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# sol02 - MIT OpenCourseWare http/ocw.mit.edu 6.055J 2.038J...

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MIT OpenCourseWare http://ocw.mit.edu 6.055J / 2.038J The Art of Approximation in Science and Engineering Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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6.055J/2.038J (Spring 2008) Solution set 2 Do the following warmups and problems. Due in class on Monday, 03 Mar 2008 . Open universe: Collaboration, notes, and other sources of information are encouraged. However, avoid looking up answers until you solve the problem (or have tried hard). That policy helps you learn the most from the problems. Bring a photocopy to class on the due date , trade it for a solution set, and figure out or ask me about any confusing points. Your work will be graded lightly: P (made a reasonable effort), D (did not make a reasonable effort), or F (did not turn in). Warmups 1. Fish tank Estimate the mass of a typical home fish tank (filled with water and fish): a useful exercise before you help a friend move who has a fish tank. By having good aim or getting lucky, I won two tiny goldfish at our elementary school’s annual fair. They lived happily for many years in our fish tank, eventually growing to 7 inches in length. The tank was perhaps 1 foot wide, 3 feet long, and 1.5 feet high, which is a volume of V 0 . 3 m × 1 m × 0 . 5 m 0 . 15 m 3 . If it filled two-thirds with water, that’s 0 . 1 m 3 of water or 100 kg! The glass tank itself has a much lower mass so the main contribution is from the water. I estimated the mass to only one digit, neglecting the mass of the empty tank. That’s no worse than the errors from the length estimates. The tank itself is an old memory (from 30 years ago) and there’s no need to overengineer the estimate. 2. Bandwidth Estimate the bandwidth (bits/s) of a 747 crossing the Atlantic filled with CDROM’s. Divide and conquer! Here’s a tree on which to fill values: bandwidth capacity (bits) of 747 number of CDROM’s cargo mass CDROM mass CDROM capacity time to cross Atlantic
Z 2 Solution set 2 / 6.055J/2.038J: Art of approximation in science and engineering (Spring 2008) First I estimate the cargo mass. A 747 can easily carry about 400 people, each person having a mass (with luggage) of, say 140 kg. The total mass is m 400 × 140 kg 6 · 10 4 kg . A special cargo plane, with no seats or other frills for passengers, probably can carry 10 5 kg. Here are the other estimates. A CDROM’s mass is perhaps one ounce or 30 g. So the number of CDROM’s is 3 · 10 6 . The capacity of a CDROM is 600 MB or about 5 · 10 9 bits. The time to cross the Atlantic is about 8 hours or 3 · 10 4 s. Now propagate the values toward the root of the tree: bandwidth (capacity/time) 5 · 10 11 s 1 capacity (bits) of 747 1 . 5 · 10 16 number of CDROM’s 3 · 10 6 cargo mass 10 5 kg CDROM mass 30 g CDROM capacity 5 · 10 9 time to cross Atlantic 3 · 10 4 s The bandwidth is 0.5 terabits per second.

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sol02 - MIT OpenCourseWare http/ocw.mit.edu 6.055J 2.038J...

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