MHB Why are the quantities equal to 0?

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The discussion revolves around the Simplex algorithm and the formation of its tableau, specifically addressing why certain quantities equal zero. Participants explore the relationship between the objective function coefficients and the tableau entries, questioning why values like \( z_1 - c_1 = 0 \) hold true. Clarifications are made regarding the interpretation of tableau components, particularly the significance of the third column in relation to the objective function. The conversation also touches on the implications of having a constant term in the objective function, as indicated by the tableau values. Overall, the thread emphasizes understanding the structure and calculations within the Simplex tableau.
evinda
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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:

  • 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)
 
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Also, suppose that we are given the following linear programming problem:

$$\max (5x_1-4x_2) \\ x_1-x_2+x_3=6 \\ 3x_1-2x_2+x_4=24 \\ -2x_1+3x_2+x_5=9 \\ x_1 \geq 0, i=1, \dots, 5$$

The first tableau is this:

$$\begin{matrix}
B & c_B & b & P_1 & P_2 & P_3 & P_4 & P_5 & \theta \\
P_3 & 0 & 6 & 1 & -1 & 1 & 0 & 0 & \\
P_4 & 0 & 24 & 3 & -2 & 0 & 1 & 0 & \\
P_5 & 0 & 9 & -2 & 3 & 0 & 0 & 1 & \\
& z & 0 & -5 & 4 & 0& 0 & 0 &
\end{matrix}$$

How did we deduce that $z_1-c_1=-5$ ?
 
Hey evinda! (Smile)

What are $c_k$ and $z_k$? (Wondering)

The objective function is $z=5x_1-4x_2 \quad\Rightarrow\quad z -5x_1 + 4x_2 = 0$.
The bottom row consists of those coefficients of $x_k$. (Nerd)
 
I like Serena said:
The objective function is $z=5x_1-4x_2 \quad\Rightarrow\quad z -5x_1 + 4x_2 = 0$.
The bottom row consists of those coefficients of $x_k$. (Nerd)

Ah I see... (Nod)

What represents the third column of the bottom row? (Thinking)
 
evinda said:
Ah I see... (Nod)

What represents the third column of the bottom row? (Thinking)

Erm... the third column of the bottom row... that seems to be $-5$.
That's the coefficient of $x_1$ in the objective function, multiplied by $-1$. (Thinking)
 
I like Serena said:
Erm... the third column of the bottom row... (Thinking)

I meant the column of b...
Do we write $0$ at the last row since there is no 4th component of $b$ or is something else meant?
 
evinda said:
I meant the column of b...
Do we write $0$ at the last row since there is no 4th component of $b$ or is something else meant?

Indeed, there is no constant in the function we want to maximize, so the corresponding column $b$ has a $0$ in its position. (Nod)
 
I like Serena said:
Indeed, there is no constant in the function we want to maximize, so the corresponding column $b$ has a $0$ in its position. (Nod)

At the second tableaux, at this position there is the number $30$...
So doesn't this mean anything? (Thinking)
 
evinda said:
At the second tableaux, at this position there is the number $30$...
So doesn't this mean anything? (Thinking)

I don't see a second tableau. :confused:

Either way, if there is a value of $30$ there, it means we have a constant in the function we want to maximize or minimize. (Nerd)
 
  • #10
I like Serena said:
I don't see a second tableau. :confused:

This is the second tableau:

$\begin{matrix}
B & c_B & b & P_1 & P_2 & P_3 & P_4 & P_5 & \theta\\
P_1 & 5 & 6 & 1 & -1 & 1 & 0 & 0 & - \\
P_4 & 0 & 6 & 0 & 1 & -3 & 1 & 0 & \frac{6}{1}\\
P_5 & 0 & 21 & 0 & 1 & 2 & 0 & 1 & \frac{21}{1} \\
& z & 30 & 0 & -1 & 5 & 0 & 0 &
\end{matrix}$
I like Serena said:
Either way, if there is a value of $30$ there, it means we have a constant in the function we want to maximize or minimize. (Nerd)
Could you explain it further to me? (Thinking)
 
  • #11
evinda said:
This is the second tableau:

$\begin{matrix}
B & c_B & b & P_1 & P_2 & P_3 & P_4 & P_5 & \theta\\
P_1 & 5 & 6 & 1 & -1 & 1 & 0 & 0 & - \\
P_4 & 0 & 6 & 0 & 1 & -3 & 1 & 0 & \frac{6}{1}\\
P_5 & 0 & 21 & 0 & 1 & 2 & 0 & 1 & \frac{21}{1} \\
& z & 30 & 0 & -1 & 5 & 0 & 0 &
\end{matrix}$

Could you explain it further to me? (Thinking)

I may be wrong about a plus or minus sign, for which you should check your course material. (Worried)
But I think the bottom row means that we want to maximize $z$ under the constraint that:
$$z - x_2 + 5x_3 = 30 \quad \Rightarrow \quad z = 30 + x_2 - 5x_3$$
So if I'm right about the signs, we want to maximize $30 + x_2 - 5x_3$.
 
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