CS 455/655: Introduction to Computer Networks
Fall 2013
Sample Final Examination
NAME:
MATTA
This exam is closed books and closed notes. Please write clearly and neatly. Be precise in
your answers; do not just re-iterate what you know about the topic. Clearly state any
assumptions you make. All questions are weighted equally. Answer all 5 questions.
Problem 1 (IP addressing):
Subnet B:
12 hosts
Subnet A:
15 hosts
Subnet C:
45 hosts
R1
To public Internet via ISP
Consider the network shown above, consisting of a single router, R1, with three subnets
A, B and C, with 15, 12, and 45 hosts, respectively, on these subnets. Assign an address
range to the hosts in subnets A, B, and C so that only a single aggregated address needs to
be advertised by R1 to the public Internet, and that the size of the advertised aggregated
address range is minimized. In a sentence or two, explain how you arrived at your
answer. (Feel free to use variable names in addresses if the precise bit values are not
important.)
We need 15+12+45 = 72 addresses, i.e. we need to 27 addresses or 7 bits. So we have a
common prefix length of 32-7=25 bits.
Aggregated CIDR address: X.Y.Z.0/25
Subnet “A” CIDR address: We need 24 addresses, or 4 bits. CIDR address: X.Y.Z.0/28
Subnet “B” CIDR address: We need 24 addresses, or 4 bits. CIDR address: X.Y.Z.16/28
Subnet “C” CIDR address: We need 26 addresses, or 6 bits. CIDR address: X.Y.Z.64/26
Note that other solutions are possible.
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NAME: MATTA
Problem 2 (Distance-vector routing):
E
A
B
Consider the simple network shown above, in which A and B periodically exchange
distance-vector routing information. Assume routers do not implement split horizon. All
links have cost 1. After reaching convergence, suppose the A—E link fails.
(a) Give a sequence of routing table updates that leads to a routing loop between A
and B, for a destination at E.
Before A advertises unreachability, node B advertises its distance of 2 to A, which A
adopts.
(b) Estimate the probability of the scenario in (a), assuming A and B send out routing
updates at random times, each at the same average rate.
Probability that B advertises before A = 0.5.
(c) Estimate the probability of a loop forming if node A broadcasts an updated
routing report after 1 second of discovering the A—E failure, and B regularly
broadcasts every 60 seconds.
Probability that B advertises within 1 second = 1/60.
Now, consider the following extended topology, and again suppose after reaching
convergence, the A—E link fails. Even if routers implement split horizon (without poison
reverse), a routing loop between B and C, for a destination at E, can form—consider the
following scenario: node A advertises a non-reachability update to B and C; before this
non-reachability update arrives at B and C, B and C had already exchanged their
distances, but these routing messages arrive at B and C after the non-reachability
message from node A had arrived.
C
E
A
B
(d) Describe how the situation in this extended scenario will evolve. Then, describe
how the situation would evolve if routers use split horizon with poison reverse.
Without Poison Reverse, a loop forms between B and C. With Poison Reverse, B and C
advertise unreachability to each other and the loop breaks.
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NAME: MATTA
Problem 3 (BGP routing):
(a) Suppose P, Q, and R are network service providers, with respective CIDR address
allocations C1.0.0.0/8, C2.0.0.0/8, and C3.0.0.0/8 (using hexadecimal dotted
notation with mask). Each provider’s customers initially receive address
allocations that are a subset of the provider’s address space.
P has the following customers:
• PA, with allocation C1.A3.0.0/16, and
• PB, with allocation C1.B0.0.0/12.
Q has the following customers:
• QA, with allocation C2.0A.10.0/20, and
• QB, with allocation C2.0B.0.0/16.
Assume there are no other providers or customers, and that each provider connects to
both of the others. Suppose customer PA switches to provider Q and customer QB
switches to provider R, resulting in the AS topology shown below. Use the CIDR
longest match rule to give the routing table for P (only as representative) that allow
PA and QB to switch without renumbering (i.e., keeping their initial address
allocations).
CIDR address
next-hop (AS name)
C2.0.0.0/8
C3.0.0.0/8
C1.A3.0.0/16
C2.0B.0.0/16
C1.B0.0.0/12
Q
R
Q // toward PA
R // toward QB
PB
PA
PB
QA
QB
Q
P
R
(b) Suppose customer PA is multi-homed, in that it has two Internet connections from
the two providers, P and Q, as shown in figure below. (PA is taking its address
assignment from P.) Q has a CIDR longest match routing entry for PA. Which
inbound (incoming) traffic to PA might flow on the Q—PA connection,
considering only traffic from domains P, Q, and H? Consider the two cases where
Q does and does not advertise PA to the world using BGP. Recall shorter AS
paths are preferred.
PA
Q does not advertise PA to the world:
Only traffic from Q.
Q advertises PA to the world:
Q
P
X
Y
H
Traffic from Q and H.
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NAME: MATTA
Problem 4 (Indirection and encapsulation):
source à
destination
protocol
type
A à B
TCP
IP header
TCP header + data
data
Consider the IP datagram carrying a TCP segment, shown above. Consider a situation
where we want to route this datagram through an intermediate router C by encapsulating
it within another datagram.
(a) Draw the datagram as it travels to C and then as it travels out of C toward the
final destination B—make sure to show the contents of the fields: source,
destination, protocol type, and data (payload).
source à
destination
source à
destination
protocol
type
data
IP header
To router C
AàC | IP-in-IP | original datagram |
protocol
type
IP header
data
From C, To destination B
| AàB | TCP | TCP segment |
(b) Describe how the datagram is processed at router C from when it arrives at C and
until it comes out of C on its way toward the final destination B.
At router C, IP demultiplexes the datagram to the “decapsulation” software, which
removes the header and forwards the original datagram to the final destination.
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NAME: MATTA
Problem 5 (CSMA/CD protocol):
(a) In CSMA/CD, after the fourth collision, what is the probability that the node
chooses K = 3? What does the result K = 3 correspond to in terms of delay in
microseconds on a 10 Mbps Ethernet?
After the 4th collision, K is chosen uniformly from 0, …, 15. The probability that K=3 is
then 1/16. K=3 corresponds to a delay of 512*3 bit times = (512*3)/10M= 153.6 µsec.
(b) Suppose nodes A and B are on the same 10 Mbps Ethernet segment, and the
propagation delay between the two nodes is 225 bit times. Suppose at time t = 0,
B starts to transmit a frame. Suppose A also transmits at some t = x, but before
completing its transmission it receives bits from B (hence, a collision occurs at
A). Assuming node A follows the CSMA/CD protocol, what is the maximum
value of x?
x = 224 bit times /10M = 22.4 µsec.
(c) Consider two nodes A and B on the same Ethernet segment, and suppose the
propagation delay between the two nodes is 225 bit times. Suppose at time t = 0,
both nodes A and B begin to transmit a frame. At what time do they detect the
collision? Assuming both nodes transmit a 48-bit jam signal after detecting a
collision1, at what time (in bit times) do nodes A and B sense an idle channel?
How many seconds is this for a 10 Mbps Ethernet?
225 bit times = 22.5 µsec.
(225 + 48) + 225 = 498 bit times = 49.8 µsec.
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Note that once an adapter detects a collision, it stops transmitting its frame but continues to transmit a 48bit “jam signal” to ensure that collision is detected by other adapters.
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