Suppose that in solving an LP, we obtain the tableau in Table 22. Although x1 can enter the basis, this LP is unbounded. Why? TABLE 22 zx1 x2x3x4rhs 1-3-20|0 0 0 1-11|0| 3 02001| 4 - Jo JO
Suppose that in solving an LP, we obtain the tableau in Table 22. Although x1 can enter the basis, this LP is unbounded. Why? TABLE 22 zx1 x2x3x4rhs 1-3-20|0 0 0 1-11|0| 3 02001| 4 - Jo JO
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Suppose that in solving an LP, we obtain the tableau in Table 22. Although \( x_1 \) can enter the basis, this LP is unbounded. Why?
**TABLE 22**
\[
\begin{array}{c|cccc|c}
z & x_1 & x_2 & x_3 & x_4 & \text{rhs} \\
\hline
1 & -3 & -2 & 0 & 0 & 0 \\
0 & 1 & -1 & 1 & 0 & 3 \\
0 & 2 & 0 & 0 & 1 & 4 \\
\end{array}
\]
**Analysis:**
- **Objective Row**: The coefficients in the objective row (\( z \)) indicate that increasing \( x_1 \) will improve the objective function since it has a negative coefficient (-3).
- **Constraints**:
- The first constraint shows that increasing \( x_1 \) can be balanced by decreasing \( x_2 \) and increasing \( x_3 \) because for every unit increase in \( x_1 \), \( x_2 \) decreases by 1 and \( x_3 \) increases by 1, staying within the feasible region.
- The second constraint allows \( x_1 \) to increase without bounds since it has a coefficient of 2 and does not interact with \( x_3 \) directly.
Thus, \( x_1 \) can increase indefinitely without violating any constraints, marking the LP as unbounded.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b1d7b4f-6196-4e5d-a87d-bf6788a6f213%2F501231d6-e309-4f69-8d4f-4236152a109b%2Fm78mndh_processed.png&w=3840&q=75)
Transcribed Image Text:Suppose that in solving an LP, we obtain the tableau in Table 22. Although \( x_1 \) can enter the basis, this LP is unbounded. Why?
**TABLE 22**
\[
\begin{array}{c|cccc|c}
z & x_1 & x_2 & x_3 & x_4 & \text{rhs} \\
\hline
1 & -3 & -2 & 0 & 0 & 0 \\
0 & 1 & -1 & 1 & 0 & 3 \\
0 & 2 & 0 & 0 & 1 & 4 \\
\end{array}
\]
**Analysis:**
- **Objective Row**: The coefficients in the objective row (\( z \)) indicate that increasing \( x_1 \) will improve the objective function since it has a negative coefficient (-3).
- **Constraints**:
- The first constraint shows that increasing \( x_1 \) can be balanced by decreasing \( x_2 \) and increasing \( x_3 \) because for every unit increase in \( x_1 \), \( x_2 \) decreases by 1 and \( x_3 \) increases by 1, staying within the feasible region.
- The second constraint allows \( x_1 \) to increase without bounds since it has a coefficient of 2 and does not interact with \( x_3 \) directly.
Thus, \( x_1 \) can increase indefinitely without violating any constraints, marking the LP as unbounded.
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