A 8-Mbps tokenring has atoken holding timer value of 10 msec. What is the longest frame (assumeheader bits are negligible) that canbe sent on this ring? A. 8000 B frame B. 80,000 B frame C. 8 x 105 bit frame D. 10,000 B frame

A 8-Mbps tokenring has atoken holding timer value of 10 msec. What is the longest frame (assumeheader bits are negligible) that canbe sent on this
ring?
A.
8000 B frame
B.
80,000 B frame
C.
8 x 105 bit frame
D.
10,000 B frame
by

1 Answer

Answer :

10,000 B frame
by

Related questions

Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits per second) over a 1 km (kilometre) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable? a. 8000 b. 10000 c. 16000 d. 20000

Description : Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits per second) over a 1 km (kilometre) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable? a. 8000 b. 10000 c. 16000 d. 20000

Last Answer : d. 20000

A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network? a. 2 Mbps b. 60 Mbps c. 120 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network? a. 2 Mbps b. 60 Mbps c. 120 Mbps

Last Answer : a. 2 Mbps

A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? a. 1 Mbps b. 2 Mbps c. 10 Mbps d. 12 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? a. 1 Mbps b. 2 Mbps c. 10 Mbps d. 12 Mbps

Last Answer : b. 2 Mbps

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

How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame? a. 10,000 bytes b. 12,000 bytes c. 15,000 bytes d. 27,000 bytes

Description : How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame? a. 10,000 bytes b. 12,000 bytes c. 15,000 bytes d. 27,000 bytes

Last Answer : b. 12,000 bytes

A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ? (A) 2 Mbps (B) 60 Mbps (C) 120 Mbps (D) 10 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ? (A) 2 Mbps (B) 60 Mbps (C) 120 Mbps (D) 10 Mbps

Last Answer : (A) 2 Mbps Explanation: In data transmission, throughput is the amount of data moved successfully from one place to another in a given period of time, and typically measured in bits per second (bps), ... second (Mbps) or gigabits per second (Gbps). Here, Throughput = 15000 x 8000/60 = 2 Mbps 

Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.b. 2 c. 2.5 d. 5

Description : Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s. b. 2 c. 2.5 d. 5

Last Answer : b. 2

Assume that we need to download text documents at the rate of 100 pages per minute. A page is an average of 24 lines with 80 characters in each line and each character requires 8 bits. Then the required bit rate of the channel is ................ (A) 1.636 Kbps (B) 1.636 Mbps (C) 2.272 Mbps (D) 3.272 Kbps

Description : Assume that we need to download text documents at the rate of 100 pages per minute. A page is an average of 24 lines with 80 characters in each line and each character requires 8 bits. Then the required bit rate of the channel is ... ...... (A) 1.636 Kbps (B) 1.636 Mbps (C) 2.272 Mbps (D) 3.272 Kbps

Last Answer : Answer: Marks given to all

Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 ms. The minimum frame size is a. 94 b. 416 c. 464

Description : Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 ms. The minimum frame size is a. 94 b. 416 c. 464

Last Answer : c. 464

A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? (A) 1 Mbps (B) 2 Mbps (C) 10 Mbps (D) 12 Mbps

Description : A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? (A) 1 Mbps (B) 2 Mbps (C) 10 Mbps (D) 12 Mbps

Last Answer : (B) 2 Mbps

The minimum frame length for 10-Mbps Ethernet is _______bytes. A) 32 B) 80 C) 128 D) none of the above

Description : The minimum frame length for 10-Mbps Ethernet is _______bytes. A) 32 B) 80 C) 128 D) none of the above

Last Answer : none of the above

The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channelcapacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol used is: a. A b. B c. C d. D

Description : The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing ... maximum utilization, when the sliding window protocol used is: a. A b. B c. C d. D

Last Answer : c. C

A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds. Assuming no frame is lost, the sender throughput is __________ bytes/second. a. 2500 b. 2000 c. 1500 d. 500

Description : A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of ... is lost, the sender throughput is __________ bytes/second. a. 2500 b. 2000 c. 1500 d. 500

Last Answer : a. 2500

If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ______. a. 64 Kbps b. 32 Mbps c. 32 Kbps d. 64 MbpS

Description : If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ______. a. 64 Kbps b. 32 Mbps c. 32 Kbps d. 64 MbpS

Last Answer : c. 32 Kbps

Suppose transmission rate of a channel is 32 kbps. If there are ‘8’ routes from source to destination and each packet p contains 8000 bits. Total end to end delay in sending packet P is _____. a. 2 sec b. 3 sec c. 4 sec d. 1 sec

Description : Suppose transmission rate of a channel is 32 kbps. If there are ‘8’ routes from source to destination and each packet p contains 8000 bits. Total end to end delay in sending packet P is _____. a. 2 sec b. 3 sec c. 4 sec d. 1 sec

Last Answer : a. 2 sec

What is the bit rate for transmitting uncompressed 800x600 pixel colour frames with 8 bits/pixel at 40 frames/second ? (A) 2.4 Mbps (B) 15.36 Mbps (C) 153.6 Mbps (D) 1536 Mbps

Description : What is the bit rate for transmitting uncompressed 800x600 pixel colour frames with 8 bits/pixel at 40 frames/second ? (A) 2.4 Mbps (B) 15.36 Mbps (C) 153.6 Mbps (D) 1536 Mbps 

Last Answer : (C) 153.6 Mbps Explanation: The number of bits/sec is just 800x600x40x8 or 153.6 Mbps.

The maximum frame length for 10-Mbps Ethernet is ________ bytes. A) 1518 B) 1500 C) 1200 D) none of the abov

Description : The maximum frame length for 10-Mbps Ethernet is ________ bytes. A) 1518 B) 1500 C) 1200 D) none of the abov

Last Answer : 1518

In IPv6,a _________address comprises 80 bits of zero, followed by16 bitsof one, followed by the 32-bit IPv4address. A) link local B) site local C) mapped D) none of the above

Description : In IPv6,a _________address comprises 80 bits of zero, followed by16 bitsof one, followed by the 32-bit IPv4address. A) link local B) site local C) mapped D) none of the above

Last Answer : mapped

Which 8-bit chip was used in many of today’s TRS-80 computers? A) Z-8000 B) Motorola 6809 C) Z-8808 D) Z-80

Description : Which 8-bit chip was used in many of today’s TRS-80 computers? A) Z-8000 B) Motorola 6809 C) Z-8808 D) Z-80

Last Answer : Answer : D

X, Y, Z started a business by investing Rs.60,000, Rs.70,000 and Rs.80,000 respectively. Find the share of each out of an annual profit of Rs.28350 A) 8000, 9367, 11800 B) 8500, 9400, 11000 C) 8308, 9530, 10346 D) 8100, 9450, 10800

Description : X, Y, Z started a business by investing Rs.60,000, Rs.70,000 and Rs.80,000 respectively. Find the share of each out of an annual profit of Rs.28350 A) 8000, 9367, 11800 B) 8500, 9400, 11000 C) 8308, 9530, 10346 D) 8100, 9450, 10800

Last Answer : Answer: D) Ratio of shares = 60,000 : 70,000 : 80,000 = 6 : 7 : 8 A’s share = Rs(28350 *6/21) = Rs.8100 B’s share = Rs(28350 * 7/21) = Rs.9450 C’s share = Rs (28350* 8/21) = Rs. 10,800

________ can beachievedbyusing multiplexing; ______ canbe achieved by using spreading. A) Efficiency; privacy and antijamming B) Privacyand antijamming;efficiency C) Privacyand efficiency; antijamming D) Efficiency and antijamming; privacy

Description : ________ can beachievedbyusing multiplexing; ______ canbe achieved by using spreading. A) Efficiency; privacy and antijamming B) Privacyand antijamming;efficiency C) Privacyand efficiency; antijamming D) Efficiency and antijamming; privacy

Last Answer : Efficiency; privacy and antijammin

Data canbe ________. A) analog B) digital C) (a) or (b) D) none of the above

Description : Data canbe ________. A) analog B) digital C) (a) or (b) D) none of the above

Last Answer : (a) or (b)

Which of the following canbe used to select the entire document ? (A) CTRL+A (B) ALT+F5 (C) SHIFT+A (D) CTRL+K (E) CTRL+H

Description : Which of the following canbe used to select the entire document ? (A) CTRL+A (B) ALT+F5 (C) SHIFT+A (D) CTRL+K (E) CTRL+H

Last Answer : CTRL+A

Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error? a. G(x) contains more than two terms b. G(x) does not divide 1+x^k, for any k not exceeding the frame length c. 1+x is a factor of G(x) d. G(x) has an odd number of terms

Description : Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error? a. G(x) contains more than two terms b. G(x) does ... exceeding the frame length c. 1+x is a factor of G(x) d. G(x) has an odd number of terms

Last Answer : c. 1+x is a factor of G(x)

How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits, and one parity bit ? a. 600 b. 800 c. 876 d. 1200

Description : How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits, and one parity bit ? a. 600 b. 800 c. 876 d. 1200

Last Answer : b. 800

Classification of protocol that interprets a frame of data as a group of successive bit combined into predetermined pattern of fixed length, usually 8 bits each.

Description : Classification of protocol that interprets a frame of data as a group of successive bit combined into predetermined pattern of fixed length, usually 8 bits each.

Last Answer : Character and Byte-oriented protocols

A certain population of ALOHA users manages to generate 70 request/sec. If the time is slotted in units of 50 msec, then channel load would be a. 4.25 b. 3.5 c. 350 d. 450

Description : A certain population of ALOHA users manages to generate 70 request/sec. If the time is slotted in units of 50 msec, then channel load would be a. 4.25 b. 3.5 c. 350 d. 450

Last Answer : b. 3.5

Bits can be sent over guided and unguided media as analog signal by ___________ a. digital modulation b. amplitude modulation c. frequency modulation d. phase modulation

Description : Bits can be sent over guided and unguided media as analog signal by ___________ a. digital modulation b. amplitude modulation c. frequency modulation d. phase modulation

Last Answer : a. digital modulation

If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ............... (A) 64 Kbps (B) 32 Mbps (C) 32 Kbps (D) 64 Mbps

Description : If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ............... (A) 64 Kbps (B) 32 Mbps (C) 32 Kbps (D) 64 Mbps

Last Answer : (C) 32 Kbps

A Huffman code: A = 1, B = 000, C = 001, D = 01 ,P(A) = 0.4, P(B) = 0.1, P(C) = 0.2, P(D) = 0.3 The average number of bits per letter is A. 8.0 bit B. 2.0 bit C. 1.9 bit D. 2.1 bit

Description : A Huffman code: A = 1, B = 000, C = 001, D = 01 ,P(A) = 0.4, P(B) = 0.1, P(C) = 0.2, P(D) = 0.3 The average number of bits per letter is A. 8.0 bit B. 2.0 bit C. 1.9 bit D. 2.1 bit

Last Answer : C. 1.9 bit

A Huffman code: A = 1, B = 000, C = 001, D = 01 ,P(A) = 0.4, P(B) = 0.1, P(C) = 0.2, P(D) = 0.3 The average number of bits per letter is a. 8.0 bit b. 2.0 bit c. 1.9 bit d. 2.1 bit

Description : A Huffman code: A = 1, B = 000, C = 001, D = 01 ,P(A) = 0.4, P(B) = 0.1, P(C) = 0.2, P(D) = 0.3 The average number of bits per letter is a. 8.0 bit b. 2.0 bit c. 1.9 bit d. 2.1 bit

Last Answer : c. 1.9 bit

In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps ? (A) 10 bps (B) 100 bps (C) 1000 bps (D) 10000 bps

Description : In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps ? (A) 10 bps (B) 100 bps (C) 1000 bps (D) 10000 bps

Last Answer : (C) 1000 bps 

Suppose that the time to do a null remote procedure call (RPC) (i.e, 0 data bytes) is 1.0 msec, with an additional 1.5 msec for every 1K of data. How long does it take to read 32 K from the file server as 32 1K RPCs? (A) 49 msec (B) 80 msec (C) 48 msec (D) 100 msec

Description : Suppose that the time to do a null remote procedure call (RPC) (i.e, 0 data bytes) is 1.0 msec, with an additional 1.5 msec for every 1K of data. How long does it take to read 32 K from the file server as 32 1K RPCs? (A) 49 msec (B) 80 msec (C) 48 msec (D) 100 msec

Last Answer : (B) 80 msec

A byte consists of a. One bit b. Four bits c. Eight bits d. Sixteen bits

Description : A byte consists of a. One bit b. Four bits c. Eight bits d. Sixteen bits

Last Answer : Eight bits

A byte consists of a. One bit b. Four bits c. Eight bits d. Sixteen bits

Description : A byte consists of a. One bit b. Four bits c. Eight bits d. Sixteen bits

Last Answer : Eight bits

A band is always equivalent to a. a byte b. a bit c. 100 bits d. none of above

Description : A band is always equivalent to a. a byte b. a bit c. 100 bits d. none of above

Last Answer : none of above

We addr redundant bits toeach block to make thelength n = k + r. The resultingn-bit blocksare called _________. A) datawords B) blockwords C) codewords D) none of the above

Description : We addr redundant bits toeach block to make thelength n = k + r. The resultingn-bit blocksare called _________. A) datawords B) blockwords C) codewords D) none of the above

Last Answer : codewords

A burst error means that two or more bits in the dataunit havechanged. A) double-bit B) burst C) single-bit D) none of the above

Description : A burst error means that two or more bits in the dataunit havechanged. A) double-bit B) burst C) single-bit D) none of the above

Last Answer : burst

The ______ technique expands the bandwidthof a signal by replacing each data bit with n bits. A) FDM B) DSSS C) FHSS D) TDM

Description : The ______ technique expands the bandwidthof a signal by replacing each data bit with n bits. A) FDM B) DSSS C) FHSS D) TDM

Last Answer : DSSS

In______ transmission, wesend1 start bit (0) at thebeginningand 1 or more stop bits (1s) at the end of each byte. A) synchronous B) asynchronous C) isochronous D) none of the above

Description : In______ transmission, wesend1 start bit (0) at thebeginningand 1 or more stop bits (1s) at the end of each byte. A) synchronous B) asynchronous C) isochronous D) none of the above

Last Answer : asynchronous

The least significant 23 bits in a 48-bit Ethernet address identify a ________. A) multicast router B) host C) multicast group D) noneof the above

Description : The least significant 23 bits in a 48-bit Ethernet address identify a ________. A) multicast router B) host C) multicast group D) noneof the above

Last Answer : multicast group

We add r redundant bits to each block to make the length n = k + r. The resulting n-bit blocks are called _________ A. datawords B. blockwords C. codewords D. none of the above

Description : We add r redundant bits to each block to make the length n = k + r. The resulting n-bit blocks are called _________ A. datawords B. blockwords C. codewords D. none of the above

Last Answer : C. codewords

In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1- bit delay in this network is equivalent to: a. 500 metres of cable. b. 200 metres of cable. c. 20 metres of cable. d. 50 metres of cable

Description : In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1- bit delay in this network is equivalent to: a. 500 metres of cable. b. 200 metres of cable. c. 20 metres of cable. d. 50 metres of cable

Last Answer : c. 20 metres of cable.

What do you call a short bit sequence sent typically 128 bits in digital signatures? A. Hash B. Public key C. Private key D. Password

Description : What do you call a short bit sequence sent typically 128 bits in digital signatures? A. Hash B. Public key C. Private key D. Password

Last Answer : A. Hash

An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are ............... and .............. (A) 4000 bauds \ sec & 1000 bps (B) 2000 bauds \ sec & 1000 bps (C) 1000 bauds \ sec & 500 bps (D) 1000 bauds \ sec & 4000 bps

Description : An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are ............... and .............. (A) 4000 bauds \ sec & ... (B) 2000 bauds \ sec & 1000 bps (C) 1000 bauds \ sec & 500 bps (D) 1000 bauds \ sec & 4000 bps

Last Answer : (D) 1000 bauds \ sec & 4000 bps Explanation: Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or equal to the bit rate. Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps

Standard Ethernet (10-Mbps) uses_______ encoding A) NRZ B) AMI C) Manchester D) differential Manchester

Description : Standard Ethernet (10-Mbps) uses_______ encoding A) NRZ B) AMI C) Manchester D) differential Manchester

Last Answer : Manchester

The IEEE 802.3 Standard defines _________ CSMA/CD as the access method for first-generation 10-Mbps Ethernet. A) 1-persistent B) p-persistent C) non-persistent D) none of the above

Description : The IEEE 802.3 Standard defines _________ CSMA/CD as the access method for first-generation 10-Mbps Ethernet. A) 1-persistent B) p-persistent C) non-persistent D) none of the above

Last Answer : 1-persistent

There are 10 users plugged into a hub running 10Mbps half-duplex.and a serverconnected to the switch running 10Mbps half- duplex as well. Find the bandwidth does each host have to the server? a. 100 kbps b. 1 Mbps c. 2 Mbps d. 10 Mbps

Description : There are 10 users plugged into a hub running 10Mbps half-duplex.and a server connected to the switch running 10Mbps half- duplex as well. Find the bandwidth does each host have to the server? a. 100 kbps b. 1 Mbps c. 2 Mbps d. 10 Mbps

Last Answer : d. 10 Mbps

Inthe _________ method, a special packet called a ______ circulates through the ring. A) reservation: control frame B) polling: poll request C) token passing: token D) none of the above

Description : Inthe _________ method, a special packet called a ______ circulates through the ring. A) reservation: control frame B) polling: poll request C) token passing: token D) none of the above

Last Answer : token passing: token

Explain the frame format for token ring and token bus

Description : Explain the frame format for token ring and token bus

Last Answer : Access method: Token passing _ Priority and reservation _ Time limits _ Monitor stations