Tricky Network Security Subcategories

Tricky Network Security

A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces 1000 frames per second?

A. 150 frames
B. 80 frames
C. 135 frames
D. 96 frames

Explanation: Frame transmission time Tfr= 200/200 kbps or 1 ms. If the system creates 1000 frames per second, or 1 frame per millisecond, then G = 1 S = G × e−2G = 0.135 (13.5 percent) This means that, Throughput =1000 × 0.135 = 135 frames.

A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces 500 frames per second?

A. 146 frames
B. 92 frames
C. 38 frames
D. 156 frames

Explanation: If the system creates 500 frames per second, or 1/2 frames per millisecond. Then G = 1/2. S = G × e−2G = 0.184 (18.4 percent) This means that Throughput = 500 × 0.184 = 92 Only 92 frames out of 500 will probably survive This is the maximum throughput case, percentage-wise.

A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces 250 frames per second?

A. 38 frames
B. 48 frames
C. 96 frames
D. 126 frames

Explanation: If the system creates 250 frames per second, or 1/4 frames per millisecond, then G = ¼. S = G × e−2G = 0.152 (15.2 percent). This means that Throughput = 250 × 0.152 = 38. Only 38 frames out of 250 will probably survive.

A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps bandwidth. Find the throughput if the system (all stations together) produces 500 frames per second.

A. 92 frames
B. 368 frames
C. 276 frames
D. 151 frames

Explanation: G=1/2 S = G × e−G = 0.303 (30.3 percent) Throughput is 500 × 0.0303 = 151.

A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame?

A. 128 bytes
B. 32 bytes
C. 16 bytes
D. 64 bytes

Explanation: The minimum frame transmission time is Tfr = 2 × Tp = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision. The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet.

In collision free protocol channel efficiency is given by-

A. D/(d + log2(n))
B. D*(d + log2(n))
C. Log2(n)
D. (d + log2(n))

Explanation: In collision free protocol channel efficiency is given by d/(d + log2(N)).

Find the distance from source A to destination D using Dijkstra algorithm.

A. A-b-c-j-d-k
B. A-b-c-g-h-d-k
C. A-e-d-k
D. A-e-g-h-d-k

Explanation: Shortest path is: A-E-G-H-D-K and the distance is 5 units.