Why don't we allow a linear programming problem to have strictly '' constraints?
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1
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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
Arpitgt
61
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up vote
1
down vote
favorite
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
Arpitgt
61
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
Arpitgt
61
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
Arpitgt
61
asked Oct 12 '16 at 17:59
Arpitgt
61
asked Oct 12 '16 at 17:59
Arpitgt
61
asked Oct 12 '16 at 17:59
Arpitgt
61
asked Oct 12 '16 at 17:59
Arpitgt
61
61
add a comment |
add a comment |
2 Answers
2
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3
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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 Hughes
61.9k24090
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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 at 18:19
dantopa
6,38132042
answered Dec 3 at 18:15
Ashley Morgan
1
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
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 Hughes
61.9k24090
add a comment |
up vote
3
down vote
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 Hughes
61.9k24090
add a comment |
up vote
3
down vote
up vote
3
down vote
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 Hughes
61.9k24090
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 Hughes
61.9k24090
answered Oct 12 '16 at 18:05
John Hughes
61.9k24090
answered Oct 12 '16 at 18:05
John Hughes
61.9k24090
answered Oct 12 '16 at 18:05
John Hughes
61.9k24090
61.9k24090
add a comment |
add a comment |
up vote
0
down vote
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 at 18:19
dantopa
6,38132042
answered Dec 3 at 18:15
Ashley Morgan
1
add a comment |
up vote
0
down vote
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 at 18:19
dantopa
6,38132042
answered Dec 3 at 18:15
Ashley Morgan
1
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 18:19
dantopa
6,38132042
answered Dec 3 at 18:15
Ashley Morgan
1
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 at 18:19
dantopa
6,38132042
answered Dec 3 at 18:15
Ashley Morgan
1
edited Dec 3 at 18:19
dantopa
6,38132042
edited Dec 3 at 18:19
dantopa
6,38132042
edited Dec 3 at 18:19
dantopa
6,38132042
6,38132042
answered Dec 3 at 18:15
Ashley Morgan
1
answered Dec 3 at 18:15
Ashley Morgan
1
answered Dec 3 at 18:15
Ashley Morgan
1
1
add a comment |
add a comment |
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