## Assignment 1: Solutions, comments, marking schemeThe good news: Several people did a nice job on this dirt simple assignment -- they got 17 out of 20 or better.The bad: Several people either are not coming to class, or don't listen, or hear but do not believe me when I say things, and/or do not read the course outline. They ignored instructions on the assignment in various ways and usually got badly burned. Three people got 0/20 (partly because of lots of deductions for not following instructions). Dirt simple assignments should be marked strictly. All one had to do here was follow instructions from class and do some drudgery. As I said on the course outline, my job is not to see if you can do the same old stuff you did in high school. Part of the purpose of this assignment was to test knowledge of the Gauss-Jordan algorithm. Many people do not know it; worse: do not know what an elem. row op. is; also refuse to write sets and n-tuples when discussing solution sets of systems. The average mark was about 12.6 out of 20. (1) (10 marks) There is some flexibility in how the equations can be written down. E.g., the first one below could be written - f1 + f5 = -50. f1 -f5 = 50 \ f1 - f2 = 30 | - f2 + f3 = 40 > (*) f3 - f4 = 25 | f4 - f5 = 35 / If one starts with this system, then the steps, in order, of the GJ algorithm as applied to the augmented matrix of (*) are: R2 --> R2 - R1, R2 --> - R2, R3 --> R3 + (new)R2, R4 --> R4 - R3, R4 --> -R4, R5 --> R5 - (new)R4. The GJ form of the augm. matrix is 1 0 0 0 -1 50 0 1 0 0 -1 20 0 0 1 0 -1 60 0 0 0 1 -1 35 0 0 0 0 0 0 (I leave out the brackets; too much typing.) A corresp. parametric description of the solution set of (*) is { all 5-tuples (50+s, 20+s, 60+s, 35+s, s) | s >= 0 }. Note that all five "flows" are nonnegative if and only if s is nonnegative. And the streets are one-way, so all five ARE nonnegative.... One might also want to say that s must be an integer (it is unusual to have s vehicles on a road when s is not a nonnegative integer). It is also interesting that for big values of s, lots of vehicles go around the traffic circle many times before leaving it (if first in are first out). One could find the largest s for which this does NOT happen (if there is such an s). I deducted 0.5 marks for each arithmetic error. I deducted 1 mark for each row reduction step that did not come next in the GJ algorithm. (I.e., for each step that was out of place in the GJ algorithm.) |

The Chief