Final Exam Solution
1.
Please explain how to perform the flag-based frame synchronization (8%) and the CRC-based
frame synchronization (7%).
Solution:
(a) Flag-based frame synchronization was developed to transfer an arbitrary number of bits
within a frame. Take an example of HDLC, the beginning and end of an HDLC frame is
indicated by the presence of an 8-bit flag. The flag in HDLC consists of the byte 01111110.
Bit stuffing prevents the occurrence of the flag inside the frame. The transmitter examines
the contents of the frame and inserts an extra 0 after each instance of five consecutive 1s.
The receiver looks for five consecutive 1s in the received sequence. Five 1s followed by a 0
indicate that the 0 is a stuffing bit, and so the bit is removed. Five consecutive 1s followed
by 10 indicate a flag. Five 1s followed by 11 indicate an error.
(b) Take an example of generic framing procedure (GFP) for the CRC-based frame
synchronization. The first two fields, PLI and cHEC, are used to delineate a frame. The
two-byte payload length indicator (PLI) gives the size in bytes of the GFP payload area and
so indicates the beginning of the next GFP frame. The two-byte cHEC field contains the
CRC-16 redundancy check bits for the PLI and cHEC fields. The cHEC can be used to
correct single errors and to detect multiple errors. The GFP receiver synchronizes to the GFP
frame boundary through a three-state process. The receiver is initially in the hunt state
where it examines four bytes at a time to see if the CRC computed over the first two bytes
equals the contents of the next two bytes. When the receiver finds a match it moves to the
pre-sync state, the receiver uses the tentative PLI field to determine the location of the next
frame boundary. Then, in the sync state the single-error correcting capability of the CRC-16
is activated so that the occurrence of an isolated error does not disrupt frame
synchronization.
2.
The following is an HDLC frame exchange between station A and station B using asynchronous
balanced mode (ABM). Please finish the diagram by completing the labeling of the frame
exchanges (12%).
1.B,I,0,0 2.A,I,0,0 3.x,I,x,x
4.x,I,x,x
5.x,REJ,x
6.x,I,x,x
7.x,I,x,x
8.x,I,x,x
(Notice that the frame (B,I,0,0) denotes station A sends the I-frame to station B with sending
sequence number 0 and the receiving sequence number 0.)
Solution:
1. B,I,0,0 2.A,I,0,0 3.A,I,1,1 4.A,I,2,1 5.A,REJ,1 6.A,I,1,1 7.A,I,2,1 8.A,I,3,1
3.
Please describe the three different medium sharing techniques, in which the pros and cons and
its applications of each technique should be included in the description (18%).
Static channelization schemes
They involve the partitioning of the medium into separate channels that are then dedicated
to particular users.
pros:
 Highly efficient for constant-bit rate traffic
 suitable when stations generate a steady stream of information that makes efficient
use of the dedicated channel.
cons:
 Inflexible in allocation of bandwidth to users with different requirements
 Inefficient for bursty traffic
 Does not scale well to large numbers of users
Preferred approach in
 Cellular telephone networks
 Terrestrial & satellite broadcast radio & TV
Channelization:
(1) FDMA
(2) TDMA
(3) CDMA
Dynamic medium access control (MAC) schemes
They involve a dynamic sharing of the medium on a per frame basis that is better matched to
situations in which the user traffic is bursty.
This primary function of MAC is to minimize or eliminate the incidence of collisions to
achieve a reasonable utilization of the medium with two basic approaches, random access and
scheduling.
Scheduling:
pros and cons:
Schedule frame transmissions (assignment) to avoid collision in shared medium
 More efficient channel utilization
 Less variability in delay
 Fairness provisioning to stations
 Increased computational or procedural complexity
approach:
(1) Reservation
(2) Polling
(3) Token passing
Random Access:
Each station transmits its reservation message randomly until the message goes through.
pros:
 Simple and quick transfer at low load
 Can accommodate large number of bursty users
cons:
 delay result from heavy traffic
approach:
(1) ALOHA
(2) CSMA
4.
(a) Please write down the maximum achievable throughputs formula for ALOHA, CSMA and
CSMA/CD and explain the meaning of parameters used in the formula (15%).
(c) Compare the maximum achievable throughputs of the random access schemes including
ALOHA, CSMA and CSMA/CD (5%).
5.
Please describe the operation of the Ethernet hub and Ethernet switch topologies using the
twisted-pair cabling (10%) and explain how the star topology improves the performance over
the bus topology (5%).
Solution:
(a) In Ethernet hub topology, the hub monitors all transmissions from the stations. When there
is only one transmission, the hub repeats the transmission on the other lines. If there is a
collision, then the hub sends a jamming signal to all the stations to implement the backoff
algorithm. In this approach the stations are in the same collision domain, where a collision
will occur if two or more stations transmit simultaneously.
In Ethernet switch topology, each input port buffers incoming transmission. The incoming
frames are examined and transferred to the appropriate outgoing ports. Each outgoing port
implements the Ethernet MAC protocol to transmit frames. In this case the group of stations
attached to the same line constitutes a collision domain. Switching LAN provides a means
of interconnecting larger numbers of stations without reaching the limit of shared
transmission medium.
(b) Using twisted pairs in a star topology can improve the performance because the loading of
each channel is lower and the system can still remain the operation even if one of the
channels is broken down. In addition, the stations implement the CSMA-CD protocol,
which can reduce the probability of collision occurrence.
6.
Describe the operation of the distributed coordination function (DCF) of the IEEE802.11
WLAN (12%), also explain the operation of the point coordination function (PCF) of the
IEEE802.11 WLAN (8%).
Solution:
(a) The distributed coordination function (DCF) is the basic access method used to support
asynchronous data transfer on a best-effort basis. The DCF is based on the CSMA-CA
protocol. High-priority frames must only wait the short interframe space (SIFS) period
before they contend for the channel. The PCF interframe (PIFS) is intermediate in duration
and is used by the PCF to gain priority access to the medium at the start of a CFP. The DCF
interframe space (DIFS) is used by the DCF to transmit data and management MDPUs.
When the data frame is transmitted, the duration field of the frame let all station in the BSS
know how long the medium will be busy. All stations hearing the data frame adjust their
NAV based on the duration field value, which includes the SIFS interval and the
acknowledgement frame following the data frame.
(b) The point coordination function (PCF) is an optional capability that can be used to provide
connection-oriented, contention-free services by enabling polled stations to transmit without
contending for the channel. The PCF function is performed by the point coordinator (PC) in
the AP within the BSS. The CFP repetition interval determines the frequency with which the
PCF occurs. At the beginning of each CFP repetition interval, the so-called target beacon
transmission times (TBTT), all stations in the BSS update their NAV to the maximum length
of the CFP. During the CFP, stations may transmit only to respond to a poll from the PC or
to transmit an acknowledgment one SIFS interval after receipt of an MPDU. At the nominal
start of the CFP, the PC senses the medium. If the medium remains idle for a PIFS interval,
the PC transmits a beacon frame to initiate the CFP. In case the CFP is slightly loaded, the
PC can foreshorten the CFP and provide the remaining bandwidth to contention-based
traffic by issuing a CF-End or CF-End+ACK control frame. This action causes all stations
that receive the frame in the BSS to reset their NAV values.