Asssignment 1 Solution

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School

University of California, Irvine *

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Course

232

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Electrical Engineering

Date

Oct 30, 2023

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Assignment 1 CS232/NetSys201/EECS248 Fall 2023 October 8, 2023 Deadline : October 12th @11:59PM on Canvas (upload your answers in PDF under Assignment 1 and please mark each problem accordingly). Turn in : A pdf file of your answers to each question. Note : You must fully compute fractions/equations in your answers to the final numbers when possible. Also show all steps taken to reach the final answer. 1 Problem 1: Consider two Exponentially distributed random variables ? 1 and ? 2 , with rate ? 1 = 8 and ? 2 = 4 , respectively. a) Compute the probability 𝑃(? 2 > 5) b) Compute the probability 𝑃(? 1 < 4) c) Compute the probability 𝑃(min(? 1 , ? 2 ) < 6) d) Compute the probability that ? 2 is smaller than ? 1 . Solution: a) 𝑃(? > 𝑡) = ℯ −𝜆? 𝑃(? 2 > 5) = ℯ −𝜆 2 ×5 = ℯ −20 b) 𝑃(? < 𝑡) = 1 − 𝑃(? > 𝑡) = 1 − ℯ −𝜆? 𝑃(? 1 < 4) = 1 − ℯ −𝜆 1 ×4 = 1 − ℯ −32 c) 𝑃(min(? 1 , ? 2, … , ? 𝑛 ) < 𝑡) = 1 − ℯ −(𝜆 1 + 𝜆 2 +⋯+𝜆 𝑛 )? 𝑃(min(? 1 , ? 2 ) < 6) = 1 − ℯ −(𝜆 1 +𝜆 2 )×6 = 1 − ℯ −72 d) 𝑃(? 2 < ? 1 ) = ? 2 ? 1 + ? 2 = 4 12 = 1 3

2 Problem 2: A router sends out 48 packets every 3 seconds on average. Suppose that the time in between two packets sent out can be modeled as an exponential r.v. a) What is the probability that a packet will be sent out in less than 4 seconds? b) Suppose that at time 𝑡 = 0 a packet was sent out, what is the probability that at time 𝑡 = 4 no further packets were sent out? c) Assume now 96 packets every 4 seconds are sent out on average. Is the probability of point (a) larger, smaller, or equal now? Solution: a) Let ?~𝐸𝑥𝑝(?) inter-leaving time between packets where ? = 48/3 = 16 𝑃(? < 4) = 1 − 𝑃(? > 4) = 1 − 𝑒 −16? = 1 − 𝑒 −64 b) 𝑃(? > 4) = 𝑒 −64 c) Let ?~𝐸𝑥𝑝(?) inter-leaving time between packets where ? = 96/4 = 24 𝑃(? < 4) = 1 − 𝑃(? > 4) = 1 − 𝑒 −24? = 1 − 𝑒 −96 As a result, the probability is larger. 3 Problem 3: A router is receiving packets from two different clients. Assume the time between the generation of two consecutive packets at each client is exponentially distributed with parameters ? 1 = 1 packets/second for node 1, and at ? 2 = 2 packets/second for node 2. a) What is the probability that the next packet will come from node 2? b) What is the probability that the router will not receive any packet in the next 2 seconds? c) Imagine that after the two seconds mentioned in part (b) elapsed, a third client enters the system and is now sending packets through the same router to which nodes 1 and 2 have been sending packets. Assume the time between the generation of two consecutive packets at each client is exponentially distributed as before for nodes 1 and 2, but with parameter ? 3 = 3 packets/second for node 3. What is the probability that the next packet will arrive from node 3? Solution: a) 𝑃(min(? 1 , ? 2 ) = ? 2 ) = ? 2 ? 1 + ? 2 = 2 3

b) 𝑃(min(? 1 , ? 2 ) > 2) = 𝑒 −2×(1+2) = 𝑒 −6 c) 𝑃(min(? 1 , ? 2 , ? 3 ) = ? 3 ) = ? 3 ? 1 + ? 2 + ? 3 = 3 6 = 1 2 4 Problem 4: Consider a router with service rate ? pkt/s, and arrival rate ? pkt/s. At time 𝑡 = 0 a packet (packet A) is being served. The next packet (packet B) arrives (according to the exponential distribution describing the arrival time) to the router buffer. a) What is the probability that Packet A leaves the router after the arrival of Packet B? b) What is the average time packet B waits in the buffer being served? c) Assuming that the incoming link to the router is part of the system, what is the average time packet B spends in the system? Solution: a) 𝑃(min(?, ?)) = ? ? + ? b) Note that transmission time and processing time are not part of the buffering time. 𝐸[𝑇 𝑏????𝑟 ] = ? ? + ? ∙ 1 ? + ? ? + ? = ? ?(? + ?) c) 𝑇 = 𝑇 1 + 𝑅 𝐸[𝑇 1 ] = 1 ? + ? 𝐸[𝑅] = ? ? + ? ∙ 2 ? + ? ? + ? ∙ 1 ? If we consider the incoming link to the router as part of the system, then the average time packet B spends in the system is: 𝐸[𝑅] = ? ? + ? ∙ 2 ? + ? ? + ? ∙ ( 1 ? + 1 ? )

Asssignment 1 Solution

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