A Multicomputer with 256 CPUs is organized as 16x16 grid. What is the worst case delay (in hops) that a message might have to take? (A) 16 (B) 15 (C) 32 (D) 30

A Multicomputer with 256 CPUs is organized as 16x16 grid. What is the worst case delay (in hops) that a message might have to take? (A) 16 (B) 15 (C) 32 (D) 30
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(D) 30
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Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milli seconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is (1) 20 (2) 10 (3) 30 (4) 40

Description : Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milli seconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is (1) 20 (2) 10 (3) 30 (4) 40

Last Answer : (2) 10

Suppose a digitized voice channel is made by digitizing 8 kHz bandwidth analog voice signal. It is required to sample the signal at twice the highest frequency (two samples per hertz). What is the bit rate required, if it is assumed that each sample requires 8 bits? (A) 32 kbps (B) 64 kbps (C) 128 kbps (D) 256 kbps

Description : Suppose a digitized voice channel is made by digitizing 8 kHz bandwidth analog voice signal. It is required to sample the signal at twice the highest frequency (two samples per hertz). What is the bit rate required, if it ... sample requires 8 bits? (A) 32 kbps (B) 64 kbps (C) 128 kbps (D) 256 kbps

Last Answer : (C) 128 kbps

Which of the following conditions does not hold good for a solution to a critical section problem ? (A) No assumptions may be made about speeds or the number of CPUs. (B) No two processes may be simultaneously inside their critical sections. (C) Processes running outside its critical section may block other processes. (D) Processes do not wait forever to enter its critical section.

Description : Which of the following conditions does not hold good for a solution to a critical section problem ? (A) No assumptions may be made about speeds or the number of CPUs. (B) No two processes ... critical section may block other processes. (D) Processes do not wait forever to enter its critical section.

Last Answer : (C) Processes running outside its critical section may block other processes. 

The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is (A) 12 bits (B) 15 bits (C) 16 bits (D) 20 bits

Description : The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is (A) 12 bits (B) 15 bits (C) 16 bits (D) 20 bits

Last Answer : (C) 16 bits Explanation: VirtualAddress = 32 bits PageSize = 4KB = 12 bits therefore : VPNTag = 20 bits, OffsetTag = 12 bits TLBEntryLength = VPNTag = 20 bits TotalTLBEntries = 64, 4-way implies ... therefore : TLBIndex = 4 bits TLBTag = TLBEntryLength - TLBIndex = 20 - 4 = 16 bits

A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively: (1) 14 and 15 (2) 14 and 29 (3) 15 and 14 (4) 16 and 32

Description : A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively: (1) 14 and 15 (2) 14 and 29 (3) 15 and 14 (4) 16 and 32

Last Answer : (3) 15 and 14

A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in response to a request? (Note that the speed of light in air is 3,00,000 km/second). (A) 133.33 m sec (B) 266.67 m sec (C) 400.00 m sec (D) 533.33 m sec

Description : A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in response to a request? (Note that the speed of light in air is 3,00,000 km/second). (A) 133.33 m sec (B) 266.67 m sec (C) 400.00 m sec (D) 533.33 m sec

Last Answer : (D) 533.33 m sec Explanation: The request has to go up and down, and the response has to go up and down. The total path length traversed is thus 160.000 km. The speed of light in air and vacuum is 300,000 km/sec, so the propagation delay alone is 160,000/300,000 sec or about 533 msec. 

Consider the fractional knapsack instance n = 4, (p1, p2, p3, p4) = (10, 10, 12, 18), (w1, w2, w3, w4) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively) (A) 40 (B) 38 (C) 32 (D) 30

Description : Consider the fractional knapsack instance n = 4, (p1, p2, p3, p4) = (10, 10, 12, 18), (w1, w2, w3, w4) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively) (A) 40 (B) 38 (C) 32 (D) 30

Last Answer : (B) 38

EBCDIC can code up to how many different characters? A) 256 B) 16 C) 32 D) 64

Description : EBCDIC can code up to how many different characters? A) 256 B) 16 C) 32 D) 64

Last Answer : Answer : A

EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Description : EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Last Answer : 256

EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Description : EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Last Answer : 256

EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Description : EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Last Answer : a. 256

EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Description : EBCDIC can code up to how many different characters? a. 256 b. 16 c. 32 d. 64

Last Answer : a. 256

Insert the missing number in the following series: 4 16 8 64 ? 256 (A) 16 (B) 24 (C) 32 (D) 20

Description : Insert the missing number in the following series: 4 16 8 64 ? 256 (A) 16 (B) 24 (C) 32 (D) 20

Last Answer : Answer: A Square root of 16 is 4, Square root of 64 is 8, Therefore Square root of 256 is 16.

7.How does traceroute work? A. It uses an ICMP destination-unreachable message to elicit the name of a router. B. It sends a specially crafted IP packet to a router to locate the number of hops from the sender to the destination network. C. It uses a protocol that will be rejected by the gateway to determine the location. D. It uses the TTL value in an ICMP message to determine the number of hops from the sender to the router.

Description : 7.How does traceroute work? A. It uses an ICMP destination-unreachable message to elicit the name of a router. B. It sends a specially crafted IP packet to a router to locate the number of hops from ... the TTL value in an ICMP message to determine the number of hops from the sender to the router.

Last Answer : D. It uses the TTL value in an ICMP message to determine the number of hops from the sender to the router.

What is the propagation delay when a signal is transmitter by an earth station to a geosynchronous satellite about 38,500 km above earth’s equator and then received by the same earth station? A. 256 msec B. 128 msec C. 300 msec D. 400 msec

Description : What is the propagation delay when a signal is transmitter by an earth station to a geosynchronous satellite about 38,500 km above earth’s equator and then received by the same earth station? A. 256 msec B. 128 msec C. 300 msec D. 400 msec

Last Answer : A. 256 msec

Suppose that the one-way propagation delay for a 100 Mbps Ethernet having 48-bit jamming signal is 1.04 micro-seconds. The minimum frame size in bits is: a. 112 b. 160 c. 208 d. 256

Description : Suppose that the one-way propagation delay for a 100 Mbps Ethernet having 48-bit jamming signal is 1.04 micro-seconds. The minimum frame size in bits is: a. 112 b. 160 c. 208 d. 256

Last Answer : d. 256

Red-black trees are one of many Search tree schemes that are "balanced” in order to guarantee that basic dynamic-set operations take ............. time in the worst case. (1) O(1) (2) O(log n) (3) O(n) (4) O(n log n)

Description : Red-black trees are one of many Search tree schemes that are "balanced” in order to guarantee that basic dynamic-set operations take ............. time in the worst case. (1) O(1) (2) O(log n) (3) O(n) (4) O(n log n)

Last Answer : (2) O(log n) 

How much space will be required to store the bit map of a 1.3 GB disk with 512 bytes block size ? (A) 332.8 KB (B) 83.6 KB (C) 266.2 KB (D) 256.6 KB

Description : How much space will be required to store the bit map of a 1.3 GB disk with 512 bytes block size ? (A) 332.8 KB (B) 83.6 KB (C) 266.2 KB (D) 256.6 KB

Last Answer : (A) 332.8 KB

There are exactly ................ different finite automata with three states x, y and z over the alphabet {a, b} where x is always the start state. (A) 64 (B) 256 (C) 1024 (D) 5832

Description : There are exactly ................ different finite automata with three states x, y and z over the alphabet {a, b} where x is always the start state. (A) 64 (B) 256 (C) 1024 (D) 5832

Last Answer : Answer: D

Which of the following information about the UNIX file system is not correct? (A) Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks. (B) An i-node contains accounting information as well as enough information to locate all the disk blocks that holds the file’s data. (C) Each i-node is 256-bytes long. (D) All the files and directories are stored in data blocks.

Description : Which of the following information about the UNIX file system is not correct? (A) Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks. (B ... Each i-node is 256-bytes long. (D) All the files and directories are stored in data blocks. 

Last Answer : (C) Each i-node is 256-bytes long.

The number of different binary trees with 6 nodes is ............. (A) 6 (B) 42 (C) 132 (D) 256

Description : The number of different binary trees with 6 nodes is ............. (A) 6 (B) 42 (C) 132 (D) 256

Last Answer : (C) 132

The versions of windows operating system like windows XP and window Vista uses following file system: (A) FAT-16 (B) FAT-32 (C) NTFS (NT File System) (D) All of the above

Description : The versions of windows operating system like windows XP and window Vista uses following file system: (A) FAT-16 (B) FAT-32 (C) NTFS (NT File System) (D) All of the above

Last Answer : (D) All of the above

Windows 32 API supports (A) 16-bit Windows (B) 32-bit Windows (C) 64-bit Windows (D) All of the above

Description : Windows 32 API supports (A) 16-bit Windows (B) 32-bit Windows (C) 64-bit Windows (D) All of the above

Last Answer : (D) All of the above

WOW32 is a: (A) Win 32 API library for creating processes and threads. (B) Special kind of file system to the NT name space. (C) Kernel - mode objects accessible through Win32 API (D) Special execution environment used to run 16 bit Windows applications on 32 - bit machines.

Description : WOW32 is a: (A) Win 32 API library for creating processes and threads. (B) Special kind of file system to the NT name space. (C) Kernel - mode objects accessible through Win32 API (D) Special execution environment used to run 16 bit Windows applications on 32 - bit machines.

Last Answer : (D) Special execution environment used to run 16 bit Windows applications on 32 - bit machines.

Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor? (A) 8x106 bytes/sec (B) 4x106 bytes/sec (C) 16x106 bytes/sec (D) 4x109 bytes/sec

Description : Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What ... (A) 8x106 bytes/sec (B) 4x106 bytes/sec (C) 16x106 bytes/sec (D) 4x109 bytes/sec 

Last Answer : (B) 4x106 bytes/sec

What is the size of the 'total length' field in IPv 4 datagram ? (A) 4 bits (B) 8 bits (C) 16 bits (D) 32 bits

Description : What is the size of the 'total length' field in IPv 4 datagram ? (A) 4 bits (B) 8 bits (C) 16 bits (D) 32 bits

Last Answer : (C) 16 bits 

A unix file system has 1-KB blocks and 4-byte disk addresses. What is the maximum file size if i-nodes contain 10 direct entries and one single, double and triple indirect entry each? (A) 32 GB (B) 64 GB (C) 16 GB (D) 1 GB

Description : A unix file system has 1-KB blocks and 4-byte disk addresses. What is the maximum file size if i-nodes contain 10 direct entries and one single, double and triple indirect entry each? (A) 32 GB (B) 64 GB (C) 16 GB (D) 1 GB

Last Answer : (C) 16 GB

The IEEE-754 double-precision format to represent floating point numbers, has a length of ........... bits. (A) 16 (B) 32 (C) 48 (D) 64

Description : The IEEE-754 double-precision format to represent floating point numbers, has a length of ........... bits. (A) 16 (B) 32 (C) 48 (D) 64

Last Answer : (D) 64

In certain city the Taxi charges comprise of a fixed charge and the charge of the distance travelled. A person paid Rs.156 for a journey of 16 km and another person paid Rs.204 for the journey of 24 km. The amount paid by a passenger who has travelled 30 km is (A) 236 (B) 240 (C) 248 (D) 256

Description : In certain city the Taxi charges comprise of a fixed charge and the charge of the distance travelled. A person paid Rs.156 for a journey of 16 km and another person paid Rs.204 for the journey of 24 km. The amount paid by a passenger who has travelled 30 km is (A) 236 (B) 240 (C) 248 (D) 256

Last Answer : Answer: B

Which of the following algorithms has lowest worst case time complexity? a) Insertion sort b) Selection sort c) Quick sort d) Heap sort

Description : Which of the following algorithms has lowest worst case time complexity? a) Insertion sort b) Selection sort c) Quick sort d) Heap sort

Last Answer : d) Heap sort

................... comparisons are necessary in the worst case to find both the maximum and minimum of n numbers. (A) 2n-2 (B) n + floor(lg n) - 2 (C) floor(3n/2) - 2 (D) 2 lg n – 2

Description : ................... comparisons are necessary in the worst case to find both the maximum and minimum of n numbers. (A) 2n-2 (B) n + floor(lg n) - 2 (C) floor(3n/2) - 2 (D) 2 lg n – 2

Last Answer : (C) floor(3n/2) - 2

Given two sorted list of size 'm' and 'n' respectively. The number of comparison needed in the worst case by the merge sort algorithm will be (A) m x n (B) max (m, n) (C) min (m, n) (D) m + n – 1

Description : Given two sorted list of size 'm' and 'n' respectively. The number of comparison needed in the worst case by the merge sort algorithm will be (A) m x n (B) max (m, n) (C) min (m, n) (D) m + n – 1

Last Answer :  (D) m + n – 1

A ball is dropped from a height of 256 feet. The acceleration due to gravity is 32 feet per second squared. How many seconds will it take for the ball to reach the ground?

Description : A ball is dropped from a height of 256 feet. The acceleration due to gravity is 32 feet per second squared. How many seconds will it take for the ball to reach the ground?

Last Answer : ANSWER: 4

If an integer occupies 4 bytes and a character occupies 1 byte of memory, each element of the following structure would occupy how many bytes ? struct name { int age; char name[30]; }; A) 30 B) 32 C) 34 D) 36

Description : If an integer occupies 4 bytes and a character occupies 1 byte of memory, each element of the following structure would occupy how many bytes ? struct name { int age; char name[30]; }; A) 30 B) 32 C) 34 D) 36

Last Answer : C) 34

Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide. (A) 30,000 (B) 32,700 (C) 33,000 (D) None of the above

Description : Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide. (A) 30,000 (B) 32,700 (C) 33,000 (D) None of the above

Last Answer : (B) 32,700

What will be the hexadecimal value in the register ax (32-bit) after executing the following instructions? Mov al, 15 Mov ah, 15 Xor al, al Mov cl, 3 Shr ax, cl Codes: (A) 0F00 h (B) 0F0F h (C) 01E0 h (D) FFFF h

Description : What will be the hexadecimal value in the register ax (32-bit) after executing the following instructions? Mov al, 15 Mov ah, 15 Xor al, al Mov cl, 3 Shr ax, cl Codes: (A) 0F00 h (B) 0F0F h (C) 01E0 h (D) FFFF h

Last Answer : (C) 01E0 h

compiler optimization, operator strength reduction uses mathematical identities to replace slow math operations with faster operations. Which of the following code replacements is an illustration of operator strength reduction? (A) Replace P + P by 2 * P or Replace 3 + 4 by 7. (B) Replace P * 32 by P<<5 (C) Replace P * 0 by 0 (D) Replace (P<<4) – P by P * 15

Description : compiler optimization, operator strength reduction uses mathematical identities to replace slow math operations with faster operations. Which of the following code replacements is an illustration of operator strength reduction? (A) Replace P + ... * P or Replace 3 + 4 by 7. (B) Replace P * 32 by P

Last Answer : (B) Replace P * 32 by P

The sum of the seek time, and the rotational delay is called the ……………….. A) reached time B) access time C) arrived time D) common time

Description : The sum of the seek time, and the rotational delay is called the ……………….. A) reached time B) access time C) arrived time D) common time

Last Answer : B) access time

............. is the time required to move the disk arm to the required track. A) Seek time B) Rotational delay C) Latency time D) Access time

Description : ............. is the time required to move the disk arm to the required track. A) Seek time B) Rotational delay C) Latency time D) Access time

Last Answer : A) Seek time

ECL is the fastest of all logic families. High Speed in ECL is possible because transistors are used in difference amplifier configuration, in which they are never driven into ............... (1) Race condition (2) Saturation (3) Delay (4) High impedance

Description : ECL is the fastest of all logic families. High Speed in ECL is possible because transistors are used in difference amplifier configuration, in which they are never driven into ............... (1) Race condition (2) Saturation (3) Delay (4) High impedance

Last Answer : Answer: 2  Explanation: Emitter-coupled logic (ECL) is the fastest of all logic families and therefore is used in applications where very high speed is essential. High speeds have become ... configuration, in which they are never driven into saturation and thereby the storage time is eliminated.

…………………. is a group of interested components working together towards a common goal by accepting inputs and producing outputs in an organized transformation process. A) System B) Network C) Team D) System Unit

Description : …………………. is a group of interested components working together towards a common goal by accepting inputs and producing outputs in an organized transformation process. A) System B) Network C) Team D) System Unit

Last Answer : A) System

The relational model is based on the concept that data is organized and stored in two-dimensional tables called ………………………A) Fields B) Records C) Relations D) Keys

Description : The relational model is based on the concept that data is organized and stored in two-dimensional tables called ………………………A) Fields B) Records C) Relations D) Keys

Last Answer : C) Relations

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

75%of a number is equal to four-fifth of another number. What is the ratio between first number and the second number? a) 4:3 b) 30:32 c) 7:9 d) 16:15 e) None of these

Description : 75%of a number is equal to four-fifth of another number. What is the ratio between first number and the second number? a) 4:3 b) 30:32 c) 7:9 d) 16:15 e) None of these

Last Answer : Let the two number be x & y. According to question, 75/100 x : 4/5 y 3/4 x : 4/5 y 15x = 16y x /y = 16/15 Answer: d)

You are given the following address: 128.16.32.13/30. Determine the subnet mask, address class, subnet address, and broadcast address. * 255.255.255.252, B,128.16.32.12, 128.16.32.15 * 255.255.255.252, C,128.16.32.12, 128.16.32.15 * 255.255.255.252, B,128.16.32.15, 128.16.32.12 * 255.255.255.248, B,128.16.32.12, 128.16.32.15

Description : You are given the following address: 128.16.32.13/30. Determine the subnet mask, address class, subnet address, and broadcast address. * 255.255.255.252, B,128.16.32.12, 128.16.32.15 * 255.255.255.252, C,128.16.32.12 ... , B,128.16.32.15, 128.16.32.12 * 255.255.255.248, B,128.16.32.12, 128.16.32.15

Last Answer : Correct answer: A * 255.255.255.248,

Insert the missing number: 16/32, 15 /33, 17/31, 14/34, ? (A) 19/35 (B) 19/30 (C) 18/35 (D) 18/30

Description : Insert the missing number: 16/32, 15 /33, 17/31, 14/34, ? (A) 19/35 (B) 19/30 (C) 18/35 (D) 18/30

Last Answer : Answer: D 16/32 then 15/33 means -1/+1, 17/31 means +2/-2, 14/34 means -3/+3 next +4/-4 i.e 18/30