Can you please go over it and let me know is my answer correct? if not could you please fix it and let me know where I di wrong? a) Effective data throughput for the link between A and the wireless access point: Propagation delay = 50 km / 200,000 km/s = 0.25 ms Transmission time = 8,000 bits / 2 Mbps = 4 ms Effective data throughput = 8,000 bits / (2 * 0.25 ms + 4 ms) = 1,000 Kbps b) Effective data throughput for the link between the wireless access point and the wireless bridge: Propagation delay = 70 km / 200,000 km/s = 0.35 ms Transmission time = 8,000 bits / 1 Mbps = 8 ms Effective data throughput = 8,000 bits / (2 * 0.35 ms + 8 ms) = 500 Kbps c) Effective data throughput for the wireless link between the wireless bridge and B, assuming a bit error rate (BER) of 1x10^-5: Propagation delay = 60 km / 200,000 km/s = 0.3 ms Transmission time = 8,000 bits / 2 Mbps = 4 ms Frame error rate (FER) = 1 - (1 - BER) (frame _size) = 1 - (1 - 1x107-5)*(8,000) = 0.057 Effective data throughput = (1 - FER) * (8,000 bits / (2 * 0.3 ms + 4 ms)) = 0.943 * 1,000 Kbps = 943 Kbps d) For Stop and Wait on the complete path between A and B, the frame transmission time includes the propagation delay and transmission time for each link, as well as the round trip time (RTT) for each wireless link. For the link between A and the wireless access point: Propagation delay = distance / speed of light = 50 km / 200,000 km/s = 0.25 ms Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms RTT=2*05s=1s Frame transmission time = Propagation delay + Transmission time + RTT =0.25ms+4 ms +1s = 1.00425 s For the link between the wireless access point and the wireless bridge: Propagation delay = distance / speed of light = 70 km / 200,000 km/s = 0.35 ms Transmission time = frame size / link data rate = 8,000 bits / 1 Mbps = 8 ms RTT=2*05s=1s Frame transmission time = Propagation delay + Transmission time + RTT =0.35ms+8ms+1s = 1.00835 s For the wireless link between the wireless bridge and B: Propagation delay = distance / speed of light = 60 km / 200,000 km/s = 0.3 ms Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms RTT=2*05s=1s Frame transmission time = Propagation delay + Transmission time + RTT=03ms+4 ms+1s= 1.0043 s The total frame transmission time for the complete path is the sum of the frame transmission times for each link: Total frame transmission time = 1.00425 s + 1.00835 s + 1.0043 s = 3.0169 s Effective data throughput = frame size / total frame transmission time = 8,000 bits / 3.0169 s = 2,651.2 bps Assuming a bit error rate (BER) of 1x10^-5, the frame error rate (FER) for the wireless links can be calculated as before: FER =1- (1 - BER)Aframe_size) = 1 - (1 - 1x10^-5)^(8,000) = 0.057 The effective data throughput taking into account the FER for each wireless link is: Effective data throughput = (1 - FER1) * (1 - FER2) * (1 - FER3) * (8,000 bits / 3.0169 s) =(1-0.057) * (1- 0.057) * (1 - 0.057) * 2,651.2 bps = 2,099.5 bps Therefore, the effective data throughput from A to B assuming Stop and Wait on the complete path between A and B and taking into account the frame error rate is 2,099.5 bps.

Can you please go over it and let me know is my answer correct? if not could you please fix it and let me know where I di wrong? a) Effective data throughput for the link between A and the wireless access point: Propagation delay = 50 km / 200,000 km/s = 0.25 ms Transmission time = 8,000 bits / 2 Mbps = 4 ms Effective data throughput = 8,000 bits / (2 * 0.25 ms + 4 ms) = 1,000 Kbps b) Effective data throughput for the link between the wireless access point and the wireless bridge: Propagation delay = 70 km / 200,000 km/s = 0.35 ms Transmission time = 8,000 bits / 1 Mbps = 8 ms Effective data throughput = 8,000 bits / (2 * 0.35 ms + 8 ms) = 500 Kbps c) Effective data throughput for the wireless link between the wireless bridge and B, assuming a bit error rate (BER) of 1x10^-5: Propagation delay = 60 km / 200,000 km/s = 0.3 ms Transmission time = 8,000 bits / 2 Mbps = 4 ms Frame error rate (FER) = 1 - (1 - BER) (frame _size) = 1 - (1 - 1x107-5)(8,000) = 0.057 Effective data throughput = (1 - FER) (8,000 bits / (2 * 0.3 ms + 4 ms)) = 0.943 * 1,000 Kbps = 943 Kbps d) For Stop and Wait on the complete path between A and B, the frame transmission time includes the propagation delay and transmission time for each link, as well as the round trip time (RTT) for each wireless link. For the link between A and the wireless access point: Propagation delay = distance / speed of light = 50 km / 200,000 km/s = 0.25 ms Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms RTT=205s=1s Frame transmission time = Propagation delay + Transmission time + RTT =0.25ms+4 ms +1s = 1.00425 s For the link between the wireless access point and the wireless bridge: Propagation delay = distance / speed of light = 70 km / 200,000 km/s = 0.35 ms Transmission time = frame size / link data rate = 8,000 bits / 1 Mbps = 8 ms RTT=205s=1s Frame transmission time = Propagation delay + Transmission time + RTT =0.35ms+8ms+1s = 1.00835 s For the wireless link between the wireless bridge and B: Propagation delay = distance / speed of light = 60 km / 200,000 km/s = 0.3 ms Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms RTT=205s=1s Frame transmission time = Propagation delay + Transmission time + RTT=03ms+4 ms+1s= 1.0043 s The total frame transmission time for the complete path is the sum of the frame transmission times for each link: Total frame transmission time = 1.00425 s + 1.00835 s + 1.0043 s = 3.0169 s Effective data throughput = frame size / total frame transmission time = 8,000 bits / 3.0169 s = 2,651.2 bps Assuming a bit error rate (BER) of 1x10^-5, the frame error rate (FER) for the wireless links can be calculated as before: FER =1- (1 - BER)Aframe_size) = 1 - (1 - 1x10^-5)^(8,000) = 0.057 The effective data throughput taking into account the FER for each wireless link is: Effective data throughput = (1 - FER1) (1 - FER2) * (1 - FER3) * (8,000 bits / 3.0169 s) =(1-0.057) * (1- 0.057) * (1 - 0.057) * 2,651.2 bps = 2,099.5 bps Therefore, the effective data throughput from A to B assuming Stop and Wait on the complete path between A and B and taking into account the frame error rate is 2,099.5 bps.

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter9: Computer Networks

Section: Chapter Questions

Problem 3VE

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Question

Can you please go over it and let me know is my answer correct? if not could you please fix it and let me know where I di wrong?

a) Effective data throughput for the link between A and the wireless access point:

Propagation delay = 50 km / 200,000 km/s = 0.25 ms

Transmission time = 8,000 bits / 2 Mbps = 4 ms

Effective data throughput = 8,000 bits / (2 * 0.25 ms + 4 ms) = 1,000 Kbps

b) Effective data throughput for the link between the wireless access point and the wireless bridge:

Propagation delay = 70 km / 200,000 km/s = 0.35 ms

Transmission time = 8,000 bits / 1 Mbps = 8 ms

Effective data throughput = 8,000 bits / (2 * 0.35 ms + 8 ms) = 500 Kbps

c) Effective data throughput for the wireless link between the wireless bridge and B, assuming a bit

error rate (BER) of 1x10^-5:

Propagation delay = 60 km / 200,000 km/s = 0.3 ms

Transmission time = 8,000 bits / 2 Mbps = 4 ms

Frame error rate (FER) = 1 - (1 - BER) (frame _size) = 1 - (1 - 1x107-5)*(8,000) = 0.057

Effective data throughput =

(1 - FER) * (8,000 bits / (2 * 0.3 ms + 4 ms)) = 0.943 * 1,000 Kbps = 943 Kbps

d) For Stop and Wait on the complete path between A and B, the frame transmission time includes the propagation delay and transmission time for each link, as well as the round trip time (RTT) for each wireless link.

For the link between A and the wireless access point:

Propagation delay = distance / speed of light = 50 km / 200,000 km/s = 0.25 ms

Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms

RTT=2*05s=1s

Frame transmission time = Propagation delay + Transmission time + RTT =0.25ms+4 ms +1s =

1.00425 s

For the link between the wireless access point and the wireless bridge:

Propagation delay = distance / speed of light = 70 km / 200,000 km/s = 0.35 ms

Transmission time = frame size / link data rate = 8,000 bits / 1 Mbps = 8 ms

RTT=2*05s=1s

Frame transmission time = Propagation delay + Transmission time + RTT =0.35ms+8ms+1s =

1.00835 s

For the wireless link between the wireless bridge and B:

Propagation delay = distance / speed of light = 60 km / 200,000 km/s = 0.3 ms

Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms

RTT=2*05s=1s

Frame transmission time = Propagation delay + Transmission time + RTT=03ms+4 ms+1s=

1.0043 s

The total frame transmission time for the complete path is the sum of the frame transmission times for each link:

Total frame transmission time = 1.00425 s + 1.00835 s + 1.0043 s = 3.0169 s

Effective data throughput =

frame size / total frame transmission time = 8,000 bits / 3.0169 s = 2,651.2 bps

Assuming a bit error rate (BER) of 1x10^-5, the frame error rate (FER) for the wireless links can be

calculated as before:

FER =1- (1 - BER)Aframe_size) = 1 - (1 - 1x10^-5)^(8,000) = 0.057

The effective data throughput taking into account the FER for each wireless link is:

Effective data throughput = (1 - FER1) * (1 - FER2) * (1 - FER3) * (8,000 bits / 3.0169 s)

=(1-0.057) * (1- 0.057) * (1 - 0.057) * 2,651.2 bps = 2,099.5 bps

Therefore, the effective data throughput from A to B assuming Stop and Wait on the complete path

between A and B and taking into account the frame error rate is 2,099.5 bps.

Question 2:
Consider the system shown in the figure below. The frames have 8,000 bits, ACKs are
negligible in size, and the frame processing time is negligible at intermediate nodes. The
speed of light in the fiber is 200,000 Km/sec.
40 Kbps
0.5 sec RTT
Wireless
3:didend:
00000000010
10 Mbps
100 Km
Fiber
40 Kbps
0.5 sec RTT
Wireless
A
B
a) Assume that the frame transmission is error-free. Find the effective data throughput from
A to B when each wireless link uses Stop and Wait.
Hint: The effective data throughput through the system is equal to that of the bottleneck link.
b) Repeat the calculation assuming now that Stop and Wait is used on the complete path
between A and B.
Assume for questions c and d that the wireless links each have a BER of 1×10-5, the ACK transmissions are error free, and the fiber link is error free.
Note: when the wireless links are not error free, the achievable throughput drops to
(1-FER) multiplied by the error free effective data throughput of the system, where FER is the frame error rate.
c) Find the Frame Error Rate (FER), then find the effective data throughput from A to B when each wireless link uses Stop and Wait.
d) When Stop and Wait is used on the complete path between A and B, the Frame is error free if there is no error during the first wireless transmission and there is
no error during the second wireless transmission.
Find the Frame Error Rate FER in this case, then find the effective data throughput from A to B assuming now that Stop and Wait is used on the complete path
between A and B.

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Follow-up Question

Dear Writer the formulas and concepts which I have applied are correct yes only calculations needed to be corrected yes?

The correct step-by-step answer is as follows:

a) To calculate the effective data throughput for the link between A and the wireless access point, we need to first calculate the propagation delay and transmission time. Using the given values, we can calculate the propagation delay as:

Propagation delay = distance / speed of light = 50 km / 200,000 km/s = 0.00025sec=0.25 ms

The transmission time can be calculated by dividing the packet size by the link data rate:

Transmission time = 8,000 bits / 2 Mbps = 0.004sec = 4 ms

Finally, the effective data throughput can be calculated using the formula:

Effective data throughput = packet size / (2 * propagation delay + transmission time)

Plugging in the values, we get:

Effective data throughput = 8,000 bits / (2 * 0.25 ms + 4 ms) = 1777.77 Kbps

b) The calculation for the effective data throughput for the link between the wireless access point and the wireless bridge is similar to part (a), but with different values. The propagation delay can be calculated as:

Propagation delay = distance / speed of light = 70 km / 200,000 km/s = 0.35 ms

The transmission time can be calculated by dividing the packet size by the link data rate:

Transmission time = 8,000 bits / 1 Mbps = 8 ms

Finally, the effective data throughput can be calculated using the formula:

Effective data throughput = packet size / (2 * propagation delay + transmission time)

Plugging in the values, we get:

Effective data throughput = 8,000 bits / (2 * 0.35 ms + 8 ms) = 919.54 Kbps

To calculate the effective data throughput for the wireless link between the wireless bridge and B, we need to take into account the bit error rate (BER). In addition to the propagation delay and transmission time, we need to calculate the frame error rate (FER), which is the probability that a frame is received with errors. The FER can be calculated as:

FER = 1 - (1 - BER)^(frame_size)

Here, we assume a packet size of 8,000 bits. Using the given values, we can calculate the propagation delay as:

Propagation delay = distance / speed of light = 60 km / 200,000 km/s = 0.3 ms

The transmission time can be calculated by dividing the packet size by the link data rate:

Transmission time = 8,000 bits / 2 Mbps = 4 ms

Using the FER formula, we get:

FER = 1−(1−1x10−5)(8,000)=−7998

Finally, the effective data throughput can be calculated using the formula:

Effective data throughput = (1 - FER) * packet size / (2 * propagation delay + transmission time)

Plugging in the values, we get:

Effective data throughput = (1 - -7998) * (8,000 bits / (2 * 0.3 ms + 4 ms)) = 13911304 Kbps

d) For Stop and Wait on the complete path between A and B, the frame transmission time includes the propagation delay and transmission time for each link, as well as the round trip time (RTT) for each wireless link.

For the link between A and the wireless access point:

Propagation delay = distance / speed of light = 50 km / 200,000 km/s = 0.25 ms

Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms

RTT = 2 * 0.5 s = 1 s=1000ms

Frame transmission time = Propagation delay + Transmission time + RTT = 0.25 ms + 4 ms + 1000 ms = 1004.25 ms =1.004s

For the link between the wireless access point and the wireless bridge:

Propagation delay = distance / speed of light = 70 km / 200,000 km/s = 0.35 ms

Transmission time = frame size / link data rate = 8,000 bits / 1 Mbps = 8 ms

RTT = 2 * 0.5 s = 1s

Frame transmission time = Propagation delay + Transmission time + RTT = 0.35 ms + 8 ms + 1000ms = 1008.35 ms=1.00835s

For the wireless link between the wireless bridge and B:

Propagation delay = distance / speed of light = 60 km / 200,000 km/s = 0.3 ms

Transmission time = frame size / link data rate = 8,000 bits / 2 Mbps = 4 ms

RTT = 2 * 0.5 s = 1 s

Frame transmission time = Propagation delay + Transmission time + RTT = 0.3 ms + 4 ms + 1000 ms = 1004.3 ms =1.004s

The total frame transmission time for the complete path is the sum of the frame transmission times for each link:

Total frame transmission time = 3.01s

Effective data throughput =

frame size / total frame transmission time = 8,000 bits / 3.01 = 2657 bps

Assuming a bit error rate (BER) of 1x10^-5, the frame error rate (FER) for the wireless links can be

calculated as before:

FER =1- (1 - BER)Aframe_size) = 1 - (1 - 1x10^-5)^(8,000) = -7998.9

The effective data throughput taking into account the FER for each wireless link is:

Effective data throughput = (1 - FER1) * (1 - FER2) * (1 - FER3) * (8,000 bits / 3.0169 s)

=(1+7998.9) * (1 +7998.9) * (1 +7998.9) * 2657 bps = 1.36033

Therefore, the effective data throughput from A to B assuming Stop and Wait on the complete path

between A and B and taking into account the frame error rate is 1.36033

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