CS 447 Network and Data Communication
Midterm Exam No.2 (SOLUTIONS)
Spring, 2003
7:30-8:45 P.M.
April 8, 2003
This question is a closed book and closed note exam. There are 6 questions in this exam.
You have 75 minutes to finish the questions. Please write your answers on separated
pieces of papers. To avoid grading problems, please staple your papers in the ascending
order in the question number. Please bring a calculator (no calculator sharing, please).
Name: _______________________________
Student ID (optional): ___________________
Question #1 (10 minutes - 10 points)
(1) What are the two potential problems for throughput in “buffered-statistical” TDM?
Mention two possible problems. – 2 points
(1)
(2)
Memory Access Speed
Buffer capacity
(2) What are the two types of switches used in a circuit-switching network? – 2 points
(1)
(2)
Blocking Switches and Non-blocking Switches
Time Division Switch and Space Division Switches
(3) What are the internal and external operations in the Internet? – 2 points


Internal operation = Datagram
External operation = virtual circuit
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(4) What are the two types of switching networks? – 2 points
(1)
(2)
Circuit Switching Networks
Packet Switching Networks
(5) Compare, circuit switching, virtual circuit packet switching and datagram packet
switching networks in terms of  bandwidth sharing of links and  message storing
at each router – 2 points
For question :
Circuit Switching Networks: bandwidth is dedicated (no one else can use it)
Virtual Circuit Packet Switching Networks: bandwidth is reserved but shared
Datagram Packet Switching Networks: every one shares the link bandwidth
For question :
Circuit Switching Networks: no store-and-forward (data as continuous flow of bits)
Virtual Circuit Packet Switching Networks: could be store-and-forward
Datagram Packet Switching Networks: store-and-forward
Question #2 (15 minutes - 20 points)
(1) What is the minimum requirement (i.e., how much transmission rate is required) for
the shared transmission cable for a statistical TDM shown below (make sure this
TDM is not a blocking statistical TDM)? – (10 minutes) 10 points

N input lines

Multiplexer
N output lines
De-Multiplexer
Shared Transmission Cable
Assumptions are:


Each slot now has a slot header (the slot header includes only the information
that indicates the destination output line – nothing else).
Assuming there are 128 input lines.
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

All the input lines have the same transmission rate of 10Mbps.
The slot includes 35 bits as user payload for each input line (excluding the
slot header)
Solution:
Since there are 128 input lines, the slot header requires log2128 bits, which is 7 bits.
Since there are 35 bits as user payload in each slot, the slot size is 35 + 7 = 42 bits. The
ratio of the payload to the overhead is 5:1. Since there are 128 input line, each of which
has transmission rate = 10 Mbps, 1280 Mbps is required only for the user payload. If the
slot header overhead is represented by X Mbps, the X will be:
1280 Mbps:X Mbps = 5:1
Solving this equation, we will get:
X = 256 Mbps.
The total transmission rate should be 1280 + 256 Mbps = 1536 Mbps.
(2) Assume that the buffered-statistical TDM will be used. If the link bandwidth of the
shared transmission cable is 900 Mbps, and each input line always transmits data at
120 Mbps, what is the requirement for the throughput of the memory buffer (in the
number of bits that have to go through the memory unit per second), if N = 10? – (5
minutes) 10 points
Shared Transmission Cable

N input lines

Multiplexer
N output lines
De-Multiplexer
Memory Buffer
Solution:
This is one of the exercise questions we worked in the classroom. Since the difference
between the total input rate and the line capacity of the shared transmission cable is 1200900 = 300 Mbps, we will have 300M bits in the memory buffer after one second. In the
next 1/3 second, the 300M bits in the memory buffer will be transmitted. While the
300M bits in the memory buffer are being transmitted, new 400M bits will arrive and
they are stored in the memory buffer. This makes 700M bit accesses in 1/3 second. In
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the next 1/3 second, the same thing repeats, which makes another 700M bit accesses in
1/3 second. This pattern just repeats. Therefore, the required throughput is 2100M
bits/second.
Question #3 (10 minutes - 10 points)
(Exercise #8.13) Ten 57.6-Kbps lines are to be multiplexed using TDM. Ignoring
overhead bits, what is the total capacity required for synchronous TDM? Assuming that
we wish to limit average line utilization of 75%, and assuming that each line is busy 45
percent of the time, what is the capacity required for statistical TDM?
Solutions:
Since there are ten 57.6-Kbps lines, the total input rate is 576Kbps. However, the
utilization of each input line is 45%, the effective total input rate is 576-Kbps  0.45,
which is 259.2–Kbps. 259.2-Kbps must be 75% (i.e., ¾) of the link capacity. Therefore
the link capacity needed is:
259.2-Kbps  4/3 = 345.6-Kbps.
Question #4 (10 minutes - 10 points)
(Exercise #9.2) Consider a simple telephone network consisting of two end offices and
the intermediate switch with a 1-MHz full-duplex trunk (trunk = a shared transmission
cable) between each end office and the intermediate switch. The average telephone is
used to make four calls per 8-hour workday, with a mean call duration of six minutes.
Ten percent of the calls are long distance (i.e., not local – they go through the
intermediate switch). What is the maximum number of telephones and end office can
support?
Solution:
The question says, each phone makes 4 calls in 8 hour and the average call duration is 6
minutes. This means each telephone set occupies a line for 24 minutes (46) in 480
minutes (= 8 hours). Therefore, one telephone cable can support up to 20 telephone sets
(280/20).
The shared trunk cable has the bandwidth of 1Mhz. Assuming that each call occupies
64KHz, the number of simultaneous calls this trunk cable support is Floor(1,000K/64K),
which is 15 (the shared trunk cable supports up to 15 simultaneous call). This means 15
64K lines are multiplexed in one trunk cable.
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We know (from the analysis in the first paragraph) that each 64KHz line supports 20
telephone lines. Since only 10% of the calls are outgoing calls, each 64KHz line in the
shared trunk supports 200 telephone sets.
20015 = 3,000
3,000 telephone sets
Question #5 (10 minutes - 15 points)
(Exercise #10.2) Define the following parameters:
N: Number of hops between two given end hosts
L: Message length in bits
B: Data rate in bps on all links
P: Fixed packet size in bits
H: Overhead bits per packet
D: Signal propagation delay
Assume that the signaling message to set up a path and the ACK message for path set up
so small that the packet transmission delay is negligible for both of them. Also assume
that there is no overhead to process the signaling message.
N = 4, L = 3200, B = 9600, P = 1024, H = 16, D = 0.001, compute the estimated time
required to complete the data transmission.
The total time is: (2ND)+(ND)+(P/BN)+(Ceiling(L/(P-H))-1) P/B. You plug in
the numbers and finish calculation.
Question #6 (20 minutes - 10 points)
This question is about TDM bus switch. The figure below shows the internal structure of
typical 33 TDM bus switch. The transmission rate of the shared bus is usually R  n
Mbps where R is the transmission rate (in Mbps) of each input (assume that all input lines
TDM Bus switch
Input Buffer
Input lines
Switch Control Unit
A
B
C
Output switch
Input switch
5
Shared Bus
X
Y
Z
Output lines
have the same transmission rate and they are always transmitting at their full transmission
rate and there are n input lines.
The length of the TDM bus (i.e., the shared cable within a TDM bus switch) will be a
significant problem if transmission rate of each input line increases. The problem is the
signal propagation delay in the bus. Even after a bit has been transmitted at an input
switch, the output switch must hold a connection until the bit signal propagates in the bus
cable to this output switch. Only after this, the switch of the next input line can start
transmit its data.
Question (a): In order to demonstrate this problem, calculate the utilization of the
transmission cable if R = 1,000 Mbps (= 1 Gbps) and n = 3. Assume that bit interleave is
used (i.e., the bus transmits a bit for each input at a time) and the transmission rate of the
bus cable is 3 Gbps. Assume that the distance of the bus cable is 10 meters and the signal
propagation speed is 300,000Km/second. Assume that the bus length is same for any
input-output port pair.
Hints: 1 bit at 1Gbps = 1ns, which is 10-9 second.
Question (b): If two-bit inter-leaving (the bus transmits two bits for an input line at a
time) is used, what will happen to the bus cable utilization (will it increase or decrease)?
Hint: this question is really easy, if you understand Question (a) above. You could
answer this question within two minutes without calculator.
Solutions for Question (a):
Bits are transmitted at each input port at 3 Gbps, which is 0.33ns. For signal to propagate
10meter cable, it will take 10/300,000,000 = 33ns. According to the definition of the
utilization:
U = 0.33/(33+0.33) = 0.33/33.33 = 0.99%.
Solutions for Question (b):
If two bits are transmitted at a time for an input line, the utilization will be:
U = (0.33  2)/(33.00 + 0.66) = 0.66/33.66 = 1.97% (it will increase).
CS 447-Network and Data Communication, Midterm Exam #2, April 8, 2003
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