Tricky Network Security — Test 1

Correct Answer: C. C. 135 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.

Correct Answer: B. B. 92 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.

Correct Answer: A. A. 38 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.

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

Correct Answer: D. 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.

Correct Answer: A. A. D/(d + log2(n))
Explanation: In collision free protocol channel efficiency is given by d/(d + log2(N)).

Correct Answer: D. D. A-e-g-h-d-k
Explanation: Shortest path is: A-E-G-H-D-K and the distance is 5 units.