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Homework 2 Solution - CIS 3360 Security in Computing Spring...

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CIS 3360 – Security in Computing Spring 2010 Solution for Homework 2 (100 points) NOTE: For all questions, please show your calculations. Homework submitting instructions: (i) Acceptable file formats: .doc, .docx or .pdf file (ii) The file name should be yourlastname_yourfirstname_HW2 (iii)Include your Full Name and your PID at the top of your document. Notes: b = bit B = Byte = 8 bits For transmission rates, K means 10 3 , M means 10 6 , G means 10 9 , so, 1Mbps = 10 6 bits/s For file sizes, K means 2 10 , M means 2 20 , G means 2 30 , so, 1 KB = 2 10 B = 1024 B m = milli = 10 -3 = 0.001 μ = micro = 10 -6 = .000001 Q1. [5+5+5] Assume that Alice is sending a word document to Bob. The size of the file is 550 KB. a) If the connection bandwidth is 1 Mbps, how much time in seconds will it take to send the entire file? Answer File size is 550 x 2 10 = 563200 bytes Since 1 byte is 8 bits, we have the file size of 563200 x 8 = 4505600 bits The bandwidth is 1000000 bps Thus, it takes 4505600 b/1000000 bps = 4.5056 seconds to send the entire file. b) At what rate should Alice be sending the file if she wanted it to reach two seconds earlier than it does in part (a)? Answer Two seconds earlier means that we want the file to be received in 4.5056-2 = 2.5056 seconds. So, if T x is the desired transmission rate, then from Alice to Bob, we have: 4505600 bits = 2.5056 sec T x in bps => T x in bps = 4505600 bits = 1798212 bps = 1.8 Mbps 2.5056 sec c) Assume that Alice is connected to Bob through satellite. If Alice started sending the file at time t , when will Bob receive the file completely? Assume that satellite propagation delay is 270 ms. Answer It takes t + time to transmit from Alice to Bob + satellite propagation delay = t + 4.5056 s + 270 ms = t + 4.7756 s
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