Solutions to Sample Questions on Link Layer

Computer Networks and Applications
Solutions to Sample Questions on Link Layer
1) 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 and
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; full duplex: TCP is
also full duplex.
2) Show by a virtue of an example other than the one in Figure 5.5 of the textbook that
two-dimensional parity checks can correct and detect a single bit error. Also, show an
example of a double-bit error that can be detected but not corrected.
Answer: Suppose we begin with the initial two-dimensional parity matrix:
0000
1111
0101
1010
With a bit error in row 2, column 3, the parity of row 2 and column 3 is now wrong in the
matrix below:
0000
1101
0101
1010
Now, suppose there is a bit error in row 2, column 2 and column 3. The parity of row 2 is
now correct! The parity of columns 2 and 3 is wrong, but we can't detect in which rows
the error occurred!
0000
1001
0101
1010
The above example shows that a double bit error can be detected (if not corrected).
3) Consider the slotted ALOHA protocol. Assume that there are a total of N nodes and
that all the nodes always have frames to transmit. Assume that each node transmits with a
probability p during each slot.
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(a) We have already seen that the efficiency of slotted ALOHA is Np(1-p)N-1.
Find the value of p that maximizes this expression.
Answer:
E ( p ) = Np(1 ! p ) N !1
Taking the derivative with respect to p,
E ' ( p ) = N (1 ! p ) N !1 ! Np( N ! 1)(1 ! p ) N !2
= N (1 ! p ) N !2 ((1 ! p ) ! p( N ! 1))
E ' ( p ) = 0 ! p* =
1
N
(b) Using the value of p found in (a), find the efficiency of slotted ALOHA by
letting N approach infinity. Hint: (1-1/N)n approaches 1/e as N approaches
infinity.
Answer:
1 N
)
1
1 N !1
1 N !1
N
E ( p*) = N (1 ! )
= (1 ! )
=
1
N
N
N
1!
N
1
1
1
lim (1 ! ) N =
Note that, lim (1 ! ) = 1
and
N
#
"
N #"
N
e
N
(1 !
Thus,
lim E ( p*) =
N "!
1
e
4) Why would the token passing protocol be inefficient if a LAN had a very large
perimeter?
Answer: When a node transmits a frame, the node has to wait for the frame to propagate
around the entire ring before the node can release the token. Thus, if L/R is small as
compared to tprop, then the protocol will be inefficient.
5) How big is the MAC address space, the IPv4 address space and the IPv6 address
space?
Answer: 248 MAC addresses; 232 IPv4 addresses; 2128 IPv6 addresses.
6) 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?
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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).
7) 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 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.
Answer: In the worst-case assume that A is transmitting the smallest possible frame of
size 64 bytes (512 bits). At t = 0 A transmits. At t = 512 bit time, A would finish
transmitting. In the worst case, B begins transmitting at time t = 224 bit time. At time
t = 224 + 225 = 449 bit time, B 's first bit arrives at A . Because 449 < 512, A aborts
before completing the transmission of the packet, as it is supposed to do.
8) Suppose that you had to design a 100 Mbps CSMA/CD protocol in which the
maximum one-way propagation delay between any two hosts is 100 x 10-6 sec. What will
you use as the minimum size of a transmitted frame if you wish the transmitting node to
detect a collision before completing the transmission of the frame?
Answer: The minimum size of the frame should be such that the transmission of a frame
will take at least 2 x 100 x 10-6 sec, the round trip propagation delay. The minimum frame
size therefore is 200 x 100 x 10-6 bits or 2500 bytes.
9) In CSMA/CD, after the fifth collision, what is the probability that a node chooses
K=4? The result K=4 corresponds to a delay of how many seconds on a 10Mbps
Ethernet?
Answer: After the 5th collision, the adapter chooses from {0, 1, 2, …, 31}. The
probability that it chooses 4 is 1/32. It waits 204.8 microseconds.
10) 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 C (that is, the adapter’s parent node)?
How would your answers change if A sent 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.
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11) Consider a network with 6 nodes connected in a star topology (all nodes directly
connect to a switch) to a central switch. Suppose that (i) A sends a frame to D, (ii) D
replies with a frame to A, (iii) C sends a frame to D, (iv) D replies with a frame to C. The
switch table is initially empty. Show the state of the switch table before and after each of
these events. For each of these events, identify the link(s) on which the transmitted frame
will be forwarded, and briefly justify your answers.
Answer:
Action
Switch Table State
Link(s) packet is
forwarded to
B, C, D, E, and F
A sends a
frame to D
Switch learns interface
corresponding to MAC
address of A
D replies with
a frame to A
Switch learns interface
corresponding to MAC
address of D
A
C sends a
frame to D
Switch learns the
interface corresponding
to MAC address of C
D
D replies with
a frame to C
Switch table state
remains the same as
before
C
Explanation
Since switch table is
empty, so switch
does not know the
interface
corresponding to
MAC address of D
Since switch already
knows interface
corresponding to
MAC address of A
Since switch already
knows the interface
corresponding to
MAC address of D
Since switch already
knows the interface
corresponding to
MAC address of C