VIPSolutionsSource: Wayne L. Winston, Operations Research: Applications andAlgorithms, 4th Edition. :Simplex Algorithm Problems Page 160 Question 6Section 4.8 Unbounded LPs Page 160 Question 6) (19 pts) 6 Show that the following LP is unbounded: minz = —x; — 3x st. xn —2p=
Question:
Source: Wayne L. Winston, Operations Research: Applications andAlgorithms, 4th Edition. :Simplex Algorithm Problems Page 160 Question 6
Section 4.8 Unbounded LPs Page 160 Question 6) (19 pts)
6 Show that the following LP is unbounded:
minz = —x; — 3x
st. xn —2p=
This is AI generated Response.
Asked by: chandu
Created at: 2025-11-13 00:21:45
Asked by: chandu
Created at: 2025-11-13 00:21:45
Thought Process:
Rendered
Raw Text
**Examining inequalities**
I'm looking at the first inequality, which suggests that if x2 is large, then x1 has to be less than or equal to 4 + 2x2. This is manageable as long as x1 isn’t excessively large. The second inequality indicates that x2 has to stay within a certain range relative to x1, specifically, x2 cannot be much greater than x1 + 3. I realize we can still move in a direction where both x1 and x2 can grow while keeping their differences in check.**Analyzing feasible solutions**
I'm examining the situation where I choose x2 = x1 + 3. I need to check how this affects the first constraint: it simplifies to -x1 - 6 <= 4, which is true for any nonnegative x1. So, along the line x2 = x1 + 3, both constraints are satisfied as long as x1 and x2 are nonnegative. I realize that as x1 increases towards infinity, so does x2. The objective z becomes -4x1 - 9, which trends toward negative infinity, indicating an unbounded solution.**Considering parametrization**
I’m looking at a parametrization where x1 = t (with t >= 0) and x2 = t + 3 (which maintains x2 >= 3). I need to ensure the constraints are met. The first check simplifies to -t - 6 <= 4, which holds true for all t >= 0. The second constraint becomes an equality. So, we have a feasible solution, but the objective leads toward negative infinity as t increases. I wonder if I can demonstrate this using the simplex method, as outlined in Winston's work. I might need to convert the LP to standard form.**Demonstrating unboundedness**
I need to provide a feasible point to show unboundedness. Let’s take t = 0, which gives x1 = 0 and x2 = 3. So, checking the constraints: for the first one, it yields -6 <= 4, which is true, and for the second, 3 <= 3, also true. Great, it's feasible! I can move in the direction d = (1,1) while keeping x2 - x1 = 3 constant. As I examine the objective function, z(t) goes to negative infinity, but I wonder if this direction maintains the first constraint?**Demonstrating unboundedness step-by-step**
First, I compute the expression for the constraints. For x1 - 2x2, substituting gives me -6 - t, which indeed decreases with t, confirming unboundedness. I’ll provide a feasible ray—x1 = t and x2 = t + 3, ensuring both constraints hold for t >= 0. Evaluating the objective function, z(t) leads to -4t - 9, which drops towards negative infinity as t increases. I can also mention the recession direction r = (1,1) to further confirm unboundedness. Let’s keep it clear and concise!
Final Answer:
Rendered
Raw Text
Did you like this answer?
Claim: The LP is unbounded below.
min z = −x1 − 3x2
s.t.
x1 − 2x2 ≤ 4
−x1 + x2 ≤ 3
x1, x2 ≥ 0
Consider the feasible ray obtained by keeping the second constraint tight:
x2 = x1 + 3. Let x1 = t ≥ 0, so x2 = t + 3.
Feasibility for all t ≥ 0:
- First constraint: x1 − 2x2 = t − 2(t + 3) = −t − 6 ≤ 4 (always true).
- Second constraint: −x1 + x2 = −t + (t + 3) = 3 (satisfied at equality).
- Nonnegativity: x1 = t ≥ 0, x2 = t + 3 ≥ 0.
Objective value along this ray:
z(t) = −t − 3(t + 3) = −4t − 9 → −∞ as t → ∞.
Thus there exists a feasible direction r = (1,1) along which the objective decreases without bound; the LP is unbounded.
Question:
give me correct structure
This is AI generated Response.
Asked by: chandu
Created at: 2025-11-13 00:35:24
Asked by: chandu
Created at: 2025-11-13 00:35:24
Final Answer:
Rendered
Raw Text
Did you like this answer?
# Unbounded Linear Program Analysis
## Problem Statement
We need to show that the following linear program (LP) is unbounded:
### Objective Function
\[
\text{min } z = -x_1 - 3x_2
\]
### Subject to Constraints
\[
\begin{align*}
1. & \quad x_1 - 2x_2 \leq 4 \\
2. & \quad -x_1 + x_2 \leq 3 \\
3. & \quad x_1, x_2 \geq 0
\end{align*}
\]
## Claim
The LP is unbounded below.
## Analysis
### Feasible Ray
Consider the feasible ray obtained by keeping the second constraint tight:
\[
x_2 = x_1 + 3
\]
Let \( x_1 = t \geq 0 \). Then:
\[
x_2 = t + 3
\]
### Feasibility Check
We need to verify that this choice satisfies all constraints for \( t \geq 0 \):
1. **First Constraint:**
\[
x_1 - 2x_2 = t - 2(t + 3) = -t - 6 \leq 4 \quad \text{(always true)}
\]
2. **Second Constraint:**
\[
-x_1 + x_2 = -t + (t + 3) = 3 \quad \text{(satisfied at equality)}
\]
3. **Nonnegativity:**
\[
x_1 = t \geq 0, \quad x_2 = t + 3 \geq 0
\]
### Objective Value Along the Ray
Calculate the objective value along this ray:
\[
z(t) = -t - 3(t + 3) = -4t - 9
\]
As \( t \to \infty \):
\[
z(t) \to -\infty
\]
### Conclusion
There exists a feasible direction \( r = (1, 1) \) along which the objective decreases without bound. Therefore, the LP is unbounded below.
Question:
add some explaination
This is AI generated Response.
Asked by: chandu
Created at: 2025-11-13 00:50:37
Asked by: chandu
Created at: 2025-11-13 00:50:37
Final Answer:
Rendered
Raw Text
Did you like this answer?
# Unbounded Linear Program Analysis
## Problem Statement
We need to show that the following linear program (LP) is unbounded:
### Objective Function
\[
\text{min } z = -x_1 - 3x_2
\]
### Subject to Constraints
\[
\begin{align*}
1. & \quad x_1 - 2x_2 \leq 4 \\
2. & \quad -x_1 + x_2 \leq 3 \\
3. & \quad x_1, x_2 \geq 0
\end{align*}
\]
## Claim
The LP is unbounded below.
## Analysis
### Feasible Ray
To demonstrate unboundedness, we examine a feasible ray that satisfies the constraints while allowing the objective function to decrease indefinitely. We can keep the second constraint tight:
\[
x_2 = x_1 + 3
\]
Let \( x_1 = t \geq 0 \). Then:
\[
x_2 = t + 3
\]
This substitution will help us analyze the behavior of the LP as \( t \) increases.
### Feasibility Check
We need to verify that this choice satisfies all constraints for \( t \geq 0 \):
1. **First Constraint:**
\[
x_1 - 2x_2 = t - 2(t + 3) = -t - 6 \leq 4
\]
This inequality holds for all \( t \geq 0 \) since \(-t - 6\) will always be less than or equal to 4.
2. **Second Constraint:**
\[
-x_1 + x_2 = -t + (t + 3) = 3
\]
This constraint is satisfied at equality, confirming that the point lies on the boundary.
3. **Nonnegativity:**
\[
x_1 = t \geq 0, \quad x_2 = t + 3 \geq 0
\]
Both variables are nonnegative as required.
### Objective Value Along the Ray
Now we calculate the objective value along this ray:
\[
z(t) = -t - 3(t + 3) = -4t - 9
\]
As \( t \to \infty \):
\[
z(t) \to -\infty
\]
This shows that the value of the objective function decreases without bound as \( t \) increases.
### Conclusion
Since we have identified a feasible direction \( r = (1, 1) \) along which the objective function decreases indefinitely, we conclude that the LP is unbounded below.
This means that there is no minimum value for the objective function within the feasible region defined by the constraints.
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :