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Consider the following Linear Programming Problem (LPP):
Maximize z = 3x1 + 2x2
Subject to x1 ≤ 4
x2 ≤ 6
3x1 + x2 ≤ 18
x2 ≥ 0, x2 ≥ 0
Maximize z = 3x1 + 2x2
Subject to x1 ≤ 4
x2 ≤ 6
3x1 + x2 ≤ 18
x2 ≥ 0, x2 ≥ 0
- a)The LPP has a unique optimal solution
- b)The LPP is infeasible
- c)The LPP is unbounded
- d)The LPP has multiple optimal solutions.
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Consider the following Linear Programming Problem (LPP):Maximize z = 3...
Explanation:
To determine the nature of the LPP, we need to solve it by using the simplex method.
Step 1: Convert the inequalities into equations by introducing slack variables.
x1 - 4x2 + x3 = 6
3x1 - x2 + x4 = 18
x2 ≥ 0, x3 ≥ 0, x4 ≥ 0
Step 2: Write the initial simplex tableau.
x1 x2 x3 x4 RHS
Z -3 -2 0 0 0
x3 1 -4 1 0 6
x4 3 -1 0 1 18
Step 3: Choose the most negative coefficient in the Z-row as the entering variable. Here, x2 has the most negative coefficient of -2.
Step 4: Determine the leaving variable by selecting the variable with the smallest non-negative ratio of RHS to the coefficient in the entering column. Here, the ratio for x3 is 6/(-4) = -1.5, which is negative, so x3 cannot be the leaving variable. The ratio for x4 is 18/(-1) = -18, which is negative, so x4 cannot be the leaving variable. Therefore, the LPP is unbounded.
Hence, the correct answer is option C, the LPP is unbounded.
To determine the nature of the LPP, we need to solve it by using the simplex method.
Step 1: Convert the inequalities into equations by introducing slack variables.
x1 - 4x2 + x3 = 6
3x1 - x2 + x4 = 18
x2 ≥ 0, x3 ≥ 0, x4 ≥ 0
Step 2: Write the initial simplex tableau.
x1 x2 x3 x4 RHS
Z -3 -2 0 0 0
x3 1 -4 1 0 6
x4 3 -1 0 1 18
Step 3: Choose the most negative coefficient in the Z-row as the entering variable. Here, x2 has the most negative coefficient of -2.
Step 4: Determine the leaving variable by selecting the variable with the smallest non-negative ratio of RHS to the coefficient in the entering column. Here, the ratio for x3 is 6/(-4) = -1.5, which is negative, so x3 cannot be the leaving variable. The ratio for x4 is 18/(-1) = -18, which is negative, so x4 cannot be the leaving variable. Therefore, the LPP is unbounded.
Hence, the correct answer is option C, the LPP is unbounded.
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Consider the following Linear Programming Problem (LPP):Maximize z = 3x1 + 2x2Subject to x1 ≤ 4 x2 ≤ 63x1 + x2 ≤ 18x2 ≥ 0, x2 ≥ 0a)The LPP has a unique optimal solution b)The LPP is infeasible c)The LPP is unbounded d)The LPP has multiple optimal solutions.Correct answer is option 'A'. Can you explain this answer?
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