SIMPLEX METHOD

Examples from the book on Inelastic Analysis of Structures solved by a simplex applet.

Example 12.5 on page 163: Optimum limit design of a frame

The corresponding LPP in the standard form is described by the simplex tableau in Table 12.4. The applet denotes the unknowns x1, x2, ... x17, and the direct transcription of the problem would read as follows:
 
min: 2 x1 + 3 x2 + 2 x3; 
c1: 0.5 x4 + 0.5 x5 + 0.5 x7 - 0.5 x8 - 0.5 x9 - 0.5 x11 = 1;
c2: -0.5 x5 +1.5 x6 + x7 + 0.5 x9 -1.5 x10 - x11 = 3;
c3: x1 - x4 - x8 - x12 = 0;
c4: x1 - x5 - x9 - x13 = 0;
c5: x2 - x5 - x9 - x14 = 0;
c6: x2 - x6 - x10 - x15 = 0;
c7: x2 - x7 - x11 - x16 = 0;
c8: x3 - x7 - x11 - x17 =0; 
x1 >= 0; 
x2 >= 0; 
x3 >= 0; 
x4 >= 0; 
x5 >= 0; 
x6 >= 0; 
x7 >= 0; 
x8 >= 0; 
x9 >= 0; 
x10>= 0; 
x11>= 0; 
x12>= 0; 
x13>= 0; 
x14>= 0; 
x15>= 0; 
x16>= 0; 
x17>= 0; 

This problem definition can be simply pasted into the input area of the applet. After clicking on the "Solve" button (with the option "Some  Feedback" changed to "Only Solution" ), we obtain the following results in the output area:
 
Value of objective function: 7.09090909090909
x1 = 0.3636363636363634
x2 = 1.2727272727272727
x3 = 1.2727272727272727
x4 = 0.3636363636363634
x5 = 0.3636363636363636
x7 = 1.2727272727272727
x8 = 0.0
x9 = 0.0
x11 = 0.0
x6 = 1.2727272727272727
x10 = 0.0
x12 = 0.0
x13 = 0.0
x14 = 0.9090909090909092
x15 = 0.0
x16 = 0.0
x17 = 0.0
Using Simplex

The results perfectly agree with those given in Table 12.5 on page 168.

Since the applet does not require the LPP to be in the standard form, it is possible to simplify the input. Variables x12, x13, ... x17 are artificial variables that were introduced in order to convert the conditions of plastic admissibility from inequalities to equalities. We can leave out these variables and write the conditions with the >= sign:
 
min: 2 x1 + 3 x2 + 2 x3; 
c1: 0.5 x4 + 0.5 x5 + 0.5 x7 - 0.5 x8 - 0.5 x9 - 0.5 x11 = 1;
c2: -0.5 x5 +1.5 x6 + x7 + 0.5 x9 -1.5 x10 - x11 = 3;
c3: x1 - x4 - x8 >= 0;
c4: x1 - x5 - x9 >= 0;
c5: x2 - x5 - x9 >= 0;
c6: x2 - x6 - x10 >= 0;
c7: x2 - x7 - x11 >= 0;
c8: x3 - x7 - x11 >=0; 
x1 >= 0; 
x2 >= 0; 
x3 >= 0; 
x4 >= 0; 
x5 >= 0; 
x6 >= 0; 
x7 >= 0; 
x8 >= 0; 
x9 >= 0; 
x10>= 0; 
x11>= 0; 

The results would be exactly the same as before. The applet does not even require the variables to be nonnegative (this is why such conditions have been written explicitly in the previous examples). Recall that the variables x4, x5, x6 and x7 are the positive parts of moments M12, M21, M32 and M45, and the variables x8, x9, x10 and x11 are negative parts of those moments. One could also represent each moment by a single variable and drop the condition that this variable is nonnegative. The condition of plastic admissibility for each moment must then be described by two inequalities. For example, the absolute value of bending moment M12 must not exceed the design moment M0cl, which is represented by variable x1. The conditions of plastic admissibility -M0cl <= M12 <= M0cl are then translated into the inequalities x1-x4>=0 and x1+x4>=0. So, another possible description of the problem would read as follows:
 
min: 2 x1 + 3 x2 + 2 x3; 
c1: 0.5 x4 + 0.5 x5 + 0.5 x7 = 1;
c2: -0.5 x5 +1.5 x6 + x7 = 3;
c3: x1 - x4 >= 0;
c4: x1 + x4 >= 0;
c5: x1 - x5 >= 0;
c6: x1 + x5 >= 0;
c7: x2 - x5 >= 0;
c8: x2 + x5 >= 0;
c9: x2 - x6 >= 0;
c10: x2 + x6 >= 0;
c11: x2 - x7 >= 0;
c12: x2 + x7 >= 0;
c13: x3 - x7 >=0; 
c14: x3 + x7 >=0;

The conditions of nonnegative x1, x2 and x3 have also been left out, since they cannot really restrict the optimal solution. This is physically clear from the physical meaning of variables x1, x2 and x3 as the design moments, and mathematically follows from the fact that the condition x1>=0 can be obtained by summing the constraints c3 and c4, etc.