Minimize c = 2x + 2y + 3z subject to x + z ≥ 340 2x + y ≥ 170 y + z ≥ 170 x ≥ 0, y ≥ 0, z ≥ 0. Step 1 Recall that for an LP problem to be standard, it needs to satisfy the following requirements. We are maximizing (not minimizing) an objective function. The constraints (apart from the requirement that each variable be nonnegative) are all ≤ constraints, with the right-hand sides nonnegative. We are to minimize the objective c = 2x + 2y + 3z, with the given constraints. x + z ≥ 340 2x + y ≥ 170 y + z ≥ 170 x ≥ 0, y ≥ 0, z ≥ 0 We will convert the minimization problem into a maximization problem by taking the negative of the objective function. All of the constraints remain unchanged. The minimization problem c = 2x + 2y + 3z converts to the maximization problem p = −2x (?)(- or +) 2y(?)(- or +) 3z. Recall that we used slack variables for all constraints given as ≤, by adding a positive value to the left-hand side to make it equal. Since the listed constraints are all ≥, we must "subtract" some nonnegative number. We will call the numbers s, t, and u, respectively and refer to the variables as a surplus variable. Rewrite the constraints and objective function in standard form. x + z − s = 2x + y − t = 170 y + z − (?) = 170
Minimize c = 2x + 2y + 3z subject to x + z ≥ 340 2x + y ≥ 170 y + z ≥ 170 x ≥ 0, y ≥ 0, z ≥ 0. Step 1 Recall that for an LP problem to be standard, it needs to satisfy the following requirements. We are maximizing (not minimizing) an objective function. The constraints (apart from the requirement that each variable be nonnegative) are all ≤ constraints, with the right-hand sides nonnegative. We are to minimize the objective c = 2x + 2y + 3z, with the given constraints. x + z ≥ 340 2x + y ≥ 170 y + z ≥ 170 x ≥ 0, y ≥ 0, z ≥ 0 We will convert the minimization problem into a maximization problem by taking the negative of the objective function. All of the constraints remain unchanged. The minimization problem c = 2x + 2y + 3z converts to the maximization problem p = −2x (?)(- or +) 2y(?)(- or +) 3z. Recall that we used slack variables for all constraints given as ≤, by adding a positive value to the left-hand side to make it equal. Since the listed constraints are all ≥, we must "subtract" some nonnegative number. We will call the numbers s, t, and u, respectively and refer to the variables as a surplus variable. Rewrite the constraints and objective function in standard form. x + z − s = 2x + y − t = 170 y + z − (?) = 170
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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Question
Minimize
c = 2x + 2y + 3z
subject tox | + | z | ≥ | 340 | ||
2x | + | y | ≥ | 170 | ||
y | + | z | ≥ | 170 |
x ≥ 0, y ≥ 0, z ≥ 0.
Step 1
Recall that for an LP problem to be standard, it needs to satisfy the following requirements.
- We are maximizing (not minimizing) an objective function.
- The constraints (apart from the requirement that each variable be nonnegative) are all ≤ constraints, with the right-hand sides nonnegative.
We are to minimize the objective
c = 2x + 2y + 3z,
with the given constraints.x + z ≥ 340 |
2x + y ≥ 170 |
y + z ≥ 170 |
x ≥ 0, y ≥ 0, z ≥ 0 |
We will convert the minimization problem into a maximization problem by taking the negative of the objective function. All of the constraints remain unchanged.
The minimization problem
c = 2x + 2y + 3z
converts to the maximization problem p = −2x (?)(- or +) 2y(?)(- or +) 3z.Recall that we used slack variables for all constraints given as ≤, by adding a positive value to the left-hand side to make it equal. Since the listed constraints are all ≥, we must "subtract" some nonnegative number. We will call the numbers s, t, and u, respectively and refer to the variables as a surplus variable.
Rewrite the constraints and objective function in standard form.
x + z − s | = |
|
2x + y − t | = | 170 |
y + z − (?)
|
= | 170 |
2x + 2y (?)(- or +) 3z + p | = | 0 |
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