Ans .
B
We can download 56 Kb in one sec. Data downloaded in 2 minutes = 56 K x 2 x 60 =6720 Kb = 6.72 Mbits
Ans .
D
The P-persistent approach combines the advantages of the other two strategies. It reduces the chance of collision and improves efficiency.The P-persistent method is used if the channel has time slots with a slot duration >= the maximum propagation time
Ans .
B
A polyalphabetic cipher is any cipher based on substitution, using multiple substitution alphabets. The Vigenère cipher is probably the best-known example of a polyalphabetic cipher, though it is a simplified special case. The Enigma machine is more complex but is still fundamentally a polyalphabetic substitution cipher..
Ans .
A
Answer : 100 Meter The maximum allowed length of a Cat 6 cable is 100 meters or 328 feet.
Ans .
D
The common gateway interface (CGI) is a standard way for a Web server to pass a Web user's request to an application program and to receive data back to forward to the user.
Ans .
C
The main issue with Distance Vector Routing (DVR) protocols is Routing Loops, since Bellman-Ford Algorithm cannot prevent loops. This routing loop in DVR network causes Count to Infinity Problem. Routing loops usually occur when any interface goes down or two-routers send updates at the same time. Counting to infinity problem: So in this example, the Bellman-Ford algorithm will converge for each router, they will have entries for each other. B will know that it can get to C at a cost of 1, and A will know that it can get to C via B at a cost of 2. If the link between B and C is disconnected, then B will know that it can no longer get to C via that link and will remove it from it’s table. Before it can send any updates it’s possible that it will receive an update from A which will be advertising that it can get to C at a cost of 2. B can get to A at a cost of 1, so it will update a route to C via A at a cost of 3. A will then receive updates from B later and update its cost to 4. They will then go on feeding each other bad information toward infinity which is called as Count to Infinity problem.
Ans .
C
The IEEE single-precision and double-precision format to represent floating-point numbers, has a length of 32 and 64 respectively.
Ans .
D
Ans is C by formula (n-1)(n-2)/2 =99x98/2=4851 (graph should be simple i.e having no parallel edges and self loops)
e.g 2 disconnected graph with 2 nodes have 0 edge (2-1)(2-2)/2=0
disconnected graph with 3 nodes have at most 1 edge (3-1)(3-2)/2=1
disconnected graph with 4 nodes have at most 3 edges between 3 nodes and 4th node as isolated vertex (4-1)(4-2)/2=3
disconnected graph with 5 nodes have at most 6 edges (5-1)(5-2)/2=6
in gen with n nodes (n-1)(n-2)/2
Ans .
D
Amortized analysis is a method for analyzing a given algorithm's time complexity, or how much of a resource, especially time or memory in the context of computer programs, it takes to execute. A splay tree is a self-adjusting binary search tree with the additional property that recently accessed elements are quick to access again. It performs basic operations such as insertion, look-up and removal in O(log n) amortized time.
Ans .
D
A maximum depth path always takes the larger part of partition In question given that always β <=0.5 So maximum partition will be (1-β ) n We have to find maximum depth d, means how many times quick sort loop will be executed!! One iteration splits the number of elements from n to (β n ) and (1-β)n So i iterations reduces the number of elements to β i n and (1-β) i n . At a leaf, there is just one remaining element, In case of maximum depth path ( let maximum depth be d) (1-β)d n = 1 (1-β)d = 1/n Take logarithm on both sides, d log ( 1-β) = -log n d = -log n/log ( 1-β) Similarly we can find minimum depth d' = -log n/log β
Ans .
C
Answer will be 4) d + 1 Notice that its just a binary tree. A binary tree can have maximum of two children, but not more. Here its asking the minimum, hence we will consider only one node at each level. Hence for the depth of d, we need d + 1 nodes. For example consider this tree. Each and every node 1 child.
Ans .
C
because in case of doubly linked list only the node to be inserted/deleted needs to be manipulated and no traversals in forward or backward direction required for the operation so it is more efficient.
Ans .
C
There are 5 state and all state accept a and last state is acceptng infinit a
Ans .
D
Correct answer would be D) (1 + 01)* (0 + λ) Strings 010010 can be generated with option (a) by using 010 twice. String 1001 can be generated with option (b). string 0100 can be generated with option (c). So except d) option all can generate strings with 00 as a substring.
Ans .
A
To solve this kind of problem, think that how many stack would you need to solve the problem. FA + 0 stack = Regular Languages FA + 1 stack = Context Free languages FA + 2 stack = Turing Machines We can implement (FA + n stack with FA + 2 stack). Hence By Turing Machine, we can solve any problem. L1 is a regular language, we do not need any stack to implement this kind of language. L2, can not be implemented without using stack. We need One stack to implement this. Hence this is CFL.
Ans .
C
Suppose we have the following grammar − G: N = {S, A, B} T = {a, b} P = {S → AB, A → aA|a, B → bB|b} The language generated by this grammar − L(G) = {ab, a2b, ab2, a2b2, ………} = {am bn | m ≥ 1 and n ≥ 1}
Ans .
A/p>
Answer A: S1 is correct but S2 is wrong! For a Context free language: Membership, Emptiness, Finiteness are decidable. ie we can make an algorithm to check whether a given CGLis infinite or not. But we cannot design an algorithm to check equivalence of two CFLS. Very important. Following are decidable for various types of languages Regular language: {Membership, Emptiness, Finiteness, Equivalence, Everything (Σ*), Ambiguity, Regularity, Disjointedness} Context free language: {Membership, Emptiness, Finiteness} Context sensitive : {Membership } Recursive: {Membership } Recursive Enumerable : None.
Ans .
A
Number of 8-bit strings beginning with 111 are 32. First 3 bits are fixed and the remaining 5 bits can be 0 or 1. So the total combinations are 25=32. Same is the case with the strings starting with 101. So total number of strings are 32+32=64.
Ans .
B
Total number of ways 12 offices can be painted = 12! But 3 of them will be green, 2 of them pink, 2 of them yellow and 5 of them white. Answer = 12! /(3!* 2 ! * 2! * 5!) = 166320
Ans .
D
Lets go one by one, (i) A graph in which there is a unique path between every pair of vertices is a tree. ==> This statement is true, Because graph can have unique path only when it does not have cycle. And according to the definition of tree, its a graph without cycel. Hence this is a valid statement. (ii) A connected graph with e = v − 1 is a tree. ==> This statement is true. Not every graph with e = v - 1, will be a tree. But if the graph is connected hence its true. (iii) A graph with e = v − 1 that has no circuit is a tree. ==> This statement is also true, Here he has not mentioned the connected thing, but mentioned that it has no circuit. It means that its conn
Ans .
C
Lets go one by one, (i) A graph in which there is a unique path between every pair of vertices is a tree. ==> This statement is true, Because graph can have unique path only when it does not have cycle. And according to the definition of tree, its a graph without cycel. Hence this is a valid statement. (ii) A connected graph with e = v − 1 is a tree. ==> This statement is true. Not every graph with e = v - 1, will be a tree. But if the graph is connected hence its true. (iii) A graph with e = v − 1 that has no circuit is a tree. ==> This statement is also true, Here he has not mentioned the connected thing, but mentioned that it has no circuit. It means that its conn
Ans .
B
A simple graph with n vertices is connected if if has more than (n−1)(n−2)/2 edges. The n vertex graph with the maximal number of edges that is still disconnected is a Kn−1 a complete graph Kn−1 with n−1 vertices has (n−12)edges, so (n−1)(n−2)/2 edges. Adding any possible edge must connect the graph, so the minimum number of edges needed to guarantee connectivity for an n vertex graph is (n−1)(n−2)/2 + 1 Hence,Option(B)More than (n−1)(n−2)/2.
Ans .
C
Ex-OR and Ex-NOR are best suited for parity checking and Parity generating. Hence,Option(C)Ex-OR and Ex-NOR.
Ans .
C
The quantification ∃!x P(x) denotes the proposition “There exists a unique x such that P(x) is true”, express the quantification using universal and existential quantifications and logical operators: ∃x P(x) ∨ ∀x∀y ((P(x) ∨ P(y)) → x = y) ∀ x P(x) ∧ ∀x∀y ((P(x) ∨ P(y)) → x = y) ∃x P(x) ∧ ∀x∀y ((P(x) ∧ P(y)) → x = y) ∃x ..
Ans .
B
Answer : Function Count Function count is used to get the number of entries in a number field that is in a range or array of numbers. And it does not depend on what Programming Language is being used .
Ans .
A
Size of 1 record = 8 + 4 = 12 Let the order be N. No. of index values per block = N - 1 (N - 1) 12 + 4 = 512 12N - 12 + 4 = 512 16N = 1009 N = 43.3333
Ans .
B
Binary Search Tree, is a node-based binary tree data structure which has the following properties: The left subtree of a node contains only nodes with keys less than the node’s key. The right subtree of a node contains only nodes with keys greater than the node’s key. The left and right subtree each must also be a binary search tree. There must be no duplicate nodes.
Ans .
B
Answer is B) Agent Discovery and Registration Mobile IP is the basic behind how wireless devices offer IP connectivity. Agent discovery and registration are two basic functions involved here. So the options is B.
Ans .
A
Answer : Before the CPU time slice expires In preemptive scheduling tasks are usually assigned with priorities. At times it is necessary to run a certain task that has a higher priority before another task although it is running. Therefore, the running task is interrupted for some time and resumed later when the priority task has finished its execution. This is called preemptive scheduling. In non-preemptive scheduling, a running task is executed till completion. It cannot be interrupted.
Ans .
D
ample for decrease in the average turn round time : Suppose we have two process P1 and P2 with burst time 20 sec and 2 sec respectively. P1 arrives at time t= 0 sec and P2 at t=2 sec. When time quantum for RR = 1 sec. Then TAT for P1 = 22 sec. and for P2 = 3 sec. So AVG TAT = (22 + 3 ) /2 sec. Now do same with time quantum for RR = 2 sec. You will get AVG TAT = (22 + 2 )/2 sec.
Ans .
A
Answer : Mutual Exclusion A mutual exclusion (mutex) is a program object that prevents simultaneous access to a shared resource. This concept is used in concurrent programming with a critical section, a piece of code in which processes or threads access a shared resource. Only one thread owns the mutex at a time, thus a mutex with a unique name is created when a program starts. When a thread holds a resource, it has to lock the mutex from other threads to prevent concurrent access of the resource. Upon releasing the resource, the thread unlocks the mutex.
Ans .
B
Answer is B) (i) and (iii) are true Compared to the processor speed, the speed of the primary memory is slow. Cache memory is a small memory which sits in between the processor and primary memory and fetches information to the processor at a much higher speed or it makes it appear so. Caching can be effective based on a property of computer programs called locality of reference. Analysis of program show that the majority of the execution time is spent around a small part of the program may be a simple loop,nested loop or a few functions. The rest of the program is accessed infrequently. There is something called temporal locality and spatial locality also which we need to know when we talk about cache. But cache memory is ideally suited for small loops. Interleaved memory is a technique for increasing the speed of RAM. Here multiple memory chips are grouped together to form what are known as banks.Each of them take turns for supplying data. An interleaved memory with "n" banks is said to be n-way interleaved. Macintosh systems are considered to be one using memory interleaving. So the answer for this question is option B.
Ans .
C
Ans .
D
Answer : Bits Check Sum is a simple error-detection scheme in which each transmitted message is accompanied by a numerical value based on the number of set bits in the message. The receiving station then applies the same formula to the message and checks to make sure the accompanying numerical value is the same. If not, the receiver can assume that the message has been garbled.
Ans .
B
In case of signed Magnitude Representation the range is from -(2^n - 1 - 1) to (2^n -1 - 1) Min no that can be represented in this system is -(2^n -1 - 1) Max no that can be represented in this system is (2^n -1 - 1) In case of 2's complement no system the range is from -2^n -1 to 2^n -1 - 1 Min no that can be represented in this system is -2^n -1 Max no that can be represented in this system is 2^n -1 - 1 As they said 2 Bytes = 16 bits we can use max no that can be represented here which is 2^n -1 - 1 2^16 -1 to 2^15 - 1
Ans .
B
Answer : ECL In electronics, emitter-coupled logic (ECL) is a high-speed integrated circuit bipolar transistor logic family
Ans .
D
Answer : Software A software interrupt is a type of interrupt that is caused either by a special instruction in the instruction set or by an exceptional condition in the processor itself. A software interrupt is invoked by software, unlike a hardware interrupt, and is considered one of the ways to communicate with the kernel or to invoke system calls, especially during error or exception handling.
Ans .
B
Ans .
A
XSTL :eXtensible Style Sheet Language Transformations is a language for transforming XML documents into other XML documents, or other formats such as HTML for web pages, plain text or into XSL Formatting Objects, which may subsequently be converted to other formats, such as PDF, PostScript and PNG.
Ans .
A
Answer A: hiding What is a view? Views are subset of table. View also has set of records in the form of rows and columns. But it is created based on the records in one or more tables. A table will have large number of data and table will be fired with specific frequently. In such case, instead of rewriting the query again and again, a name is given to the query and it will be called whenever it is required. Hence view is also called as named query or stored query.
Ans .
D
Answer : Dijikstra Vector Interior Gateway Routing Protocol (IGRP) is a distance vector interior routing protocol (IGP) developed by Cisco. It is used by routers to exchange routing data within an autonomous system. IGRP is a proprietary protocol. The Routing Information Protocol (RIP) is one of the oldest distance-vector routing protocols which employ the hop count as a routing metric. RIP prevents routing loops by implementing a limit on the number of hops allowed in a path from source to destination.
Ans .
A
Decision Tree Classifier is a classification technique. Classification technique is a systematic approach to build classification models from an input data set. For example, decision tree classifiers, rule-based classifiers, neural networks, support vector machines, and naive Bayes classifiers are different technique to solve a classification problem.
Ans .
B
A Crosstab query is a type of select query which allows data to view as both horizontally and vertically so that the data can be more compact and easier to read.
Ans .
D
the Thomas write rule is a rule in timestamp-based concurrency control. It can be summarized as ignore outdated writes.
Ans .
D
The Global Positioning System (GPS), originally Navstar GPS, is a space-based radionavigation system owned by the United States government and operated by the United States Air Force. It is a global navigation satellite system that provides geolocation and time information to a GPS receiver anywhere
Ans .
B
the Thomas write rule is a rule in timestamp-based concurrency control. It can be summarized as ignore outdated writes.
Ans .
B
We can assume single level paging and TLB look and page table look happens sequentially meaning TLB look time is to be considered even for TLB miss. On a TLB miss we need to access main memory for getting the physical address for the page table before doing the actual memory access for the data. Effective access time = Hit ratio*Time during hit + Miss Ratio * Time During Miss =0.7*(30+100) + 0.3 (30+100+100) =91+69 =160 ns Hence,Option(D)160ns is the correct choice.
Ans .
B
ans is B 143 ms and 123 ms at clock time 160 millisecond only request number 1 2 3 4 are there to be serviced request number 5 (with track no 75) will be ready after 15 millisecond starting head position 65 now for SSTF 12 40 75 85 100 it will serve 85(since 75 is not come yet) first then 75 then 100 then 40 and finally 12 so total movements =20+10+25+60+28=143 total time 143x1 ms=143ms now for look 12 40 75 85 100 direction towards highest no of tracks it will serve 85(since 75 is not come yet) first then 100 then 75 then 40 and finally 12 so total movements =20+15+25+35+28=123 total time 123x1 ms=123ms
Ans .
(Wrong question)
Ans .
A
Answer A: Without pipeline one task needs tn time. n tasks need ntn time. With pipeline, First task needs k cycles to finish. So time will be k tp Other n-1 tasks needs tp time only to finish. Total time = (k+ n -1 ) tp Speed up = T pipeline / T without pipeline = n t n / (k+n-1) t p