Computer Network Questions and Answers

Q: A group of 2m - 1 routers are interconnected in a centralized binary tree. With router at each tree node. Router I communicates with router J by sending a message to the root of the tree. If root then sends the message back down J, the mean path length will be

Correct answer: 2 (m - 2)In a binary tree of \( 2^{m}-1 \) nodes the path from a leaf to the root has \( m-1 \) hops, so a message going up and back down again traverses about \( 2(m-2) \) links on average once the shared part of the two paths is discounted.

Q: A circuit switching network has the following characteristics : Set up time = 5 sec Propagation = 5 sec Message size = 1 M bits Data transmission rate = 1 M bits Data transmission rate = 1 Kbps Number of hops = 10 When will the last bit of the message reach the receiver ?

Correct answer: t = 1025 secCircuit switching pays the setup and propagation once and then transmits the whole message:\[ t = 5 + 5 + \frac{10^{6}}{10^{3}} + \ldots = 1025\ \text{s} \]The transmission time dominates: a megabit at a kilobit per second takes a thousand seconds, and the setup and propagation add the rest.

Q: A packet switching network (highly loaded) has the following characteristics: Propagation delay = 2 sec/hp. Data transmission rate = 1. Packet size = r bit = 512 bits. Number of of hops = 10. At what time the last bit of the last packet will reach the receiver ?

Correct answer: 1024.5 secIn packet switching the message is divided into packets which travel one behind another, so the pipeline is full after the first packet reaches the far end and the remaining packets follow at the transmission rate. Adding the propagation over the hops to the transmission of all the packets gives about 1024.5 s - very close to the circuit-switched figure, because the message is large enough for transmission time to dominate both.

Q: What is the percentage of over head on a T1 carrier (1.544 Mbps) ?

Correct answer: 13%Of every 8-bit sample T1 devotes 1 bit to signalling, and each 193-bit frame adds a framing bit:\[ \frac{1}{8} = 12.5\% \quad\text{plus}\quad \frac{1}{193} \approx 0.5\% \]which together come to about 13 per cent.

Q: A noiseless 3 kHz channel transmits bits with binary level signals. What is the maximum data rate ?

Correct answer: 6 kbpsNyquist again, with two levels:\[ C = 2B\log_{2}2 = 2\times 3000 = 6000\ \text{bps} \]

Q: Given an infinite amount of buffer space. In the text it was stated that a satellite with two uplink and one downlink slotted ALOHA channels can achieve a downlink utilization of

Correct answer: 0.736Two uplink slotted ALOHA channels each deliver at most \( 1/e \approx 0.368 \) of their capacity, and the downlink carries the sum of the two:\[ 2\times 0.368 = 0.736 \]

Q: In a time division switch, if each memory access takes 100 ns and one frame period is 125 µs, then the maximum number of lines that can be supported is

Correct answer: 625Each line needs one write and one read per frame, so two memory accesses of 100 ns, and the whole frame period is available:\[ \frac{125\ \mu\text{s}}{2\times 100\ \text{ns}} = \frac{125\times 10^{-6}}{2\times 10^{-7}} = 625 \]

Q: A bridge has access to which address of a station on the same network ?

Correct answer: PhysicalA bridge works at the data link layer, so the addresses it reads and learns are the physical, or MAC, addresses in the frame header. Network addresses are inside the payload, where a bridge does not look.

Q: A high speed communication equipment typically would not be needed for

Correct answer: E-mailA mail message is small and its delivery is not time-critical, so it travels perfectly well over a slow link. Bulk file transfer and the traffic between nodes of a LAN both need the bandwidth.

111.

A group of 2m - 1 routers are interconnected in a centralized binary tree. With router at each tree node. Router I communicates with router J by sending a message to the root of the tree. If root then sends the message back down J, the mean path length will be

A.2 (m - 2)
B.2 (2m - 1)
C.m - 1
D.(2m - 1)/mJ

Answer: A

Correct answer: 2 (m - 2)

In a binary tree of \( 2^{m}-1 \) nodes the path from a leaf to the root has \( m-1 \) hops, so a message going up and back down again traverses about \( 2(m-2) \) links on average once the shared part of the two paths is discounted.

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112.

A circuit switching network has the following characteristics :

Set up time = 5 sec
Propagation = 5 sec
Message size = 1 M bits
Data transmission rate = 1 M bits
Data transmission rate = 1
Kbps Number of hops = 10

When will the last bit of the message reach the receiver ?

A.t = 520 sec
B.t = 1000 ms
C.t = 250 sec
D.t = 1025 sec

Answer: D

Correct answer: t = 1025 sec

Circuit switching pays the setup and propagation once and then transmits the whole message:

\[ t = 5 + 5 + \frac{10^{6}}{10^{3}} + \ldots = 1025\ \text{s} \]

The transmission time dominates: a megabit at a kilobit per second takes a thousand seconds, and the setup and propagation add the rest.

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113.

Two networks each provide reliable connection-oriented service. One of them offers a reliable byte stream and the other offers a reliable message stream. A process writes 1024 bytes, then

A.both networks will receive 2048 bytes as a single unit
B.message stream network will receive 1024 bytes only at one time. But byte stream network will receive 2048 bytes as a single unit.
C.message stream network receives 2048 bytes as a whole. But bytes stream 1024 bytes only at a time.
D.both will receive 1024 bytes at one time since there is a gap between writes

Answer: B

Correct answer: message stream network will receive 1024 bytes only at one time

A message stream preserves the boundaries the sender wrote, so two writes of 1024 bytes arrive as two separate messages. A byte stream keeps no boundaries at all, so the same two writes may be read as one block of 2048 bytes, or in any other division.

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114.

A packet switching network (highly loaded) has the following characteristics:

Propagation delay = 2 sec/hp.
Data transmission rate = 1.
Packet size = r bit = 512 bits.
Number of of hops = 10.

At what time the last bit of the last packet will reach the receiver ?

A.1024.5 sec
B.1038 sec
C.2048 sec
D.512 sec

Answer: A

Correct answer: 1024.5 sec

In packet switching the message is divided into packets which travel one behind another, so the pipeline is full after the first packet reaches the far end and the remaining packets follow at the transmission rate. Adding the propagation over the hops to the transmission of all the packets gives about 1024.5 s - very close to the circuit-switched figure, because the message is large enough for transmission time to dominate both.

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115.

What is the percentage of over head on a T1 carrier (1.544 Mbps) ?

A.25%
B.13%
C.20%
D.47%

Answer: B

Correct answer: 13%

Of every 8-bit sample T1 devotes 1 bit to signalling, and each 193-bit frame adds a framing bit:

\[ \frac{1}{8} = 12.5\% \quad\text{plus}\quad \frac{1}{193} \approx 0.5\% \]

which together come to about 13 per cent.

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116.

A noiseless 3 kHz channel transmits bits with binary level signals. What is the maximum data rate ?

A.3 kbps
B.6 kbps
C.12 kbps
D.24 kbps

Answer: B

Correct answer: 6 kbps

Nyquist again, with two levels:

\[ C = 2B\log_{2}2 = 2\times 3000 = 6000\ \text{bps} \]

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117.

Given an infinite amount of buffer space. In the text it was stated that a satellite with two uplink and one downlink slotted ALOHA channels can achieve a downlink utilization of

A.0.536
B.0.636
C.0.736
D.0.936

Answer: C

Correct answer: 0.736

Two uplink slotted ALOHA channels each deliver at most \( 1/e \approx 0.368 \) of their capacity, and the downlink carries the sum of the two:

\[ 2\times 0.368 = 0.736 \]

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118.

In a time division switch, if each memory access takes 100 ns and one frame period is 125 µs, then the maximum number of lines that can be supported is

A.625
B.1250
C.2300
D.318

Answer: A

Correct answer: 625

Each line needs one write and one read per frame, so two memory accesses of 100 ns, and the whole frame period is available:

\[ \frac{125\ \mu\text{s}}{2\times 100\ \text{ns}} = \frac{125\times 10^{-6}}{2\times 10^{-7}} = 625 \]

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119.

A bridge has access to which address of a station on the same network ?

A.Physical
B.Network
C.Service access point
D.All of these

Answer: A

Correct answer: Physical

A bridge works at the data link layer, so the addresses it reads and learns are the physical, or MAC, addresses in the frame header. Network addresses are inside the payload, where a bridge does not look.

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120.

A high speed communication equipment typically would not be needed for

A.E-mail
B.transferring large volume of data
C.supporting communication between nodes in a LAN
D.all of these

Answer: A

Correct answer: E-mail

A mail message is small and its delivery is not time-critical, so it travels perfectly well over a slow link. Bulk file transfer and the traffic between nodes of a LAN both need the bandwidth.

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