edited by
3,076 views

2 Answers

Best answer
8 votes
8 votes

Packet Size = 1KB, Bandwidth = $10^9$ bps, Propagation Time = 15 ms
$\text{Transmission Time} = \frac{\text{Packet Size}}{\text{Bandwidth}} \\= \frac{1 \text{ KB}}{ 10^9 \text{ bps}} \\ = \frac{1024 \times 8 }{10^9 \text{ bps}} \approx 0.008 \text{ ms}.$

Sender Utilization = Fraction of Time Sender is busy (Same as link utilization which is the fraction of time the link is carrying useful data)

In Stop-and-wait, sender can retransmit only when ACK arrives for the send data which requires 1 transmission time for the data packet, 1 propagation delay for the data packet to reach the receiver, 1 propagation delay for the ACK to reach the sender. (ACK normally being small its transmission time can be ignored unless given in question). So,

$\text{Utilization} = \frac{\text{Transmission Time}}{\text{ Transmission Time } + 2 \times \text{Propagation Time}}  \\= \frac{0.008}{0.008 + 2 \times 15} \\= \frac{0.001 }{30.008} \\= 0.000266  = 0.026 \%.$
 

selected by
3 votes
3 votes
Here in the given L= 1* 1024 * 8 bits, Tp =15ms= 15000 micro sec

channel capacity = BW (units in b/s) so,BW= 10^9 b/s

Now,Tt =  L/B =1*1024*8 / 10^9 =8.1 micro sec

Utilisation is 1 / 1+2a where a is Tp/Tt

1+2a= 3704.7 , 1 / 1+2a =1 / 3704.7  = 0.02%
Answer:

Related questions

0 votes
0 votes
2 answers
3
Shivam_j asked Oct 16, 2022
672 views
Class B network on the internet has a subnet mask of 255.255.119.0 what is maximum possible hosts per subnet. Assuming Classfull Addressing Scheme