Suppose packets are being routed from one computer to another and a router has a
ID: 3221765 • Letter: S
Question
Suppose packets are being routed from one computer to another and a router has a choice of sending the packet over two different links - Link A takes either 1, 2, 3, or 4 hops with equal probability before getting to the destination. Each hop takes 2 milliseconds. Link B takes 7, 8, or 9 hops with equal probability, but each of these hops only takes 1 millisecond. Let X be a random variable that represents the number of milliseconds it takes to transmit the packet from one computer to the other. a. What is the probability mass function (pmf) of X over Link A? b. What is the expected time E[X] for a packet sent over Link A? c. What is the expected time E[X] for a packet sent over Link B? d. If the router decides to route a packet over Link A or Link B with equal probability, what is the pmf of X for both links?Explanation / Answer
a.X : Number of milliseconds it takes to transmit the packet from one computer to the other
Probability mass function of X over Link A.
The possible value of X = 2milliseconds
Link A can take any of the four hops 1,2,3,4 with equal probability for transmitting the packet from one computer to another. And each hop take 2 milliseconds
Probability of taking Hop1 to transmit the packet in 2 milliseconds = 1/4
Probability of taking Hop2 to transmit the packet in 2 milliseconds= 1/4
Probability of taking Hop3 to transmit the packet in 2 milliseconds= 1/4
Probability of taking Hop4 to transmit the packet in 2 milliseconds = 1/4
Probability of transmitting the packet from one computer to another in 2 milliseconds = Probability of taking Hop1 or Hop2 or Hop3 or Hop3 to transmit the packet in 2 milliseconds = Probability of taking Hop1 to transmit the packet in 2 milliseconds+Probability of taking Hop2 to transmit the packet in 2 milliseconds+Probability of taking Hop3 to transmit the packet in 2 milliseconds+Probability of taking Hop4 to transmit the packet in 2 milliseconds =1/4+1/4+1/4+1/4 =1
The possible value of X = 2milliseconds
Probability of transmitting the packet from one computer to another in 2 milliseconds = 1
b.Expected value of X over Link A = x P(X=x) = 2 x 1 =2
c. Similarly for pmf X over Link B
X =1 milli scond ; P(X=1) =1
E(X) over link B =x P(X=x) = 1x1 =1
d.
Probability of routing packet thru Link A =1/2
Probability of routing packet thru Link B =1/2
X : Number milliseconds
X : Number of milli seconds if packet routed thru Link 'A' = 2 milli seconds; Probability of routing thru link 'A' = P(X=2) =1/2
X : Number of milliseconds if packet route thru Link 'B' =1 millisecond ; Probability of routing thru link'B' =P(X=1) =1/2
Probability mass function X for both links
X=1 ; P(X=1) =1/2
X= 2 ;P(X=2) =1/2
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