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What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20ns and the propagation time is 30ns?
The efficiency of a Stop and Wait ARQ protocol is determined by the ratio of the time it takes to send a frame to the total time taken for a round trip of a single frame plus the acknowledgment. In simpler terms, it's the ratio of the useful time spent transmitting data over the total time spent incRead more
The efficiency of a Stop and Wait ARQ protocol is determined by the ratio of the time it takes to send a frame to the total time taken for a round trip of a single frame plus the acknowledgment. In simpler terms, it’s the ratio of the useful time spent transmitting data over the total time spent including waiting for acknowledgments.
Given:
– Transmission Time (Tt) = 20ns (nanoseconds)
– Propagation Time (Tp) = 30ns
The efficiency ((E)) of Stop and Wait protocol can be calculated using the formula:
[
E = frac{Tt}{Tt + 2Tp}
]
Substituting the given values:
[
E = frac{20}{20 + 2(30)} = frac{20}{80} = frac{1}{4} = 0.25
]
Therefore, the efficiency of the Stop and Wait protocol, given the provided transmission and propagation times, is 0.25 or 25%.
See lessWhat will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20ns and the propagation time is 30ns?
In the Stop and Wait protocol, the efficiency is primarily determined by the ratio of the time taken to transmit a frame to the total time taken for the round-trip of a frame plus the time taken for the acknowledgment to come back. This can be simplified as the ratio of the transmission time (Tt) toRead more
In the Stop and Wait protocol, the efficiency is primarily determined by the ratio of the time taken to transmit a frame to the total time taken for the round-trip of a frame plus the time taken for the acknowledgment to come back. This can be simplified as the ratio of the transmission time (Tt) to the sum of the transmission time and twice the propagation time (since the signal has to travel to the receiver and then the acknowledgment has to travel back to the sender), i.e., Tt + 2Tp.
Given that the transmission time (Tt) for a frame is 20 ns and the propagation time (Tp) is 30 ns, the efficiency (η) can be calculated as follows:
[ η = frac{T_t}{T_t + 2 * T_p} ]
Substituting the given values:
[ η = frac{20}{20 + 2 * 30} = frac{20}{80} = frac{1}{4} ]
[ η = 0.25 ] or 25%
Thus, the efficiency of the Stop and Wait protocol under the given conditions would be 25%.
See lessIn serial data transmission, every byte of data is padded with a ‘0’ in the beginning and one or two ‘1’ s at the end of byte because
In serial data transmission, padding a byte of data with a '0' at the beginning and one or two '1's at the end serves several important purposes: 1. Frame Synchronization: The added bits, particularly the '0' at the beginning and the '1's at the end, help in establishing and maintaining the synchronRead more
In serial data transmission, padding a byte of data with a ‘0’ at the beginning and one or two ‘1’s at the end serves several important purposes:
1. Frame Synchronization: The added bits, particularly the ‘0’ at the beginning and the ‘1’s at the end, help in establishing and maintaining the synchronization between the sender and receiver. This makes it easier for the receiver to identify the start and end of each byte in the continuous stream of data.
2. Error Detection: The structured pattern of padding bits can also assist in error detection. By expecting a specific pattern at the beginning and the end of each byte, any deviation from this pattern can signal a transmission error.
3. Signal Integrity: The inclusion of these bits can also help in maintaining signal integrity over the data transmission path by providing regular transitions between ‘0’ and ‘1’ states. This is particularly important in some transmission mediums where a long sequence of similar bits (all ‘0’s or all ‘1’s) can cause the receiver to lose track of the bit boundaries.
4. Bit Stuffing: In protocols that use bit stuffing, adding specific bits ensures that the actual data does not accidentally mimic control signals. In some protocols, a long sequence of ‘1’s might be interpreted as a control signal, so breaking up such sequences with ‘0’s ensures that data is not misinterpreted as a signal or delimiter.
Every protocol may implement these principles differently based on its specific requirements for
See lessMatch the following: (P) SMTP (1) Application layer (Q) BGP (2) Transport layer (R) TCP (3) Data link layer (S) PPP (4) Network layer (5) Physical layer
P - 1) Application layerQ - 4) Network layerR - 2) Transport layerS - 3) Data link layer
P – 1) Application layer
Q – 4) Network layer
R – 2) Transport layer
S – 3) Data link layer
See lessIn the slow start phase of the TCP congestion control algorithm, the size of the congestion window
In the slow start phase of the TCP congestion control algorithm, the size of the congestion window increases exponentially. Specifically, the congestion window size starts at a small value, typically one or two segments, and then doubles every round trip time (RTT), as long as acknowledgments continRead more
In the slow start phase of the TCP congestion control algorithm, the size of the congestion window increases exponentially. Specifically, the congestion window size starts at a small value, typically one or two segments, and then doubles every round trip time (RTT), as long as acknowledgments continue to be received and no packet loss is detected. This exponential growth occurs because for every acknowledgment received, the congestion window size increases by one segment, effectively doubling the window size each RTT. This exponential increase continues until a packet loss is detected or the congestion window size reaches a threshold value, known as the slow start threshold (ssthresh), at which point the congestion control algorithm transitions from the slow start phase to the congestion avoidance phase.
See lessConsider different activities related to email. m1:Send an email from a mail client to mail server m2:Download an email from mailbox server to a mail client m3:Checking email in a web browser Which is the applicable level protocol user in each activity?
For the activities related to email that you've mentioned, the applicable level protocols are as follows:- m1: Sending an email from a mail client to a mail server typically involves the Simple Mail Transfer Protocol (SMTP). SMTP is used for the transmission of emails from the source address to theRead more
For the activities related to email that you’ve mentioned, the applicable level protocols are as follows:
– m1: Sending an email from a mail client to a mail server typically involves the Simple Mail Transfer Protocol (SMTP). SMTP is used for the transmission of emails from the source address to the destination.
– m2: Downloading an email from mailbox server to a mail client can involve the Post Office Protocol version 3 (POP3) or the Internet Message Access Protocol (IMAP). POP3 is used when the mail client connects to the mail server to download emails to the local device, whereas IMAP is used to access and manage emails directly on the mail server, allowing for more versatile handling of emails across multiple devices.
– m3: Checking email in a web browser doesn’t use the traditional email protocols (SMTP, POP3, IMAP) directly from the client side; instead, it involves Hypertext Transfer Protocol (HTTP) or Hypertext Transfer Protocol Secure (HTTPS). These protocols are used for secure communication between the user’s web browser and the web-based email service (webmail server). The behind-the-scenes action of receiving and sending emails through the web interface still utilizes SMTP, POP3/IMAP as relevant, but this is abstracted from the user.
In summary:
– m1: SMTP for sending emails.
– m2: POP3 or IMAP for downloading emails.
– m3: HTTP or HTTPS for checking emails in a web browser.
See lessSending a file from your personal computer's primary memory or disk to another computer is called
Uploading
Uploading
See lessIf a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is __________
The data rate can be calculated using the formula:[ text{Data rate} = frac{text{Total number of characters}}{text{Time in seconds}} ]Given that the total number of characters is 50,000 and the time taken to send these characters is 40 seconds, we can plug these values into our formula:[ text{Data raRead more
The data rate can be calculated using the formula:
[ text{Data rate} = frac{text{Total number of characters}}{text{Time in seconds}} ]
Given that the total number of characters is 50,000 and the time taken to send these characters is 40 seconds, we can plug these values into our formula:
[ text{Data rate} = frac{50,000 , text{characters}}{40 , text{seconds}} ]
[ text{Data rate} = 1,250 , text{characters per second} ]
See lessIf 5 files are transferred from server A to client B in the same session. The number of TCP connection between A and B is
The number of TCP connections between server A and client B for transferring 5 files in the same session would be 1. TCP establishes a connection before transferring data and maintains that connection until the session ends. Assuming these files are transferred sequentially or even concurrently (givRead more
The number of TCP connections between server A and client B for transferring 5 files in the same session would be 1. TCP establishes a connection before transferring data and maintains that connection until the session ends. Assuming these files are transferred sequentially or even concurrently (given the capabilities of modern TCP implementations) in the same session, only a single TCP connection is necessary.
See lessIn a packet switching network, if the message size is 48 bytes and each packet contains a header of 3 bytes. If 24 packets are required to transmit the message, the packet size is ________
In a packet switching network, where the message is 48 bytes and each packet includes a 3-byte header, to calculate the packet size given that 24 packets are required for transmission, we take the following steps: 1. Determine the total amount of data being sent (including headers).2. Find the sizeRead more
In a packet switching network, where the message is 48 bytes and each packet includes a 3-byte header, to calculate the packet size given that 24 packets are required for transmission, we take the following steps:
1. Determine the total amount of data being sent (including headers).
2. Find the size of each packet.
Firstly, it’s important to note that there seems to be a misunderstanding or typo in the question as presented. If the message size is mentioned to be 48 bytes and 24 packets are required, it implies a division of the total data into these packets. However, the description as is suggests an unrealistic scenario regarding typical packet-switched networking practices where 24 packets would convey much more than 48 bytes even with additional header data per packet.
However, proceeding under the assumption we’re trying to work out the packet size given a typical scenario (and interpreting the 48-byte message size as a mistake or miscommunication), here’s how it would generally be approached:
Given:
– Each packet has a 3-byte header.
– 24 packets are required.
Without clear total message size that fits the context, another way to look at this is perhaps you’re asking how to calculate the size of a packet when parts of its structure are known (header versus payload):
– Each packet has some payload plus the 3-byte header.
If you’re looking for the packet size including the header when 24 packets are transmitted, but based on typical networking principles where the
See less