- #1
evinda
Gold Member
MHB
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Hello! (Wave)
I am looking at the general form of the Simplex algorithm with the use of tableaux.
$\overline{x_0}$ is a basic non degenarate feasible solution and thus the columns $P_1, \dots, P_m$ are linearly independent.
The first step is to create a $(m+1) \times (n+4)$ matrix as follows:
Remark
The value of $z_k$ is the dot product of the second and the $(3+k)$-th column.Why does it hold that $z_1-c_1=0, \dots, z_m-c_m=0$ ?
Isn't it $z_1-c_1=c_1 \cdot 1 + c_2 \cdot 0+ \dots + c_m \cdot z_m=c_1$? Or am I wrong? (Thinking)
I am looking at the general form of the Simplex algorithm with the use of tableaux.
$\overline{x_0}$ is a basic non degenarate feasible solution and thus the columns $P_1, \dots, P_m$ are linearly independent.
The first step is to create a $(m+1) \times (n+4)$ matrix as follows:
- At the first column we write the basic columns: $P-1, \dots, P_m$.
- At the second column we write the values of the corresponding coefficients of the objective funtion.
- At the third column we write the initial basic feasible non degenerate solution $\overline{x_0}$.
- At the next $n$ columns we write the elements of the columns of the matrix $A$.
- The last column remains empty for now.
- At the last row we write the value $z_0$ of the objective function that corresponds to the solution $\overline{x_0}$ and also the values of the differences $z_k-c_k, k=1, \dots, n$
Remark
The value of $z_k$ is the dot product of the second and the $(3+k)$-th column.Why does it hold that $z_1-c_1=0, \dots, z_m-c_m=0$ ?
Isn't it $z_1-c_1=c_1 \cdot 1 + c_2 \cdot 0+ \dots + c_m \cdot z_m=c_1$? Or am I wrong? (Thinking)
