%% Math 464 Linear Optimization (Spring 2023) %% Lecture 14, 02/23/2023 %% Iterations of the simplex method >> format rat >> format compact %% Setting up the problem >> A = [1 1 -1 0 0; 3 1 0 -1 0; 3 2 0 0 1]; >> b = [2 4 10]'; >> c = [2 1 0 0 0]'; >> [m,n]=size(A); %% Starting at B(10/3, 0) %% Iteration 1 >> Bind = [1 3 4]; % for B(10/3, 0) >> B = A(:,Bind); Binv = inv(B); >> cB = c(Bind) cB = 2 0 0 >> cP = (c' - cB'*Binv*A)' cP = 0 -1/3 0 0 -2/3 %% Reduced cost is NOT >= 0, so current bfs is not optimal %% We select j=2 (2nd basic direction) >> j=2; d = zeros(n,1); d(j)=1; dB = -Binv*A(:,j); d(Bind) = dB d = -2/3 1 1/3 -1 0 %\$ Initializing the current bfs >> x = zeros(n,1); xB = Binv*b; x(Bind) = xB x = 10/3 0 4/3 6 0 %% Notice that d(1) and d(4) are < 0, and hence %% candidates for the min-ratio test >> [theta, l] = min( [-x(1)/d(1) -x(4)/d(4)] ) theta = 5 l = 1 >> x_ptB = x; % just saving a copy of the bfs >> x = x + theta*d x = 0 5 3 1 0 %% We're now at C(0,5) %% Iteration 2 >> Bind = [2 3 4]; >> B = A(:,Bind); Binv = inv(B); >> cB = c(Bind) cB = 1 0 0 >> cP = (c' - cB'*Binv*A)' cP = 1/2 0 0 0 -1/2 %% bfs not optimal; only option is to choose j=5 to proceed >> j=5; d = zeros(n,1); d(j)=1; dB = -Binv*A(:,j); d(Bind) = dB d = 0 -1/2 -1/2 -1/2 1 >> [theta, l] = min( [-x(2)/d(2) -x(3)/d(3) -x(4)/d(4)] ) theta = 2 l = 3 >> x = x + theta*d x = 0 4 2 0 2 %% We're now at D(0,4); %% Iteration 3 >> Bind = [2 3 5]; >> B = A(:,Bind); Binv = inv(B); >> cB = c(Bind) cB = 1 0 0 >> cP = (c' - cB'*Binv*A)' cP = -1 0 0 1 0 %% bfs not optimal; only choice is j=1 to proceed >> j=1; d = zeros(n,1); d(j)=1; dB = -Binv*A(:,j); d(Bind) = dB d = 1 -3 -2 0 3 >> [theta, l] = min( [-x(2)/d(2) -x(3)/d(3)] ) theta = 1 l = 2 >> x = x + theta*d x = 1 1 0 0 5 %% We're now at E(1,1) %% Iteration 4 >> Bind = [2 1 5]; %% Note that x(1) entered in place of x(3) >> B = A(:,Bind); Binv = inv(B); >> cB = c(Bind) cB = 1 2 0 >> cP = (c' - cB'*Binv*A)' cP = 0 0 1/2 1/2 0 %% c' >= 0 ==> current bfs is optimal! >> x x = 1 1 0 0 5 %% z* = min (c^T x) is the optimal objective function >> zSt = c'*x zSt = 3 >>