If the partition skew of a relation is such that half the tuples reside on a single node, what is the maximum speedup we can expect with 10 nodes for a parallel projection?

Suppose we have a byte addressable computer that has a 32-byte cache with 8 bytes per block. the memory address is 8 bits long. the system accesses memory addresses (in hex) in this exact order: 6e, b9, 17, e0, 4e, 4f, 50, 91, a8, ab, ad, 93, and 94. (a) assuming the cache is direct mapped, what memory addresses will be in cache block 2 after the last address has been accessed? (b) assuming the cache is direct mapped, what is the hit ratio for the entire memory reference sequence given, assuming the cache is initially empty? (c) assuming the cache is 2-way set associative with a lru replacement policy, what is the hit ratio?

Pp 4.1 design and implement a class called sphere that contains instance data that represents the sphere’s diameter. define the sphere constructor to accept and initialize the diameter and include getter and setter methods for the diameter. include methods that calculate and return the volume and surface area of the sphere (see pp 3.5 for the formulas). include a tostring method that returns a one-line description of the sphere. create a driver class called multisphere, whose main method instantiates and updates several sphere objects.

This question involves a class named textfile that represents a text file. public class textfile { private string filename; private string filename; private arraylist words; // constructors not shown // postcondition: returns the number of bytes in this file public int filesize() { } // precondition: 0 < = index < words.size() // postcondition: removes numwords words from the words arraylist beginning at // index. public void deletewords(int index, int numwords) { } // precondition: 0 < = index < = words.size() // postcondition: adds elements from newwords array to words arraylist beginning // at index. pub lic voidaddwords(int index, string[] newwords) { } // other methods not shown } complete the filesize() method. the filesize() is computed in bytes. in a text file, each character in each word counts as one byte. in addition, there is a space in between each word in the words arraylist, and each of those spaces also counts as one byte. for example, suppose the words arraylist stores the following words: { mary had a little lamb; its fleece was white as snow. } the filesize() method would compute 4 + 3 + 1 + 6 + 5 + 4 + 6 + 3 + 5 + 2 + 5 as the sum of the lengths of each string in the arraylist. the value returned would be this sum plus 10, because there would also be 10 spaces in between the 11 words. complete the filesize() method below: // postcondition: returns the number of bytes in this file public int filesize() { }