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Chapter 5 Review Questions
1. The transportation mode, e.g., car, bus, train, car.
2. 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
3. Framing: there is also framing in IP and TCP; link access; reliable delivery: there is
5. 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.
7. In polling, a discussion leader allows only one participant to talk at a time, with each
8. When a node transmits a frame, the node has to wait for the frame to propagate
9. 248 MAC addresses; 232 IPv4 addresses; 2128 IPv6 addresses.
10. C’s adapter will process the frames, but the adapter will not pass the datagrams up the
11. 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
13. The three Ethernet technologies have identical frame structures.
15. In 802.1Q there is a 12- bit VLAN identifier. Thus 212 = 4,096 VLANs can be
supported.
16. We can string the N switches together. The first and last switch would use one port
Chapter 5 Problems
Problem 1
1 1 1 0 1
Problem 2
Suppose we begin with the initial two-dimensional parity matrix:
0 0 0 0
1 1 1 1
0 1 0 1
Problem 3
01001100 01101001
+ 01101110 01101011
------------------------------
Problem 4
a) To compute the Internet checksum, we add up the values at 16-bit quantities:
00000001 00000010
00000011 00000100
00000101 00000110
Problem 5
Problem 6
Problem 7
a) Without loss of generality, suppose ith bit is flipped, where 0<= i <= d+r-1 and
assume that the least significant bit is 0th bit.
Problem 8
a)
1
)1()(
N
pNppE
21
)1)(1()1()('
NN
pNNppNpE
))1()1(()1(
2
NpppN
N
N
ppE 1
*0)('
b)
Problem 9
12
*0)('
N
ppE
Problem 10
a) A’s average throughput is given by pA(1-pB).
Problem 11
a) (1 – p(A))4 p(A)
where, p(A) = probability that A succeeds in a slot
p(A) = p(A transmits and B does not and C does not and D does not)
= p(A transmits) p(B does not transmit) p(C does not transmit) p(D does
not transmit)
= p(1 – p) (1 – p)(1-p) = p(1 – p)3
Hence, p(A succeeds for first time in slot 5)
= (1 – p(A))4 p(A) = (1 – p(1 – p)3)4 p(1 – p)3
Problem 12
Problem 13
The length of a polling round is
)/(
poll
dRQN
.
Problem 14
a), b) See figure below.
c)
1. Forwarding table in E determines that the datagram should be routed to interface
2. The adapter in E creates and Ethernet packet with Ethernet destination address 88-
3. Router 2 receives the packet and extracts the datagram. The forwarding table in
this router indicates that the datagram is to be routed to 198.162.2.002.
33-33-33 and source address of 55-55-55-55-55-55 via its interface with IP
address of 198.162.2.003.
192.168.3.002
88
-88-88-88-88-88
77
-77-77-77-77-77
55
-55-55-55
-55
44-44-44-44-44-44
33
-33-33-33-33-33
LAN
Router 1
LAN
C
192.168.1.002
22
-22-22-22-22-22
192.168.1.003
E
Router 2
LAN
d) ARP in E must now determine the MAC address of 198.162.3.002. Host E sends out
Problem 15
a) No. E can check the subnet prefix of Host F’s IP address, and then learn that F is on
the same LAN. Thus, E will not send the packet to the default router R1.
Ethernet frame from E to F:
Source IP = E’s IP address
b) No, because they are not on the same LAN. E can find this out by checking B’s IP
address.
Ethernet frame from E to R1:
Source IP = E’s IP address
c) Switch S1 will broadcast the Ethernet frame via both its interfaces as the received
ARP frame’s destination address is a broadcast address. And it learns that A resides
B won’t send ARP query message asking for A’s MAC address, as this address can
be obtained from A’s query message.
Problem 16
Lets call the switch between subnets 2 and 3 S2. That is, router R1 between subnets 2 and
3 is now replaced with switch S2.
a) No. E can check the subnet prefix of Host F’s IP address, and then learn that F is on
the same LAN segment. Thus, E will not send the packet to S2.
Ethernet frame from E to F:
Source IP = E’s IP address
b) Yes, because E would like to find B’s MAC address. In this case, E will send an ARP
query packet with destination MAC address being the broadcast address.
This query packet will be re-broadcast by switch 1, and eventually received by Host
B.
Ethernet frame from E to S2:
Source IP = E’s IP address
c) Switch S1 will broadcast the Ethernet frame via both its interfaces as the received
ARP frame’s destination address is a broadcast address. And it learns that A resides
on Subnet 1 which is connected to S1 at the interface connecting to Subnet 1. And, S1
will update its forwarding table to include an entry for Host A.
Yes, router S2 also receives this ARP request message, and S2 will broadcast this
query packet to all its interfaces.
Problem 17
Wait for 51,200 bit times. For 10 Mbps, this wait is
Problem 18
At
0 t
A
transmits. At
576 t
,
A
would finish transmitting. In the worst case,
B
Problem 19
Time,
t
Event
0
A
and
B
begin transmission
245
A
and
B
detect collision
Problem 20
a) Let
Y
be a random variable denoting the number of slots until a success:
1
)1()(
m
mYP
EE
,
where
E
is the probability of a success.
1
1
)1(
)1(1
N
N
pNp
pNp
x
b)
Maximizing efficiency is equivalent to minimizing
x
, which is equivalent to maximizing
c)
efficiency
1
1
)
1
1(
)
1
1(1
N
N
N
N
k
k
d) Clearly,
1 ek
k
approaches 1 as
fok
.
Problem 21
133.333.333.002
88
-88-88-88-88-88
77
-77-77-77-77-77
55-55-55-55-55
44-44-44-44-44-44
33
-33-33-33-33-33
LAN
Router 1
LAN
111.111.111.002
22
-22-22-22-22-22
Router 2
LAN
122.222.222.004
C
E
i) from A to left router: Source MAC address: 00-00-00-00-00-00
Destination MAC address: 22-22-22-22-22-22
Problem 22
i) from A to switch: Source MAC address: 00-00-00-00-00-00
Destination MAC address: 55-55-55-55-55-55
Source IP: 111.111.111.001
Problem 23
Problem 24
Each departmental hub is a single collision domain that can have a maximum throughput
Problem 25
Problem 26
Action
Switch Table State
Link(s) packet is
forwarded to
Explanation
B sends a
frame to E
Switch learns interface
corresponding to MAC
A, C, D, E, and F
Since switch table is
empty, so switch
Problem 27
a) The time required to fill
8L
bits is
b) For
,500,1 L
the packetization delay is
c) Store-and-forward delay
R
L408
d) Store-and-forward delay is small for both cases for typical link speeds. However,
packetization delay for
1500 L
is too large for real-time voice applications.
Problem 28
The IP addresses for those three computers (from left to right) in EE department are:
The IP addresses for those three computers (from left to right) in CS department are:
111.111.2.1, 111.111.2.2, 111.111.2.3. The subnet mask is 111.111.2/24.
The router’s interface card that connects to port 1 can be configured to contain two sub-
111.111.2.0 is associated with VLAN 12. This means that each frame that comes from
subnet 111.111.1/24 will be added an 802.1q tag with VLAN ID 11, and each frame that
comes from 111.111.2/24 will be added an 802.1q tag with VLAN ID 12.
Suppose that host A in EE department with IP address 111.111.1.1 would like to send an
802.1q tag.
Problem 29
D
R3
R4
R5
0
1
0
0
A
in out out
label label dest interf.
1
0
in out out
label label dest interf.
in out out
0
Problem 30
D
R3
R4
1
0
0
R6
in out out
label label dest interf.
1
0
0
Problem 31
Your computer first uses DHCP to obtain an IP address. You computer first creates a
special IP datagram destined to 255.255.255.255 in the DHCP server discovery step, and
puts it in a Ethernet frame and broadcast it in the Ethernet. Then following the steps in
the DHCP protocol, you computer is able to get an IP address with a given lease time.
A DHCP server on the Ethernet also gives your computer a list of IP addresses of first-
hop routers, the subnet mask of the subnet where your computer resides, and the
addresses of local DNS servers (if they exist).
HTTP request message will be segmented and encapsulated into TCP packets, and then
further encapsulated into IP packets, and finally encapsulated into Ethernet frames. Your
computer sends the Ethernet frames destined to the first-hop router. Once the router
receives the frames, it passes them up into IP layer, checks its routing table, and then
sends the packets to the right interface out of all of its interfaces.
Problem 32
a) Each flow evenly shares a link’s capacity with other flows traversing that link, then the
80 flows crossing the B to access-router 10 Gbps links (as well as the access router to
Problem 33
a) Both email and video application uses the fourth rack for 0.1 percent of the time.
b) Probability that both applications need fourth rack is 0.001*0.001 = 10-6.
