Write the dual for the following linear program
$begingroup$
Write the dual for the following linear program:
max($3x_1 + 8x_2$)
subject to
$x_1 + 4x_2$ ≤ 20
$x_1 + x_2$ ≥ 7
$x_1$ ≥ -1
$x_1$ ≤ 5
The posted solution is as follows, but does not show the steps.
Where did I go wrong?
linear-programming
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
$endgroup$
add a comment |
$begingroup$
Write the dual for the following linear program:
max($3x_1 + 8x_2$)
subject to
$x_1 + 4x_2$ ≤ 20
$x_1 + x_2$ ≥ 7
$x_1$ ≥ -1
$x_1$ ≤ 5
The posted solution is as follows, but does not show the steps.
Where did I go wrong?
linear-programming
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
$endgroup$
$begingroup$
Welcome to MathStackExchange. I please urge you to rewrite the pictures using MathJax, here is a guide, as it reallu helps reading the questions and also helps anyone struggling with a similar problem to find your question in the future.
$endgroup$
– Mefitico
Dec 10 '18 at 4:02
add a comment |
$begingroup$
Write the dual for the following linear program:
max($3x_1 + 8x_2$)
subject to
$x_1 + 4x_2$ ≤ 20
$x_1 + x_2$ ≥ 7
$x_1$ ≥ -1
$x_1$ ≤ 5
The posted solution is as follows, but does not show the steps.
Where did I go wrong?
linear-programming
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
$endgroup$
Write the dual for the following linear program:
max($3x_1 + 8x_2$)
subject to
$x_1 + 4x_2$ ≤ 20
$x_1 + x_2$ ≥ 7
$x_1$ ≥ -1
$x_1$ ≤ 5
The posted solution is as follows, but does not show the steps.
Where did I go wrong?
linear-programming
linear-programming
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
asked Dec 10 '18 at 3:56
Leonardo LopezLeonardo Lopez
1033
1033
$begingroup$
Welcome to MathStackExchange. I please urge you to rewrite the pictures using MathJax, here is a guide, as it reallu helps reading the questions and also helps anyone struggling with a similar problem to find your question in the future.
$endgroup$
– Mefitico
Dec 10 '18 at 4:02
add a comment |
$begingroup$
Welcome to MathStackExchange. I please urge you to rewrite the pictures using MathJax, here is a guide, as it reallu helps reading the questions and also helps anyone struggling with a similar problem to find your question in the future.
$endgroup$
– Mefitico
Dec 10 '18 at 4:02
$begingroup$
Welcome to MathStackExchange. I please urge you to rewrite the pictures using MathJax, here is a guide, as it reallu helps reading the questions and also helps anyone struggling with a similar problem to find your question in the future.
$endgroup$
– Mefitico
Dec 10 '18 at 4:02
$begingroup$
Welcome to MathStackExchange. I please urge you to rewrite the pictures using MathJax, here is a guide, as it reallu helps reading the questions and also helps anyone struggling with a similar problem to find your question in the future.
$endgroup$
– Mefitico
Dec 10 '18 at 4:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Check your definition of $z_1$ and $z_2$.
$$z_1 = x_1 + 1 iff x_1 = z_1-1$$
$$z_2 = -x_2+5 iff x_2 = -z_2+5$$
Hence $$x_1+4x_2 le 20$$ becomes
$$(z_1-1)+4(-z_2+5) le 20$$ which is equivalent to
$$z_1 -4z_2 le 1.$$
Do the same thing for the second inequality.
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
If inequality is greater than you don’t multiply by $-1$, but add the constant. Note:
$$x_1ge -1 Rightarrow underbrace{x_1+1}_{z_1}ge 0;\
x_2le 5Rightarrow -x_2ge -5 Rightarrow underbrace{-x_2+5}_{z_2}ge 0.$$
Note that your conversion to dual is correct.
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033424%2fwrite-the-dual-for-the-following-linear-program%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Check your definition of $z_1$ and $z_2$.
$$z_1 = x_1 + 1 iff x_1 = z_1-1$$
$$z_2 = -x_2+5 iff x_2 = -z_2+5$$
Hence $$x_1+4x_2 le 20$$ becomes
$$(z_1-1)+4(-z_2+5) le 20$$ which is equivalent to
$$z_1 -4z_2 le 1.$$
Do the same thing for the second inequality.
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
Check your definition of $z_1$ and $z_2$.
$$z_1 = x_1 + 1 iff x_1 = z_1-1$$
$$z_2 = -x_2+5 iff x_2 = -z_2+5$$
Hence $$x_1+4x_2 le 20$$ becomes
$$(z_1-1)+4(-z_2+5) le 20$$ which is equivalent to
$$z_1 -4z_2 le 1.$$
Do the same thing for the second inequality.
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
Check your definition of $z_1$ and $z_2$.
$$z_1 = x_1 + 1 iff x_1 = z_1-1$$
$$z_2 = -x_2+5 iff x_2 = -z_2+5$$
Hence $$x_1+4x_2 le 20$$ becomes
$$(z_1-1)+4(-z_2+5) le 20$$ which is equivalent to
$$z_1 -4z_2 le 1.$$
Do the same thing for the second inequality.
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Check your definition of $z_1$ and $z_2$.
$$z_1 = x_1 + 1 iff x_1 = z_1-1$$
$$z_2 = -x_2+5 iff x_2 = -z_2+5$$
Hence $$x_1+4x_2 le 20$$ becomes
$$(z_1-1)+4(-z_2+5) le 20$$ which is equivalent to
$$z_1 -4z_2 le 1.$$
Do the same thing for the second inequality.
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 10 '18 at 4:05
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
add a comment |
add a comment |
$begingroup$
If inequality is greater than you don’t multiply by $-1$, but add the constant. Note:
$$x_1ge -1 Rightarrow underbrace{x_1+1}_{z_1}ge 0;\
x_2le 5Rightarrow -x_2ge -5 Rightarrow underbrace{-x_2+5}_{z_2}ge 0.$$
Note that your conversion to dual is correct.
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
$endgroup$
add a comment |
$begingroup$
If inequality is greater than you don’t multiply by $-1$, but add the constant. Note:
$$x_1ge -1 Rightarrow underbrace{x_1+1}_{z_1}ge 0;\
x_2le 5Rightarrow -x_2ge -5 Rightarrow underbrace{-x_2+5}_{z_2}ge 0.$$
Note that your conversion to dual is correct.
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
$endgroup$
add a comment |
$begingroup$
If inequality is greater than you don’t multiply by $-1$, but add the constant. Note:
$$x_1ge -1 Rightarrow underbrace{x_1+1}_{z_1}ge 0;\
x_2le 5Rightarrow -x_2ge -5 Rightarrow underbrace{-x_2+5}_{z_2}ge 0.$$
Note that your conversion to dual is correct.
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
$endgroup$
If inequality is greater than you don’t multiply by $-1$, but add the constant. Note:
$$x_1ge -1 Rightarrow underbrace{x_1+1}_{z_1}ge 0;\
x_2le 5Rightarrow -x_2ge -5 Rightarrow underbrace{-x_2+5}_{z_2}ge 0.$$
Note that your conversion to dual is correct.
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
answered Dec 10 '18 at 4:31
farruhotafarruhota
20k2738
20k2738
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033424%2fwrite-the-dual-for-the-following-linear-program%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Welcome to MathStackExchange. I please urge you to rewrite the pictures using MathJax, here is a guide, as it reallu helps reading the questions and also helps anyone struggling with a similar problem to find your question in the future.
$endgroup$
– Mefitico
Dec 10 '18 at 4:02