The efficiency of a Stop and Wait (SAW) protocol is given by the formula:

[ text{Efficiency} = frac{T_t}{T_t + 2 times T_p} ]

Where:

– (T_t) = Transmission time

– (T_p) = Propagation time

Given:

– Transmission time ((T_t)) = 20 ns

– Propagation time ((T_p)) = 30 ns

Plugging the values into the formula:

[ text{Efficiency} = frac{20}{20 + 2 times 30} ]

[ text{Efficiency} = frac{20}{20 + 60} ]

[ text{Efficiency} = frac{20}{80} ]

[ text{Efficiency} = 0.25 ] or 25%

Hence, the efficiency of the Stop and Wait protocol given the conditions is 25%.

To calculate the packet size in a packet switching network, we need to consider both the size of the data and the size of the header in each packet. Given that each packet contains a header of 3 bytes and that 24 packets are required to transmit the entire message, we can calculate the packet size as follows:

Let’s denote:

– (H) as the header size per packet,

– (N) as the total number of packets,

– (D) as the data size per packet,

– (T) as the total message size,

– (P) as the packet size (which includes both the header and the data).

From the provided information:

– (H = 3) bytes (header size),

– (N = 24) (number of packets),

– (T = 48) bytes (total message size).

The total size of the message is given, but it seems there might be a confusion in the question as it presents the message size being less than what would be expected given the number of packets and the header size.

However, based on the usual way to find the packet size:

– (P = D + H), where (D) must be calculated or clarified.

For the calculation or approach mentioned in the question (assuming the total message size as transmitted data without headers), the total data that can be transmitted (excluding headers) is:

– Total transmitted data ((T)) minus the total size of all headers ((

For a generator polynomial G(x) used in CRC (Cyclic Redundancy Check) to detect an odd number of bit errors, the polynomial must satisfy the condition that it includes the factor (x + 1). Specifically, G(x) has the ability to detect all odd numbers of bit errors if and only if (G(x)) is divisible by (x + 1). This is because (x + 1) represents a polynomial that, in binary, corresponds to the pattern 11 (i.e., an error of two bits). Multiplying this by any polynomial of even weight (even number of 1s) will result in a polynomial of even weight that captures odd numbers of errors. This property ensures that any error pattern with an odd number of flipped bits will result in a non-zero remainder when divided by (G(x)), thereby indicating the presence of errors.

To determine the number of packets Station A will transmit to Station B, given the sliding window protocol with a window size of 3, and a go-back-n error control strategy where every 5th packet that A transmits gets lost, let’s walk through the process.

The sliding window protocol allows Station A to send up to 3 packets before needing an acknowledgment (ACK) for the first packet in the window. With go-back-n, if a packet is lost, all packets sent after the lost packet must be retransmitted, even if they were received successfully.

Since every 5th packet gets lost:

1. The first four packets (1, 2, 3, 4) are sent successfully. Packet 5 is lost.
2. Station A realizes packet 5 is lost and hence retransmits packets 5, 6, and 7.
3. Packet 5 (on the second try, transmission number 6 overall) is successfully transmitted, but now the 10th packet from the start of the process, which is packet 8 this time (since we are counting retransmissions in our total), is lost.
4. Station A retransmits packets 8, 9, and receives ACKs without loss.

Let’s count the transmissions including retransmissions:

– Original sequence of transmissions: 1, 2, 3, 4, (5 lost), 6, 7, 8

– First re

To determine the maximum duration for which the computer can transmit at the full 10Mbps given the conditions of the token bucket, we need to calculate how long the initial tokens (16 Megabits) plus the tokens being added at a rate of 2Mbps can sustain a 10Mbps transmission.

Here’s a step-by-step calculation:

1. Token Consumption Rate: The network aims to transmit at 10Mbps, but tokens are being added at a rate of 2Mbps. Thus, the net token consumption rate when transmitting at full speed is (10Mbps – 2Mbps = 8Mbps).

2. Initial Token Bucket Capacity: The token bucket starts fully filled with 16 Megabits.

3. Duration Calculation: To find out how long the computer can sustain a 10Mbps transmission, we calculate the duration for which the initial tokens plus the tokens added can sustain this consumption rate. Since the consumption rate is 8Mbps, and you have an initial capacity of 16 Megabits, you divide the capacity by the consumption rate.

[ text{Duration} = frac{text{Initial Capacity}}{text{Consumption Rate}} = frac{16 text{ Megabits}}{8 text{ Mbps}} ]

[ text{Duration} = 2 text{ seconds} ]

Therefore, the computer can transmit at the full 10Mbps for a maximum duration of 2 seconds before the tokens in the