.

Connectionless unacknowledged service (CUAS) and connectionless acknowledged service (CLAS) are
two types of communication services provided by network protocols. The main difference between
these two services lies in the level of reliability and error handling.

Connectionless unacknowledged service is the simplest type of connectionless service. In CUAS, the
sender simply transmits a data unit (DU) to the receiver without any guarantee of delivery. The receiver
does not send any acknowledgments to the sender, and the sender does not wait for any
acknowledgments before sending the next DU. This type of service is very fast, but it is also unreliable.
Packets can be lost or corrupted without the sender or receiver being aware of it.
Connectionless acknowledged service is a more reliable type of connectionless service. In CLAS, the
sender transmits a DU to the receiver and waits for an acknowledgment from the receiver before
sending the next DU. If the acknowledgment is not received within a certain time period, the sender will
retransmit the DU. This type of service is slower than CUAS, but it is more reliable. Packets are less likely
to be lost or corrupted because the sender will retransmit any packets that are not acknowledged by the
receiver.
The protocols that provide these services also differ. CUAS protocols, such as UDP (User Datagram
Protocol), are very simple and do not require any additional control messages. CLAS protocols, such as
TCP (Transmission Control Protocol), are more complex and require additional control messages to
manage the acknowledgments.

2/PPP stands for Point-to-Point Protocol. It is a full-duplex protocol, which means that data can be
transferred in both directions simultaneously. PPP is commonly used for dial-up internet access and for
connecting two computers directly together.

HIDIC stands for High-level Data Link Control with Internet Checksum. It is a half-duplex protocol, which
means that data can only be transferred in one direction at a time. HIDIC is commonly used for
connecting embedded devices together, such as sensors and actuators.

Sure. PPP and HIDIC are both protocols for transferring data between devices, but they have some key
differences.

PPP stands for Point-to-Point Protocol. It is a full-duplex protocol, which means that data can be
transferred in both directions simultaneously. PPP is commonly used for dial-up internet access and for
connecting two computers directly together.

HIDIC stands for High-level Data Link Control with Internet Checksum. It is a half-duplex protocol, which
means that data can only be transferred in one direction at a time. HIDIC is commonly used for
connecting embedded devices together, such as sensors and actuators.

Here is a table that summarizes the key differences between PPP and HIDIC:

Feature PPP HIDIC

Full-duplex Yes No

Commonly used for Dial-up internet access, connecting two computers directly together Connecting
embedded devices together

Transport layer protocol IP UDP

Error correction CRC-16 None

Here is an example of how PPP and HIDIC might be used in different scenarios:

You want to connect your home computer to the internet using a dial-up modem. In this case, you
would use PPP to establish a connection to your ISP.

You want to connect two sensors together to transmit data. In this case, you would use HIDIC to create a
reliable connection between the two sensors.

3/Here are the answers to your questions:

a. How many subnets are needed?

The topology has 4 LANs, so we need at least 4 subnets.

b. How many bits must be borrowed to support the number of subnets in the topology table?

We can borrow 3 bits to support 8 subnets.

c. How many subnets does this create?

Borrowing 3 bits creates 8 subnets.

d. How many usable hosts does this create per subnet?

Borrowing 3 bits leaves 28 bits for host addresses, which is 256 hosts per subnet.

Note: If your answer is less than the 25 hosts required, then you borrowed too many bits.
4/a. How long does
it take to download
the file over a 32
kilobit/second
modem?

A 1 megabyte file is
equal to 8,000,000
bits. A 32
kilobit/second
modem can transfer data at a rate of 32,000 bits per second. Therefore, it would take 8,000,000 /
32,000 = 250 seconds to download the file.

b. How long does it take to take to download the file over a 1 megabit/second modem?

A 1 megabit/second modem can transfer data at a rate of 1,000,000 bits per second. Therefore, it would
take 8,000,000 / 1,000,000 = 8 seconds to download the file.

c. Suppose data compression is applied to the text file. How much do the transmission times in parts (a)
and (b) change?

Data compression can significantly reduce the size of a text file. For example, a lossless compression
algorithm could reduce the size of the file by a factor of 2 or 3. In this case, the download time would be
reduced to 125 seconds for a 32 kilobit/second modem and 2 seconds for a 1 megabit/second modem.

5/Organization A

First IP address: 198.16.0.0

Last IP address: 198.16.3999.255

Mask: 24 bits (/24)

Organization B

First IP address: 198.16.4000.0

Mask: 24 bits (/24)

Organization C

First IP address: 198.16.7000.0

Last IP address: 198.16.9999.255

Mask: 24 bits (/24)

Organization D

First IP address: 198.16.10000.0

Last IP address: 198.16.17999.255

Mask: 23 bits (/23)

The reason for the different masks is that we need to ensure that each organization has a unique
subnet. For example, if we used a mask of 24 bits for all four organizations, then organizations A and B
would have overlapping subnets.
1/Connection-oriented acknowledged service (COAS) and connectionless acknowledged service (CLAS)
are two types of communication services provided by network protocols. The main difference between
these two services lies in the way that they handle the delivery of data.

Connection-oriented acknowledged service (COAS) is a reliable service that guarantees the delivery of
data. In COAS, the sender and receiver establish a connection before any data is transferred. This
connection provides a reliable channel for the transfer of data. The sender sends data to the receiver,
and the receiver acknowledges the receipt of each data packet. If a data packet is lost or corrupted, the
receiver will not acknowledge it, and the sender will retransmit the packet.

Connectionless acknowledged service (CLAS) is a less reliable service that does not guarantee the
delivery of data. In CLAS, the sender and receiver do not establish a connection before any data is
transferred. Each data packet is sent independently, and the receiver does not acknowledge the receipt
of each data packet. If a data packet is lost or corrupted, the receiver will not know about it, and the
sender will not retransmit the packet.
The protocols that provide these services also differ. COAS protocols, such as TCP (Transmission Control
Protocol), are more complex than CLAS protocols, such as UDP (User Datagram Protocol). COAS
protocols need to maintain state information about the connection, such as the sequence numbers of
the data packets that have been sent and received. CLAS protocols do not need to maintain any state
information, which makes them simpler and faster.

2/The distance between the earth and the moon is 375,000 km.

The speed of light is 3 x 10^8 meters/second.

The data rate is 1.5 Mbps, which is 1,500,000 bits per second.

The round-trip time (RTT) for a signal to travel from the earth to the moon and back is:

RTT = 2 * distance / speed of light = 2 * 375,000 km / 3 x 10^8 meters/second = 250 milliseconds

The smallest possible frame size is the amount of data that can be transmitted in one RTT. This is equal
to the data rate multiplied by the RTT:

frame size = data rate * RTT = 1,500,000 bits/second * 250 milliseconds = 468750 bits

This means that the smallest possible frame size is 468750 bits. Any smaller frame size would result in a
gap between frames that would be longer than the RTT, which would prevent continuous transmission.
4/. How does the protocol need to be modified to accommodate this change?

The Selective Repeat ARQ protocol needs to be modified in two ways to accommodate this change:

The sender needs to keep track of the next m frames that it expects to receive an acknowledgment for.

The receiver needs to send an acknowledgment message that contains a list of the next m frames that it
expects to receive.

The sender can keep track of the next m frames that it expects to receive by using a sliding window. The
sliding window is a list of frame numbers that the sender has sent, but has not yet received an
acknowledgment for. The size of the sliding window is m.

The receiver can send an acknowledgment message that contains a list of the next m frames that it
expects to receive by simply listing the frame numbers of the next m frames in the sequence.

b. What is the effect of the change on protocol performance?

The change to Selective Repeat ARQ that allows the receiver to send a list of the next m frames that it
expects to receive can improve protocol performance in two ways:

It can reduce the number of acknowledgment messages that need to be sent.

It can reduce the number of retransmissions that need to be made.

If the receiver sends a list of the next m frames that it expects to receive, then the sender only needs to
send an acknowledgment message when one of the frames in the list is not received. This can reduce
the number of acknowledgment messages that need to be sent by up to m-1.

If the receiver sends a list of the next m frames that it expects to receive, then the sender only needs to
retransmit the frames that are not in the list. This can reduce the number of retransmissions that need
to be made by up to m-1.

In general, the change to Selective Repeat ARQ that allows the receiver to send a list of the next m
frames that it expects to receive can improve protocol performance by reducing the number of
acknowledgment messages and retransmissions that need to be made.
5/Here is the detailed explanation:

The generator polynomial g(x) = x^3 + x + 1 has a degree of 3. This means that the codeword must be 3
bits longer than the information sequence.

The information sequence 1001 is 4 bits long. This means that the codeword must be 7 bits long.

We can find the codeword by performing polynomial arithmetic on the information sequence and the
generator polynomial.

g(x) = x^3 + x + 1

1001 = 1*x^3 + 0*x^2 + 0*x + 1

(g(x))*(1001) = (x^3 + x + 1)*(1*x^3 + 0*x^2 + 0*x + 1)

= x^6 + x^4 + x^2 + x + 1

The first three bits of the result, x^6 + x^4 + x^2, are all 0. This is because the information sequence is
only 4 bits long, and the first three bits of the generator polynomial are 0.

The last four bits of the result, x + 1, are the codeword. In this case, the codeword is 3, 1, 1, 3.