1 If you still are having problems getting your program to work, send me for help. You will need to use it later in the term to start Programming Project #2 and to find answers to problems. Assignment #2 is now available: Due Thurs. Oct. 4 at the beginning of class. Mon. Oct. 8 is a holiday (Thanksgiving): late deadline is Tues. Oct. 9 at 3:30pm, you can slip late submissions under my office door.

Alan Turing Celebration Lecture Biological Evolution as a Form of Learning Leslie Valiant, CC and Applied Math., Harvard Wed. Oct 10, 3:30pm, ECS 124 CSC 445/545: This is the same time as our class and it is the class immediately before the midterm. If a vote is unanimous, we can reschedule the midterm.

The primal: Maximize c T x subject to A x ≤ b, x ≥ 0. Writing this out: Maximize c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n ≤ b 1 a 21 x 1 + a 22 x a 2n x n ≤ b 2... a m1 x 1 + a m2 x a mn x n ≤ b m x 1, x 2,..., x n ≥ 0

Theorem: For every primal feasible solution x= (x 1, x 2,..., x n ) and for every dual feasible solution y= (y 1, y 2,..., y m ), c 1 x 1 + c 2 x c n x n ≤ b 1 y 1 + b 2 y b m y m That is, the objective function values for dual feasible solutions always give upper bounds on the objective function values for primal feasible solutions.

Proof: Because the constraints of the dual are (and x ≥ 0): a 11 y 1 + a 21 y a m1 y m ≥ c 1 a 12 y 1 + a 22 y a m2 y m ≥ c 2... a 1n y 1 + a 2n y a mn y m ≥ c n we have that S= (a 11 y 1 + a 21 y a m1 y m )x 1 + (a 12 y 1 + a 22 y a m2 y m )x (a 1n y 1 + a 2n y a mn y m ) x n ≥ c 1 x 1 + c 2 x c n x n

We have that S= (a 11 y 1 + a 21 y a m1 y m )x 1 + (a 12 y 1 + a 22 y a m2 y m )x (a 1n y 1 + a 2n y a mn y m ) x n ≥ c 1 x 1 + c 2 x c n x n Regroup the terms on the LHS so they are in terms of x i ’s instead of y j ’s: S= (a 11 x 1 + a 12 x a 1n x n )y 1 + (a 21 x 1 + a 22 x a 2n x n )y (a m1 x 1 + a m2 x a mn x n ) y m

S= (a 11 x 1 + a 12 x a 1n x n )y 1 + (a 21 x 1 + a 22 x a 2n x n )y (a m1 x 1 + a m2 x a mn x n ) y m From the primal problem: a 11 x 1 + a 12 x a 1n x n ≤ b 1 a 21 x 1 + a 22 x a 2n x n ≤ b 2... a m1 x 1 + a m2 x a mn x n ≤ b m Therefore since y ≥0, S ≤ b 1 y 1 + b 2 y b m y m

Summing it up, we have shown that c 1 x 1 + c 2 x c n x n ≤ S ≤ b 1 y 1 + b 2 y b m y m so by transitivity, this proves what we were trying to show: c 1 x 1 + c 2 x c n x n ≤ b 1 y 1 + b 2 y b m y m Important observation: it is critical for preserving the direction of the inequalities that the solutions x and y are feasible and hence non-negative.

A linear programming problem can have an optimal solution, or it can be infeasible or unbounded. Thought question: which combinations are possible?