Why does the Lagrange multiplier $lambda$ change when the equality constraint is scaled? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$f$ does not have extrema at the Lagrange multiplier valueSimple Lagrange Multiplier Problem, not working outCan lagrange multiplier(Kuhn tucker multipliers?) change in corner solution?Why is the lagrange multiplier a constant?How to solve the following optimzation use lagrange multiplier method?LASSO relationship between Lagrange multiplier and constraint and why it doesn't matterMaximizing a bivariate quadratic form with Lagrange's methodWhen is there a symmetry between constraint and objective function in Lagrange multipliers?Computing the Lagrange multiplier in constrained optimization problemWhy is the Lagrange multiplier considered to be a variable?
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Why does the Lagrange multiplier $lambda$ change when the equality constraint is scaled?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$f$ does not have extrema at the Lagrange multiplier valueSimple Lagrange Multiplier Problem, not working outCan lagrange multiplier(Kuhn tucker multipliers?) change in corner solution?Why is the lagrange multiplier a constant?How to solve the following optimzation use lagrange multiplier method?LASSO relationship between Lagrange multiplier and constraint and why it doesn't matterMaximizing a bivariate quadratic form with Lagrange's methodWhen is there a symmetry between constraint and objective function in Lagrange multipliers?Computing the Lagrange multiplier in constrained optimization problemWhy is the Lagrange multiplier considered to be a variable?
$begingroup$
Consider the problem
$$beginarrayll textmaximize & x^2+y^2 \ textsubject to & dfracx^225 + dfracy^29 = 1endarray$$
Solving this using the Lagrange multiplier method, I get
$$x = pm5, qquad y = 0, qquad lambda = 25$$
However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?
optimization lagrange-multiplier constraints qcqp
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
asked Apr 7 at 6:29
PGuptaPGupta
1746
$endgroup$
add a comment |
$begingroup$
Consider the problem
$$beginarrayll textmaximize & x^2+y^2 \ textsubject to & dfracx^225 + dfracy^29 = 1endarray$$
Solving this using the Lagrange multiplier method, I get
$$x = pm5, qquad y = 0, qquad lambda = 25$$
However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?
optimization lagrange-multiplier constraints qcqp
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
asked Apr 7 at 6:29
PGuptaPGupta
1746
$endgroup$
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
Apr 7 at 6:34
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
Apr 7 at 6:58
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
Apr 7 at 7:14
$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
Apr 7 at 7:17
add a comment |
$begingroup$
Consider the problem
$$beginarrayll textmaximize & x^2+y^2 \ textsubject to & dfracx^225 + dfracy^29 = 1endarray$$
Solving this using the Lagrange multiplier method, I get
$$x = pm5, qquad y = 0, qquad lambda = 25$$
However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?
optimization lagrange-multiplier constraints qcqp
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
asked Apr 7 at 6:29
PGuptaPGupta
1746
$endgroup$
Consider the problem
$$beginarrayll textmaximize & x^2+y^2 \ textsubject to & dfracx^225 + dfracy^29 = 1endarray$$
Solving this using the Lagrange multiplier method, I get
$$x = pm5, qquad y = 0, qquad lambda = 25$$
However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?
optimization lagrange-multiplier constraints qcqp
optimization lagrange-multiplier constraints qcqp
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
asked Apr 7 at 6:29
PGuptaPGupta
1746
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
asked Apr 7 at 6:29
PGuptaPGupta
1746
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
edited Apr 7 at 20:10
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked Apr 7 at 6:29
PGuptaPGupta
1746
asked Apr 7 at 6:29
PGuptaPGupta
1746
asked Apr 7 at 6:29
PGuptaPGupta
1746
1746
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
Apr 7 at 6:34
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
Apr 7 at 6:58
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
Apr 7 at 7:14
$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
Apr 7 at 7:17
add a comment |
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
Apr 7 at 6:34
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
Apr 7 at 6:58
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
Apr 7 at 7:14
$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
Apr 7 at 7:17
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
Apr 7 at 6:34
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
Apr 7 at 6:34
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
Apr 7 at 6:58
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
Apr 7 at 6:58
1
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
Apr 7 at 7:14
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
Apr 7 at 7:14
$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
Apr 7 at 7:17
$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
Apr 7 at 7:17
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$
Thus, from $$fracpartial fpartial x=2x+18lambda x=0$$ and
$$fracpartial fpartial y=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac19$ or $lambda=-frac125.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac19(9x^2+25y^2-225)=25-frac169y^2leq25,$$ where the equality occurs for $y=0.$
If we consider $f(x,y)=x^2+y^2+lambdaleft(fracx^225+fracy^29-1right)$ so we'll get $lambda=-25$
and we'll get the same answer of course.
It happens because $-frac19cdot225=-25$ and
$$x^2+y^2-frac19(9x^2+25y^2-225)=x^2+y^2-25left(fracx^225+fracy^29-1right).$$
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
$endgroup$
– PGupta
Apr 7 at 7:05
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
Apr 7 at 7:13
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
Apr 7 at 7:19
add a comment |
$begingroup$
If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag1 nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.
Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
$endgroup$
$begingroup$
Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
$endgroup$
– PGupta
Apr 7 at 7:36
1
$begingroup$
@PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
$endgroup$
– littleO
Apr 7 at 8:05
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
Apr 7 at 9:00
add a comment |
$begingroup$
Writing the equation $$fracx^225+fracy^29=1$$ in the form
$$y^2=9-frac925x^2$$ you will have the objective function
$$f(x)=frac1625x^2+9$$
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$
Thus, from $$fracpartial fpartial x=2x+18lambda x=0$$ and
$$fracpartial fpartial y=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac19$ or $lambda=-frac125.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac19(9x^2+25y^2-225)=25-frac169y^2leq25,$$ where the equality occurs for $y=0.$
If we consider $f(x,y)=x^2+y^2+lambdaleft(fracx^225+fracy^29-1right)$ so we'll get $lambda=-25$
and we'll get the same answer of course.
It happens because $-frac19cdot225=-25$ and
$$x^2+y^2-frac19(9x^2+25y^2-225)=x^2+y^2-25left(fracx^225+fracy^29-1right).$$
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
$endgroup$
– PGupta
Apr 7 at 7:05
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
Apr 7 at 7:13
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
Apr 7 at 7:19
add a comment |
$begingroup$
Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$
Thus, from $$fracpartial fpartial x=2x+18lambda x=0$$ and
$$fracpartial fpartial y=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac19$ or $lambda=-frac125.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac19(9x^2+25y^2-225)=25-frac169y^2leq25,$$ where the equality occurs for $y=0.$
If we consider $f(x,y)=x^2+y^2+lambdaleft(fracx^225+fracy^29-1right)$ so we'll get $lambda=-25$
and we'll get the same answer of course.
It happens because $-frac19cdot225=-25$ and
$$x^2+y^2-frac19(9x^2+25y^2-225)=x^2+y^2-25left(fracx^225+fracy^29-1right).$$
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
$endgroup$
– PGupta
Apr 7 at 7:05
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
Apr 7 at 7:13
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
Apr 7 at 7:19
add a comment |
$begingroup$
Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$
Thus, from $$fracpartial fpartial x=2x+18lambda x=0$$ and
$$fracpartial fpartial y=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac19$ or $lambda=-frac125.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac19(9x^2+25y^2-225)=25-frac169y^2leq25,$$ where the equality occurs for $y=0.$
If we consider $f(x,y)=x^2+y^2+lambdaleft(fracx^225+fracy^29-1right)$ so we'll get $lambda=-25$
and we'll get the same answer of course.
It happens because $-frac19cdot225=-25$ and
$$x^2+y^2-frac19(9x^2+25y^2-225)=x^2+y^2-25left(fracx^225+fracy^29-1right).$$
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$
Thus, from $$fracpartial fpartial x=2x+18lambda x=0$$ and
$$fracpartial fpartial y=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac19$ or $lambda=-frac125.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac19(9x^2+25y^2-225)=25-frac169y^2leq25,$$ where the equality occurs for $y=0.$
If we consider $f(x,y)=x^2+y^2+lambdaleft(fracx^225+fracy^29-1right)$ so we'll get $lambda=-25$
and we'll get the same answer of course.
It happens because $-frac19cdot225=-25$ and
$$x^2+y^2-frac19(9x^2+25y^2-225)=x^2+y^2-25left(fracx^225+fracy^29-1right).$$
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
edited Apr 7 at 7:11
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
answered Apr 7 at 7:00
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
$begingroup$
Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
$endgroup$
– PGupta
Apr 7 at 7:05
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
Apr 7 at 7:13
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
Apr 7 at 7:19
add a comment |
$begingroup$
Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
$endgroup$
– PGupta
Apr 7 at 7:05
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
Apr 7 at 7:13
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
Apr 7 at 7:19
$begingroup$
Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
$endgroup$
– PGupta
Apr 7 at 7:05
$begingroup$
Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
$endgroup$
– PGupta
Apr 7 at 7:05
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
Apr 7 at 7:13
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
Apr 7 at 7:13
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
Apr 7 at 7:19
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
Apr 7 at 7:19
add a comment |
$begingroup$
If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag1 nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.
Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
$endgroup$
$begingroup$
Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
$endgroup$
– PGupta
Apr 7 at 7:36
1
$begingroup$
@PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
$endgroup$
– littleO
Apr 7 at 8:05
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
Apr 7 at 9:00
add a comment |
$begingroup$
If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag1 nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.
Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
$endgroup$
$begingroup$
Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
$endgroup$
– PGupta
Apr 7 at 7:36
1
$begingroup$
@PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
$endgroup$
– littleO
Apr 7 at 8:05
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
Apr 7 at 9:00
add a comment |
$begingroup$
If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag1 nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.
Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
$endgroup$
If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag1 nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.
Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
answered Apr 7 at 7:26
littleOlittleO
30.5k649111
30.5k649111
$begingroup$
Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
$endgroup$
– PGupta
Apr 7 at 7:36
1
$begingroup$
@PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
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– littleO
Apr 7 at 8:05
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Great, got it. Thanks a lot!
$endgroup$
– PGupta
Apr 7 at 9:00
add a comment |
$begingroup$
Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
$endgroup$
– PGupta
Apr 7 at 7:36
1
$begingroup$
@PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
$endgroup$
– littleO
Apr 7 at 8:05
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
Apr 7 at 9:00
$begingroup$
Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
$endgroup$
– PGupta
Apr 7 at 7:36
$begingroup$
Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
$endgroup$
– PGupta
Apr 7 at 7:36
1
1
$begingroup$
@PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
$endgroup$
– littleO
Apr 7 at 8:05
$begingroup$
@PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
$endgroup$
– littleO
Apr 7 at 8:05
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
Apr 7 at 9:00
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
Apr 7 at 9:00
add a comment |
$begingroup$
Writing the equation $$fracx^225+fracy^29=1$$ in the form
$$y^2=9-frac925x^2$$ you will have the objective function
$$f(x)=frac1625x^2+9$$
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Writing the equation $$fracx^225+fracy^29=1$$ in the form
$$y^2=9-frac925x^2$$ you will have the objective function
$$f(x)=frac1625x^2+9$$
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Writing the equation $$fracx^225+fracy^29=1$$ in the form
$$y^2=9-frac925x^2$$ you will have the objective function
$$f(x)=frac1625x^2+9$$
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
Writing the equation $$fracx^225+fracy^29=1$$ in the form
$$y^2=9-frac925x^2$$ you will have the objective function
$$f(x)=frac1625x^2+9$$
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
edited Apr 7 at 7:18
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Apr 7 at 6:37
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
79k42867
add a comment |
add a comment |
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$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
Apr 7 at 6:34
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
Apr 7 at 6:58
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
Apr 7 at 7:14
$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
Apr 7 at 7:17