/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 730 According to the Fundamental The... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 730

According to the Fundamental Theorem of linear programming, if either a linear program or its dual has no feasible point, then the other one has no solution. Illustrate this assertion with an example.

Short Answer

Expert verified
We constructed a linear program (LP) and its dual linear program (DLP). After analyzing the LP, we found it has no feasible region, as no point satisfies all three constraints simultaneously. According to the Fundamental Theorem of linear programming, if the LP has no feasible point, then its DLP must have no solution. True to the theorem, we found that there is no combination for y_1, y_2, and y_3 that satisfy all DLP constraints simultaneously, proving the assertion.

Step by step solution

01

Create a linear program

: Let's create a linear program (LP). Maximize: \(z = 2x_1 + 3x_2\) Subject to: \(x_1 + x_2 鈮� 1\) \(x_1 - x_2 鈮� 3\) \(-2x_1 + 3x_2 鈮� 1\) \(x_1, x_2 \geq 0\)
02

Create the dual of the linear program

: Now, let's construct the dual linear program (DLP) for the given LP. Minimize: \(w = y_1 + 3y_2 + y_3\) Subject to: \(y_1 + y_2 - 2y_3 鈮� 2\) \(y_1 - y_3 + 3y_3 鈮� 3\) \(y_1, y_2, y_3\geq 0\)
03

Check for feasible region in the linear program

: Now, let's check whether the given LP has a feasible region (in which an optimal solution could exist) by finding the intersection of the constraints. Plot the constraints: \(x_1 + x_2 = 1\), \(x_1 - x_2 = 3\), and \(-2x_1 + 3x_2 = 1\). Upon analyzing the plots, we can see that no point satisfies all three constraints simultaneously, indicating that the LP does not have a feasible region.
04

Check the dual linear program for a solution

: According to the Fundamental Theorem of linear programming, since the given LP has no feasible point, its DLP should have no solution: Check the DLP constraints: \(y_1 + y_2 - 2y_3 = 2\), \(y_1 - y_3 + 3y_3 =3\), and \(y_1, y_2, y_3\geq 0\). Upon attempting to solve the DLP constraints, we find that there is no combination of values for y_1, y_2, and y_3 that satisfy all constraints simultaneously. Therefore, the DLP does not have a solution. Thus, the given example illustrates the Fundamental Theorem of linear programming asserting that if either a linear program or its dual has no feasible point, the other one also has no solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In order to produce 1000 tons of non-oxidizing steel for engine valves, at least the following units of manganese, chromium and molybdenum, will be needed weekly-: 10 units of manganese, 12 units of chromium, and 14 units of molybdenum ( 1 unit is 10 pounds). These metals are obtainable from dealers in non-ferrous metals, who, to attract markets make them available in cases of three sizes, \(S, M\) and \(L .\) One \(S\) case costs \(\$ 9\) and contains 2 units of manganese, 2 units of chromium and 1 unit of molybdenum. One \(\mathrm{M}\) case costs \(\$ 12\) and contains 2 units of manganese, 3 units of chromium, and 1 unit of molybdenum. One \(\mathrm{L}\) case costs \(\$ 15\) and contains 1 unit of manganese, 1 unit of chromium and 5 units of molybdenum. How many cases of each kind should be purchased weekly so that the needed amounts of manganese, chromium and molybdenum are obtained at the smallest possible cost? What is the smallest possible cost?

Use the simplex algorithm to solve the following linear programming problem: Maximize $$ z=4 x_{1}+8 x_{2}+5 x_{3} $$ subject to $$ \begin{aligned} &x_{1}+2 x_{2}+3 x_{3} \leq 18 \\ &x_{1}+4 x_{2}+x_{3} \leq 6 \\ &2 x_{1}+6 x_{2}+4 x_{3} \leq 15 \\ &x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0 \end{aligned} $$

Maximize \(\mathrm{x}_{0}=3 \mathrm{x}_{1}+9 \mathrm{x}_{2}\) subject to: $$ \begin{gathered} \mathrm{x}_{1}+4 \mathrm{x}_{2} \leq 8 \\ \mathrm{x}_{1}+2 \mathrm{x}_{2} \leq 4 \\ \mathrm{x}_{1}, \mathrm{x}_{2} \geq 0 \end{gathered} $$ Use the simplex technique to solve.

Maximize \(3 \mathrm{x}_{1}+\mathrm{x}_{2}+3 \mathrm{x}_{3}\) subject to: $$ \begin{aligned} & 2 \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3} \leq 2 \\ \mathrm{x}_{1}+2 \mathrm{x}_{2}+3 \mathrm{x}_{3} \leq 5 \\ 2 \mathrm{x}_{1}+2 \mathrm{x}_{2}+\mathrm{x}_{3} \leq 6 \\ \mathrm{x}_{1} \geq 0, & \mathrm{x}_{2} \geq 0, & \mathrm{x}_{3} \geq 0 \end{aligned} $$ Find the optimal solution to this linear program by using the simplex method.

Find an optimal solution to this maximization problem using the simplex technique. $$ \text { maximize } \mathrm{x}_{0}=3 \mathrm{x}_{1}+2 \mathrm{x}_{2}+5 \mathrm{x}_{3} $$ subject to: $$ \begin{aligned} &\mathrm{x}_{1}+2 \mathrm{x}_{2}+\mathrm{x}_{3} \leq 430 \\ &3 \mathrm{x}_{1}+2 \mathrm{x}_{3} \leq 460 \\ &\mathrm{x}_{1}+4 \mathrm{x}_{2} \leq 420 \\ &\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3} \geq 0 \end{aligned} $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.