Why don't we allow a linear programming problem to have strictly '' constraints?
$begingroup$
I am new to linear programming and I have been asked this question
"Why don't we allow a linear programming problem to have strictly '<' or '>' constraints?"
But unable to answer it.
Kindly provide me an explanation on this.
linear-programming
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
$endgroup$
add a comment |
$begingroup$
I am new to linear programming and I have been asked this question
"Why don't we allow a linear programming problem to have strictly '<' or '>' constraints?"
But unable to answer it.
Kindly provide me an explanation on this.
linear-programming
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
$endgroup$
add a comment |
$begingroup$
I am new to linear programming and I have been asked this question
"Why don't we allow a linear programming problem to have strictly '<' or '>' constraints?"
But unable to answer it.
Kindly provide me an explanation on this.
linear-programming
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
$endgroup$
I am new to linear programming and I have been asked this question
"Why don't we allow a linear programming problem to have strictly '<' or '>' constraints?"
But unable to answer it.
Kindly provide me an explanation on this.
linear-programming
linear-programming
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
asked Oct 12 '16 at 17:59
ArpitgtArpitgt
61
61
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Consider the linear program on $mathbb R$ consisting of one constraint: $x < 1$, with the function to be optimized being $f(x) = x$. What's the optimum? At what point is it achieved?
Answer: There's no optimum. Normally, it'd be at $x = 1$, but that just barely fails to meet the constraint. But for any $x$ less than $1$, there's a better solution, namely $(1+x)/2$.
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
$endgroup$
add a comment |
$begingroup$
Because of The Divisibility Assumption.
The Divisibility Assumption requires that each decision variable is allowed to assume fractional values. For example, the Divisibility Assumption implies that it is acceptable to produce $1.5$ or $1.63$ of a product or service.
edited Dec 3 '18 at 18:19
dantopa
6,46942243
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Consider the linear program on $mathbb R$ consisting of one constraint: $x < 1$, with the function to be optimized being $f(x) = x$. What's the optimum? At what point is it achieved?
Answer: There's no optimum. Normally, it'd be at $x = 1$, but that just barely fails to meet the constraint. But for any $x$ less than $1$, there's a better solution, namely $(1+x)/2$.
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
$endgroup$
add a comment |
$begingroup$
Consider the linear program on $mathbb R$ consisting of one constraint: $x < 1$, with the function to be optimized being $f(x) = x$. What's the optimum? At what point is it achieved?
Answer: There's no optimum. Normally, it'd be at $x = 1$, but that just barely fails to meet the constraint. But for any $x$ less than $1$, there's a better solution, namely $(1+x)/2$.
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
$endgroup$
add a comment |
$begingroup$
Consider the linear program on $mathbb R$ consisting of one constraint: $x < 1$, with the function to be optimized being $f(x) = x$. What's the optimum? At what point is it achieved?
Answer: There's no optimum. Normally, it'd be at $x = 1$, but that just barely fails to meet the constraint. But for any $x$ less than $1$, there's a better solution, namely $(1+x)/2$.
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
$endgroup$
Consider the linear program on $mathbb R$ consisting of one constraint: $x < 1$, with the function to be optimized being $f(x) = x$. What's the optimum? At what point is it achieved?
Answer: There's no optimum. Normally, it'd be at $x = 1$, but that just barely fails to meet the constraint. But for any $x$ less than $1$, there's a better solution, namely $(1+x)/2$.
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
answered Oct 12 '16 at 18:05
John HughesJohn Hughes
63.3k24090
63.3k24090
add a comment |
add a comment |
$begingroup$
Because of The Divisibility Assumption.
The Divisibility Assumption requires that each decision variable is allowed to assume fractional values. For example, the Divisibility Assumption implies that it is acceptable to produce $1.5$ or $1.63$ of a product or service.
edited Dec 3 '18 at 18:19
dantopa
6,46942243
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
$endgroup$
add a comment |
$begingroup$
Because of The Divisibility Assumption.
The Divisibility Assumption requires that each decision variable is allowed to assume fractional values. For example, the Divisibility Assumption implies that it is acceptable to produce $1.5$ or $1.63$ of a product or service.
edited Dec 3 '18 at 18:19
dantopa
6,46942243
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
$endgroup$
add a comment |
$begingroup$
Because of The Divisibility Assumption.
The Divisibility Assumption requires that each decision variable is allowed to assume fractional values. For example, the Divisibility Assumption implies that it is acceptable to produce $1.5$ or $1.63$ of a product or service.
edited Dec 3 '18 at 18:19
dantopa
6,46942243
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
$endgroup$
Because of The Divisibility Assumption.
The Divisibility Assumption requires that each decision variable is allowed to assume fractional values. For example, the Divisibility Assumption implies that it is acceptable to produce $1.5$ or $1.63$ of a product or service.
edited Dec 3 '18 at 18:19
dantopa
6,46942243
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
edited Dec 3 '18 at 18:19
dantopa
6,46942243
edited Dec 3 '18 at 18:19
dantopa
6,46942243
edited Dec 3 '18 at 18:19
dantopa
6,46942243
6,46942243
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
answered Dec 3 '18 at 18:15
Ashley MorganAshley Morgan
1
1
add a comment |
add a comment |
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