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Leviafan [203]
3 years ago
10

An upper-layer packet is split into 10 frames, each of which has an 80% chance of arriving undamaged. If no error control is don

e by the data link protocol, how many times must the message be sent on average to get the entire thing through
Computers and Technology
2 answers:
nikdorinn [45]3 years ago
6 0

Answer:

The expected or average no. of transmissions = 9.32

Explanation:

Chance of each frame to arrive undamaged = 80%

Total number of frames = 10

Probability of getting whole message to arrive undamaged = p = 0.80¹⁰

p = 0.1073

q = 1 - p = 1 - 0.1073 = 0.8927

where q is the probability of not getting whole message to arrive undamaged

The expected or average no. of transmissions = ∑ kp(1-p)^{k-1}

where k is from 0 to ∞

The above series can be reduced to p/(1-q)²

The expected or average no. of transmissions = p/(1-q)²

The expected or average no. of transmissions = 0.1073/(1-0.8927)²

Therefore, the expected or average no. of transmissions = 9.32

Arisa [49]3 years ago
5 0

Answer: The message must be sent 9.313 approximately 9 times to get the entire data through.

Explanations: To find how many times the message must be sent on average to avoid error control in data link later.

E = 1/P

E = The average number of attempt before successful transmission.

P= Total probability of transmission without error.

STEP1 : FIND TOTAL PROBABILITY;

Since it each frame has a probability of 80% to be successful.

For each frame p = 80/100 = 0.8

For the 10 frame; total probability

P= (0.8)^10 = 0.1074

STEP2: FIND THE AVERAGE NUMBER OF OF TRIAL BEFORE A SUCCESSFUL TRANSMISSION WITHOUT ERROR;

Using equation above

E = 1/P

E= 1 ÷ 0.1074 = 9.313

Therefore they must be an average of 9.313 approximately 9 trials before a successful transmission without error.

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