Assignment 1: Solutions, comments, marking scheme

The 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 problem
clearly says to solve the system by applying the GJ algorithm.

I deducted 2 marks each time someone did something to a
matrix that was not even an elementary row operation. I warned
people in class about pulling this sort of thing.

I deducted two marks if set brackets did not appear in the answer,
two marks if an ordered 5-tuple did not appear, and 1 mark if 
nothing was written down about the parameter (that it is arbitrary,
or so -- or, better: nonnegative but otherwise arbitrary). I
warned people about all these things in class too.

I deducted 2 marks usually if someone row reduced and stopped
at a matrix that was not in r. r-e. form. I deducted 0.5 marks
for each of the first few times people did not use one of the
two notations I said in class I would accept for indicating row
operations.


(2) (10 marks)

After writing down the augmented matrix of (*), one 
performs the following row ops. in the order given:

R2 --> R2 - 2 R1,
R3 --> R3 - 3 R1,
R2 --> (1/3) R2,
R3 --> R3 - 7 (new)R2,
R3 --> (3/7) R3,
R2 --> R2 + (4/3) R3,
R1 --> R1 - 3 R3,     (ok to do this one before the previous one,
                       but I prefer this order - sweep up, just as
                       we swept down when sweeping UNDER the steps)
R1 --> R1 + 2 R2.

The resulting matrix is

1  0  0   15/7   11
0  1  0   -4/7  -16  (with square brackets I don't want to type).
0  0  1  -10/7  -12

The corresponding param. description of the solution set of (*) is

{ (11 - (15/7)s, (4/7)s - 16, (10/7)s -12, s) | s is arbitrary }.





The Chief