Consider the following idealized WiMAX scenario. The downstream sub-frame (see F
ID: 3770734 • Letter: C
Question
Consider the following idealized WiMAX scenario. The downstream sub-frame (see Figure 6.17) is slotted in time, with N downstream slots per sub-frame, with all time slots of equal length in time. There are four nodes, A, B, C, and D, reachable from the base station at rates of 10 Mbps, 5 Mbps, 2.5 Mbps, and 1 Mbps, respectively, on the downstream channel. The base station has an infinite amount of data to send to each of the nodes, and can send to any one of these four nodes during any time slot in the downstream sub- frame.
1)What is the maximum rate at which the base station can send to the nodes, assuming it can send to any node it chooses during each time slot? Is your solution fair? Explain and define what you mean by “fair.”
2)If there is a fairness requirement that each node must receive an equal amount of data during each downstream sub- frame, what is the average transmission rate by the base station (to all nodes) during the downstream sub- frame? Explain how you arrived at your answer.
3)Suppose that the fairness criterion is that any node can receive at most twice as much data as any other node during the sub- frame. What is the average transmission rate by the base station (to all nodes) during the sub-frame? Explain how you arrived at your answer.
Explanation / Answer
a)10Mbps if it only the transmits to node A.The solution is not fair.since only the "A" getting as served.
it is fair means for each of the four nodes are should be alloted equal number of slots.
b)For the fairness requirement such that the each node receives an equal amount of the data during each downstream the sub-frame.
let n1,n2,n3 ,n4 respectively to represent the number os slots that A,B,C,D
now
The data transmitted to A in 1(first) slot=10t Mbits
hence the total amount of the data transmitted to A is =10t n1
The total number of data transmitted B,C,D are equal to 5t n2,2.5t n3, and t n4
respectively.
Now the fulfill given fairness requirement .
10t n1=5t n2=2.5t n3=t n4
hence the values are
n2=2n1
n3=4n1
n4=10n1
The total number of slots is N
n1+n2+n3+n4=N
n1+2n1+4n1+10n1=N
17 n1=N
n1=N/17
hence n2=2N/17
n3=4N/17
n4=10N/17
The avarage transmission rate is given by
(10t n1+5t n2+2.5t n3+t n4)/tN
(10N/17+5*2N/17+2.5*4N/17+1*10N/17)/N
40/17 =2.35 Mbps
c)Let the node A receives twice as much the data as nodes as B,C,D during the sub frame.
10t n1=2 * 5tn2=2 * 2.5tn3=2*tn4
n2=n1
n3=2n1
n4=5n1
again
n1+n2+n3+n4=N
n1+n1+2n1+5n1=N
n1=N/9
The average transmission is
(10t n1+5t n2+2.5t n3+t n4)/tN
=25/9=2.78 Mbps
The Nodes B,C,D receive the twice as much the data as any other the nodes the different values for the average transmission rate
can be calculated.
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