Channel has data rate of 4 Kbps and propagation delay of 20mswhat range of frame size do stop and wait give an efficiency of atleast 50?

Answer 1

You need to use two formulas - firstly a= propagation delay/transmission time. We'll denote this by tprop and ttrans.

tprop is given in the question, it is 20 x 10-3
ttrans is Distance/Data Rate

so ttrans = L / 4 x 103

ttrans = 80/L

(becasue the powers of ten cancel each other out so you just multiply 20 x 4.
we use powers of ten because we need the same format as the propogation delay

a = 80/L

2. Use the formula 1/1+2a >= .5 = 1/1+ 160/L>=.5


First 1>= .5(1+160/L)
second 1>= .5 + 80/L
third 80/L >= 1 x .5
fourth .5 >= 80/L
L >= 80/.5
L >= 160