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Consider the use of 10 K-bit size frames on a 10 Mbps satellite channel with 270 ms delay. What is the link utilization for stop-and-wait ARQ technique assuming P = 10-3?

### My doubt - In frame size 10 K-bit, what should you take value of K? i.e K as 1000 bits or K = 1024 bits? I'm more inclined about taking K = 1024 bits, but the question's solution has K = 1000 bits!

Given solution - Link utilization = (1-P) / (1+2a)
Where a = (Propagation Time) / (Transmission Time)
Propagation time = 270 msec
Transmission time = (frame length) / (data rate)
= (10 K-bit) / (10 Mbps)
= 1 msec
Hence, a = 270/1 = 270
Link utilization = 0.999/(1+2*270) ≈0.0018 =0.18%

edited | 148 views
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In bandwidth we take 1kbps as  1000 bps

In frame size if 1 Kb is given , it's equivalent to 1024 bit
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@magma, see solution!

Given solution - Link utilization = (1-P) / (1+2a)
Where a = (Propagation Time) / (Transmission Time)
Propagation time = 270 msec
Transmission time = (frame length) / (data rate)
= (10 K-bit) / (10 Mbps)
= 1 msec
Hence, a = 270/1 = 270
Link utilization = 0.999/(1+2*270) ≈0.0018 =0.18%
+1

data rate given as = 10 Mbps

Tt = 10 x 2 10 / 10 x 106   = 1024 / 106  = 1024 x 10-6 sec

Tp =  270 x  10 -3  sec

Capacity of a channel = 2*Tp* BW

= 2*   270 x  10 -3  * 10 x 106

=  5400 x 10 3 bit

No of packets sent = 5400 x 10 3   /  10 x 210

=  527.34  = 527 packet

you can send 527 packet

but In stop and wait we only sent 1 packet

therefore link utilization : 1/527 = 0.001897 = 0.18 %

just go with the concept

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@magma y u r using 2*tp*b (since it is point to point link half duplex must be used)

In storage K= 1024 like in Operating System and CO

in transmission K = 1000 like in Computer Networks

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K = 1000 like in Computer Networks  even for the packet size/length?

Packet contains data.

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Both used but i think because we are using MKS system so better if you go with K=1000.

+1 vote