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R1. What are some of the possible services
that a link-layer protocol can offer to the
network layer? Which of these link-layer
services have corresponding services in IP?
In TCP?
Answer :
Framing: there is also framing in IP and
TCP;
link access;
reliable delivery: there is also reliable
delivery in TCP;
flow control: there is also flow control in
TCP;
error detection: there is also error detection
in IP and TCP;
error correction;
Half/full duplex: TCP is also full duplex.
R2. If all the links in the Internet were to
provide reliable delivery service, would the
TCP reliable delivery service be redundant?
Why or why not?
Answer :
Although each link guarantees that an IP
datagram sent over the link will be received
at the other end of the link without errors, it
is not guaranteed that IP datagrams will
arrive at the ultimate destination in the
proper order. With IP, datagrams in the
same TCP connection can take different
routes in the network, and therefore arrive
out of order. TCP is still needed to provide
the receiving end of the application the byte
stream in the correct order. Also, IP can
lose packets due to routing loops or
equipment failures.
R3. In section 5.3, we listed four
desirable characteristics of a
broadcast channel. Which of these
characteristic does slotted ALOHA
have ? Which of these characteristic
does token passing have?
Answer :
Slotted Aloha: 1, 2 and 4 (slotted
ALOHA is only partially decentralized,
since it requires the clocks in all nodes
to be synchronized). Token ring: 1, 2,
3, 4.
R4. Suppose two nodes start to
transmit at the same time a packet of
length L over a broadcast channel of
rate R. Denote the propagation delay
between the two nodes an dprop. Will
there be a collision if dprop< L/R? why
or why not?
Answer :There will be a collision in the
sense that while a node is transmitting
it will start to receive a packet from the
other node.
R7. Suppose nodes A, B, and C each
attach to the same broadcast LAN
(through their adapters). If A sends
thousands of IP datagrams to B with
each encapsulating frame addressed
to the MAC address of B, will C’s
adapter process these frames? If so,
will C’s adapter pass the IP datagrams
in these frames to the network layer C?
how would your answers change if A
sends frames with the MAC broadcast
address?
Answer :C’s adapter will process the
frames, but the adapter will not pass
the datagrams up the protocol stack. If
the LAN broadcast address is used,
then C’s adapter will both process the
frames and pass the datagrams up the
protocol stack.
R8.How big is the MAC address space?
The IPv4 address space? The IPv6
address space?
Answer :
248 MAC addresses;
232 IPv4 addresses;
2128 IPv6 addresses.
R9. Why is an ARP query sent within a
broadcast frame? Why is an ARP
response sent within a frame with a
specific destination MAC address?
Answer :
An ARP query is sent in a broadcast frame
because the querying host does not which
adapter address corresponds to the IP
address in question. For the response, the
sending node knows the adapter address to
which the response should be sent, so
there is no need to send a broadcast frame
(which would have to be processed by all
the other nodes on the LAN).
P1. Suppose the information content
of a packet is the bit pattern
1110110010001010 and an even parity
scheme is being used. What would the
value of the field containing the parity
bits be for the case of a twodimensional parity scheme? Your
answer should be such that a
minimum-length checksum field is
used.
Answer:
The rightmost column and bottom row are for parity
bits.
11101
11000
10001
10100
00000
P2. Suppose the information portion of
a packet (D in Figure 5.4 ) contains 10
bytes consisting of the 8-bit unsigned
binary representation of the integers 0
through 9. Compute the Internet
checksum for this data.
Answer:
P3. Consider the previous problem,
but instead of containing the binary
of the numbers 0 through 9
suppose these 10 bytes contain
a. The binary representation of the
numbers 1 through 10.
b. The ASCll representation of the
letters A through J (uppercase).
c. The ASCll representation of the
letters a through j (lowercase).
Compute the Internet checksum for
this data.
Answer:
P6 .Consider the 4-bit generator, G,
shown in Figure 5.8,and suppose
that D has the value
a. 10010001
b. 10100011
c. 01010101
What is the value of R?
Answer:
• If we divide 1001 into
10010001000 we get 10000001,
with a remainder of R = 001.
b) If we divide 1001 into we get
10100011000 we get 10110101,
with a remainder of R = 101.
c) If we divide 1001 into 01010101000
we get 010111101, with a
remainder of R =110.
P12. Consider three LANs interconnect by two routers, as
shown in Figure 5.38.
C
A
E
B
F
Subnet 1
D
Subnet 2
Subnet
3
a.
b.
c.
d.
e.
Redraw the diagram to include adapters.
Assign IP addresses to all the interfaces. For
Subnet 1 use addresses of the form
111.111.111.xxx; for Subnet 2 uses addresses
of the form 122.122.122.xxx; and for Subnet 3
use addresses of the form 133.133.1333.xxx.
Assign MAC addresses to all of the adapters.
Consider sending an IP datagram from Host A
to Host F. Suppose all of the ARP tables are
up to date. Enumerate all the steps, as done
for the single-router example in Section5.4.2.
Repeat (d), now assuming that the ARP table
in the sending host is empty (and the other
tables are up to date).
a), b), c) See figure below.
e) ARP in A must now determine the LAN
address of 111.111.111.002. Host A sends out an
ARP query packet within a broadcast Ethernet
frame. The first router receives the
query packet and sends to Host A an ARP
response packet. This ARP response packet is
carried by an Ethernet frame with Ethernet
destination address 00-00-00-00-00-00.
P14. Recall that with the CSMA/CD
protocol, the adapter waits K*512 bits
times after a collision, where K is
drawn randomly. For K = 100, how
long does the adapter wait until
returning to Step 2 for a 1 Mbps
Ethernet? For a 10Mbps Ethernet?
Answer :
Wait for 51,200 bit times. For 10 Mbps,
this wait is
For 100 Mbps, the wait is 512 μ sec.
P15. Suppose nodes A and B are on the same 10
Mbps Ethernet bus, and the propagation delay
between the two nodes is 225 bit times. Suppose A
and B send frames at the same time, the frames
collide, and then A and B choose different values of
K in the CSMA/CD algorithm. Assuming no other
nodes are active, can the retransmissions from A
and B collide? For our purposes, it suffices to work
out the following example. Suppose A and B begin
transmission at t = 0 bit times. They both detect
collisions at t = 225 bit times. They finish
transmitting a jam signal at t = 225+48=273 bit
times. Suppose KA =0 and KB = 1. At what time
does B schedule its retransmission? At what time
does A begin transmission? (Note :The nodes must
wait for an idle channel after returning to Step 2 –
see protocol. ) At what time does A’s signal reach B?
does B refrain from transmitting at its scheduled
time.
Answer :
P16. Suppose nodes A and B are on the same
10Mbps Ethernet bus, and the propagation delay
between the two nodes is 225 bit times. Suppose
node A begins transmitting a frame and, before it
finishes, node B begins transmitting a frame. Can A
finish transmitting before it detects that B has
transmitted? Why or why not? If the answer is yes,
then A incorrectly believes that its frame was
successfully transmitted without a collision. Hint:
Suppose at time t = 0 bit times, A begins to
transmitting a frame. In the worst case, A transmits
a minimum-sized frame of 512+64 bit times. So A
would finish transmitting the frame at t = 512 +64
bit times. Thus ,the answer is no, if B’s signal
reaches A before bit time t= 512+64 bits. In the
worst case, when does B’s signal reach A?
Answer :
At t = 0 A transmits. At t = 576 , A would
finish transmitting. In the worst case, B
begins transmitting at time t = 224 . At time t
= 224 + 225 = 449 B 's first bit arrives at
A . Because 449 < 576 , A aborts before
completing the transmission of the packet,
as it is supposed to do. Thus A cannot finish
transmitting before it detects that B
transmitted. This implies that if A does not
detect the presence of a host, then no other
host begins transmitting while A is
transmitting.
P19. Suppose two nodes, A and B, are attached to opposite
ends of a 900 m cable, and that they each have one frame of
1,000 bits (including all headers and preambles) to send to
each other. Both nodes attempt to transmit at time t= 0.
suppose there are four repeaters between A and B, each
inserting a 20-bit delay. Assume the transmission rate is 10
Mbps, and CSMA/CD with backoff intervals of multiples of
512 bits is used. After the first collision, A draws K =0 and B
draws K =1in the exponential backoff protocol. Ignore the jam
signal and the 96-bit time delay.
a. What is the one-way propagation delay (include repeater
delays) between A and B in seconds? Assume that the signal
propagation speed is 2*108 m/sec .
b. At what time ( in seconds) is A’s packet completely
delivered at B ?
c. Now suppose that only A has a packet and that the
repeaters are replaced with switches. Suppose that each
switch has a 20-bit processing delay in addition to a storeand-forward delay. At what time, in seconds ,is A’s packet
delivered at B?
Answer:
P21.Suppose now that the leftmost router in Figure 5.38 is
replaced by a switch. Hosts A, B, C, and D and the right
router are all star-connected into this switch. Give the source
and destination MAC address in the frame encapsulating this
IP datagram as the frame is transmitted (i) from A to the
switch, (ii) from the switch to the right router, (iii) from the right
router to E. Also give the source and the destination IP
addresses in the IP datagram encapsulated within the frame
at each of these points in time.
C
A
E
B
F
Subnet 1
D
Subnet 2
Subnet
3
Answer:
P23. Consider Figure 5.26. suppose that all links
are 10 Mbps. What is the maximum total aggregate
throughput that can be achieved among the 14 end
systems in this network? Why?
1Gbps
1Gbps
1Gbps
100Mbps
(fiber)
100Mbps
(fiber)
Mixture of 10 Mbps,
100Mbps,1Gbps,
Cat 5 cable
100Mbps
(fiber)
Electrical Engineering
Computer Engineering
Computer Science
Answer:
If all the 14 nodes send out data at the
maximum possible rate of 10 Mbps, a
total aggregate throughput of 14*10 =
140 Mbps is possible
P24. Suppose the three departmental switches in
Figure 5.26 are replaced by hubs. All links are 10
Mbps. What is the maximum total aggregate
throughput that can be achieved among the 14 end
system in this network? Why?
1Gbps
1Gbps
1Gbps
100Mbps
(fiber)
100Mbps
(fiber)
Mixture of 10 Mbps,
100Mbps,1Gbps,
Cat 5 cable
100Mbps
(fiber)
Electrical Engineering
Computer Engineering
Computer Science
Answer:
Each departmental hub is a single collision
domain that can have a maximum
throughput of 10 Mbps. The links
connecting the web server and the mail
server has a maximum throughput of 10
Mbps. Hence, if the three collision domains
and the web server and mail server send
out data at their maximum possible rates of
10 Mbps each, a maximum total aggregate
throughput of 50 Mbps can be achieved
among the 14 end systems.