Give short answers to the following questions.
Q1. In Data Link Control briefly explain the procedure known as byte stuffing. If data from an upper layer consists
of the data given in Figure 4.1, explain how byte stuffing would be implemented to ensure that the data is not
misinterpreted.
Flag
ESC
Figure 4.1
Q2. How the inefficiency of Stop-and-Wait protocol is overcome in sliding window protocol?
Q3. What is piggybacking? What is its advantage?
Q4. The timer of a system using the Stop-and-Wait ARQ Protocol has a time-out of 4 ms. Draw the flow diagram
similar to Figure 11.11 in lecture notes for four data frames if the round trip delay is 3 ms. Assume that the first
data frame is damaged and fourth one is lost, but all other data and control frames are not lost or damaged.
Q5. Define random access and list three protocols in this category.
Q6. A network with one primary and four secondary stations uses polling. The size of a data frame is 50 bytes. The
size of the poll, ACK, and NAK frames are 5 bytes each. Each station has 3 frames to send. How many total
bytes are exchanged if there is no limitation on the number of frames a station can send in response to a poll?
Q7. In Figure 4.2, the date rate is 100 Mbps, the distance between station A and C is 4,000 m, and the propagation
speed is 2 × 10°m/s. Station A starts sending a long frame at time t, = 0; station C starts sending a long frame
at time ty = 4 us. The size of the frame is long enough to guarantee the detection of collision by both stations.
Find:
a.
The time when station C hears the collision (tg).
b.
The time when station A hears the collision (t.).
C.
The number of bits station A has sent before detecting the collision.
d.
The number of bits station C has sent before detecting the collision.
Transmission
First bit of A
First bit of C
A’s collision
detection
and abortion
Collision
occurs
Cs collision
detection and
abortion
Time
Figure 4.2 Collision of the first bit in CSMA/CD.
Q8. What is the type of the following destination addresses?
a.
4B:35:20:41:15:1C
b.
78:FF:41:20:51-FF
FF:FF:FF:FF:FF:FF
09. How does the Ethernet address 78:FF:41:20:51:FF appear on the line in binary?
Q10. An Ethernet MAC sublayer receives 1,529 bytes of data from the upper layer.
Can the data be encapsulated in one frame?
b.
If not, how many frames need to be sent?
What is the size of the data in each frame?
