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

The number of times that the message must be sent on average to get the entire thing through is; 10 times

The question did not tell us anything about datalink protocols.

Now, let us assume that there will be a successful transmission of the whole fragmented message with the single condition that all the frames arrive intact.

Thus;

P(successful transmission of the message) = P(all the frames arrive intact); p = 0.8¹⁰

Now, probability of unsuccessful transmission of the message;

q = 1 - p

Now, the Mean value of number of tries is simply the Expected value of k.

The formula for that is;

Σ[k][P(k)]

where k is from 1 to ∞

The binomial expansion of that formula gives us;

1p + 2qp + 3q²p + 4q³p

This can be reduced to;

p(1 + 2q + 3q² + 4q³ )

This can be further reduced to;

p/(1 - q)²

Recall that; q = 1 - p

Thus;

p/(1 - q²) = p/(1 - (1 - p))²

⇒ 1/p

Put 0.8¹⁰ for p to get;

1/0.8¹⁰ = 9.31

Thus, mean number of tries will be approximately 10.

Read more about expected value at; link

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 = ∑

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

deudtgc hdb d gd gsyjs bgwgsvsvd bsbeve,r3hfnfb. dv c cncnd.