Algorithms
October 3, 2023JNU PhD CS 2019
October 3, 2023JNU PhD CS 2019
| Question 1 |
An upper-layer packet is split into 10 frames, each of which has an 80 percent chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through (in transmissions)?
| 18.6 | |
| 27.9 | |
| 9.3 | |
| None of the above |
Question 1 Explanation:
Total frames=10
Each frame has a chance of 80% undamage which is 0.8 of getting through, the chance of the whole message getting through is (0.8)10, which is about 0.107. Call this value p.
The expected number of transmissions for an entire message is then
Now use a = 1-p to get E = 1/p. Thus, it takes an average of 1/0.107, or about 9.3 transmissions.
Each frame has a chance of 80% undamage which is 0.8 of getting through, the chance of the whole message getting through is (0.8)10, which is about 0.107. Call this value p.
The expected number of transmissions for an entire message is then
Now use a = 1-p to get E = 1/p. Thus, it takes an average of 1/0.107, or about 9.3 transmissions.
Correct Answer: C
Question 1 Explanation:
Total frames=10
Each frame has a chance of 80% undamage which is 0.8 of getting through, the chance of the whole message getting through is (0.8)10, which is about 0.107. Call this value p.
The expected number of transmissions for an entire message is then
Now use a = 1-p to get E = 1/p. Thus, it takes an average of 1/0.107, or about 9.3 transmissions.
Each frame has a chance of 80% undamage which is 0.8 of getting through, the chance of the whole message getting through is (0.8)10, which is about 0.107. Call this value p.
The expected number of transmissions for an entire message is then
Now use a = 1-p to get E = 1/p. Thus, it takes an average of 1/0.107, or about 9.3 transmissions.