Sheet 3 Model Answer (1)

8/13/2019 Sheet 3 Model Answer (1)

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Assignment 2Advanced Database Management (CP3020)

1. Consider the SQL !er"#

SELECT$.%name& $.Lname& S.%name& S.Lname

FROM $MPL'$$ AS $& $MPL'$$ AS S

WHERE

$.S!er*ssn + S.Ssnin ,hich& -or each emo"ee (-rom the coman" schema given in

"o!r te/tboo)& retrieves the emo"ees -irst and ast name and

the -irst and ast name o- his or her immediate s!ervisor.

Ass!ming that the $MPL'$$ reation is#

$MPL'$$ (%name#string& Lname#string& Ssn#string&

date#date& Address#string& Se/#char& Saar"#rea&

S!er*ssn#string& Dno#integer) and the sie o- each -ied is as

-oo,s#

%name 20 b"tes

Lname 20 b"tes

Ssn 10 b"tes

date 4 b"tes

Address 50 b"tes

Se/ 1 b"te

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Saar" 5 b"tes

S!er*ssn 10 b"tes

Dno 1 b"te

6hat is& records are -i/ed and each record contains 115 b"tes.

(a)Dra, t,o !er" trees that can reresent this !er". 7nder ,hat

circ!mstances ,o!d "o! !se each o- "o!r !er" trees8

(b)Dra, the initia (canonica) !er" tree -or this !er"& and then

sho, (ste b" ste& ,ith some 9!sti-ications:e/anations) ho,

the !er" tree is otimied b" the agorithm o!tined in Chater

1;.

(c) Comare (in terms o- the n!mber o- b"tes that m!st be

trans-ered) "o!r o,n !er" trees o- art (a) and the initia and -ina

!er" trees o- art (b).

2. Consider schedules S1, S2, S3, and S4 belo.

S1 ! r1"#$, r2"%$,1"#$, r3"#$, c1,2"%$, r2"& $,3"#$,2"& $, r3"& $, c2,3"& $, c3'

S2 ! r1"#$, r2"%$, r2"& $,1"#$, r3"#$, c1,2"& $,3"#$, c3, r2"#$,2"%$,2"#$, c2'

S3 ! r1"#$, r2"%$,1"#$, r3"#$,2"%$, r2"& $,3"#$,2"& $, r3"& $,3"& $'

S4 ! r1"#$, r2"%$, r2"& $,1"#$, r3"#$,2"& $,3"#$, r2"#$,2"%$,2"#$'

"a$ (e)er*ine he)her each schedule S1 and S2 is s)ric), cascadeless, reco+erable,

or nonreco+erable.

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Solu)ion!

S)ric)ness is a s)roner re-uire*en) )han cascadeless, because

addi)ional li*i)a)ions on )he ri)in beha+ior are i*osed. S)ric)ness

i*lies cascadeless. Casacdeless i*lies reco+erabili)/.

The schedules o0 S1 and S2 are as 0ollos!

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S1 S2

T1 T2 T3 T1 T2 T3

r"#$ r"#$

r"%$ r"%$

"#$ r"&$

r"#$ "#$

c1 r"#$

"%$ c1

r"&$ "&$

"#$ "#$

"&$ c3

r"&$ r"#$

c2 "%$

"&$ "#$

c3 c2


4/12

Schedule S1 is recoverable.

Since there is no such a situation that one transaction reads a valuewritten by another, but commits before that one. Its recoverable.Since r(Y) in T3 is deendent on w(Y) in T! and before it committed.Its not cascadeless. Thus its not strict.

Schedule S! is strict"#nly T! reads a value (r($)) written by one of the other transactions(w($)), namely T3. This value has already been committed. ScheduleS! therefore is casacdeless, and is therefore also recoverable. S! is

also strict, because no ob%ect is read or written, that has beenudated by a transaction that has not committed or aborted yet.n shor

Schedule Reco+erable Cascadeless S)ric)

S1 &es o o

S2 &es &es &es

"b$ (e)er*ine he)her or no) each schedule S3 and S4 is con0lic) serialiable. lso

0or each serialiable schedule, de)er*ine )he e-ui+alen) serial schedule.

The schedules S3 is as 0ollo!

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5

S3

T1 T2 T3

r"#$

r"%$

"#$

r"#$

"%$

r"&$

"#$

"&$

r"&$

"&$


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S3 is con0lic) serialiable.

T1 T2 E-ui+alen) serial schedules

T3

The schedules S4 is as 0ollo!

6

S4

T1 T2 T3

r"#$

r"%$

r"&$

"#$

r"#$

"&$

"#$

r"#$

"%$

"#$

T1&&&&T3T!&&&&T3


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S3 is con0lic) serialiable.

T1 T2 E-ui+alen) serial schedules

T3

"c$ l/ )he basic )i*es)a* orderin alori)h* )o

7

T1&&&&T3&&&T!T1&&&&T!


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schedules S3 and S4. (e)er*ine he)her or no) )he alori)h*

allos )he e5ecu)ion o0 )he schedules, and discuss cascadin

rollbac6 "i0 an/$.

Hin)s! each oera)ion )a6es one )i*e uni), and )i*es)a* o0

each )ransac)ion is )he

)i*e associa)ed )o i)s 0irs) oera)ion. For e5a*le,

)i*es)a*s o0 )ransac)ions T1, T2,

and T3 in schedule S4 are 1, 2, and 7 "resec)i+el/$.

S3! Ti*es)a*s T181, T282, T384.

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9

S3

T1 T2 T3

r"#$

r"%$

"#$

r"#$

"%$

r"&$

"#$

"&$

r"&$

"&$

Set '&timestam($) 1

Set '&timestam () !

Set *&timestam ($) 1

Set '&timestam ($) +

Set *&timestam () !

Set '&timestam (Y) !

Set *&timestam ($) +

Set *&timestam (Y) !

Set '&timestam (Y) +

Set *&timestam (Y) +


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A transactions are e/ec!ted s!ccess-!" b" the timestam

ordering agorithm.

S3! Ti*es)a*s T181, T282, T387.

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S4

T1 T2 T3

r"#$

r"%$

r"&$

"#$

r"#$

"&$

"#$

r"#$"abor)$

"%$

"#$

Set '&timestam($) 1

Set '&timestam () !

Set '&timestam (Y) !

Set *&timestam ($) 1

Set '&timestam ($)

Set *&timestam (Y) !

Set *&timestam ($)

Set '&timestam ($) !(cascadin- rollbac)


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Since r(X) in T2 reads X written by T3, T3 is abort.

And there is a chain rollback.

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Sheet 3 Model Answer (1)