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Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milliseconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is ______

  1. 20
  2. 10
  3. 30
  4. 40
in Others by Veteran (105k points) | 3k views

5 Answers

+5 votes

Bottleneck  bandwidth = 64 kbps=64x103 bps

Round trip delay = 40 ms =40x10-3 sec

Total data $40\times 64\times 10^3 \times 10^{-3} bits  = 40\times \dfrac{64}{8}  bytes =320 bytes$

1 packet size = 32 byte

No. of packets 320/32  = 10

Hence (2 ) is the ans

by Boss (49.3k points)
edited by
0

Why you multiply delay with rate

Here is the link which has different way to solve same kind of problem, that seems correct

http://www.geeksforgeeks.org/computer-networks-set-11/

+4 votes
transmission delay = Packet Length / Bandwidth = 32 Byte/ (64/8 KByte);

transmission delay = 4 ms;

1 Packet = Transmission delay

?(N) = Transmission delay + RTT( 2 * Propagation delay )

 No of packets (N) = (Transmission delay + RTT) / Transmission delay = 1 + 2a, where a = Propagation delay / transmission delay

N = 1+ ((2 *  20)/4)

Therefore,

1+10 = 11 Packets.

Usually Some author ignore "1" in "1+2a" formulae..

Hence, Answer can be 10 or 11
by (119 points)
0 votes

Total data 40x64x103 x10-3 =  =320 bytes.

No. of packets 320/32  = 10

ANS: 2

by Active (3.8k points)
0 votes
Length=32*8bits

Bandwidth=128*10^3bps

Transmission time =frame length/ Bandwidth

So ,Transmission time=4ms

RTT=40ms

No of frames to achieve 100% utilization= RTT/Transmission time

So 40/2=20

20 is the optimal window size
by (17 points)
0 votes
$\underline{\textbf{Answer:}\Rightarrow}\;\text{(D)}40$

$\underline{\textbf{Explanation:}\Rightarrow}$

Round Trip propogation delay $=80\;\text{ms}$

Frame Size $=32\times 8\;\text{bits}$

Transmission Time $=\mathbf{\dfrac{L}{B}} = \dfrac{32 \times 8}{128} \;\text{ms} = 2\;\text{ms}$

Let $\mathbf n$ be the window size.

Utilization $=\mathbf{\dfrac{n}{1+2a}}$, where $\mathrm {a = \dfrac{Propogation\; time}{transmission\; time}} = \dfrac{\mathrm n}{1+\dfrac{80}{2}}$

For maximum Utilization Efficiency $ = 1$

$\Rightarrow 1 = \dfrac{\mathrm n}{1+\dfrac{80}{2}}\\ \Rightarrow\mathrm n = 41$

Which is close to option $\mathbf D$

$\therefore\;\mathbf D$ is the correct answer.
by Boss (19k points)
edited by
0
why u took bandwidth as $128$ ? It's given as $64 kbps$ right ?
Answer:

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