by L.N.Vaserstein

## Spring 2000    |   Fall 2000   |   Fall 2001 Roe | Spring 2001

 Math 484.1, Linear Programming |  outlines & grading textbook: class notes. MWF  10:10-11, 102 McAllister office hours:  MWF 8:30-10:00 am. 205 MB Current Topic:   Transp.problem Updated last:  April 27, 2001
Number of students  : 37.

|13  | 30 | 10 |    30 | 19       | 49|    16 | 10 |   10 |  50 | 30  | 25 | 15 | 15 | 50 ||  -5      |10 |15| -7  || 385  points
| q1 | h1 | q2   |  h2  |  q3    |  m1 |  q4  || h3 | q5 | m2 | h4  | h5 | q6 | q7| m3 | bonus | q8  |h6 |bonus | ||100%

Grades: 2 WPs,  4 Fs,  D, 2 Cs, C+,3 B-s,5 Bs, 2 B+s, 6 A-s,,10 As.
The best students: Todd and Jason can get free pictures of the class and free lunches.
----------------------

Homework due March 16, 30 points: solve your diet problem.

Homework due M, Jan 15, 30 points: Ex 2.6 on page 18. For this problem, you can choose your gender and weight.
Late homeworks will be acccepted with partial credit. Grading:
-2 pts for missing store name, -2 points for missing date, -3 pts for  messing up the units for the variables,
-3 points for missing protein.

Quiz 2: 3 points for solving the system in the case a =/ 1,-1. In this case, there is one
solution. There are no solutions when a = 1 or -1.
The student with name like B.Neunahe (the second row in class on Jan 17): please, contact me.

Homework 2, 30 pts, due W., Jan.24, page 18, Exercise 2.7 (solve  the  blending problem in Example 2.2).

Quiz 3 on Jan 24:   0=1 implies everything. 5 points for solving the system in the case a /= -½ . Wnen
a = -½  there are solutions with y = 1.  Students doing almost anything in this case were given the full credit.

We do convex sets on M, Jan 29 an Midterm 1 on F, Febr 2.

Febr. 7. It is not OK to replace  f = b  by   f -z <= b, z >= 0.

Solution to q4. Using  standard tricks, we replace  f=x-z by -f, inntroduce a slack variable
w= x+z-3 >=0, and  replace z by z' - z".
Standard form: -f=-x+z'-z" -> min, 2x-3y+z'-z" =1, x+z'-z"-w=3, x,y,w,z',z" >=0.
Note that z=z'-z" is not a part of the original problem or the standard form. It is  a part of connection
between two problems.
Canonical form: -f=-x+z'-z" -> min, 2x-3y+z'-z" <=1, -2x+3y-z'+z" <=-1, x+z'-z"<=3, x,y,z',z" >=0.

Another way: We solve the equation for z   (z= 1 -2x+3y) and exclude it from the problem.
Standard form:  -f = -3x+3y +1 -> min, x - 3y +w = -2, x,y,w >= 0.
Canonical form:  -f = -3x+3y +1 -> min, x - 3y  <= -2, x,y >= 0.

This trick requires more work, but gives smaller problems. We can easily solve the last problem with
2 variables by graphical method. After this, we use z= 1 -2x+3 y to get an answer for the original problem.

Solution to q5:
 x y z 1 1 1 1 -1 =u -1 1 -1 2 =v 1 2 1 0 -> min

m3. Grading: -1pt  for  not computing the total cost.  About -1 pt for each 10% deviation  from min.

q8:  8 pts for finding minmax =1 and maxmin=0.
h6:  The value of game is 3/7. A few students got it exactly.  -5 points  for not writing any strategy
for a player.

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