Question

An upper-layer packet is split into 10 frames, each of which has an 80% 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

Answer 1

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

Answer 2

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.