By Andrew S. Tanenbaum
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Additional resources for Computer Networks (4th Edition) Solutions Manual
Unless the gigabit line is amazingly cheap, it is probably not worth having for this application. 18. Here are three reasons. First, process IDs are OS-specific. Using process IDs would have made these protocols OS-dependent. Second, a single process may establish multiple channels of communications. A single process ID (per process) as the destination identifier cannot be used to distinguish between these channels. Third, having processes listen on well-known ports is easy, but well-known process IDs are impossible.
For each possible 56-bit key, decrypt the first ciphertext block. If the resulting plaintext is legal, try the next block, etc. If the plaintext is illegal, try the next key. 9. The equation 2n = 1015 tells us n, the number of doubling periods needed. Solving, we get n = 15 log2 10 or n = 50 doubling periods, which is 75 years. Just building that machine is quite a way off, and Moore’s law may not continue for 75 more years. 42 PROBLEM SOLUTIONS FOR CHAPTER 8 10. The equation we need to solve is 2256 = 10n .
If these are then broken up into lines of 80 bytes, 52 such lines will be needed, adding 52 CRs and 52 LFs. The total length will then be 4200 bytes. 14. , =FF, this sequence will be mistakenly interpreted as an escape sequence. The solution is to encode the equal sign itself, so all equal signs always start escape sequences. 15. Some examples and possible helpers are application/msexcel(Excel), application/ppt (PowerPoint), audio/midi (MIDI sound), image/tiff (any graphics previewer), video/x-dv (QuickTime player).
Computer Networks (4th Edition) Solutions Manual by Andrew S. Tanenbaum