databasehomeworkhelp.com_ Database System Assignment Help (1).pptx

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Problem 1.
To reduce the number of plans the query optimizer must consider, the Selinger
Optimizer employs a number of heuristics to reduce the search space. List three:
Solution.
a. Push selections down to the leaves.
b. Push projections down. – Consider only left deep plans.
c. Join tables without join predicates (cross product joins) as late as possible.
Problem 2.
Give one reason why the REDO pass of ARIES must use physical logging.
Solution.
The pages on disk may be in any state between “no updates applied” and “all updates
applied.” Logical redo cannot be used on an indeterminate state, whereas physically
redo
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logging just overwrites the data on disk with the logged values. In other words, physical
logging is idempotent whereas logical logging is not.
II Optimistic Concurrency Control
Problem 3.
For the following transaction schedules, indicate which transactions would commit and
which would abort when run using the parallel validation scheme described in the Kung
and Robinson paper on Optimistic ConcurrencyControl. Also give a brief justification for
your answer.You may assume that if a transaction aborts, it does not execute its write
phase, rolling back instantly at the end of the validation phase.

Solution.
Transactions that commit:T1,T2 orT1
Transactions that abort: {} orT2
Justification:
There are two correct solutions:
T1 andT2 both commit.T1 commits because it validates first.T2 will also commit
because while its write set intersectsT1’s write set,T1 finishes its write phase beforeT2
starts its write phase (condition 2 in section 3.1 of the paper).
T1 commits,T2 aborts.T1 commits because it validates first. In the pseudocode from
the paper (section 5),T1 is executing the line marked “write phase” whenT2 starts its
validation phase.T2 findsT1 in the “finish active” set, and aborts due to a write/write
conflict.This is sort of a “false” conflict, since it would be safe forT2 to commit.
However, the parallel algorithm assumes thatT1 is still writing at the time thatT2
detects the conflict, so it must abort.
Problem 4.

Solution
Transactions that commit:T1,T3
Transactions that abort:T2
Justification:
T1 andT3 commit,T2 aborts.T1 commits because it validates first.T2 must abort
becauseT1’s write set intersectsT2’s read set, andT2 was started reading beforeT1
finished writing.T3 then commit because although it has a conflict withT2, when it
begins the validation algorithm,T2 has already aborted (it finishes aborting at the end of
the validation phase).

III Schema Design
Consider a relational table:
( professor
professor name, professor id,
student name, student office id,
student designated refrigerator id, refrigerator owner id,
refrigerator id, refrigerator size, secretary name,
secretary id, secretary office )
Suppose the data has the following properties:
A. Professors and secretaries have individual offices, students share offices.
B. Students can work for multiple professors.
C. Refrigerators are owned by one professor.
D. Professors can own multiple refrigerators.
E. Students can only use one refrigerator.
F. The refrigerator the student uses must be owned by one of the professors they work
for.
G. Secretaries can work for multiple professors.
H. Professors only have a single secretary.

Problem 5.
[Put this table into 3rd normal form by writing out the decomposed tables; designate keys
in your tables by underlining them. Designate foreign keys by drawing an arrow from a
foreign key to the primary key it refers to. Note that some of the properties listed above
may not be enforced (i.e., guaranteed to be true) by a 3NF decomposition.
Solution.

Problem 6.
Which of the eight properties (A–H) of the data are enforced (i.e., guaranteed to be true)
by the 3NF decomposition and primary and foreign keys you gave above
Solution:
B,C,D,E,G,H; the schema does not enfore that profs/secretaries have individual offices
or that students use the fridge of the professor they work for.
Problem 7.
What could a database administrator do to make sure the properties not explicitly
enforced by your schema are enforced by the database?
Solution.
Use triggers or stored procedures to ensure that these constraints are checked when
data is inserted or updated.

IV External Aggregation
Suppose you are implementing an aggregation operator for a database system, and you
need to support aggregation over very large data sets with more groups than can fit into
memory.
Your friend Johnny Join begins writing out how he would implement the algorithm.
Unfortunately, he stayed up all night working on Lab 3 and so trails off in an incoherent
stream of obscenities and drool before finishing his code. His algorithm begins as follows:
input :TableT, aggregation function f name, aggregation field agg f , group by field gby
f /* AssumeT has |T| pages and your system has |M| pages of memory
/* Partition phase
n |T/M| ← ;
Allocate bu f s, n pages of memory for output buffers ;
Allocate f iles, an array of n files for partitions ;
foreach record r ∈T do
h ← hash(r.gby f) ; /* h ∈ [1 . . . n] */
Write r into page bu f s[h] ;
if page bu f s[h] is full then
Add bu f s[h] to file f iles[h], and set the contents of bu f s[h] to empty ;
end

end
/* Aggregate phase
foreach f ∈ f iles do
/*Your code goes here.
End
Problem 8.
Relative to |T|, how big must |M| be for the algorithm to function correctly? Justify your
answer with a sentence or brief analytical argument.
Solution.
Because bu f s should fit in memory, n ≤ |M|.T
The number of hash buckets is computed as n = | M | . | |T
Plugging in for n, we get that | | M | | ≤ |M|.
Therefore, |T| ≤ |M| 2 .
It follows that |M| ≥ √ |T|.
Problem 9.
Assuming the aggregate phase does only one pass over the data, and that f name = ’AVG’,

describe briefly what should go in the body of the for loop (in place of “Your code goes
here”).You may write pseudocode or a few short sentences. Suppose the system provides
a function emit(group,val) to output the value of a group.
Solution.
All tuples with the same gby f will end up in the same partition, but there may (very likely
will) be multiple gby f values per partition.
The basic algorithm should be something like:
H = new HashTable // stores (gbyf, ) pairs
for tuple t in f:
if (( = H.get(t.gbyf)) == null)
sum = sum + t.aggf cnt = cnt + 1
T.put(t.gbyf,)
for (group, ) inT:
emit(group,sum/cnt)
Any in-memory aggregation would work here, so if you prefer to sort the values in
memory and then bucket them as is done in the next question, that is also an acceptable
answer.

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