Lab/Tutorial :
(a) (1-persistent CSMA)
CCN2238 : Data Communications and Networking
Session 5 : MAC Protocols
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(Solution)
Q1. The finish time of each packet is as follows:
(b) (1-persistent CSMA/CD)
(c) (Non-persistent CSMA)
SEHH2238 Computer Networking
Tutorial 5
Q2. Let us find the relationship between the minimum frame size and the data rate. We know that Tfr = (frame size) / (data rate) = 2 × Tp = 2 × distance / (propagation speed)
or (frame size) = [2 × (distance) / (propagation speed)] × (data rate)] or (frame size) = K × (data rate)
This means that minimum frame size is proportional to the data rate (K is a constant). When the data rate is increased, the frame size must be increased in a network with a fixed length to continue the proper operation of the CSMA/CD. In Example 12.5, we mentioned that the minimum frame size for a data rate of 10Mbps is 512 bits. We calculate the minimum frame size based on the above proportionality relationship
Data rate = 10 Mbps Data rate = 100 Mbps Data rate = 1 Gbps Data rate = 10 Gbps
→ minimum frame size = 512 bits
→ minimum frame size = 5120 bits
→ minimum frame size = 51,200 bits → minimum frame size = 512,000 bits
Given t1 = 0s and t2 = 3μs
a) t3–t1 = Duration of A sends the first bit to C gets the first bit
= Distance / Propagation Speed = 2000 / 2×108 = 10 μs t3 = 10 μs + t1 = 10μs
b) t4–t2 = Duration of C sends the first bit to A gets the first bit = Distance / Propagation Speed = 2000 / 2×108= 10 μs
t4 = 10 μs + t2 = 13μs
c) Duration of data transmission in A = t4 – t1 = 13 – 0 = 13 μs No. of bits sent at A = 13μs x 10Mbps =130 bits
d) Duration of data transmission in C = t3 – t2 = 10 – 3 = 7 μs
Q4. Repeat Q3 Given t1 =
b) t4–t2 t4
bits sent at C = 7μs x 10Mbps =70 bits
if the data rate is 100 Mbps. 0s and t2 = 3μs
= Duration of A sends the first bit to C gets the first bit = Distance / PropagationSpeed = 2000 / 2×108= 10 μs = 10 μs + t1 = 10μs
= Duration of C sends the first bit to A gets the first bit = Distance / PropagationSpeed = 2000 / 2×108= 10 μs = 10 μs + t2 = 13μs
c) Duration of data transmission in A = t4 – t1 = 13 – 0 = 13 μs No. of bits sent at A = 13μs x 100Mbps =1300 bits
d) Duration of data transmission in C = t3 – t2 = 10 – 3 = 7 μs No. of bits sent at C = 7μs x 100Mbps =700 bits
SEHH2238 Computer Networking Tutorial 5 Page 2
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