Q.Consider a generalized processor sharing (GPS) system with a processing capaci
ID: 1766145 • Letter: Q
Question
Q.Consider a generalized processor sharing (GPS) system with a processing capacity of 2 packets/sec serving 4 buffers and respective weights of w1=0.3, w2=0.2, w3=0.4 and w4=0.1. Assume a scenario where packets arrive simultaneously as follows: 4 packets at buffer 1, 3 packets at buffer 2, 2 packets at buffer 3 and 2 packets at buffer 4. Find the following: a) Serving time for all packets in the system b) First buffer to become empty and the time that it occurs c) Second buffer to become empty and the time that it occurs.
Explanation / Answer
a) Serving time for all packets = Total number of packets/ Rate of packets sevrving per sec
= 11/2 = 5.5 sec
b) First buffer to become empty
Buffer 3 has highest weight of 0.4
Buffer 3 time = 2/(2 x 0.4) = 2.5 sec
c) Next heighest weight is buffer 1
Buffer1 will run at weight 0.3 for 2.5sec and thereafter remaining packet at 0.7 wight
Packets transferred in 2.5 sec for buffer 1 = 2.5 x 0.3 x2 =1.5
Remaining packets = 4-1.5 = 2.5
Remaining packets at 0.7 weight = 2.5/(2x 0.7) = 1.7857
Buffer 1 gets empty after (2.5 + 1.7857) = 4.2857 sec
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