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PE NWC203c by PH - abc
Networking (Trường Đại học FPT)
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PE NWC203c
NWC203c Shape Spring 2022
Q1: Explain the difference between connectionless unacknowledged service and
connectionless acknowledged service. How do the protocols that provide these services
differ?
[Answer]
Connection Less:
Connectionless service comes with a single free-standing data unit for all transmissions.
→ In this, each unit contains all of the protocols that control information necessary for
delivery perspective, but this also contains no provision for sequencing or flow control.
+ Acknowledged:
This is achieved by the use of ACK and NAK control messages.
→ These types of protocols are well suited for communication over the network,
where high layers are very sensitive to loss and can have a significant probability
of error in these underlying networks.
Example: HDLC, which offers for unnumbered acknowledgment service(setup and release).
+ Unacknowledge:
→ This comes with a very simpler version and provides faster communication for
networks, which are inherently reliable or provide service to a higher layer, that
can tolerate loss in the information, or which has built-in error control/recovery
feature.
=========================================================================
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Q2. Explain the difference between connection-oriented acknowledged service and
connectionless acknowledged service. How do the protocols that provide these services differ?
[Answer]
Less & Oriented:
+ Connection-oriented:
→ In this type of service, a setup phase will be initialized between sender and
receiver, to establish a context for transferring the information
→ This connection is provided to the sender for all SDUs.
→ This service requires a stateful protocol, which is used to keep track of
sequence numbers, and timers.
+ ConnectionLess:
→ Here, there will be no prior context provided for transferring the information
between sender and receiver.
→ The sender will pass its SDU to an underlying layer without any notice.
→ And in this, the sender requires an acknowledgment of SDU delivery.
→ The protocols are very different in these services
→ this service also does not require transmitting protocols to track the
acknowledgment of PDU.
→ After receiving the PDU, the receiver needs to send acknowledgment, If not
received
in time, then it will return failure.
=========================================================================
Q3: Explain the differences between PPP and HDLC.
[Answer]
HDLC is a short form of High-level Data Link Control that does the data encapsulation.
PPP is an acronym for Point-to-Point Protocol that can be used by different devices without
any data format change.
A few major differences are as below:
+ For communication through HDLC, a bit-oriented protocol is used for point-to-point
links as well as for multipoint link channels. However, PPP uses a byte-oriented protocol
for point-to-point links at the time of communication.
+ HDLC does the encapsulation for synchronous media only whereas PPP can do the
encapsulation for synchronous as well as for asynchronous media.
+ HDLC can be used only for CISCO devices whereas PPP can be easily used for other
devices.
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Q4: 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.
[Answer]
Go-Back-N
Selective Repeat
Maximum Send Window Size in
Default HDLC Frame
Maximum Send Window Size in
Extended HDLC Frame
7
4
127
64
The round trip propagation delay
2tprop = 2 = 2 = 2.5s
We have: =
Go-Back-N:
If N = 7 → = 2.5 s → = 535715 bits
If N = 127 → = 2.5 s → = 29528 bits
Selective Repeat:
If N = 4 → = 2.5 s → = 973500 bits
If N = 64 → = 2.5 s → = 58594 bits
=========================================================================
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Q5: 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?
[Answer]
R = 1.5 * 106 bps, and nf =2000 bits.
The distance that the information must travel is the earth to satellite distance → d ≈ 36,000 km.
The speed of light c: 3 x 108.
Tprop = = = 120 ms
Tf = = = 1.33 ms
We can use either Go-Back-N or Selective Repeat ARQ.
The default window size is N = 7
The maximum information rate is achieved with no error, and hence, no retransmission.
Tcycle = Tf + 2Tprop = 1.33 + 2 * 120 = 241.33 ms
n = N * nf = 7 * 2000 = 14,000 bits
Rmax = = = 58 kbps
If the extended sequence numbering option (7-bit) is used,
the maximum send window size would be N = 127 → n = 127 * 2000 = 254000 bits
Rmax = = = 1.052 Mbps
=========================================================================
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Q6: 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.
[Answer]
Find out the probability of the k packets that have reached the T- second.
N=60 and p=0.1
The average number for the arrivals of the packets can be given as Np = 6
The average number of packets received through the first line as below:
→ The average number of packets received through the first line = 4.59 packets
The average number of packets transmitted through the second line per T second can be
obtained as below:
The average number of packets transmitted through the second line = 6 - 4.59 = 1.41 packets
Therefore, it will transmit 1.41 packets on average per T second from the second line.
=========================================================================
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Q7: 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.
[Answer]
a,
Message size is important because in real-time signals of voice it is necessary to transfer
a fixed packet size of that holds no more than 20ms of the speech signal. The handling of
arbitrary message size is not as important as long as the desired packet size for voice can be
handled.
Sequencing is essential because each packet needs to arrive in the same sequence that it
was generated. Reliability is moderately important since voice transmission can tolerate a
certain level of loss and error.
Pacing and flow control are not as important because the synchronous nature of the voice
signal implies that the end systems will be matched in speed.
Timing, for real-time voice transfer, is important because this adaptation function helps
to control the jitter in the delivered signal.
Addressing is only during the connection setup phase if we assume some form of virtual
circuit packet switching method.
Privacy, integrity, and authentication have traditionally not been as important as the other
issues discussed above.
b,
If the underlying network is reliable then the end-to-end approach is better because the
probability of error is very low so processing at the edge suffices to provide acceptable
performance.
If the underlying network is unreliable then the hop-by-hop approach may be required.
For example, if the probability of error is very high, as in a wireless channel, then error
recovery at each hop may be necessary to make effective communication possible.
=========================================================================
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Q8: 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?
[Answer]
a,
The sender in the stop-and-wait protocol described in the chapter retransmits a frame
when an acknowledgment is not received in time. The modified protocol says that the frame is
retransmitted every time the sender or receiver sees an error.
Therefore, the only difference is that frames are retransmitted sooner. So, the protocol
will work correctly.
b, No. The state transition diagram will stay the same.
c, The error recovery process will be faster with this modified protocol.
=========================================================================
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Q9: 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.
[Answer]
Bytes each to be transmitted in several PDUs in order to exchange messages of any size.
A single PDU must include all small messages.
Peer processes must exchange information that permits messages to be reassembled at
the recipient. The message length, for example, could be included in the first PDU. A message
end-of-message marker could be included in the last PDU. In connection-oriented networks,
sequence numbers can be used to detect loss, while in connectionless networks, they can be
used to aid in message reconstruction. Finally, because variable-size PDUs are allowed, the
PDU size must be specified in the PDU header.
In this instance, each PDU must be identified with a stream ID in addition to all of the
header information specified in (b), so that the receiver may treat each stream separately while
reassembling messages.
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.
[Answer BY PH]
a. Develop an approach that allows the peer processes to exchange messages of arbitrary
size.
To exchange messages of arbitrary size, large messages must be segmented into parts of M-H
bytes each in length to be transmitted in multiple PDUs. Small messages must be placed in a
single PDU.
b. What essential control information needs to be exchanged between the peer processes?
The peer processes need to communicate information that allows for the reassembly of
messages at the receiver. For example, the first PDU may contain the message length. The last
PDU may contain and end-of-message marker. Sequence numbers may also be useful to detect
loss in connection oriented networks and to help in reconstruction of the messages in
connectionless networks. Lastly, since variable size PDUs are permitted, the size of the PDU
must be transmitted in the PDU header.
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?
In this case, in addition to all of the header information mentioned in b), each PDU must be
labeled with a stream ID, so that the receiver can treat each stream independently when
reassembling messages.
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Q10: 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 errorsWe conclude
that it is extremely unlikely that the file will arrive error free.
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?
[Answer]
a,
Note for n large and p very small, (1 − p)n ≈ e − np.
P[no error in the entire file] = (1 – p)n ≈ e – np
for n > 1, p < 1 = e-8 = 3.35 * 10-4
We conclude that it is extremely unlikely that the file will arrive error free.
b,
A subblock of length is received without error with probability:
P[no error in subblock] =
A block has no errors if all subblocks have no errors
So P[no error in block] = P[no errors in subblock]N = = (1 – p)n
So simply dividing the blocks does not help.
c, We assume the following:
t0 = tf + tACK = +
(tprop ≈ 0).
P[nt = i ] = P[one success after i – 1 failure] = (1 – Pf) Pf i – 1
ttotal | i transmissions = i * t0
E[ttotal] =
Here, nf = n → na , Thus t0 ≈ tf =
E[ttotal] = = = = 23881s = 6.633 hours!
The file gets through, but only after many retransmissions.
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Q11:
In this activity, you are given the network address of 192.168.100.0/24 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?
[Answer]
a, Based on the topology, how many subnets are needed?
→5
b, How many bits must be borrowed to support the number of subnets in the topology table?
→3
c, How many subnets does this create?
→8
d, How many usable hosts does this create per subnet?
→ 25 – 2 = 30
=========================================================================
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Q12: Five stations (S1-S5) are connected to an extended LAN through transparent bridges
(B1-B2), 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.
[Answer]
Address
S1
S3
S4
S2
S5
Port
1
2
2
1
2
Addres
s
S5
S2
S3
S1
S4
Port
1
2
2
1
2
=========================================================================
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Q13: 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)
Destinatio
n
Cost
Next Hop
[Answer]
Iteration
Initial
1
2
3
4
5
N
{4}
{ 2, 4 }
{ 2, 3, 4 }
{ 2, 3, 4, 5 }
{ 2, 3, 4, 5, 6 }
{1, 2, 3, 4, 5, 6}
D1
5
4
4
4
4
4
D2
1
1
1
1
1
1
D3
2
2
2
2
2
2
D5
3
3
3
3
3
3
D6
∞
3
3
3
3
3
b, Find the set of associated routing table entries (Destination, Next Hop, Cost)
Destinatio
n
1
2
3
5
6
Cost
Next Hop
2
2
3
5
3
5
1
2
3
3
=========================================================================
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Q14:
You are a network technician assigned to install a new network for a customer. You must
create multiple subnets out of the 192.168.12.0/24 network address space to meet the following
requirements:
-
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 192.168.12.0/24. 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 192.168.12.0 with the chosen subnet mask.
Subnet
Address
Prefix
Subnet Mask
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[Answer]
a, How many host addresses are needed in the largest required subnet?
NNumbeof subnets required = 4 which implies that the network is divided into four parts
The given IP address i.e 192.168.12.0 is a Class C IP address.
Number of host addresses required = 26 - 2 = 62
→ Therefore the max number of hosts possible in each subnet is 62 .
b, What is the minimum number of subnets required?
Two subnet are required for LAN-A and LAN-B and two subnets are needed to be left for
future use.
→ Therefor the total number of subnets are 4 .
c, The network that you are tasked to subnet is 192.168.12.0/24. What is the /24 subnet mask in
binary?
Subnet mask for any network is obtained by changing the net id bits to 1's and host id
bits to 0's.
Since the given network is a Class C network so the number of netid bits are 24 and the
number of host id bits are 8 and there are two bits reserved for subnet id bits.
→ The subnet mask for network in binary: 11111111.11111111.11111111.11000000.
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.
1.) In the network mask, what do the ones and zeros represent?
→ In the nerwork maslk the ones represent the net id bits and the zeroes represent the
host id bits.
2.) 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 192.168.12.0 with the chosen subnet mask.
Subnets for LAN-A are :
1. 192.168.12.00000000
2. 192.168.12.00000001
3. 192.168.12.00000010
4. 192.168.12.00000011
Last network for the subnet will be 192.168.12.00111111
Subnets for LAN-B are :
1. 192.168.12.01000000
2. 192.168.12.01000001
3. 192.168.12.01000010
4. 192.168.12.01000011
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......
Last network for the subnet will be 192.168.12.01111111
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 192.168.12.0 with the chosen subnet mask.
Subnet address
Prefix
Subnet Mark
192.168.12.0
/26
255.255.255.192
192.168.12.64
/26
255.255.255.192
192.168.12.128
/26
255.255.255.192
192.168.12.192
/26
255.255.255.192
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Q15: 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?
[Answer]
a) 2 things are needed to be changed:
The frame header needs to be modified to recieve the list of frames and Since the
reciever explicitly indicates which frames are needed to be transmitted.
Change in transmitter operation is needed. If the recieved list contains m oldest frames
that are yet to be recieved , then it can be used to skip retransmission of frames that have
already been received.
b)
Performance will surely increase if the error rate is high or delay is high. A single frame
can ask for the retransmission of several frames.
The complexity of the protocol will surely increase relative to the unchanged Selective
repeat ARQ
Other two subnets are :
Starting address of third subnet is : 192.168.12.10000000
Last address of the third subnet is : 192.168.12.10111111
Starting address of forth subnet is : 192.168.12.11000000
Last address of the forth subnet is : 192.168.12.11111111
=========================================================================
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More:
0
1
2
3
4
5
6
U
U,T
U,T,W
UTWV
UTWVX
UTWVXY
U
0,U
-
W
~
3,U
3,U
-
V
~
3,U
3,U
3,U
-
T
~
2,U
-
Y
~
~
9,T
9,T
9,T
9,T
-
X
~
~
~
9,W
6,V
-
Z
~
~
~
~
~
14,X
14,X
=> Z -> X -> V -> U
A
A, C
A,C,B
ACBD
ACBDE
ACBDEG
ACBDEGF
A
0,A
-
B
~
4,A
4,A
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C
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~
~
~
9,B
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E
~
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10,C
10,C
10,C
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F
~
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14,D
14,D
14,D
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G
~
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11, G
11,G
-
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z
~
~
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15, G
15,G
step
0
1
2
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4
5
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lOMoARcPSD|28579846
Các Dạng của Các Kỳ Trước:
1. Let g1(x) = x3 + x + 1 and let g2(x) = x3 + x2 + 1. Consider the information bits
(1,1,0,1,1,1).
a. Find the codeword corresponding to these information bits if g1(x) is used as the
generating polynomial.
b. Find the codeword corresponding to these information bits if g2(x) is used as the
generating polynomial.
[Answer]
a,
G1(x) is used as the generating polynomial
G1(x) = x + 1 = 11 → n = 2
Step 1: Add n-1 bit 0 to the information bits → Information bits: 1101110
Step 2:
1101110
11
11
100101
00011
11
0010
11
01
R(x) = 01
Q(x) = 100101
Step 3: Add R(x) to the information bits → Codeword: 1101111
b,
G2(x) is used as the generating polynomial
G2(x) = x3 + x2 + 1 = 1101 → n=4
Step 01: Add n-1 bit 0 to the information bits → Information bits: 110111000
Step 02:
1101111000
1101
1101
100010
00001100
1101
00010
R(x) = 010
Q(x) = 100010
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Step 3: Add R(x) to the information bits → Codeword: 110111010
2. Let g1(x) = x + 1 and let g2(x) = x3 + x2 + 1. Consider the information bits (1,1,0,1,1,0).
a. Find the codeword corresponding to these information bits if g1(x) is used as the
generating polynomial.
b. Find the codeword corresponding to these information bits if g2(x) is used as the
generating polynomial.
[Answer]
a,
G1(x) is used as the generating polynomial
G1(x) = x + 1 = 11 → n=2
Step 01: Add n-1 bits 0 to the information bits → Information bits: 1101100
Step 02:
1101100
11
11
100100
00011
11
0000
00
0
R(x) = 0
Q(x) = 100100
Step 03: Add R(x) to the information bits → Codeword: 1101100
b,
G2(x) is used as the generating polynomial
G2(x) = x3 + x2 + 1 = 1101 → n = 4
Step 01: Add n-1 bits 0 to the information bits → Information bits: 110110000
Step 02:
110110000 1101
1101
100011
00001000
1101
01010
1101
0111
R(x) = 111
Q(x) = 100011
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Step 03: Add R(x) to the information bits → Codeword: 110110111
2. Let g1(x) = x + 1 and let g2(x) = x3 + x2 + 1. Consider the information bits (1,1,0,1,1,0).
a. Find the codeword corresponding to these information bits if g1(x) is used as the
generating polynomial.
b. Find the codeword corresponding to these information bits if g2(x) is used as the
generating polynomial.
[Answer]
a,
G1(x) is used as the generating polynomial
G1(x) = x + 1 = 11 → n = 2
Step 01: Add n-1 bits 0 to the information bits → Information bits: 1101100
Step 02:
1101100
11
11
100100
00011
11
0000
00
0
R(x) = 0
Q(x) = 100100
Step 03: Add R(x) to the information bits → Codeword: 1101100
b,
G2(x) is used as the generating polynomial
G2(x) = x3 + x2 + 1 = 1101 → n=4
Step 01: Add n-1 bits 0 to the information bits → Information bits: 110110000
Step 02:
110110000 1101
1101
100011
00001000
1101
01010
1101
0111
R(x) = 111
Q(x) = 100011
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Step 03: Add R(x) to the information bits → Codeword: 110110111
3. Consider the 7-bit generator, G=10111, and suppose that D has the value 1010100001.
What is the value of R? Show your all steps to have result.
[Answer]
G = 10111 → The polynomial expression of G: x4 + x2 + x + 1
D = 1010100001
Here, the degree of the expression is 4 → r = 4.
Thus, D + r → 10101000010000
10101000010000 10111
10111
1001011001
00010000
10111
0011101
10111
010100
10111
00011000
10111
01111
Therefore, the value of R = 1111
===================================================================
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4. Consider the 7-bit generator, G=10011, and suppose that D has the value 1010101010.
What is the value of R? Show your all steps to have result.
[Answer]
G = 10011 → The polynomial expression of G: x4 + x + 1
D = 1010101010
Here, the degree of the expression is 4 → r = 4.
Thus, D + r → 10101010100000
10101010100000 10011
10011
1011011100
0011001
10011
010100
10011
0011110
10011
011010
10011
010010
10011
0000100
Therefore, the value of R = 0100
===================================================================
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8. 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
[Answer]
G(x)=x3+x+1 = 1011
The information sequence: 1001
Here, the degree of the expression is 3 → r = 3.
Thus, the information sequence → 1001000
Using polynomial arithmetic we obtain:
1001000
1011
1011
101
001000
1011
00110
→ Codeword: 1001110
====================================================================
9. Let g(x)=x3+x+1. Consider the information sequence 1011. Find the codeword
corresponding to the preceding information sequence. Using polynomial arithmetic we
obtain
[Answer]
G(x) = x3+x+1 = 1011
The information sequence: 1011
Here, the degree of the expression is 3 → r = 3.
Thus, the information sequence → 1011000
Using polynomial arithmetic we obtain:
1011000
1011
1011
1000
0000000
→ Codeword: 1011000
====================================================================
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14. Sender A wants to send 100111010011110 to receiver B. This transmission uses CRC
algorithm for error detection with generator polynomial bits string is 10110. What is bits
string will be transmitted on the medium. Show your all steps to have result
[Answer]
D: 100111010011110
Generator polynomial: 10110 → n = 5
Step 1: Add n-1 bits 0 to the bit stream → D: 1001110100111100000
Step 2:
1001110100111100000
10110
10110
101000010100010
0010110
10110
0000010011
10110
0010111
10110
000010000
10110
001100
R(x) = 1100
Step 3: Add R(x) to the information bits → Codeword: 1001110100111101100
Bits string will be transmitted on the medium is: 1001110100111101100
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15. A bit stream 10011101 is transmitted using the standard CRC method. The generator
polynomial is x3 + 1. Show the actual bit string transmitted. Suppose the third bit from the
left is inverted during transmission. Show that this error is detected at the receivers end.
[Answer]
Bit stream: 10011101
Generator polynomial: x3+ 1 = 1001 → n = 4
Step 1: Add n-1 bits 0 to the bit stream → bit stream: 10011101000
Step 2:
10011101000
1001
1001
10001100
00001101
1001
01000
1001
000100 (Remainder)
R(x) = 100
Step 3: Actual frame transmitted: 10011101000 – 100 = 10011101100 (modulo 2 subtraction)
Now suppose the third bit from the left is garbled and frame is received as 10111101100.
Hence on dividing this by the polynomial generator we get a remainder of 100 which shows
that an error has occurred. Had the received frame been error free we would have got a
remainder of zero.
10111101100
1001
1001
10101000
001011
1001
001001
1001
0000100
→ Remainder indicating error = 100
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Sender A wants to send 100111010011011 to receiver B. This transmission uses CRC
algorithm for error detection with generator polynomial bits string is 10111. What is bits
string will be transmitted on the medium. Show your all steps to have result.
[Answer]
Generator polynomial is 10111 = x4 + x2 + x1 + x0 → r = 4
Step 1: Add 0000 to data bits string. It will be 1001110100110110000
Step 2:
1001110100110110000
10111
10111
101010001000110
0010010
10111
0010110
10111
000010110
10111
000011100
10111
010110
10111
000010
R(x) = 0010
Step 03: add R(x) to data bits string → bits string: 1001110100110110010
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5. Consider the following network Figure 1. With the indicated link costs, use Dijkstra’s
shortest-path algorithm to compute the shortest path from u to all network nodes. Show
how the algorithm works by computing a table.
[Answer]
u
u, x
u, x, v
u, x, v, y
u, x, v, y, w
u
-
v
2, u
2, u
2,U
-
w
5, u
4, x
4, x
4, x
-
x
1, x
-
U -> X
y
~
2, x
-
z
~
~
5, y
5, y
5, y
-> Y -> Z
====================================================================
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6. A router has the following CIDR entries in its routing table:
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?
[Answer]
a, The bit pattern of 135.46.63.10 is: 10000111.00101110.00111111.00001010
Taking the first 22 bits of the above IP as network address (it is equivalent of making the last
10 bit zero), we have: 10000111.00101110.00111100.00000000 = 135.46.60.10
→ It matches the network address of the 2rd row. The packet will be forwarded to
interface 1.
b, The bit pattern of 135.46.57.14 is 10000111.00101110.00111001.00001110
Taking the first 22 bits of the above IP as network address (it is equivalent making the last
10 bits zero), we have 10000111.00101110.00111000.00000000 = 135.46.56.0
→ It matches the network address of the first row. The packet will be forwarded to interface
0.
c, The bit pattern of 135.46.52.2 is 10000111.00101110.00111001.00001110
Taking the first 22 bits of the above IP address as network address, we have 135.45.52.0.
It dose not matche the network addresses of the first three rows.
→ The packet will be forwarded to default gateway which is Router 2.
d, The bit pattern of 192.53.40.7 is 11000000.00110101.00101000.00000111
Taking the first 23 bits of the above IP address as network address, we have 192.53.40.0.
→ It matches the network address of the third row. The packet will be forwarded to
Router 1.
e, The bit pattern of 192.53.56.7 is 11000000.00110101.00111000. 00000111
Taking the first 23 bits of the above IP address as network address, we have 192.53.56.0.
→ It does not matche the network addresses of the first three rows. The packet will be
forwarded to default gateway which is Router 2.
====================================================================
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7. Suppose two hosts, A and B, are separated by 30,000 kilometers and are connected by a
direct link of R = 3 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 900,000 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?
[Answer]
a.
The distance between two hosts A and B = 30,000 km = 3 * 107 m
Transmission rate(R) of the direct link between A and B = 3Mbps = 3 * 106 bps
Propagation Speed(S) of the link between A and B = 2.5 * 108 meters/sec
Propagation delay: dprog =
band-width delay product: R * dprog = 3 * 106 * 0.12 = 360000 bits
Therefore, band-with delay product is: 360000 bits
b.
Size of the file = 900 000 bits = 9 * 105 bits
Transmission rate(R) of the direct link between A and B = 3Mbps = 3 * 106 bps
The band-width delay product: R * dprog = 3 * 106 * 0.12 = 360000 bits
Therefore, the maximum number of bits at a given time will be 360000 bits
====================================================================
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10. A packet switch receives a packet and determines the outbound link to which the
packet should be forwarded. When the packet arrives, one other packet is halfway done
being transmitted on this outbound link and four other packets are waiting to be
transmitted. Packets are transmitted in order of arrival.
Suppose all packets are 2,500 bytes and the link rate is 3 Mbps. What is the queuing delay
for the packet? More generally, what is the queuing delay when all packets have length L,
the transmission rate is R, x bits of the currently-being-transmitted packet have been
transmitted, and n packets are already in the queue?
[Answer]
Packet length = L
Transmission rate = R
Currently transmitted packet = x bits
Waiting queue = n packets
Formula for Queuing delay:
Queuing delay =
Given data:
L = 2500 bytes = 8 * 2500 = 20000 bits
R = 3 Mbps. = 3 * 106 bps/s
X = bytes = 8 * 1250 = 10000 bits
n=4
Queuing delay =
Thus, the queuing delay: 0.03s
====================================================================
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11. Suppose a header consists of four 16-bit words: (11111111 11111110,
11111111 00000000, 11110000 11110000, 11000000 11000001). Find the Internet checksum
for this code
[Answer]
A0 = 11111111 11111110 = 215 → 21 = 65534
A1 = 11111111 00000000 = 215 → 28 = 65280
A2 = 11110000 11110000 = 215 + 214 + 213 + 212 + 27 +26 + 25 + 24 = 61680
A3 = 11000000 11000001 = 215 + 214 + 27 + 26 + 20 = 49345
X = (A0 + A1 + A2 + A3) % 65534 = 241839 % 65534 = 45237
A4 = -X % 65534 = -45237 % 65534 = 20297
So, 20297 is 0100111101001001 binary
→ The Internet checksum = 01001111 01001001
====================================================================
8. Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111 00000000,
11110000 11110000, 11000000 11000000). Find the Internet checksum for this code
[Answer]
A0 = 11111111 11111111 = 215 → 20 = 65535
A1 = 11111111 00000000 = 215 → 28 = 65280
A2 = 11110000 11110000 = 215 + 214 + 213 + 212 + 27 +26 + 25 + 24 = 61680
A3 = 11000000 11000000 = 215 + 214 + 27 + 26 = 49344
X = (A0 + A1 + A2 + A3) % 65535 = 241839 % 65535 = 45234
A4 = -x % 65535 = -45234 % 65535 = 20301
So, 20301 is 0100111101001101 binary
→ The Internet checksum = 01001111 01001101
====================================================================
8. Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111 00000000,
11110000 11110000, 11000000 11000001). Find the Internet checksum for this code
[Answer]
A0 = 11111111 11111111 = 215 → 20 = 65535
A1 = 11111111 00000000 = 215 → 28 = 65280
A2 = 11110000 11110000 = 215 + 214 + 213 + 212 + 27 +26 + 25 + 24 = 61680
A3 = 11000000 11000001 = 215 + 214 + 27 + 26 + 20 = 49345
X = (A0 + A1 + A2 + A3) % 65535 = 241840 % 65535 = 45235
A4 = -x % 65535 = -45234 % 65535 = 20300
So, 20300 is 0100111101001100 binary
→ The Internet checksum = 100111101001100
9. Consider a packet of length 2,000 bytes that propagates over a link of distance 3,500 km
with propagation speed of 2,5 · 108 m/s, and transmission rate 2 Mbps?
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a. How long does the packet propagation take?
b. Does this propagation delay depend on the packet length?
c. Does this propagation delay depend on the transmission rate?
[Answer]
Length: 2000 bytes = 16000 bits
Distance: 3,500 km = 35 * 105 m
Propagation Speed: 2,5 * 108 m/s
Rate: 2Mbps = 2 * 106 bit/s
a.
Transmission delay = = = 8 ms
Propagation delay = = = 14 ms
Therefore, the total time = 8ms + 14ms = 22ms
b. No, the propagation delay only depend on D,S not L,R → the delay depend on packet length
is not true.
c. No, the propagation delay only depend on D,S not L,R → the delay depends on transmission
rate is not true.
====================================================================
9. Consider a packet of length 1,000 bytes that propagates over a link of distance 2,500 km
with propagation speed of 2,5 · 108 m/s, and transmission rate 2 Mbps?
a. How long does the packet propagation take?
b. Does this propagation delay depend on the packet length?
c. Does this propagation delay depend on the transmission rate?
[Answer]
a.
Length: 1000 bytes = 8000 bits
Distance: 2,500 km = 25 * 105 m
Propagation Speed: 2,5 * 108 m/s
Rate: 2Mbps = 2 * 106 bit/s
Transmission delay = = = 4 ms
Propagation delay = = = 10 ms
Therefore, the total time = 4ms + 10ms =14ms
b. No, the propagation delay only depend on D,S not L,R → the delay depend on packet length
is not true.
c. No, the propagation delay only depend on D,S not L,R → the delay depends on transmission
rate is not true.
10. Suppose Host A wants to send a large file to Host B. The path from Host A to Host B
has three links, of rates R1 = 250 kbps, R2 = 3 Mbps, and R3 = 2 Mbps.
a. Assuming no other traffic in the network, what is the throughput for the file transfer?
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b. Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how
long will it take to transfer the file to Host B?
[Answer]
a, Consider given data:
R1 = 250 kbps
R2 = 3 Mbps
R3 = 2 Mbps
The throughput for the file transfer = min {R1, R2, R3}
min {250 kbps, 3 Mbps, 2 Mbps} = 250 kbps
Thus, the throughput for the file transfer = 250 kbps
b, Consider given data:
The file size = 4 million bytes = 32000000 bits
- From (a) Throughput for the file transfer=250 Kbps = 250 000 bps
- Time to transfer the file to Host B = = 128 s
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10. Suppose Host A wants to send a large file to Host B. The path from Host A to Host B
has three links of rates R1 = 500 kbps, R2 = 2 Mbps, and R3 = 1 Mbps.
a. Assuming no other traffic in the network, what is the throughput for the file transfer?
b. Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how
long will it take to transfer the file to Host B?
c. Repeat (a) and (b), but now with R2 reduced to 100 kbps.
[Answer]
a) Consider given data:
R1 = 500 kbps
R2 = 2 Mbps,
R3 = 1 Mbps
The throughput for the file transfer = min {R1, R2, R3}
min {500 kbps, 2 Mbps, 1 Mbps} = 500 kbps
So, the throughput for the file transfer = 500 kbps
b) Consider given data:
The file size = 4 million bytes = 32000000 bits
- From (a), Throughput for the file transfer = 500 Kbps = 500000 bps
- Time to transfer the file to Host B = = 64 s
c) Repeat (a) and (b), but now with R2 reduced to 100 kbps.
Consider given data: R1 = 500 kbps, R2=100 kbps, and R3 = 1 Mbps
The throughput for the file transfer = min {R1, R2, R3}
min {500 kbps, 100 kbps, 1 Mbps} = 100 kbps
So, the throughput for the file transfer = 100 kbps
- From (a), Throughput for the file transfer = 100 Kbps = 100000 bps
- Time to transfer the file to Host B = = 320 s
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11. 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 = 200 bytes, 1000 bytes, 2000 bytes.
[Answer]
TCP/IP over Ethernet allows data frames with a payload size up to 1460 bytes.
Therefore, L = 200, L=1000 and L= 2000 are within this limit
The message overhead includes:
• TCP: 20 bytes of header
• IP: 20 bytes of header
• Ethernet: total 18 bytes of header and trailer.
Therefore, the total message overhead is 58 bytes
Therefore
L = 200 bytes,
( ) * 100 = 77,5%
L = 1000 byte,
() * 100 = 94.5%
L = 2000 bytes,
() * 100 = 97,1%
====================================================================
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11. 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
[Answer]
TCP/IP over Ethernet allows data frames with a payload size up to 1460 bytes.
Therefore, L = 100, 500 and 1000 bytes are within this limit.
The message overhead includes:
• TCP: 20 bytes of header
• IP: 20 bytes of header
• Ethernet: total 18 bytes of header and trailer
The total message overhead is 58 bytes
Therefore:
L = 100 bytes,
( ) * 100 = 63,3% efficiency
L = 500 byte,
( ) * 100 = 89,6% efficiency
L = 1000 bytes,
( ) * 100 = 94,5% efficiency
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12. Suppose the size of an uncompressed text file is 1 megabyte
a. How long does it take to download the file over a 35 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?
If we assume a maximum compression ratio of 1:6, then we have the following times for
the 35 kilobit and 1 megabit lines respectively:
[Answer]
a,
Size text file = 1 megabyte = 1 *1024 * 1024 * 8 = 8388608 bits
Speed = 35 kilobit/second = 35 * 1000 = 35000 bits/s
Time to download the file over = = 239.67 s
b,
Size text file = 1 megabyte = 1 *1024 * 1024 * 8 =8388608 bits
Speed = 1 megabit/second = 1000000 bits/s
Time to download the file over = = 8.38 s
c,
T(35k) = = 39.9 s
T(1M) = = 1.39 s
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12. Suppose the size of an uncompressed text file is 1 megabyte
a. How long does it take to download the file over a 35 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?
If we assume a maximum compression ratio of 1:6, then we have the following times for
the 35 kilobit and 1 megabit lines respectively:
[Answer]
a,
Size text file = 1 megabyte = 1 *1024 * 1024 * 8 = 8388608 bits
Speed = 35 kilobit/second = 35 * 1000 = 35000 bits/s
Time to download the file over = = 239.67 s
b,
Size text file = 1 megabyte = 1 *1024 * 1024 * 8 =8388608 bits
Speed = 1 megabit/second = 1000000 bits/s
Time to download the file over = = 8.38 s
c,
T(35k) = = 39.9 s
T(1M) = = 1.39 s
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13. 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
request also rejected?
[Answer]
a. In a three-way handshake procedure, one must ensure the selection of the initial sequence
number is always unique. If station B receives an old SYN segment from A, B will
acknowledge the request based on the old sequence number. When A receives the
acknowledgment segment from B, A will find out that B received a wrong sequence number. A
will discard the acknowledgment packet and reset the connection
b. If an old SYN segment from A arrives at B, followed by an old ACK segment from A to a
SYN segment from B, the connection will also be rejected. Initially, when B receives an old
SYN segment, B will send a SYN segment with its own distinct sequence number set by itself.
If B receives the old ACK from A, B will notify A that the connection is invalid since the old
ACK sequence number does not match the sequence number previously defined by B.
Therefore, the connection is rejected
====================================================================
A university has 150 LANs with 100 hosts in each LAN.
(a) Suppose the university has one Class B address. Design an appropriate subnet
addressing scheme.
(b) Design an appropriate CIDR addressing scheme.
[Answer]
a) Class B's address has 14 bits for the network ID and 16 bits for the host ID. We need to
decide how many bits to allocate to the subnet id versus the host id for designing an appropriate
subnet addressing scheme. We can choose either 7 bits or 8 bits to identify the hosts.ideTCP
b) The CIDR addressing scheme involves generating a prefix length that indicates the length of
the network mask.
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Suppose that a group of computers is connected to an Ethernet LAN. If the computers
communicate only with each other, does it make sense to use IP protocol in the
computers? Should the computers run TCP directly over Ethernet? How is addressing
handled?
[Answer]
IP convention can be utilized since the IP convention is a lot of necessities for tending to also as
directing information on the Internet. Since the computer can't run principles-based TCP
without IP, TCP uses IP addresses. There is a need to utilize an individual custom streaming
convention. that depends on Ethernet or another link layer as a lower layer.
===================================================================
(1) The figures below show the TCP/UDP communication pattern diagrams. Which
diagram works for TCP? Why?
(2) Fill the missing steps (blank boxes) in both diagrams for TCP/UDP correspondingly.
[Answer]
1) Diagram (b) will work for TCP since it is a network protocol that shows the details of how
data is sent as well as received.
2) Missing steps: (a) bind, connect, (b) bind>listen>accept, connect.
====================================================================
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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?
[Answer]
a) To convert arbitrary sizes, large contents must be split into bytes of each length that will be
transmitted in multiple PDUs.
b) Peer-to-peer processes move information allowing messages to be aggregated.
c) A PDU must be attached to the sequence ID, to process each sequence independently when a
message is selected. This stream ID can be dodged if the source and target work for them to
process a content transfer at a certain time.
====================================================================
Consider a network link that has distance of 100 meters, and signal traverses at the speed
of light in cable 2.5 x 10^8 meters per second. The link has transmission bandwidth of 100
megabits/second (100 x 10^6 bits per second). The packet size is 400 bits. What is the
signal propagation delay?
[Answer]
d = 100 megabits / second
(Cách đổi đơn vị: 100 megabits / second = 100*103 kilobits / seconds, đổi ngược lại thì chia cho
1000)
v = 2.5 * 108 meters / second
Signal Propagation = = = 4 *10-7 s
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Consider a network link that has distance of 100 meters, and signal traverses at the speed
of light in cable 2.5 x 108 meters per second. The link has transmission bandwidth of 100
megabits/second (100 x 106 bits per second). The packet size is 400 bits. What is the packet
transmission delay?
[Answer]
L = 400 bits
R = 100 *106 bit/s
Packet transmission = = = 4 *10-6 s
====================================================================
Three users are sharing a common link of 1 Mb/s. User A is downloading large files and is
connected to the common link via a slow access link at x Mb/s, user B is connected via a
100 Mb/s link, but the application she is using requires at most x Mb/s, and finally user C
is connected via a 1 Gb/s link and is downloading a movie that can take up any amount of
bandwidth available to it. What is the max-min fair allocation for these three flows at the
common link?
[Answer]
Two Cases:
Case 1: If x < Mb/s then A and B get x Mb/s, C gets (1-2x) Mb/s.
Otherwise: A, B, and C get Mb/s
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