Patterson Chap 2 Q 25

Question

25. Suppose you are designing a sliding window protocol for a 1-Mbps point-to-point link to the stationary satellite revolving around the Earth at an altitude of 3 × 10⁴ km. Assuming that each frame carries 1 KB of data, what is the minimum number of bits you need for the sequence number in the following cases? Assume the speed of light is 3 × 10⁸ m/s. (a) RWS=1 (b) RWS=SWS

Answer 1

Official Solution =>

25. One-way latency of the link is 100 ms. (Bandwidth)× (roundtrip
delay) is about 125 pps × 0.2 sec, or 25 packets. SWS should be
this large.
(a) If RWS= 1 , the necessary sequence number space is 26.
Therefore, 5 bits are needed.
(b) If RWS= SWS, the sequence number space must cover twice
the SWS, or up to 50. Therefore, 6 bits are needed.
 

Comments

@Shubham+Pandey+2 in above mentioned 1KB of data what should I take ?    $ 2^{10}  $  or $ {10}^{3} $

---

I have a few doubts in solving this question.

B = 1 Mbps = 10^6 b/s

d = 3*10^4 km = 3*10^7 m

L = 1 kB = 2^13 b

v = 3*10^8 m/s

Tt = transmission time = L/B = (2^13/10^6) s

Tp = propagation time = d/v = 0.1 s

1 < WS <= (1 +2a)  ( For sliding window, WS > 1)
For max efficiency, WS = (1 +2a) = 1 + 2*Tp/Tt = 25.4

Doubt 1: Is Bandwidth = 2^20 or 10^6?

Doubt 2: 1. If  (1+2a) is not a natural number, do we take floor or ceil?

Doubt 3: For finding minimum sequence number, do we assume that efficiency should be high? Because if efficiency can be low, then minimum will be (2+1)=3 (WS=2 because in sliding window WS>1).

Thanks.

---

B means bytes

b means bits

Answer 2

tp=3*10^7/3*10^8=0.1 s

RTT=0.2 s

bandwidth delay product=10^6*0.2 s

sender window size=10^6*0.2/8*10^3=25

If rws=1 then total sequence no=25+1=26 so 5 bits

if sws=rws

total sequence no=50

so no of bits=6