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I Short Answer
1. To reduce the number of plans the query optimizer must consider, the Selinger Opti mizer
employs a number of heuristics to reduce the search space. List three:
– Push selections down to the leaves.
– Push projections down.
– Consider only left deep plans.
– Join tables without join predicates (cross product joins) as late as possible.
2. Give one reason why the REDO pass of ARIES must use physical logging.
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 logging just
over writes the data on disk with the logged values. In other words, physical logging is
idempotent whereas logical logging is not.
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II Optimistic Concurrency Control
Read Validate Write
T1
Read set: {A,B} Write set: {B,C}
T2
Read set: {A,D} Write set: {A,B}
Transactions that commit: T1,T2 or T1 Transactions that abort: {} or T2
Justification:
There are two correct solutions:
T1 and T2 both commit. T1 commits because it validates first. T2 will also commit because while its write set intersects T1’s write
set, T1 finishes its write phase before T2 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” when T2 starts its validation phase. T2 finds T1 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 for T2 to commit. However, the parallel algorithm assumes that T1 is
still writing at the time that T2 detects the conflict, so it must abort.
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 Concurrency
Control. 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.
3.
Time
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4.
Read Validate Write
T1
Read set: {A,B} Write set: {B,C}
T2
Read set: {A,B} Write set: {B,D}
T3
Read set: {D} Write set: {D}
Transactions that commit: T1,T3
Transactions that abort:T2
Justification:
T1 and T3 commit, T2 aborts. T1 commits because it validates first. T2 must abort because T1’s
write set intersects T2’s read set, and T2 was started reading before T1 finished writing. T3 then
commit because although it has a conflict with T2, when it begins the validation algorithm, T2 has
already aborted (it finishes aborting at the end of the validation phase).
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III Schema Design
Consider a relational table:
Professor(
professor name, professor id, professor office id,
student 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.
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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.
prof: pid pname poffice psec
student: sid sname soffice sfridge
sec: secid secname secoffice
fridge: fridgeid fridgesize fridgeowner
sworksfor: sid profid
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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?
B,C,D,E,G,H; the schema does not enfore that profs/secretaries have individual offices or that stu dents
use the fridge of the professor they work for.
7.What could a database administrator do to make sure the properties not explicitly en
forced by your schema are enforced by the database?
Use triggers or stored procedures to ensure that these constraints are checked when data is in serted
or updated
IV External Aggregation
Suppose you are implementing an aggregation operator for a database system, and you need
to support aggre gation 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:
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��
��
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.
Because bu f s should ﬁt in memory, n ≤ |M|.
The number of hash buckets is computed as n =
.
|T |
|M|
|T |
Plugging in for n, we get that
|M|
≤
|M|.
Therefore, |T| ≤ |M|2.
It follows that |M| ≥
√
|T
|.
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.
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, <sum,cnt>) pairs for tuple t in
f :
i f ((<sum,cnt> = H.get(t.gbyf)) == null)
<sum,cnt> = <0,0> sum = sum + t.aggf cnt
= cnt + 1
T.put(t.gbyf,<sum,cnt>) for (group, <sum,cnt>)
in T:
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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10. Describe an alternative to this hashing-based algorithm. Your answer shouldn’t require
more than a sentence or two.
Remember that the tuples do not fit in memory.
One solution would be to use an external memory sort on the table, where the sort key for
each tuple is the gbyf field.
After sorting, sequentially scan the sorted file’s tuples, keeping a running average for each gbyf
grouping (they will appear in contiguous chunks in the file). The average should be calculated
on each aggf field.
Other solutions included using external memory indices, such as B+Trees, on gbyf to group the
tuples on which to calculate an average.
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V Cost Estimation
Suppose you are given a database with the following tables and sizes, and that each
data page holds 100 tuples, that both leaf and non-leaf B+Tree pages are dense-packed
and hold 100 keys, and that you have 102 pages of memory. Assume that the buffer
pool is managed as described in the DBMIN Paper (“An Evaluation of Buffer
Management Strategies for Relational Database Systems.”, VLDB 1985.)
Table Size, in pages
T1 100
T2 1000
T3 5000
11. Estimate the minimum number of I/Os required for the following join operations. Ig nore the
difference between random and sequential I/O. Assume that B+Trees store pointers to records in heap
files in their leaves (i.e., B+Tree pages only store keys, not tuples.)
•Nested loops join between T1 and T2, no indices.
T1 fits in memory. Put it as the inner and read all of T1 once. Then scan through T2 as the outer. Total: |T
1| + |T2| = 1100 I/Os.
•Grace hash join between T2 and T3, no indices.
Grace join hashes all of T2, T3 in one pass, outputting records as it goes. It then scans through the hashed output to do
the join. Therefore it reads all tuples twice and writes them once for a total cost of: 3(|T2| + |T3|) = 18000 I/Os.
•Index nested loops join between a foreign key of T2 and the primary key of T3, with a B+Tree index on
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T3 and no index on T2.
Put T2 as the outer and T3 as the inner. Assume the matching tuples are always on a single B+Tree page
and that selectivity is 1 due to foreign-key primary-key join. Cache the upper levels of the B+Tree. T3 is
500,000 tuples, so it needs 3 levels (top level = 100 pointers, second level = 1002 = 10, 000, third level
1003 = 1, 000, 000 pointers). Cache the root plus all but one of the level 2 pages (100 pages). Read all of
T2 once (one page at a time). For each tuple in T2, we do a lookup in the B+Tree and read one B+Tree leaf
page (at level 3), then follow the pointer to fetch the actual tuple from the heap ﬁle (using the other page of
memory).
Total cost is: 1000(|T2|) + 100(top o f B + Tree) + {T2} × No.BTree lookups
For 99/100 of the B+Tree lookups, two levels will be cached, so {T2} × No.BTree lookups is:
99/100 ∗(2 × |T2|) = 99/100 ∗2 × 100000 = 198000 pages.
For 1/100 of the B+Tree lookups, only the root level will be cached:
1/100 ∗(3 × |T2|) = 1/100 ∗3 × 100000 = 3000 pages.
So the total is :
1000 + 100 + 198000 + 3000 = 202100 I/Os
We were ﬂexible with the exact calculation here due to the complexity introduced by not being able to
completely cache both levels of the B+Tree.
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LSN TID PrevLsn Type Data
1 1 - SOT -
2 1 1 UP A
3 1 2 UP B
4 2 - SOT -
5 2 4 UP C
6 - - CP dirty, trans
7 3 - SOT -
8 2 5 UP D
9 3 7 UP E
10 1 3 COMMIT -
11 2 8 UP B
12 2 8 CLR B
13 3 7 CLR E
VI ARIES with CLRs
Suppose you are given the following log ﬁle.
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