# LP Problem and Solutions ## Given LP Minimize \( z = 5x_1 8x_2 \) to: - \( x_1 + 2x_2 \geq 8 \) - \( x_1 + x_2 \geq 5 \) - \( x1 \geq \) - \( x_2 \geq \) --- ## a) Draw the feasible region and identify basic feasible solutions ### Step 1: Rewrite constraints as equalities for boundary lines: 1. \( x_1 + 2x_2 = 8 \) 2. \( x_1 + x_2 = 5 \) ### Step 2: Plot the lines: - Line 1: \( x_1 + 2x_2 = 8 \) - Line 2: \( x_1 + x_2 = 5 \) ### Step 3: Determine the feasible region: Since inequalities are \( \geq \), the feasible region is where the region is **above or on** the lines: - \( x_1 + 2x_2 \geq 8 \) - \( x_1 + x_2 \geq 5 \) ### Step 4: Find intersection points (vertices): - Intersection of lines: Solve: \[ x_1 + 2x_2 = 8 \\ x_1 + x_2 = 5 \] Subtract second from first: \[ (x_1 + 2x_2) - (x_1 + x_2) = 8 - 5 \Rightarrow x_2 = 3 \] Plug into \( x_1 + x_2 = 5 \): \[ x_1 + 3 = 5 \Rightarrow x_1 = 2 \] **Vertex 1:** \( (2, 3) \) - Intersection of each line with axes: Line 1: \( x_1 + 2x_2 = 8 \) - \( x_2 = \Rightarrow x_1 = 8 \) - \( x_1 = \Rightarrow 2x_2 = 8 \Rightarrow x_2=4 \) Line 2: \( x_1 + x_2 = 5 \) - \( x_2= \Rightarrow x_1=5 \) - \( x_1= \Rightarrow x_2=5 \) ### Step 5: Identify the feasible vertices: - **Vertex A:** \( (2, 3) \) - **Vertex B:** Intersection of \( x_1 + 2x_2=8 \) with \( x_2= \): \( (8, ) \) - **Vertex C:** Intersection of \( x_1 + x_2=5 \) with \( x_2= \): \( (5, ) \) - **Vertex D:** Intersection of \( x_1 + 2x_2=8 \) with \( x_2=4 \): \( x_1 + 2(4)=8 \Rightarrow x_1= \), so \( (, 4) \) ### Step 6: Check which vertices satisfy the inequalities: - \( (8,) \): \( x_1+2x_2=8+=8 \geq 8 \), OK; \( x_1 + x_2=8+=8 \geq 5 \), OK - \( (5,) \): \( 5+=5 \geq 8? \) No. So NOT in feasible region. - \( (,4) \): \( +8=8 \geq 8 \), OK; \( +4=4 \geq 5? \) No. So NOT in feasible region. - \( (2,3) \): \( 2+6=8 \geq 8 \), OK; \( 2+3=5 \geq 5 \), OK **Feasible vertices:** - \( (2,3) \) - \( (8,) \) --- ## b) Direction of Unboundedness ### Step 1: Plot the feasible region and observe: - The feasible region is unbounded in the direction where \( x_1 \) and/or \( x_2 \) increase indefinitely, satisfying the constraints. ### Step 2: Find the direction vector: - For LP with constraints \( \geq \), the region extends infinitely outward. - The **direction of unboundedness** can be checked by examining the feasible region's boundary lines and the inequalities. ### Step 3: Verify the direction: - Consider the vector \( d = (d_1, d_2) \) for which moving along \( d \) keeps us in the feasible region: Check the constraints under movement: \[ \text{For } x + \lambda d: \] - \( (x_1 + \lambda d_1) + 2(x_2 + \lambda d_2) \geq 8 \) - \( (x_1 + \lambda d_1) + (x_2 + \lambda d_2) \geq 5 \) which simplifies to: \[ \text{Constraints on } d: \] \[ d_1 + 2d_2 \geq \\ d_1 + d_2 \geq \] ### Step 4: Find a feasible direction: Choose \( d \) such that: \[ d_1 + 2d_2 \geq \\ d_1 + d_2 \geq \] For example, pick \( d_2=1 \): - \( d_1 + 2(1) \geq \Rightarrow d_1 \geq -2 \) - \( d_1 + 1 \geq \Rightarrow d_1 \geq -1 \) Choose \( d_1 = \): - Then, both inequalities are satisfied: \[ + 2(1)=2 \geq \\ + 1=1 \geq \] **Thus, the direction vector**: \[ d = (, 1) \] indicates the LP is **unbounded** in the direction of increasing \( x_2 \). **Conclusion:** The LP is unbounded in the direction \( (, 1) \), which satisfies the constraints for all sufficiently large \( x_2 \). --- ## c) Applying the Two-Phase Simplex Method ### Phase 1: - Convert inequalities to equalities by adding artificial variables. ### Step 1: Rewrite constraints with artificial variables: Let \( s_1, s_2 \) be artificial variables: \[ x_1 + 2x_2 - s_1 = 8 \\ x_1 + x_2 - s_2 = 5 \] with \( s_1, s_2 \geq \). ### Step 2: Set up Phase 1 LP: Minimize \( w = s_1 + s_2 \) Subject to the above constraints and \( x_1, x_2, s_1, s_2 \geq \). ### Step 3: Find initial basic feasible solution: - \( s_1=8, s_2=5 \), with \( x_1=x_2= \). ### Step 4: Optimize \( w \): - Since \( s_1, s_2 \) are positive, the initial solution is feasible but not optimal. - The goal is to find a basic feasible solution with \( s_1=s_2= \). ### Phase 2: - Once \( s_1=s_2= \), the solution to the original LP can be obtained. - Given the earlier analysis, the LP is unbounded, so Phase 2 will indicate unboundedness unless the artificial variables can be eliminated. ### **Summary:** - The feasible region is unbounded in the direction of increasing \( x_2 \), with vertices at \( (2,3) \) and \( (8,) \). - The LP's objective function decreases indefinitely along this unbounded direction, confirming the problem is unbounded. --- **Note:** For detailed graphical plotting, a graph is recommended to visualize the feasible region and check the unboundedness visually.