%% Math 364: Sensitivity Analysis using matrix form 
%% 	     of the tableau simplex method.
%% Matlab session from Lecture 23 on 11/05/2024
%%
%% Comments added at "% " for increased readability

% We find the optimal tabluea for the Farmer Jones LP directly using
% the knowledge of the optimal basis
%
% Original Farmer Jones LP:
% max  z =  30 x1 + 100x2
% s.t. 	       x1 +    x2 <=  7 (land avail)
% 	     4 x1 + 10 x2 <= 40 (labr hrs)
%	     10x1      	  >= 30 (min corn) % **NOT** scaled by 10
%	       x1,     x2 >=  0 (non neg)


 
>> A = [ 1 1 1 0 0 0; 4 10 0 1 0 0; 10 0 0 0 -1 1]
A =
     1     1     1     0     0     0
     4    10     0     1     0     0
    10     0     0     0    -1     1
>> b= [7 40 30]'
b =
     7
    40
    30

% We can use M=10,000 for the big-M

>> M = 10000; c = [30 100 0 0 0 -M]' 
c =
          30
         100
           0
           0
           0
      -10000

% x1, x2, s1 are basic in the optimal tableau

>> basis = [1 2 3] 
basis =
     1     2     3
>> cB = c(basis)
cB =
    30
   100
     0
>> n = 6; m=3;
>> cN = c(setdiff(1:n,basis))
cN =
           0
           0
      -10000

% initialize the starting tableau

>> T = [ [1 -c' 0]; [zeros(m,1) A b]]
T =
           1         -30        -100           0           0           0       10000           0
           0           1           1           1           0           0           0           7
           0           4          10           0           1           0           0          40
           0          10           0           0           0          -1           1          30
>> format rat

% We find the basis matrix B, and use it to form the elementary matrix
% whose inverse represents all EROs to convert T0 to the optimal
% tableau.


>> B = A(:,basis)
B =
       1              1              1       
       4             10              0       
      10              0              0
      
>> E = [1 -cB'; zeros(m,1) B]
E =
       1            -30           -100              0       
       0              1              1              1       
       0              4             10              0       
       0             10              0              0

% The elementary matrix is E^{-1}
>> inv(E)
ans =
       1              0             10             -1       
       0              0              0              1/10    
       0              0              1/10          -1/25    
       0              1             -1/10          -3/50    

% Let's find the optimal tableau

>> Topt = inv(E)*T
Topt =
       1              0              0              0             10              1           9999            370       
       0              1              0              0              0             -1/10           1/10           3       
       0              *              1              0              1/10           1/25          -1/25          14/5     
       0              0              0              1             -1/10           3/50          -3/50           6/5

% The "*" entry is essentially zero - it's listed as an extremely
% small number close to zero due to numerical tolerances.

>> Topt(3,2)
ans =
      -1/18014398509481984
>>