INSTITUTO SUPERIOR TÉCNICO Administração e optimização de Bases de Dados

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1 INSTITUTO SUPERIOR TÉCNICO Administração e optimização de Bases de Dados Exam 1 - solution 16 June The duration of this exam is 2,5 Hours. You can access your own written materials, but the exam is to be done individually. You are not allowed to use computers, tablets, nor mobile phones. The maximum grade of the exam is 20 pts. Write your answers below the questions. Write your number and name at the top of each page. Present all calculations performed. After the exam starts, you can leave the room one hour after delivering the exam. The following table is be used by instructors, ONLY: SUM

2 1. (4 vals) Indexing 1.1. (2,5 pts) Suppose that we are using extendable hashing on a file that contains records with the following search-key values: Search-key values Hash value Ronaldo Messi Hernandez Iniesta Lahm Ibrahimovic Rooney Neymar Show the extendable hash structure for this file, if the hash values for each search key are as shown in the table above, and if buckets can hold up to three records. Use the least significant bits of the hash value. 2

3 1.2. (1 pt) Suppose that you have a sorted file and want to construct a dense primary B+ tree index on this file. a) One way to accomplish this task is to scan the file, record by record, inserting each one using the B+ tree insertion procedure. What performance and storage utilization problems are there with this approach? b) Explain how the bulk-loading algorithm improves upon this scheme. a) This approach is likely to be quite expensive, since each entry requires us to start from the root and go down to the appropriate leaf page. Even though the index level pages are likely to stay in the buffer pool between successive requests, the overhead is still considerable. Also, according to the insertion algorithm, each time a node splits, the data entries are redistributed evenly to both nodes. This leads to a fixed page utilization of 50%. b) The bulk loading algorithm has good performance and space utilization compared with the repeated inserts approach. Since the B+ tree is grown from the bottom up, the bulk loading algorithm allows the administrator to pre-set the amount each index and data page should be filled. This allows good performance for future inserts, and supports some desired space utilization (0,5 pt) What is the difference between a B+-tree index and a B+-tree file organization? Indicate one advantage of the each schema. In a B+ tree file organization, the leaf nodes store records instead of pointers to records. B+ tree indices solve the problem of index file degradation while B+ tree file organization solves the data file degradation problem. 3

4 2. (4 pts) Query Processing and Optimization 2.1. (2,5 pts) Consider performing a natural join between the following two relations: Client(Name,ID) ClientDetails(ID,Property,Value) Assume that the Client tuples are stored contiguously on 2000 disk blocks and that the ClientDetails tuples are stored contiguously on 400 blocks. Each block of Client or ClientDetails holds up to 50 tuples. There are 102 memory blocks available. Compute the I/O cost for each of the following join algorithms, justifying your result. Ignore the I/O cost of writing the output to disk. Unless stated otherwise, the tuples in the relations are not sorted. a) Merge join, sorted relations (i.e., assume that both relations are sorted). b) Merge join, unsorted relations (i.e., assume that both relations are unsorted). c) Index join. Assume that there is an index on the ID column of Client. We read a block of ClientDetails and, for each tuple in this block, we use the index to find all matching tuples of Client. Each of these Client tuples is read into memory and joined with the tuples from ClientDetails. We repeat the process for all blocks of ClientDetails. Assume that the index is entirely in memory, and assume that, on average, each tuple of ClientDetails matches 4 tuples of Client. d) Hash join. Assume Client as the build relation. a) Cost merge join = bclient + bclientdetails = = 2400 b) Cost merge join + cost sorting Client + cost sorting ClientDetails Cost sorting Client = bclient(2*ceiling(log M-1 (bclient/m)) + 1) = = 2000 * (2*ceiling(log 101 (2000/102)) + 1) = = 2000 * (2*1 + 1) = 6000 Cost sorting ClientDetails = 400 * (2*ceiling(log 101 (400/102)) + 1) = = 400 * (2*1 + 1) = 1200 Cost = = 9600 c) bclientdetails + nbtuplesclientdetails * c, where c is the cost of a selection using an index c = h + b, where h is the height of the index, i.e., the number of blocks to go from the root until the leaves in this case, it is zero because the index is entirely in memory, so there is no need to transfer blocks from disk to memory. b is the number of blocks containing records with the specified search key, i.e. the number of Client tuples that match each ClientDetails tuple. nbtuplesclientdetails = 400 * 50 =

5 Cost = * 4 = d) bclient/m = 2000/102 = 19.6 < 102 so there is no need for partitioning Cost = 3* (bclientdetails + bclient) = 3 *( ) = (1 pt) Consider the following database relations: Client(Name,Address,ClientID) ClientSubcriptions(ClientID,SubscriptionType) ClientID: Foreign Key(Client) The relation Client has 200 tuples and the relation ClientSubscriptions has 600 tuples. Answer the following questions: a) Estimate the number of tuples of Client X ClientSubscriptions. b) Consider the selection: σ ClientID=2 (Client X ClientSubscriptions). Estimate the number of tuples returned by the selection. a) ClientID is primary key of Client and foreign key in ClientSubscriptions referencing Client, so the number of tuples in the result of the join is the same as the number of tuples in ClientSubscriptions = 600 b) σ ClientID=2(Client X ClientSubscriptions) = 600 / V(ClientID, Client X ClientSubscriptions) = 600/200 = (0,5 pt) What would change in the answer to question 2.2.b) if the selection condition was ClientID>2? The number of tuples would be n/2 =600/2 = 300 because we do not have any statistical information about the maximum and minimum values that the attribute ClientID can take. 5

6 3.(4 pts) Transactions and Concurrency Control 3.1. (2,5 pts) Consider a multi-granularity locking system, with lock modes IX, X, IS, S and SIX. The objects are arranged in the following hierarchy: relation / \ / \ block A block B / \ / \ / \ / \ a1 a2 a3 b1 b2 Assume there are two active transactions T1 and T2, and there are no lock upgrades. a) Transaction T1 has already obtained the following locks: IS lock on the relation, S on B. Transaction T2 wants to modify b2 (and nothing else) while T1 is active. What lock does T2 need to get? Which of these locks can T2 get at this point? b) Transaction T1 has already obtained the following locks: IS lock on the relation, S on A. Transaction T2 wants to read b2 and modify a1 (and nothing else) while T1 is active. What locks does T2 need to get? Which of these locks can T2 get at this point? c) Transaction T1 has already obtained the following locks: IX lock on the relation, IX on A, X on a1. Transaction T2 wants to read a2 and modify a3 (and nothing else) while T1 is active. What locks does T2 need to get? Which of these locks can T2 get at this point? d) Transaction T1 has already obtained the following locks: IX lock on the relation, IX on A, X on a1. Transaction T2 wants to read a2 and modify a1 (and nothing else) while T1 is active. What locks does T2 need to get? Which of these locks can T2 get at this point? a) T2 needs to get: IX(relation); IX(B); X(b2) Can get: IX(relation) b) T2 needs to get: IX(relation); IX(A); IS(B); X(a1), S(b2) Can get: IX(relation); IS(B); S(b2) c) T2 needs to get: IX(R); IX(A); S(a2); X(a3) Can get all d) T2 needs to get: IX(relation); IX(A); S(a2); X(a1) Can get all except X(a1) 6

7 3.2. (1 pt) Consider the following two transactions: T1: R1(X) W1(X) R1(Y) W1(Y) T2: R2(Y) W2(Y) Which of the following schedules would be allowed under the 2-Phase Locking protocol? Justify your answer. Assume that the L lock actions in the schedules are exclusive. a) L1(X) L1(Y) L2(Y) R1(X) W1(X) R1(Y) W1(Y) R2(Y) W(Y) U2(Y) U1(Y) U1(X) b) L1(Y) L1(X) R1(X) W1(X) R1(Y) W1(Y) U1(X) U1(Y) L2(Y) R2(Y) W2(Y) U2(Y) Schedule a) is not possible, because T2 can only get a lock on Y (L2(Y)) after T1 releases the lock it ot on Y. Schedule b) is possible because each transaction first obtains the locks and when it starts to release the locks, it does not obtain anymore (0,5 pt) Consider the following schedule for three concurrent transactions and indicate whether it is possible under the timestamp-based protocol. Justify. T1 T2 T R(A) R(C) W(A) W(B) W(B) W(C) If we consider TS(T1) < TS(T2) < TS(T3) When T2:W(A) TS(T2) < Read_TS(A) = TS(T3) so the operation is rejected and T2 is rolled back 7

8 4. (4 pts) Recovery Management 4.1. (2,5 pts) Briefly answer the following questions regarding the ARIES recovery algorithm: a) If the system fails repeatedly during recovery, what is the maximum number of log records that can be written (as a function of the number of update and other log records written before the crash) before restart completes successfully? Justify. b) What is the oldest log record we need to retain? Justify. a) Let us take the case where each log record is an update record of an uncommitted transaction and each record belongs to a different transaction. This means there are n records of n different transactions, each of which has to be undone. During recovery, we have to add a CLR record of the undone action and an end transaction record after each CLR. Thus, we can write a maximum of 2n log records before restart completes. b) The oldest begin checkpoint referenced in any fuzzy dump or master log record (1 pt) In the context of the ARIES algorithm, explain the purpose of the checkpoint mechanism. Explain how does the frequency of checkpoints affect: (i) The system's performance when no failure occurs. (ii) The time it takes to recover from a system crash. (iii) The time it takes to recover from a disk crash (i.e., a crash on stable storage). Checkpointing is done with log-based recovery schemes such as ARIES to reduce the time required for recovery after a crash. If there is no checkpointing, then the entire log must be searched after a crash, and all transactions have to be undone/redone from the log. If checkpointing had been performed, then most of the log-records prior to the checkpoint (i.e., those associated to committed transactions, that are already reflected in stable storage) can be ignored at the time of recovery. Another reason to perform checkpoints is to clear log-records from stable storage as it gets full. Since checkpoints cause some loss in performance while they are being taken (i.e., they decrease system performance when no failure occurs), their frequency should be reduced if fast recovery is not critical. If we need fast recovery from a system crash, checkpointing frequency should be increased. Checkpoints have no effect on recovery from a disk crash (i.e., the equivalent of checkpoints for recovery from disk crashes are the archival dumps). 8

9 4.3. (0,5 pt) How does the recovery manager ensure atomicity of transactions? How does it ensure durability? The Recovery Manager ensures atomicity of transactions by undoing the actions of transactions that do not commit. It ensures durability by making sure that all actions of committed transactions survive system crashes and media failures. 9

10 5. (4 pts) Miscellaneous 5.1. (1 pt) Discuss how the SQL Server DBMS supports data indexing, trying to answer the following particular questions: a) Are clustered indexes on non-key attributes supported? b) Are composite indexes supported? Besides composite indexes, is there any other way to implement the idea of "covering indexes" through non-clustered indexes? c) Can the non-clustered indexes be sparse, or do they have to be dense? d) How are tuple locators (i.e., the pointers to the actual tuples associated to the index keys) implemented in the case of non-clustered indexes (i.e., what is the relationship between clustered and non-clustered indexes)? e) Why should clustered index keys use as few columns as possible? a) In SQL Server 2012, one can create non-clustered indexes on a single attribute, or on multiple attributes of a given relation (i.e., composite indexes), which may be keys or not. b) Besides composite indexes, SQL Server also supports the idea of having several attributes, besides the index key, included in the leaves of the indexing structure. This is particularly useful for defining indexes that can cover all attributes involved in a particular query, thus avoiding the access to the actual tuples. c) As in all DBMS, non-clustered indexes have to be dense, so that each index key is associated to the corresponding "tuple locator". d) In SQL Server, the structure of the "tuple locators" depends on whether the data pages from the corresponding relation are stored in a heap or a clustered table (i.e., depending on whether the table has a clustered index or not). For a heap, a row locator is a pointer to the row. For a clustered table, the row locator is the clustered index key. e) In SQL Server, it is beneficial to define clustered index keys with as few columns as possible. If a large clustered index key is defined, any non-clustered indexes that are defined on the same table will be significantly larger, because the nonclustered index entries contain the clustering key (1 pt) Explain the two general strategies through which parallelism can be used in DBMS processing. Clearly indicate how DBMSs can make use of these two general strategies (i.e., what are the typical operations where each of the strategies can be employed). The two general strategies are pipeline parallelism (i.e., many machines each doing one step in a multi-step process) and partition parallelism (i.e., many machines doing the same thing to different pieces of data). Partitioning can be exploited in the context of intra-operator parallelism (e.g., get all machines working to compute a given operation, for instance a scan, a sort, or a join). Pipelining can be exploited in the context of inter-operator parallelism (e.g., each operator of a given complex query may run concurrently on a different site). 10

11 5.3.(1 pt) Consider a database storing information about the expenses and revenues associated to the employees of a given company, specifically (i) the monthly revenues generated by each employee, and (ii) the yearly salary for each employee. Consider also the following two equivalent SQL queries, which return the employees that are paid yearly the same value that they generate to the company in revenues: SELECT * FROM Employees WHERE salary = 12*revenue; SELECT * FROM Employees WHERE salary/12 = revenue; Explain which query is better (i.e., in which situations is one query better than the other, in terms of execution efficiency). Justify your answer. First, it is important to remember that even if there are indexes on a table, but the query has an algebraic expression in the corresponding index key, the index will likely not be used. With this in mind, we should try to use the query most likely to explore existing indexes. * If there is an index on one of the tables, then the query where the corresponding attribute is not using an arithmetic operation should be used. * If there are indexes on both tables, and if there's a clustering index on the larger table, we should try to use it. * If there are indexes on both tables, but only a non-clustering index on the larger table, then we should: (i) if the small table has (much) less records than the larger table has pages, then we should use use the non-clustering index on the large table; (ii) otherwise, we should use the index on the small table (1 pt) Explain the difference between a system crash and a "disaster." Indicate what kinds of strategies are used in database management systems for handling both types of problems. In a system crash, the CPU goes down and disk may also crash, but stable-storage at the site is assumed to survive system crashes. In a disaster, everything at a site is destroyed. Recovery algorithms such as ARIES are typically used in DBMSs for recovering from system crashes. For recovering from "disasters, stable storage needs to be distributed and, in the context of DBMSs, different types of backup strategies can also be used to recover from disasters. 11