EE 122 1st Term Exam
Date: October 9, 2002
Name:
SID:
ee122 login:
Day/Time of section you attend:
Problem Points
1
/10
2
/10
3
/20
4
/20
5
/20
6
/20
Total
/100
1. Question 1 (10 pt)
Use no more than three sentences to answer the questions below:
(a) State the “end-to-end arguments”. (4 pt)
(b) Error detection techniques detect corrupt bits. Give an example of an error that
can be detected by the transport layer that the link layer cannot. (3 pt)
(c) Given that the transport layer can detect errors that the data link layer cannot, why
would you implement error detection in the lower layers? (3 pt)
Solution:
(a) A
lower layer should not implement functionality that can be correctly and completely
implemented only by a higher layer. The only exception is when partial functionality in a
lower layer optimizes performance significantly, and doesn't adversely affect applications
at the higher layer that do not require this functionality.
(b) The transport layer can detect errors that occur after a packet has passed the data link
layer. For example, a packet can be corrupted in the buffer or the switching backplane of
a router.
Another example is when a stronger error detection technique is used at the transport
layer. Since the transport layer is implemented only at end-hosts and not on every router
along the path, typically more computational resources are available at the transport
layer.
Note: You have to give an example that some how involves corrupt bits. If you talk about
lost or out-of-order packets, you receive partial credit only.
(c) Implementing error detection at a lower layer can improve performance. The transport
layer can only detect errors at the destination, and hence the packet with the error is
carried over the entire path wasting bandwidth. The retransmission has to be done over
the entire path too.
Another example is that different data link layer technologies have different error
characteristics, so a high error rate technology (wireless) can have strong error detection,
while a low error rate technology (optical) can have weak error detection. We can thus
avoid the overhead of having strong error detection over the entire path.
2. Question (10 pt)
Compare packet switching and circuit switching using the following criteria: (a)
forwarding cost, (b) bandwidth utilization, and (c) ability to handle resource reservations.
Briefly explain your reasoning.
Solution:
(a) Packet switching has a higher per-packet forwarding cost because
 The router has to examine (and possibly rewrite) the header of every packet
it forwards.
 The router has to maintain routing tables which may have a large number of
entries.
 The router needs to do a longest prefix matching for every packet,
The per-packet forwarding cost is much lower in circuit switching once the
connection has been established.
(b) In packet switching, different users use statistical multiplexing to share the
bandwidth on any link. In the absence of congestion and packet drops, no
bandwidth is wasted and the utilization is typically high.
In circuit switching, bandwidth may be wasted if an established connection is idle.
For applications with a high peak to average sending rate ratio, this wastage can be
high.
(c) Circuit switching inherently requires resources to be reserved for a particular
connection. On the other hand, in packet switching the bandwidth available to a
particular user (flow) depends on the sending rate of other flows. We need
additional scheduling mechanisms at the router to support reservations to a
particular flow.
3. Question (20 pt)
We want to transfer a file of size d bits. Each link has bandwidth b bps and fixed delay f
sec. All the routers on the path are store-and-forward. We use packets of size p bits,
which includes headers of size h bits. We always pad the last packet so that it is full.
There is no setup time for the transfer. Packets are sent continuously and are not lost.
There is no queueing delay and processing overhead.
(a)
(b)
(c)
(d)
How many packets are sent? (5 pt)
What is the delay to transfer the file across one link? (5 pt)
What is the delay for one packet to arrive at a destination n links away? (5 pt)
What is the delay to transfer the file across n links? (5 pt)
Solution:
(a) You have to send all of your data, only p – h of your data fits in a packet, and you
have to round up because the last packet may not be full, so the number of packets

  d 
file _ size
is 

.
 data _ per _ packet   p  h 
(b)
sender
receiver
f
packets
p/b
total _ delay  transmissi on _ delay  fixed _ delay
number _ of _ packets * packet _ size

 fixed _ delay
bandwidth
 d 
 p  h p
 f

b
links
(c)
receiver
sender
f
packet
p/b
f
p/b
f
p/b
delay  (transmissi on _ delay _ per _ link  fixed _ delay _ per _ link ) * num _ links
p

   f n
b

(d)
receiver
sender
f
p/b
f
p/b
f
p/b
p/b
p/b
delay  delay _ of _ first _ packet  delay _ for _ remaining _ packets _ after _ first
p

delay    f n  (number _ of _ packets  1) * transmissi on _ delay _ per _ packet
b

p
   d   p
   f n   
 1
b
   p  h   b
Alternatively, you can think about it as
delay  total _ fixed _ delay _ across _ all _ links 
total _ transmissi on _ delay _ for _ all _ packets 
initial _ delay _ for _ first _ packet _ to _ travel _ across _ all _ links _ except _ first
 d p
p
delay  fn  
 n  1 (this simplifies to the same as above)

b
 p  h b
4. Question (20 pt)
Assume two end-hosts communicate using the sliding window protocol. Assume the
receiver window is always smaller than the sender’s window and the size of the receiver
window is w bits. Let C be the link capacity between the two end-hosts in bps, and RTT
be the roundtrip time between the two end-hosts in sec. What is the maximum throughput
achieved by the two end-hosts?
Note: Assume every bit is acknowledged.
Answer: There are two cases (see figure below)
case a: RTT > w/C, throughput = w/RTT
case b: RTT <= w/C, throughput = C
receiver
sender
receiver
sender
RTT
RTT
w/C
w/C
(a)
(b)
5. Question (20 pt)
Consider the network below. Assume that the network implements the Distance Vector
routing protocol.
(a) Assume that the routing protocol has converged. Write down the routing table of each
node. (5 pt)
(b) Assume the cost of the link between nodes A and B increases from 1 to 3. Write down
the routing table of each node at every step until routing tables converge. Assume that
both A and B see that the change of the link (A, B)’s cost at the same time, and that
exchanges of routing information and routing table updates are synchronous (i.e., they
happen at the same time at all nodes) (5 pt)
(c) Assume the cost of the link between nodes A and B increases from 1 to 50. How
many iterations does it take the routing protocol to converge? (5 pt)
(d) Give one solution to reduce the number of iterations at point (c). Explain. (5 pt)
A
1
B
1
20
C
1
D
a) See first row below
b) See the last three rows below
DA
B
C
DB
A
C
DC
A
B
D
DD
C
B
1
3
A
1
21
A
1
21
3
A
2
C
3
1
C
2
20
B
2
20
4
B
3
D
4
2
D
3
21
D
3
23
1
C
1
1st iteration
DA
B
C
DB
A
C
DC
A
B
D
DD
C
B
3
3
A
3
21
A
1
21
3
A
2
C
5
1
C
4
20
B
2
20
4
B
3
D
6
2
D
5
21
D
3
23
1
C
1
2nd iteration
DA
B
C
DB
A
C
DC
A
B
D
DD
C
B
3
3
A
3
21
A
1
23
3
A
2
C
7
1
C
4
20
B
4
20
4
B
3
D
8
2
D
5
21
D
3
25
1
C
1
D
DD
C
3rd iteration
DC
A B
DA
B
C
DB
A
C
B
3
5
A
3
21
A
1
23
3
A
2
C
7
1
C
4
20
B
4
20
4
B
5
D
8
2
D
5
21
D
3
25
1
C
1
4th iteration
DA
B
C
DB
A
C
DC
A
B
D
DD
C
B
3
5
A
3
21
A
1
23
3
A
2
C
7
1
C
4
20
B
4
20
6
B
5
D
8
2
D
5
21
D
3
25
1
C
1
c) There are 20 iterations (i.e., the second highest cost of a link). See below:
DA
B
C
DB
A
C
DC
A
B
D
DD
C
B
1
3
A
1
21
A
1
21
3
A
2
C
3
1
C
2
20
B
2
20
4
B
3
D
4
2
D
3
21
D
3
23
1
C
1
1st iteration
DA
B
C
DB
A
C
DC
A
B
D
DD
C
B
50
3
A
50
21
A
1
21
3
A
2
C
52
1
C
51
20
B
2
20
4
B
3
D
53
2
D
52
21
D
3
23
1
DA
B
C
DB
A
C
DC
A
B
C
1
From now on, only the entries
in red will change!
D
DD C
B
50
3
A
50
21
A
1
41
3
A
2
C
70
1
C
51
20
B
4
20
4
B
3
D
71
2
D
52
21
D
3
41
1
C
1
D
DD
C
2nd iteration
3rd iteration
DC
A B
DA
B
C
DB
A
C
B
50
5
A
50
21
A
1
41
3
A
2
C
70
1
C
51
20
B
4
20
4
B
5
D
71
2
D
52
21
D
3
41
1
C
1
B
D
DD
C
4th
iteration
DC
A
DA
B
C
DB
A
C
B
50
5
A
50
21
A
1
41
3
A
2
C
70
1
C
51
20
B
6
20
6
B
5
D
71
2
D
52
21
D
3
41
1
C
1
…
20th
iteration
DA
B
C
DB
A
C
DC
A
B
D
DD
C
B
50
21
A
50
21
A
1
41
3
A
2
C
70
1
C
51
20
B
22
20
22
B
21
D
71
2
D
52
21
D
3
41
1
C
1
Note: Full credit was awarded for only specifying the number of iterations. Also we gave
full credit to students that were slightly off in estimating the number of iterations, e.g.,
19, 21.
d) Poisoned reverse: a node sends to its neighbor the route learnt from that node with a
cost of infinity.
6. Question (20 pt)
Consider a wireless network consisting of n nodes where nodes are placed on a line at the
coordinates 0, d, 2*d, …, (n-1)*d. Assume the range of transmission of each node is d +
e, where 0 < e < d, and that n is even.
(a) What is the maximum number of packets that can be simultaneously transmitted in
such a network? (10 pt)
(b) What is the maximum number of packets that can be simultaneously transmitted in a
network with n2 nodes arranged in an n x n matrix, where the distance between any two
nodes on both the x and y axis is d? (10 pt)
For full credit it is enough to give an example that achieves the maximum number of
transmitted packets for each case. There is a 10 points bonus if you prove that your
examples indeed achieve the maximum number of transmitted packets in all cases.
Note: A node cannot send and receive at the same time, and a node cannot receive if it is
within the transmission range of two or more senders.
Solution:
(a) The maximum number of packets is n/2; see the example below. Each arrow
represents a packet that is transmitted from the sender to the receiver. Note that two
neighbor nodes can send successfully two packets as long as the receivers are at a
distance larger than d + e
(b) There are two possible cases:
(b.1) 0  e  ( 2  1) * d ; the maximum number of packets is n2/2 (see example below)
(b.2) ( 2  1) * d  e  d ; the maximum number of packets is n2/3 (see example below)
Proof (for bonus points):
(a) Let S be the number of senders that successfully send a packet, and let R be the
number of receivers that successfully receive a packet. Since a node cannot send and
receive at the same time, it follows that S + R <= n. Finally, for each successfully
transmitted packet there is exactly one sender and one receiver, it follows that the
maximum number of transmitted packets is at most n/2.
(b.1) Same as (a).
(b.2) Let S be the number of senders, R the number of receivers, and I the number of idle
nodes, i.e., a node that can neither send nor receive a packet. The crux of the proof is to
show that we can associate an idle node with each pair of nodes that communicate. Since
transmitting one packet involves exactly one sender and one receiver, the total number of
transmitted packets is min(S, R). Since according to the above statement we have min(S,
R) <= I, and by definition I + S + R = n2, it follows that min(S,R) <= n2/3.
To prove that we can associate with each transmission an idle node, we define a simple
rule: To each pair of nodes (A, B) that communicate we associate node C that is a
neighbor of both A and B. Ties are broken according to the lowest y-coordinate first, and
then according to the lowest x-coordinate. The following figure shows all possible cases.
B
c
A
(a)
A
c
B
(b)
B
B
A
A
c
(c)
c
(d)
Note two things. First, node C can neither send nor receive other packets than the one
exchanged between A and B. This holds true irrespective of which is the sender, i.e., A or
B. Second, a node C cannot be associated to more than one transmission (this can be
easily verified by taking case by case).