA GR,adr

[,XR]

Shifts GR (exceptforthesign bit) leftorright bythenumberofbitsspecified bytheeffectiveaddress.When a leftshiftisperformed, those bitsthat are

Compareresult FR bitvalue(GR) > (effective address) 00(GR) = (effective address) 01(GR) < (effective address) 10


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Shift Right Arithmetic SRA GR,adr

[,XR]

leftvacant bytheshiftoperation arefilledwithzeroes. When a rightshiftisperformed, those bits

that are leftvacant bytheshiftoperation arefilledwiththesamevalue asthesign bit.Sets FRin accordancewiththevalueoftheshiftoperationresult.

Shift Left Logical SLL GR,adr

[,XR]

Shifts GR (includingthesign bit) leftorright bythenumberofbitsspecified bytheeffectiveaddress.Those bitsthat are leftvacant bytheshiftoperation arefilledwithzeroes.Sets FRin accordancewiththevalueoftheshiftoperationresult.

Shift Right Logical SRL GR,adr

[,XR]

(6) BranchinstructionsJumpon PlusorZero JPZ adr[,XR] Branchestotheeffective address, dependingon

thevalueofFR. Ifcontrol doesnot branchto a

new address, executioncontinueswiththenextinstruction.J

umpon MInus JMI adr[,XR]

Jumpon Non Zero JNZ adr[,XR]

Jumpon ZEro JZE adr[,XR]

unconditional JuMP JMP adr[,XR] Branchesunconditionallytotheeffective address.

(7) StackoperationinstructionsPUSH effective address PUSH adr[,XR] After subtracting (by means of "address

su

btr

action

")

one

from

SP,this

instruction

stores

theeffective addressin address (SP).POP up POP GR Aftersettingthecontentsofaddress (SP)in GR,

thisinstruction adds (bymeansof"addressaddition")oneto SP.

Instruction ValueofFRinorderto branch JPZ 00, 01JMI 10JNZ 00, 10JZE 01


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(8) Call andreturninstructionsCALL subroutine CALL adr[,XR] After subtracting (by means of "address

subtraction") one from SP, this instructionstores

the value resulting from adding (by means of

"address addition")twotothecurrentvalueofthe

PC in address (SP), and then branches to the

effective address.RETurnfromsubroutine RET Afterfetchingthecontentsofaddress (SP), this

instruction adds (bymeansof"address addition")onetothe SP andthen branchestothe addressthatisindicated bythepreviouslyfetchedcontents.(Thefetchedcontents aresetinthe PC.)

1.3 Character SetCOMET supports a JIS X0201 Romaji/katakana charactersetthatuses 8-bitcodes. Partofthe

charactersetisshowninthetable below. Ifanycharactersnot listedinthis areneededtosolveone

oftheproblems, the bitconfigurationforthatcharacterwill begivenintheproblem.

[ExplanationofTable]

Eight bits are used to represent one

character; the upper four bits indicate the

column in the table, and the lower four bits

indicate the row. For example, the

hexadecimal codes for the space character,

"3," "F," and "Z" are 20, 33, 46, and 5A,

respectively.

The characters that correspond to the

hexadecimal codes 21 to 5F, 60 to 7E, and A1

to DF are called "graphics characters." This

table omits the characters corresponding to

codes 60 through 7E (defined as lower-case

letters) andcodes A1 through DF (defined as

kana, etc.).

Graphics characters can be displayed

(printed) as characters on an output device

such as a monitorora printer.2. Specifications of the CASL Assembly Language

Assembly language for the COMET is

called "CASL." ThespecificationsforCASL

aredescribed below.2.1 Instruction Types

Column


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Pseudoinstruction FunctionSTART Definesthetopofa program.

Defines thestarting addressforexecutionofa program.Defines the entry for linking with other

programs.END Definestheendofa program. DS Allocates an area. DC Defines a constant.

CASL consists of four pseudo instructions(START,END ,DS,and DC),threemacroinstructions (IN,OUT,and EXIT) and machine language instructions

(COMET instructions).

(1) Pseudo Instructions

Pseudo instructions are used for tasks

such as assembler control, definition of

constants, and creating data needed to link

theprogram. The functions of thepseudo

instructions are listedinthetable atright.

(2) Macroinstructions

Macro instructions use a pre-defined

groupofinstructions andoperanddata togenerate a group of instructions that

performs a desiredfunction.

CASLprovides macro instructionsforperforming I/O, terminating a program, etc. Because

machine language I/O instructions, etc., arenotdefined in CASL, suchprocessing is left to the

operatingsystem. When a macro instruction is

encountered, CASL generates a group of

instructions that calls the operating system.

Assumethatthenumberofwordsinthegroupof

instructionsthatisgeneratedisundefined.

When a macroinstructionisexecuted, thecontentsofthe GRregisters aresaved butthecontentsofFR

areundefined.

Thefunctionsofthemacroinstructions are listedinthetable atright

(3) Machine languageinstructions

There are 23 machine languageinstructions, as

describedinsection 1.2, "Instructions."

Macroinstruction FunctionIN InputOUT OutputEXIT Terminatesprogramexecution.


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2.2 Instruction Format

Pseudo instructions, macro instructions, and machine language instructions all have a label

field, anopcodefield, anoperandfield, and a commentfield. Eachfieldisdefi ned below.Label field The label consistsofuptosix characters, startingfromthefirstcharacter.

Opcodefield When there is no label: Starts from the second or any other subsequent

characterposition.

Whenthereis a label: Startsfrom anycharacterpositionfollowing at least

one space

character afterthe label field.

Operandfield Startsfrom anycharacterpositionfollowing at leastonespacecharacterafter

theopcodefield, uptothe 72ndcharacter. Thisfieldcannot becontinuedon

thenext lineintheprogram.

Commentfield When a semicolon (;) isencountered in a line, everything thatfollowsuntiltheendofthe lineistreated as a comment. (Thisdoesnot applytosemicolons

(;)encounteredin a characterstringfora DC instruction, however.)

Notethatifthefirstnon -spacecharacterin a lineis a semicolon, theentire line

istreated as a comment.

Anycharacter (thatis allowed bytheprocessingsystem)can bewrittenin a

commentfield.

The format of each

instructionin CASL isshownin

thetable atright.When "blank" isindicatedfor

a field, that field must not be

described.

(1) Registerspecification

In the operand field of a machine language

Label Instructioncode Operand Howtoreadlabel START Executionstart address STARTBlank END Blank END[label] DC Constant Define Constant[label] DS Lengthofarea inwords Define Storage [label] IN Input area,

inputstring length INput

[label] OUT Output area,

outputstring length

OUTput

[label] EXIT Blank EXIT[label] Machine languageinstruction (Referto 1.2, "Instructions.")

Specification byregistersymbol

General-purposeregisternumber

GR0 0GR1 1GR2 2GR3 3GR4 4


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instruction, thenumbers 0 through 4 areused to

specify a general register, butthesymbolsshown

inthetable atrightcan also beused.

(2) Label field

"label" inthe label fieldrepresents a label. A label canconsistofuptosix characters; thefirst

charactermust be anupper-case letter. Subsequentcharactersmay beeitherupper-case lettersor

numbers. The labels assigned to DC, DS, IN, OUT, EXIT andmachine language instructions

indicate the address of the first word of the associated area or instruction word (or group of

instructions, inthecaseofa macroinstruction).

The label that is assigned to a START instruction can bereferenced as anentry name from

anotherprogram.

2.3 Pseudo Instructions

(1) START [Executionstart address]

Thispseudoinstructionindicatesthetopofaprogram. Inotherwords, thispseudoinstruction

must bewritten atthevery beginningofa program.

The label namethatisdefinedwithinthisprogramspecifiestheexecutionstart address. Ifthe

label isomitted, execution begins atthestartoftheprogram.

(2) END

Thispseudo instruction indicates the end of theprogram. Thispseudo instruction must be

written attheveryendoftheprogram.(3) DC Constant

Thispseudo instructionstoresconstantdata thathas beenspecified as a constant. There are

fourtypesofconstants: decimal constants, hexadecimal constants, characterconstants and address

constants.


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Typeof

constant

Format

DescriptionofinstructionOpco

de

Operand

Decimal

constant

DC n This instructionstores the decimal value specified by "n" as

onewordofbinarydata. Ifnisoutsideoftherangeof32768

to 32767, justthe lower16 bitsofn arestored.

Hexadeci

mal

constant

DC #h Assume "h" is a four-digit hexadecimal number.

(Hexadecimal notationuses 0 through 9 and A through F.) This

instructionstoresthehexadecimal valuespecified by "h" asone

wordofbinarydata. (0000 ehe FFFF)

Characte

r

constant

DC 'character

string'

This instructionstoreseach character in the character string,

startingfromthe leftendofthestring, inthe lowereight bitsof

consecutivewords. Inotherwords, thefirstcharacterisstored

in bits 8 through 15 of the first word, the second character is

storedin bits 8 through 15 ofthesecondword, andsoon, sothat

the character data for each character in the string is stored

sequentiallyinmemory. Bits 0 through 7 ofeachword arefilled

withzeroes. Spaces and anyofthegraphicscharacters (referto

1.3, "Character Set") can be written in a character string.

Apostrophes (')cannot bewrittenincharacterstrings, however.

A character string can not have a length of zero (i.e., the

characterstringcannot beempty).

Address

constant

DC Label name This instructionstores an addressvaluecorresponding to the

label name asonewordofbinarydata. Ifthe label nameisnot

defined within this program, the assembler refrains from

determining the address and entrusts that task to theoperating

system. (Referto 3.1, "Undefined Labels.") (4) DS Area word length

Thispseudoinstruction allocates an area ofthespecifiedword length.

Theword lengthofthe area isspecified by a decimal number (u 0). If"0" isspecifiedforthe

word lengthofan area, the area isnot allocated, butthe label nameinthe label fieldisvalid. 2.4 Macro Instructions

(1) IN Input area, inputcharacterlength

This macro instruction inputs one record of data (character data) to the input area from a

previously assigned input device. (For details on assigning an input device, refer to 3.1(3),

"Input/outputdevice assignment.")

Write label names for the input area and the input character length operands in theoperand


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field.

Theinput area operandshould bethe label nameofan 80-wordworkarea (whichissufficient

for 80 characters). The input data is input in this area (beginning at the starting address), one

characterperword. Zeroes arestored in bits 0 through 7 ofeachword (in thesamemanner as

characterconstantswiththe DC instruction).

The lengthofthedata thatwasinput (thenumberofcharactersintheinputrecord)isstored as

binary data in the area (one word) specified by the input character length operand. No record

delimitercode (such as a linereturncode, whenusing a keyboard)isstored.

If the inputdata is less than 80 characters long, thepreviousdata (thedata thatwas already

there beforethismacrowasexecuted)is left asisintheremainingportionoftheinput area. Ifthe

inputdata exceeds 80 characters, theexcesscharacters areignored.

"0" and "1" arestored astheinputcharacter lengthinthefollowinginstances:

0: Wheninputting anemptyrecord (forexample, whenonlythereturncodewasinputfromthetypewriterterminal).

1: When EOF (endoffile)wasdetected (whenusing a cardreader, etc.)

(2) OUT Output area, outputcharacterlength

This macro instruction outputs data stored in the output area as one record of data to the

previously assigned output device. (For details on assigning an output device, refer

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