Foundation of Network 13

Foundation of Network 1
Foundation of Network
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Foundation of Network 2
Answer to question 1
The rate of data transmission= 10kbps
The frame size, n=898 bytes
The information bytes per single frame, k = 870 bytes
Propagation delay, td= 25ms
The acknowledgement size, ta= 8 bytes
The roundtrip processing delay, tp = 10ms
Tf= (898*8)/8000=0.898
Ta= (8*8)/8000=0.008
Tp=(10*8)/8000=0.01
Td=(25*8)/8000=0.025
Tf+ta+tp+td=0.898+0.008+0.01+0.025=0.941
The throughput k = (870*8)/0.941=6960/0.941=7396.38 bytes
=7.396 kbps
A=td/tf
A=0.025/0.898=0.0278
1/ (1+2a)= 1/(1+0.0556)
= 94.73%

Foundation of Network 3
Answer to question 2
Bit error rate= 0.00001
If a bit is error free it’s probability= 1-0.00001= 0.99999.
Frame length being free of error= (0.99999) 1024= 0.989812
The probability of Error free frame is: 1- 0.9898
P=0.0102
Answer to question 3
P-persistence In the case of idle media, the transmission
probability which is Prop; the time unit
delay probability will be (1-p). The time
unit is always similar to the maximum
delay of the propagation. Also, if there is a
busy media, the naval will continue up to
the point the channel becomes idle. When
the time unit delays, then one moves to
the first step.
1-Persistence During the period of ready frame, the
sender will check if the bus is busy or idle.
Due to the existence fo the delay of
propagation, the frame has never been
serviced.
Non-persistence In this case, 1-persistent and non-

Foundation of Network 4
persistent is combined as they try to
minimize the collisions which happen in
these two algorithms. Here, the station has
the time managing slots for the
transmitted messages. These messages
have to pick the random number still and
be transmitted only when the number is
less than the probability.
Answer to question 4
4a) Answer
First convert the megabytes per second to bytes per second
1Mbps= 100,000,000bps
If the 40% is being utilized,
It will be (40/100)*100000000=40,000,000bps
In every second, the transmitted number of information is 40,000,000 bytes
4b) Answer
From the 40,000,000 bytes transmitted per second, first convert this to Megabytes
40,000,000/1,000,000= 40Mbps
The frame number= (the average bits per frames number)/(S* time used for transmission)
The frames number= (800*8)/ (40*4)