Name: ________Jeffrey Absher____________
TDC462
DATA COMMUNICATIONS
Final Exam
Summer I 2002 Quarter
due by
July 17, 2002
Wednesday, 11:59:59pm
Total
Overall Grade Points
Total Points
100
40
Points
Remarks:
1. Read the questions carefully and clearly state any assumptions that you make.
2. You must show all of your work, including all the intermediate steps of a calculation,
to get ANY credit.
3. Write clearly and eligibly, unreadable answers will be marked wrong.
4. Due date is July 17th by 11:59:59pm.
5. No late submission will be accepted.
6. Please turn it in via DLWeb.
7. You have got 4 additional points on the exam, which may be considered as extra
credit. That is, the total is 104 while 100 is full credit for the final.
8. Please answer them individually. Not to scare you but, in the very recent past, I
caught 8 students in cheating in take-home finals. So, please…
(8 Points) 1. Synchronization is an important protocol function. What are the 4
most commonly used techniques to implement this function? Explain each briefly.
a. Synchronization for synchronous transmission can occur at the Bit level or
at the Block Level.
At the bit level, the data blocks are transmitted without having a start or a
stop bit.
i)
You can use Manchester or another biphase encoding method that
includes an extra mid-bit transition. This extra transition is used to
synchronize the clocks between the sender and the receiver.
ii)
You can use an external clock to synchronize sender and receiver
over short distances. (or one side can act as the clock)
b. At the block level for synch transmission
i)
You can use preambles and postambles or expected patterns to
synchronize the clocks between sender and receivers so the both know
the start and end of each block.
c. For Asynchronous transmission, each sent character is synchronized by
itself, when the receiver notices a transition, it listens to the medium for the
next character length, and the transmitter and receiver re-synch for each
character, if the transmission is continuous, there is a minimum stopelement time between characters. This is good for sparse transmissions, but
has a high overhead for dense data transmissions.
(8 Points) 2. Explain how digital and analog data are transmitted over T-1 carrier
(DS-1 Frame format). Derive the actual user data rate in Mbps for analog transmission,
digital transmission with robbed bit signaling and digital transmission with clear
channel.
A DS1 consists of 24 64000 bps channels TDM’ed into 8000 frames/sec of 193 bits
(24 * 8bits+1 framing bit) per frame providing a gross data rate of 1.544 Mbps.
T1’s use AMI meaning that a pulse (high or low) indicates a 1 and that a non-pulse
(middle) is a zero; they must maintain a 12.5% 1’s density and there may be no
more than 15 consecutive zero’s.
a. For Analog Transmission, The analog data for each channel is transformed
to and from digital data with ADC’s/DAC’s (PCM). The first bit of each
193-bit frame, and the 7th bit of every 6th frame is used as overhead for
framing control and synchronization. Resulting in a net user data rate of
8000 frames/sec * (192 bits/frame * 5/6 + 1/6 * (192 – 24) bits/frame)
= 1.504 Mbps
b. For Digital with robbed bit signaling, the result is that one cannot use 64000
bps for the DS0’s. The maximum speed for a DS0 that is multiplexed into a
T1 w/ RBS is 56000 bps, so 24 channels*56000 bits/(sec channel) = 1.344
Mbps
c. For Digital Clear Channel (B8ZS encoding) There are no bits robbed, and
the T1’s 1’s density is maintained by sending a special-case byte for an allzero’s byte that violates the expected polarity. This results in the full
64kbps of each channel being usable for a net user datarate of 64000
bits/(sec channel) * 24 channels = 1.536 Mbps
(8 Points) 3. Two directly connected nodes (A and B) use a sliding-window
protocol with a 3-bit sequence numbers in the DLC protocol field. They employ GoBack-N ARQ mechanism with a window size of 6. Show the window positions of A
and B after each of the following events, except for part (f) for which just answer the
question.
a. Before any transmission
A:0,1,2,3,4,5 B:0,1,2,3,4,5
b. Station-A sends frames 0, 1, 2, 3 and 4.
A:5 B:0,1,2,3,4,5
c. B receives 0, 1, 2 and 3, but can only process 0, 1 and 2 and sends ACK for these 3
frames.
A:5 B:4,5,6,7,0
d. A receives the ACK from B
A:5,6,7,0 B:4,5,6,7,0
e. A transmits frames 5, and 6.
A:7,0 B:4,5,6,7,0
f. B receives 5 and 6. What should B do at this point?
At minimum it should send a REJ4 and discard frames 5 and 6, it may also want
to send an ACK3 after processing frame 3.
(8 Points) 4. Utilization of the CSMA system depends on the propagation time and
transmission time. Explain in details under what conditions a better utilization will
result in by means of the two aforementioned parameters.
If transmission time is larger (frames are larger), and propagation time is kept
small (maximum distance shorter), then utilization of the CSMA medium is better.
The posit [propagation time << transmission time] allows for each potential
transmitting station to know almost immediately that a transmission has
commenced and that the medium is in use, so they do not attempt to transmit and
cause a collision. A small propagation time makes the possible collision window
very small. A large transmission time means that there are fewer windows per unit
time. Both of these will result in fewer collisions, Fewer collisions results in more
data transmitted per unit time. The one problem is if the transmission time is
enlarged too much, then utilization could go back down again because stations will
have to wait too long for other stations to relinquish the shared medium. Also
increasing the transmission time could mean that transmission slots could get too
large and then the medium is unused for too long while a station must wait for the
next slot to start.
(8 Points) 5. Critique the following statement: “So long as we keep the offered load
below the network capacity, additional traffic will always increase the throughput of the
system.”
Throughput is generally defined as the data transfer rate. (data transferred/time)
so the statement is correct, if the network can handle additional traffic (because
the load < capacity) then that additional traffic would increase throughput. But,
the statement is almost trivial because you are merely increasing throughput by
having artificial bounds on your offered load, thus setting up a 1-to-1 ratio of your
product (offered load) to your raw material (network capacity). This is not the
manner that you would want to increase throughput, instead, proper management
of resources and statistically determining loads based on time of day (and
rescheduling jobs) could increase the throughput. As well data compression could
help. So I would not only be wary of this statement, but also of the person that said
it and of people that felt that it had import.
(8 Points) 6. Is there a minimum and maximum frame size requirements for IEEE
802.3 Ethernet standard? Explain why or why not.
Yes, There is both a maximum and a minimum. The minimum (64 bytes) is because
you do not want to finish sending a frame before you know whether a collision has
occurred or not, therefore the minimum is based on the maximum distance between
any 2 nodes and the type of medium connecting the nodes. Having larger frames
and shorter propagations also increases media utilization. The 802.3 frame
maximum is due to the Length Field of the 802.3 spec being the old type field of
Ethernet II, and that the type-field of Ethernet II only contained valid values above
1536. Hence this field can be used by network devices to auto detect Ethernet II or
802.3.
(8 Points) 7. What is the period of the signal represented by the following sine wave
function?
s(t )  sin( 2 10 6 t )  13 sin( 32 10 61 t )  15 sin( 52 10 6 t )  17 sin( 72 10 6 t )
The Period (based on fundamental frequency) is ‘t’ when 106t/2=2
106t/2=2
106t/2=2
106t=4
t=4/106
t = 0.000004
And the fundamental frequency would be 1/(4E-6) or 250000hz.
(8 Points) 8. A data link uses 1000 bit frames, has a data rate of 10 Kbps, and has a
distance of 10 km. Assume that the velocity of the signal is 100 km/s. What is the link
utilization (U) if stop-and-wait flow control is used?
Utilization = tframe/(2 tprop + tframe)
U= 1/(1 + 2a)
a = (d/V)/(L/R)
a = (10000 m/100000 m/s) / (1000 bits / 10000 bits/sec)
a=1
Utilization = 1/ (1 + 2) = 1/3 = 0.333
(8 Points) 9. Problem 10.20 on page 341 (Figure 10.20b only).
fromYtoX
A
B
C
D
E
F
G
H
I
J
K
A
X
1,A
2,B
4,H
1,A
5,C
2,E
3,G
X
6,D
5,D
B
1,B
X
1,B
4,H
2A
4,C
2,C
3,G
X
5,C
5,F
C
2,B
1,C
X
3,H
2,G
3,C
1,C
2,G
X
4,C
4,F
D
4,E
4,C
3,G
X
3,G
2,K
2,H
1,D
X
2,D
1,D
E
1,E
2,A
2,G
3,H
X
5,C
1,E
2,G
X
5,D
4,D
F
5,B
4,C
3,F
2,K
5,G
X
4,C
3,D
X
4,D
1,F
G
2,E
2,C
1,G
2,H
1,G
4,C
X
1,G
X
4,D
3,D
H
3,E
3,C
2,G
1,H
2,G
3,K
1,H
X
X
3,D
2,D
I
X
X
X
X
X
X
X
X
X
X
X
J
6,E
5,C
4,J
2,J
5,G
4,K
4,H
3,D
X
X
3,D
K
5,E
5,C
4,F
1,K
4,G
1,K
3,H
2,D
X
3,D
X
(8 Points) 10. Suppose that you are working for an ISP (Internet Service Provider)
as a network engineer. Would you want to maximize or minimize your company’s
network utilization? Explain your reasoning.
Per-subscriber I want to minimize my network utilization, so I want to have each
subscriber use as little of the network resources as possible (if they only need 128
kbps, then allocate on1y 128 kbps to their connection). But, for things that I pay
for such as a T1 or a FR circuit, I want to maximize its utilization over time. So I
want every “monthly cost” resource to be maximized (maximize overhead cost
utilization), while I want every per-unit cost product to have minimum cost
(minimize production costs).
(8 Points) 11. What are two functions of HEC field of the ATM header? Explain
which layer uses it and how.
a. The HEC is the Header Error Control Field of the ATM frame (cell). It is 8
bits and is first used as a CRC for ATM to detect and correct errors on the
ATM header (not the payload), this occurs at the “ATM layer” of the ATM
stack.
b. Also, the physical layer of ATM can use the HEC field if “cell-based
physical layer” is used during initial startup and synch. The receiver goes
through a set of states (FSM) to try to find a valid HEC field, and based on
thresholds of correct or incorrect consecutive HECs determines whether it
is synchronized with the sender.
(8 Points) 12. Explain how binary exponential backoff algorithm works.
For the ‘nth’ attempt at transmitting this frame:
a. Wait until the medium is free, attempt to transmit a frame
b. If a collision is detected, all active transmitters immediately send a “JAM”
signal for 32 bits, and then stop transmitting.
c. Each transmitter then generates a random number between 0 and 2min(n,10)1 and waits that many “slottimes.”
d. Then increase the ‘n’-count and go back to step a unless n > 16
(8 Points) 13. Define the following terms: Admission Control, Usage Parameter
Control, Traffic policing, Traffic shaping, AIMD.
a. Admission control – Simply not allowing the initial “call” to go through to
avoid congestion. (all circuits are busy, please try your call again later) This
is used in ATM during the initial VCC/VPC request.
b. Usage parameter control – On VCC’s and VPCs, if the peak cell rate
(within the agreed delay variation parameters) exceeds the agreed-upon
peak cell rate then the exceeding cells may be discarded or tagged as low
priority. Similarly for sustained cell rates within burst tolerance
parameters.
c. Traffic policing – Generic term to describe Usage Parameter Control and
similar methods of congestion control that discard cells that exceed the
agreed-upon limits.
d. Traffic shaping – in contrast to Traffic policing, Traffic shaping attempts to
“smooth out” the bursts of network traffic by adding buffers to hold cells
that exceed agreed-upon limits until they can be delivered successfully
under the limits. Of course this could fail because the buffers could fill up.
e. AIMD - additive increase and multiplicative decrease. A concept of slowly
(additively) approaching an unknown threshold (such as optimal window
size) and upon an error or reaching the threshold, dropping back below
that threshold quickly. AIMD is used in the ‘congestion-avoidance’ phase of
TCP end-to-end congestion control.