Changes in Primal and Dual Solutions When Scaled
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Suppose that we have an LP in the standard form
$$text{(P) min }c^Tx: Ax=b, xle 0$$
and it's dual would be
$$text{(D) max }b^Tx: Axle c$$
Let $overline{x}$ and $overline{y}$ be the optimal solutions to (P) and (D) respectively.
How do these solutions change if we multiply the constraint $Ax=b$ by a scalar? I think we'd have the following
$$text{(P$alpha$) min }c^Tx: alpha Ax=b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}-{0}$
My initial thoughts are that
1. The optimal value's of (P), (D), (P$alpha$), and (D$alpha$) won't change since the feasible region wont' change.
2. The optimal solutions will just be $overline{x}/alpha$ and $overline{y}/alpha$ for (P$alpha$) and (D$alpha$) respectively.
Yet, this seems wrong.
I'm rather new to Linear Programming and duality and would love an explanation of why my intuition is/isn't correct.
linear-programming
asked Nov 26 at 12:45
CandyKing312312
63
add a comment |
up vote
1
down vote
favorite
Suppose that we have an LP in the standard form
$$text{(P) min }c^Tx: Ax=b, xle 0$$
and it's dual would be
$$text{(D) max }b^Tx: Axle c$$
Let $overline{x}$ and $overline{y}$ be the optimal solutions to (P) and (D) respectively.
How do these solutions change if we multiply the constraint $Ax=b$ by a scalar? I think we'd have the following
$$text{(P$alpha$) min }c^Tx: alpha Ax=b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}-{0}$
My initial thoughts are that
1. The optimal value's of (P), (D), (P$alpha$), and (D$alpha$) won't change since the feasible region wont' change.
2. The optimal solutions will just be $overline{x}/alpha$ and $overline{y}/alpha$ for (P$alpha$) and (D$alpha$) respectively.
Yet, this seems wrong.
I'm rather new to Linear Programming and duality and would love an explanation of why my intuition is/isn't correct.
linear-programming
asked Nov 26 at 12:45
CandyKing312312
63
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that we have an LP in the standard form
$$text{(P) min }c^Tx: Ax=b, xle 0$$
and it's dual would be
$$text{(D) max }b^Tx: Axle c$$
Let $overline{x}$ and $overline{y}$ be the optimal solutions to (P) and (D) respectively.
How do these solutions change if we multiply the constraint $Ax=b$ by a scalar? I think we'd have the following
$$text{(P$alpha$) min }c^Tx: alpha Ax=b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}-{0}$
My initial thoughts are that
1. The optimal value's of (P), (D), (P$alpha$), and (D$alpha$) won't change since the feasible region wont' change.
2. The optimal solutions will just be $overline{x}/alpha$ and $overline{y}/alpha$ for (P$alpha$) and (D$alpha$) respectively.
Yet, this seems wrong.
I'm rather new to Linear Programming and duality and would love an explanation of why my intuition is/isn't correct.
linear-programming
asked Nov 26 at 12:45
CandyKing312312
63
Suppose that we have an LP in the standard form
$$text{(P) min }c^Tx: Ax=b, xle 0$$
and it's dual would be
$$text{(D) max }b^Tx: Axle c$$
Let $overline{x}$ and $overline{y}$ be the optimal solutions to (P) and (D) respectively.
How do these solutions change if we multiply the constraint $Ax=b$ by a scalar? I think we'd have the following
$$text{(P$alpha$) min }c^Tx: alpha Ax=b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}-{0}$
My initial thoughts are that
1. The optimal value's of (P), (D), (P$alpha$), and (D$alpha$) won't change since the feasible region wont' change.
2. The optimal solutions will just be $overline{x}/alpha$ and $overline{y}/alpha$ for (P$alpha$) and (D$alpha$) respectively.
Yet, this seems wrong.
I'm rather new to Linear Programming and duality and would love an explanation of why my intuition is/isn't correct.
linear-programming
linear-programming
asked Nov 26 at 12:45
CandyKing312312
63
asked Nov 26 at 12:45
CandyKing312312
63
asked Nov 26 at 12:45
CandyKing312312
63
asked Nov 26 at 12:45
CandyKing312312
63
asked Nov 26 at 12:45
CandyKing312312
63
63
add a comment |
add a comment |
1 Answer
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We would have
$$text{(P$alpha$) min }c^Tx: alpha Ax=color{blue}alpha b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }color{blue}{alpha}b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}setminus {0}$
The primal solution would remain the same since the feasible set remains the same. The corresponding dual solution would be $frac1{alpha}bar{y}$.
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
For (P$alpha$) why would you multiply b by $alpha$? Wouldn't that make $alpha Ax=alpha b$ equivalent to $Ax=b$?
– CandyKing312312
Nov 26 at 12:59
That is what the question meant right? Multiplying the constraint $Ax=b$ by a scalar. not just multiplying each row of $A$ by $alpha$. Yes, they are equivalent.
– Siong Thye Goh
Nov 26 at 13:05
My mistake, I meant $overline{A}_i=alpha*A_i$. Where $overline{A}$ represents the modified primal constraint
– CandyKing312312
Nov 26 at 13:07
1
you probably mean $alpha in mathbb{R}setminus{0}$
– LinAlg
Nov 26 at 17:00
add a comment |
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1 Answer
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We would have
$$text{(P$alpha$) min }c^Tx: alpha Ax=color{blue}alpha b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }color{blue}{alpha}b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}setminus {0}$
The primal solution would remain the same since the feasible set remains the same. The corresponding dual solution would be $frac1{alpha}bar{y}$.
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
For (P$alpha$) why would you multiply b by $alpha$? Wouldn't that make $alpha Ax=alpha b$ equivalent to $Ax=b$?
– CandyKing312312
Nov 26 at 12:59
That is what the question meant right? Multiplying the constraint $Ax=b$ by a scalar. not just multiplying each row of $A$ by $alpha$. Yes, they are equivalent.
– Siong Thye Goh
Nov 26 at 13:05
My mistake, I meant $overline{A}_i=alpha*A_i$. Where $overline{A}$ represents the modified primal constraint
– CandyKing312312
Nov 26 at 13:07
1
you probably mean $alpha in mathbb{R}setminus{0}$
– LinAlg
Nov 26 at 17:00
add a comment |
up vote
2
down vote
We would have
$$text{(P$alpha$) min }c^Tx: alpha Ax=color{blue}alpha b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }color{blue}{alpha}b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}setminus {0}$
The primal solution would remain the same since the feasible set remains the same. The corresponding dual solution would be $frac1{alpha}bar{y}$.
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
For (P$alpha$) why would you multiply b by $alpha$? Wouldn't that make $alpha Ax=alpha b$ equivalent to $Ax=b$?
– CandyKing312312
Nov 26 at 12:59
That is what the question meant right? Multiplying the constraint $Ax=b$ by a scalar. not just multiplying each row of $A$ by $alpha$. Yes, they are equivalent.
– Siong Thye Goh
Nov 26 at 13:05
My mistake, I meant $overline{A}_i=alpha*A_i$. Where $overline{A}$ represents the modified primal constraint
– CandyKing312312
Nov 26 at 13:07
1
you probably mean $alpha in mathbb{R}setminus{0}$
– LinAlg
Nov 26 at 17:00
add a comment |
up vote
2
down vote
up vote
2
down vote
We would have
$$text{(P$alpha$) min }c^Tx: alpha Ax=color{blue}alpha b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }color{blue}{alpha}b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}setminus {0}$
The primal solution would remain the same since the feasible set remains the same. The corresponding dual solution would be $frac1{alpha}bar{y}$.
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
We would have
$$text{(P$alpha$) min }c^Tx: alpha Ax=color{blue}alpha b, xle 0$$
and it's dual would be
$$text{(D$alpha$) max }color{blue}{alpha}b^Tx: alpha Axle c$$
Where $alpha in mathbb{R}setminus {0}$
The primal solution would remain the same since the feasible set remains the same. The corresponding dual solution would be $frac1{alpha}bar{y}$.
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
edited Nov 26 at 17:18
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
answered Nov 26 at 12:54
Siong Thye Goh
97.5k1463116
97.5k1463116
For (P$alpha$) why would you multiply b by $alpha$? Wouldn't that make $alpha Ax=alpha b$ equivalent to $Ax=b$?
– CandyKing312312
Nov 26 at 12:59
That is what the question meant right? Multiplying the constraint $Ax=b$ by a scalar. not just multiplying each row of $A$ by $alpha$. Yes, they are equivalent.
– Siong Thye Goh
Nov 26 at 13:05
My mistake, I meant $overline{A}_i=alpha*A_i$. Where $overline{A}$ represents the modified primal constraint
– CandyKing312312
Nov 26 at 13:07
1
you probably mean $alpha in mathbb{R}setminus{0}$
– LinAlg
Nov 26 at 17:00
add a comment |
For (P$alpha$) why would you multiply b by $alpha$? Wouldn't that make $alpha Ax=alpha b$ equivalent to $Ax=b$?
– CandyKing312312
Nov 26 at 12:59
That is what the question meant right? Multiplying the constraint $Ax=b$ by a scalar. not just multiplying each row of $A$ by $alpha$. Yes, they are equivalent.
– Siong Thye Goh
Nov 26 at 13:05
My mistake, I meant $overline{A}_i=alpha*A_i$. Where $overline{A}$ represents the modified primal constraint
– CandyKing312312
Nov 26 at 13:07
1
you probably mean $alpha in mathbb{R}setminus{0}$
– LinAlg
Nov 26 at 17:00
For (P$alpha$) why would you multiply b by $alpha$? Wouldn't that make $alpha Ax=alpha b$ equivalent to $Ax=b$?
– CandyKing312312
Nov 26 at 12:59
For (P$alpha$) why would you multiply b by $alpha$? Wouldn't that make $alpha Ax=alpha b$ equivalent to $Ax=b$?
– CandyKing312312
Nov 26 at 12:59
That is what the question meant right? Multiplying the constraint $Ax=b$ by a scalar. not just multiplying each row of $A$ by $alpha$. Yes, they are equivalent.
– Siong Thye Goh
Nov 26 at 13:05
That is what the question meant right? Multiplying the constraint $Ax=b$ by a scalar. not just multiplying each row of $A$ by $alpha$. Yes, they are equivalent.
– Siong Thye Goh
Nov 26 at 13:05
My mistake, I meant $overline{A}_i=alpha*A_i$. Where $overline{A}$ represents the modified primal constraint
– CandyKing312312
Nov 26 at 13:07
My mistake, I meant $overline{A}_i=alpha*A_i$. Where $overline{A}$ represents the modified primal constraint
– CandyKing312312
Nov 26 at 13:07
1
1
you probably mean $alpha in mathbb{R}setminus{0}$
– LinAlg
Nov 26 at 17:00
you probably mean $alpha in mathbb{R}setminus{0}$
– LinAlg
Nov 26 at 17:00
add a comment |
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