Câu 1: Explain the difference between connectionless unacknowledged service and
connectionless acknowledged service. How do the protocols that provide these services
differ?
Câu 2. Explain the difference between connection-oriented acknowledged service and
connectionless acknowledged service. How do the protocols that provide these services
differ?
Câu 3: Explain the differences between PPP and HDLC.
Câu 4:
A 1.5 Mbps communications link is to use HDLC to transmit information to the moon.
What is the smallest possible frame size that allows continuous transmission? The
distance between earth and the moon is approximately 375,000 km, and the speed of light
is 3 x 108 meters/second.
The smallest possible frame size that allows continuous transmission is the size of the
round-trip propagation delay. The round-trip propagation delay is the time it takes for a
signal to travel from Earth to the Moon and back.
The distance between Earth and the Moon is 375,000 km, so the round-trip propagation
delay is 2 * 375,000 km / 3 x 10^8 meters/second = 250 milliseconds.
The data rate of the communications link is 1.5 Mbps, so the smallest possible frame size
is 1.5 Mbps * 250 milliseconds = 375,000 bits.
In bytes, the smallest possible frame size is 375,000 bits / 8 bits/byte = 46,875 bytes.
Therefore, the smallest possible frame size that allows continuous transmission is 46,875
bytes.
To determine the smallest possible frame size for continuous transmission over a 1.5
Mbps communications link using HDLC to the moon, we need to consider the round-trip
time (RTT) for the signal to travel from Earth to the moon and back.
Given that the distance between Earth and the moon is approximately 375,000 km and
the speed of light is 3 x 10^8 meters/second, we can calculate the RTT as follows:
RTT = (2 * distance) / speed of light
RTT = (2 * 375,000 km) / (3 x 10^8 m/s)
RTT = 2.5 seconds
To achieve continuous transmission, the frame transmission time should be less than the
RTT.
The frame transmission time can be calculated using the formula:
Frame transmission time = frame size / link speed
Since the link speed is given as 1.5 Mbps, we can substitute the values into the formula:
Frame transmission time = frame size / 1.5 Mbps
To ensure continuous transmission, the frame transmission time should be less than the
RTT of 2.5 seconds. Therefore, we can set up the following inequality:
frame size / 1.5 Mbps < 2.5 seconds
Simplifying the inequality, we can solve for the minimum frame size:
frame size < 1.5 Mbps * 2.5 seconds
frame size < 3.75 Mbps * seconds
frame size < 3.75 * 10^6 bits
Therefore, the smallest possible frame size that allows continuous transmission over the
1.5 Mbps communications link using HDLC to the moon is less than 3.75 * 10^6 bits.
Để xác định kích thước khung nhỏ nhất có thể để truyền liên tục qua liên kết truyền thông
1,5 Mb/giây sử dụng HDLC tới mặt trăng, chúng ta cần xem xét thời gian khứ hồi (RTT)
để tín hiệu truyền từ Trái đất tới mặt trăng và ngược lại.
Cho rằng khoảng cách giữa Trái đất và mặt trăng là khoảng 375.000 km và tốc độ ánh
sáng là 3 x 10^8 mét/giây, chúng ta có thể tính RTT như sau:
RTT = (2 * khoảng cách) / tốc độ ánh sáng
RTT = (2 * 375.000 km) / (3 x 10^8 m/s)
RTT = 2,5 giây
Để đạt được truyền liên tục, thời gian truyền khung phải nhỏ hơn RTT.
Thời gian truyền khung có thể được tính bằng công thức:
Thời gian truyền khung = kích thước khung / tốc độ liên kết
Vì tốc độ liên kết được cho là 1,5 Mbps, nên chúng ta có thể thay thế các giá trị vào công
thức:
Thời gian truyền khung hình = kích thước khung hình / 1,5 Mbps
Để đảm bảo truyền liên tục, thời gian truyền khung phải nhỏ hơn RTT là 2,5 giây. Do đó,
ta có thể thiết lập bất đẳng thức sau:
kích thước khung hình / 1,5 Mbps < 2,5 giây
Đơn giản hóa sự bất bình đẳng, chúng ta có thể giải quyết kích thước khung hình tối
thiểu:
kích thước khung hình < 1,5 Mbps * 2,5 giây
kích thước khung hình < 3,75 Mbps * giây
kích thước khung hình < 3,75 * 10^6 bit
Do đó, kích thước khung nhỏ nhất có thể cho phép truyền liên tục qua liên kết truyền
thông 1,5 Mb/giây sử dụng HDLC tới mặt trăng là nhỏ hơn 3,75 * 10^6 bit.
Câu 5: Suppose HDLC is used over a 1.5 Mbps geostationary satellite link. Suppose that
250-byte frames are used in the data link control. What is the maximum rate at which
information can be transmitted over the link?
The maximum rate at which information can be transmitted over the link is 299,800 bits
per second.
The data rate of the link is 1.5 Mbps, which is equal to 1.5 * 10^6 bits per second.
However, the overhead of the HDLC protocol is 250 * 8 = 2000 bits per frame. This
means that the maximum rate at which information can be transmitted over the link is 1.5
* 10^6 - 2000 = 299,800 bits per second.
In bytes, the maximum rate at which information can be transmitted over the link is
299,800 / 8 = 37,475 bytes per second.
Câu 6:
Suppose that a multiplexer receives constant-length packet from N = 60 data sources.
Each data source has a probability p = 0.1 of having a packet in a given T-second period.
Suppose that the multiplexer has one line in which it can transmit eight packets every T
seconds. It also has a second line where it directs any packets that cannot be transmitted
in the first line in a T-second period. Find the average number of packets that are
transmitted on the first line and the average number of packets that are transmitted in the
second line.
The average number of packets that are transmitted on the first line is given by:
E[x_1] = np = 60 * 0.1 = 6
where n is the number of data sources and p is the probability of a data source having a
packet in a given T-second period.
The average number of packets that are transmitted in the second line is given by:
E[x_2] = np(1 - p/m) = 6 * 0.1 * (1 - 0.1/8) = 0.133333
where m is the capacity of the first line.
Therefore, the average number of packets that are transmitted on the first line is 6 and
the average number of packets that are transmitted in the second line is 0.133333.
Câu 7:
Consider the transfer of a single real-time telephone voice signal across a packet network.
Suppose that each voice sample should not be delayed by more than 20 ms.
a. Discuss which of the following adaptation functions are relevant to meeting the
requirements of this transfer: handling of arbitrary message size; reliability and
sequencing; pacing and flow control; timing; addressing; and privacy, integrity
and authentication.
b. Compare a hop-by-hop approach to an end-to-end approach to meeting the
requirements of the voice signal.
a.
The following adaptation functions are relevant to meeting the requirements of this transfer:
Handling of arbitrary message size: The voice signal is a continuous signal, so it needs to
be divided into small packets. The packets can be of different sizes, so the network needs to
be able to handle arbitrary message sizes.
Reliability and sequencing: The voice signal is a real-time signal, so it is important that the
packets are delivered reliably and in the correct order.
Pacing and flow control: The network needs to be able to pace the delivery of the packets
so that the voice signal does not become too delayed.
Timing: The network needs to be able to keep track of the timing of the packets so that the
voice signal is not played back out of order.
Addressing: The network needs to be able to address the packets so that they can be
delivered to the correct destination.
Privacy, integrity and authentication: The network needs to be able to protect the voice
signal from unauthorized access, modification, and replay.
b.
A hop-by-hop approach to meeting the requirements of the voice signal would involve each
hop in the network handling the adaptation functions independently. This approach would
be simple to implement, but it would not be very reliable. If a packet is lost or delayed at
one hop, the other hops would not be able to recover it.
An end-to-end approach to meeting the requirements of the voice signal would involve the
network providing end-to-end guarantees for the adaptation functions. This approach would
be more reliable, but it would be more complex to implement.
The best approach to meeting the requirements of the voice signal would depend on the
specific network and the requirements of the application. If the network is reliable and the
application does not require very low latency, then a hop-by-hop approach may be
sufficient. However, if the network is not reliable or the application requires very low
latency, then an end-to-end approach may be necessary.
Câu 8 :
Consider the Stop-and-Wait protocol as described. Suppose that the protocol is modified
so that each time a frame is found in error at either the sender or receiver, the last
transmitted frame is immediately resent.
a. Show that the protocol still operates correctly.
b. Does the state transition diagram need to be modified to describe the new
operation?
c. What is the main effect of introducing the immediate-retransmission feature?
a.
The Stop-and-Wait protocol works by sending a frame, waiting for an
acknowledgement, and then sending the next frame. If the acknowledgement is not
received, the frame is resent.
The immediate-retransmission feature modifies the protocol so that the frame is
resent as soon as an error is detected. This means that the sender does not have to
wait for the acknowledgement before resending the frame.
The protocol will still operate correctly with the immediate-retransmission feature. If
a frame is received in error, the receiver will send a negative acknowledgement. The
sender will then immediately resent the frame.
b.
The state transition diagram does not need to be modified to describe the new
operation. The only difference is that the sender will now enter the "Resend frame"
state as soon as an error is detected.
c.
The main effect of introducing the immediate-retransmission feature is to reduce the
number of frames that are lost. This is because the frame is resent as soon as an error
is detected, so there is less time for the frame to be lost in the network.
The immediate-retransmission feature also improves the throughput of the protocol.
This is because the sender does not have to wait for the acknowledgement before
resending the frame, so the sender can send more frames in a given period of time.
Câu 9:
Suppose that two peer-to-peer processes provide a service that involves the transfer of
discrete messages. Suppose that the peer processes are allowed to exchange PDUs that
have a maximum size of M bytes including H bytes of header. Suppose that a PDU is not
allowed to carry information from more than one message.
a. Develop an approach that allows the peer processes to exchange messages of arbitrary
size.
b. What essential control information needs to be exchanged between the peer processes?
c. Now suppose that the message transfer service provided by the peer processes is shared
by several message source-destination pairs. Is additional control information required,
and if so, where should it be placed?
a.
To allow the peer processes to exchange messages of arbitrary size, we can use a
technique called fragmentation. This technique breaks the message into smaller pieces,
called fragments, that are each smaller than the maximum PDU size. The fragments are
then sent as separate PDUs.
The receiver reassembles the fragments into the original message. The fragmentation and
reassembly process is handled by the peer processes.
b.
The essential control information that needs to be exchanged between the peer processes
includes:
The size of the message.
The number of fragments.
The sequence number of each fragment.
c.
If the message transfer service provided by the peer processes is shared by several
message source-destination pairs, then additional control information is required. This
additional control information includes:
The source and destination of the message.
The type of message.
The priority of the message.
This additional control information is needed to ensure that the messages are routed to the
correct destination and that the messages are processed in the correct order.
The additional control information can be placed in the header of the PDU. The header of
the PDU can be up to H bytes long, so there is enough space to include the additional
control information.
Câu 10:
A 1 Mbyte file is to be transmitted over a 1 Mbps communication line that has a bit error
rate of p = 10-6.
a. What is the probability that the entire file is transmitted without errors? Note for n
large and p very small, (1 − p)n ≈ e-np.
b. The file is broken up into N equal-sized blocks that are transmitted separately.
What is the probability that all the blocks arrive correctly without error? Does
dividing the file into blocks help?
c. Suppose the propagation delay is negligible, explain how Stop-and-Wait ARQ can
help deliver the file in error-free form. On the average how long does it take to
deliver the file if the ARQ transmits the entire file each time?
.
Câu 11:
In this activity, you are given the network address of 192.168.1.0/24 to subnet and
provide the IP addressing for the Packet Tracer network. Each LAN in the network
requires at least 25 addresses for end devices, the switch and the router. The
connection between R1 to R2 will require an IP address for each end of the link.
a. Based on the topology, how many subnets are needed?
b. How many bits must be borrowed to support the number of subnets in the topology
table?
c. How many subnets does this create?
d. How many usable hosts does this create per subnet?
Note: If your answer is less than the 25 hosts required, then you borrowed too
many bits.
Câu 12:
Five stations (S1-S5) are connected to an extended LAN through transparent bridges (B1B2), as shown in the following figure. Initially, the forwarding tables are empty. Suppose
the following stations transmit frames: S1 transmits to S5, S3 transmit to S2, S4 transmits
to S3, S2 transmits to S1, and S5 transmits to S4. Fill in the forwarding tables with
appropriate entries after the frames have been completely transmitted.
Câu 13:
Consider the network in Figure.
a) Use the Dijkstra algorithm to find the set of shortest paths from node 4 to other
nodes.
Iteration
N
D1 D2 D3 D5 D6
Initial
b) Find the set of associated routing table entries (Destination, Next Hop, Cost)
Destination
D1
D2
D3
D5
D6
Cost
4
1
2
3
3
Next Hop
D2
D2
D3
D5
D3
14)
You are a network technician assigned to install a new network for a customer. You
must create multiple subnets out of the 192.168.1.0/24 network address space to meet
the following requirements:
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The first subnet is the LAN-A network. You need a minimum of 50 host IP
addresses.
The second subnet is the LAN-B network. You need a minimum of 40 host
IP addresses.
You also need at least two additional unused subnets for future network
expansion.
Note: Variable length subnet masks will not be used. All of the device subnet masks
should be the same length.
Answer the following questions to help create a subnetting scheme that meets the stated
network requirements:
a. How many host addresses are needed in the largest required subnet?
b. What is the minimum number of subnets required?
c. The network that you are tasked to subnet is 192.168.1.0/24. What is the /24
subnet mask in binary?
d. The subnet mask is made up of two portions, the network portion, and the host
portion. This is represented in the binary by the ones and the zeros in the subnet
mask.
In the network mask, what do the ones and zeros represent?
e. When you have determined which subnet mask meets all of the stated network
requirements, derive each of the subnets. List the subnets from first to last in the
table. Remember that the first subnet is 192.168.0.0 with the chosen subnet mask.
Câu 15:
Suppose that Selective Repeat ARQ is modified so that ACK messages contain a list of
the next m frames that it expects to receive.
Solutions follow questions:
a. How does the protocol need to be modified to accommodate this change?
b. What is the effect of the change on protocol performance?
Q.16. (2 marks)
Suppose the size of an uncompressed text file is 1 megabyte
Note: Explain your answer in details.
a. How long does it take to download the file over a 32 kilobit/second modem?
b. How long does it take to take to download the file over a 1 megabit/second
modem?
c. Suppose data compression is applied to the text file. How much do the transmission
times in parts (a) and (b) change?
Q17. (2 marks)
Let g(x)=x3+x+1. Consider the information sequence 1001. Find the codeword
corresponding to the preceding information sequence. Using polynomial arithmetic we obtain
Note: Explain your answer in details.
Q.18. (2 marks)
A router has the following CIDR entries in its routing table:
Address/mask Next hop
135.46.56.0/22 Interface 0
135.46.60.0/22 Interface 1
192.53.40.0 /23 Router 1
default Router 2
(a) What does the router do if a packet with an IP address 135.46.63.10 arrives?
(b) What does the router do if a packet with an IP address 135.46.57.14 arrives?
a.
The router will forward the packet with an IP address 135.46.63.10 to interface 1. This is
because the destination IP address matches the first routing table entry, which has a subnet
mask of /22. The subnet mask of /22 means that the first 22 bits of the IP address must
match for the packet to be routed to the interface. The first 22 bits of the IP address
135.46.63.10 match the first 22 bits of the subnet mask 135.46.60.0, so the packet will be
forwarded to interface 1.
b.
The router will forward the packet with an IP address 135.46.57.14 to interface 0. This is
because the destination IP address does not match any of the first two routing table entries.
The default routing table entry will then be used, which routes all packets to Router 2.
Câu 19:
A Large number of consecutive IP address are available starting at 198.16.0.0.
Suppose four organizations, A, B, C, D request 4000, 2000, 4000, and 8000
addresses, respectively. For each of these organizations, give:
1. the first IP address assigned
2. the last IP address assigned
3. the mask in the w.x.y.z/s notation
The start address, the ending address, and the mask are as follows:
Here are the details for each organization:
Organization A
Start address: 198.16.0.0
Last IP address assigned: 198.16.39.255
Mask: 255.255.252.0
The mask of 255.255.252.0 means that the first 23 bits of the IP address must match for the packet
to be routed to organization A. The first 23 bits of the IP address 198.16.0.0 match the first 23 bits
of the subnet mask 255.255.252.0, so all packets with an IP address in the range 198.16.0.0 to
198.16.39.255 will be routed to organization A.
Organization B
Start address: 198.16.40.0
Last IP address assigned: 198.16.63.255
Mask: 255.255.254.0
The mask of 255.255.254.0 means that the first 22 bits of the IP address must match for the packet
to be routed to organization B. The first 22 bits of the IP address 198.16.40.0 match the first 22
bits of the subnet mask 255.255.254.0, so all packets with an IP address in the range 198.16.40.0
to 198.16.63.255 will be routed to organization B.
Organization C
Start address: 198.16.64.0
Last IP address assigned: 198.16.95.255
Mask: 255.255.252.0
The mask of 255.255.252.0 means that the first 23 bits of the IP address must match for the packet
to be routed to organization C. The first 23 bits of the IP address 198.16.64.0 match the first 23
bits of the subnet mask 255.255.252.0, so all packets with an IP address in the range 198.16.64.0
to 198.16.95.255 will be routed to organization C.
Organization D
Start address: 198.16.96.0
Last IP address assigned: 198.16.127.255
Mask: 255.255.255.0
The mask of 255.255.255.0 means that all 32 bits of the IP address must match for the packet to
be routed to organization D. The first 32 bits of the IP address 198.16.96.0 match the first 32 bits
of the subnet mask 255.255.255.0, so all packets with an IP address in the range 198.16.96.0 to
198.16.127.255 will be routed to organization D.
Câu 20:
(2 marks) Suppose an application layer entity wants to send an L-byte message to its
peer process, using an existing TCP connection. The TCP segment consists of the
message plus 20 bytes of header. The segment is encapsulated into an IP packet that has
an additional 20 bytes of header. The IP packet in turn goes inside an Ethernet frame that
has 18 bytes of header and trailer. What percentage of the transmitted bits in the physical
layer correspond to message information, if L = 100 bytes, 500 bytes, 1000 bytes.
Câu 21:
(2 marks) Consider the three-way handshake in TCP connection setup.
(a) Suppose that an old SYN segment from station A arrives at station B, requesting a
TCP connection. Explain how the three-way handshake procedure ensures that the
connection is rejected.
(b) Now suppose that an old SYN segment from station A arrives at station B, followed a
bit later by an old ACK segment from A to a SYN segment from B. Is this connection
Câu 22:
(2 marks) Suppose a header consists of four 16-bit words: (11111111 11111111,
11111111 00000000, 11110000 11110000, 11000000 11000000). Find the Internet
checksum for this code.
The Internet checksum is a 16-bit checksum that is used to verify the integrity of IP
packets. The checksum is calculated by adding the 16-bit words in the IP header, and then
taking the one's complement of the sum.
The header consists of four 16-bit words:
0xFFFF
0x0000
0xAAFF
0xCCCC
The checksum is calculated as follows:
0xFFFF + 0x0000 + 0xAAFF + 0xCCCC = 0x155D
The one's complement of 0x155D is 0xEAAB.
Therefore, the Internet checksum for this code is 0xEAAB.
Note: SV có thể làm cách khác nhưng kế t quả đúng vẫn đươ ̣c tính điể m
Câu 23:
(2 marks)
Consider the 7-bit generator, G=10011, , and suppose that D has the value 1001010101.
What is the value of R? Show your all steps to have result.
Note: Explain your answer in details
The value of R is 1001010010.
Here are the steps:
The generator polynomial is G=10011. This means that the
polynomial that is used to generate the codewords is x^3 +
x + 1.
The dataword is D=1001010101. This means that the
dataword is a binary sequence with 7 bits.
The remainder R is calculated by dividing the dataword D
by the generator polynomial G.
R = D - G * Quotient
The quotient is the number of times that the generator
polynomial G divides evenly into the dataword D. The
remainder is the remainder that is left after the division.
In this case, the quotient is 1. This means that the
generator polynomial G divides evenly into the dataword D
once. The remainder is therefore the last 3 bits of the
dataword D, which is 1010.
Therefore, the value of R is 1001010010.
Câu 24:
(2 marks)
Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a
direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 108
meters/sec.
a. Calculate the bandwidth-delay product, R _ dprop.
b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent
continuously as one large message. What is the maximum number of bits that will be in
the link at any given time?
Note: Explain your answer in details
a.
Calculate the bandwidth-delay product, R _ dprop.
The bandwidth-delay product is calculated as the product of the link capacity (R) and
the propagation delay (dprop). In this case, we have:
Bandwidth-delay product = R * dprop = 2 Mbps * (20,000 km / 2.5 x 10^8 m/s) =
500,000,000 bits
b.
Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is
sent continuously as one large message. What is the maximum number of bits that will
be in the link at any given time?
The maximum number of bits that will be in the link at any given time is the sum of the
file size (800,000 bits) and the bandwidth-delay product (500,000,000 bits). This is
because the file is being sent continuously as one large message, so there will be a delay
between the time the first bit is sent and the time the last bit is received. During this
delay, the link will be filled with the file data.
Maximum number of bits in link = 800,000 bits + 500,000,000 bits = 500,800,000 bits
In other words, at any given time, there will be at most 500,800,000 bits in the link. This
is the maximum amount of data that can be transmitted by the sender before waiting for
acknowledgment.
Note: The bandwidth-delay product is an important concept in networking because it
limits the maximum throughput of a link. If the file size is larger than the bandwidthdelay product, then the sender will have to stop and wait for acknowledgments before
sending more data. This can lead to a decrease in throughput and an increase in
latency.
Note:
Students have to follow the steps and complete the tasks in details in order to
have the results. If the students only write the result, that is, that result is not
marked or recorded.
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Students do examination on word file and answer by English