Solve struggles to solve equation with variable constraint

up vote
3
down vote

favorite

I am trying to find x,y set satisfying FOC with the constraint:

0<x<=y<1

Here is my code, which does not give anything.

Dbuy = 1 - (x/y) - x*Log[y/x];
Dwait = (x/y) - x;
revx = D[(Dbuy*x + Dwait*CCC*y), x]
revy = D[(Dbuy*x + Dwait*CCC*y), y]
revsln=Solve[revx==0, revy==0,x,y]

When I use FindRoot function instead of Solve for a fixed value of CCC (between 0 and 1, for example 0.4), it gives me numerical values (0.322, 0.581) if I add

x,0.01, y,0.01

But I want some form of closed-form solution for

0<=C<=1

What do you guys think?

edited 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

Ovunc

234

New contributor

1

Well, do you have any reason to expect a closed-form solution exists at all?
â€“Â AccidentalFourierTransform
2 hours ago

I feel that the root for the constraint 0<=x<=y<=1 has a closed-form. Is there a way to check it? Or is there a way to add a constraint to "Solve" function? Or is there a way to find a closed-form approximation?
â€“Â Ovunc
2 hours ago

1

In 11.3 I get a "This system cannot be solved with the methods available to Solve." message. Probably means the system cannot be solved with the methods available to Solve, no? Even if I drop the floating CCC I get Root objects parametrized near to a value.
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

up vote
3
down vote

favorite

I am trying to find x,y set satisfying FOC with the constraint:

0<x<=y<1

Here is my code, which does not give anything.

Dbuy = 1 - (x/y) - x*Log[y/x];
Dwait = (x/y) - x;
revx = D[(Dbuy*x + Dwait*CCC*y), x]
revy = D[(Dbuy*x + Dwait*CCC*y), y]
revsln=Solve[revx==0, revy==0,x,y]

When I use FindRoot function instead of Solve for a fixed value of CCC (between 0 and 1, for example 0.4), it gives me numerical values (0.322, 0.581) if I add

x,0.01, y,0.01

But I want some form of closed-form solution for

0<=C<=1

What do you guys think?

edited 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

Ovunc

234

New contributor

1

Well, do you have any reason to expect a closed-form solution exists at all?
â€“Â AccidentalFourierTransform
2 hours ago

I feel that the root for the constraint 0<=x<=y<=1 has a closed-form. Is there a way to check it? Or is there a way to add a constraint to "Solve" function? Or is there a way to find a closed-form approximation?
â€“Â Ovunc
2 hours ago

1

In 11.3 I get a "This system cannot be solved with the methods available to Solve." message. Probably means the system cannot be solved with the methods available to Solve, no? Even if I drop the floating CCC I get Root objects parametrized near to a value.
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

up vote
3
down vote

favorite

I am trying to find x,y set satisfying FOC with the constraint:

0<x<=y<1

Here is my code, which does not give anything.

Dbuy = 1 - (x/y) - x*Log[y/x];
Dwait = (x/y) - x;
revx = D[(Dbuy*x + Dwait*CCC*y), x]
revy = D[(Dbuy*x + Dwait*CCC*y), y]
revsln=Solve[revx==0, revy==0,x,y]

When I use FindRoot function instead of Solve for a fixed value of CCC (between 0 and 1, for example 0.4), it gives me numerical values (0.322, 0.581) if I add

x,0.01, y,0.01

But I want some form of closed-form solution for

0<=C<=1

What do you guys think?

edited 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

Ovunc

234

New contributor

I am trying to find x,y set satisfying FOC with the constraint:

0<x<=y<1

Here is my code, which does not give anything.

Dbuy = 1 - (x/y) - x*Log[y/x];
Dwait = (x/y) - x;
revx = D[(Dbuy*x + Dwait*CCC*y), x]
revy = D[(Dbuy*x + Dwait*CCC*y), y]
revsln=Solve[revx==0, revy==0,x,y]

When I use FindRoot function instead of Solve for a fixed value of CCC (between 0 and 1, for example 0.4), it gives me numerical values (0.322, 0.581) if I add

x,0.01, y,0.01

But I want some form of closed-form solution for

0<=C<=1

What do you guys think?

equation-solving mathematical-optimization

edited 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

Ovunc

234

New contributor

edited 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

Ovunc

234

New contributor

edited 2 hours ago

b3m2a1

25k254147

edited 2 hours ago

b3m2a1

25k254147

edited 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

Ovunc

234

New contributor

asked 2 hours ago

Ovunc

234

asked 2 hours ago

Ovunc

234

New contributor

Ovunc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

1

Well, do you have any reason to expect a closed-form solution exists at all?
â€“Â AccidentalFourierTransform
2 hours ago

I feel that the root for the constraint 0<=x<=y<=1 has a closed-form. Is there a way to check it? Or is there a way to add a constraint to "Solve" function? Or is there a way to find a closed-form approximation?
â€“Â Ovunc
2 hours ago

1

In 11.3 I get a "This system cannot be solved with the methods available to Solve." message. Probably means the system cannot be solved with the methods available to Solve, no? Even if I drop the floating CCC I get Root objects parametrized near to a value.
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

1

Well, do you have any reason to expect a closed-form solution exists at all?
â€“Â AccidentalFourierTransform
2 hours ago

I feel that the root for the constraint 0<=x<=y<=1 has a closed-form. Is there a way to check it? Or is there a way to add a constraint to "Solve" function? Or is there a way to find a closed-form approximation?
â€“Â Ovunc
2 hours ago

1

In 11.3 I get a "This system cannot be solved with the methods available to Solve." message. Probably means the system cannot be solved with the methods available to Solve, no? Even if I drop the floating CCC I get Root objects parametrized near to a value.
â€“Â b3m2a1
2 hours ago

Well, do you have any reason to expect a closed-form solution exists at all?
â€“Â AccidentalFourierTransform
2 hours ago

I feel that the root for the constraint 0<=x<=y<=1 has a closed-form. Is there a way to check it? Or is there a way to add a constraint to "Solve" function? Or is there a way to find a closed-form approximation?
â€“Â Ovunc
2 hours ago

In 11.3 I get a "This system cannot be solved with the methods available to Solve." message. Probably means the system cannot be solved with the methods available to Solve, no? Even if I drop the floating CCC I get Root objects parametrized near to a value.
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
3
down vote

Until someone comes up with a method that yields an analytical solution, you can use NArgMax:

argmax[c_?NumericQ] := NArgMax[Dbuy*x + Dwait*c*y, 
 0 <= c <= 1, 0 <= x <=1, 0<= y < 1, x <= y, x, y]
ParametricPlot[Evaluate@t, t, argmax[t], t, 0, 1, 
 PlotRange -> 0, 1, 0, 1, Frame -> True]

enter image description here

edited 1 hour ago

answered 2 hours ago

kglr

167k8188390

are these the optimal (x,y) combinations for 0<=c<=1?
â€“Â Ovunc
5 secs ago

add a commentÂ |Â

up vote
2
down vote

Here's a way to get a sense for the kinds of things for want. We'll solve this (exactly!) for a large number of values for CCC, plot them, and extract meaning out of that.

Here's the basic code I'll fux with:

blurg[CCC_] :=
 Block[
 
 ccc = Rationalize[CCC],
 Dbuy,
 Dwait,
 revx,
 revy,
 x, y
 ,
 Dbuy = 1 - (x/y) - x*Log[y/x];
 Dwait = (x/y) - x;
 revx = D[(Dbuy*x + Dwait*ccc*y), x]; 
 revy = D[(Dbuy*x + Dwait*ccc*y), y];
 Solve[revx == 0, revy == 0, x, y, Reals]
 ]

solns =
 blurg /@ Rest@Subdivide[0, 1, 100];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

You can see you always have two solutions and they converge down to those end-points at CCC=1 and apparently diverge as CCC->0

You can also look at the symbolic solutions returned but they will be nasty. Lots of complicated Root objects

Here's another fun example:

solns =
 blurg /@ Rest@Subdivide[0, .01, 25];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

This shows the divergence over that side of the solutions.

edited 1 hour ago

answered 1 hour ago

b3m2a1

25k254147

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

Until someone comes up with a method that yields an analytical solution, you can use NArgMax:

argmax[c_?NumericQ] := NArgMax[Dbuy*x + Dwait*c*y, 
 0 <= c <= 1, 0 <= x <=1, 0<= y < 1, x <= y, x, y]
ParametricPlot[Evaluate@t, t, argmax[t], t, 0, 1, 
 PlotRange -> 0, 1, 0, 1, Frame -> True]

enter image description here

edited 1 hour ago

answered 2 hours ago

kglr

167k8188390

are these the optimal (x,y) combinations for 0<=c<=1?
â€“Â Ovunc
5 secs ago

add a commentÂ |Â

up vote
3
down vote

Until someone comes up with a method that yields an analytical solution, you can use NArgMax:

argmax[c_?NumericQ] := NArgMax[Dbuy*x + Dwait*c*y, 
 0 <= c <= 1, 0 <= x <=1, 0<= y < 1, x <= y, x, y]
ParametricPlot[Evaluate@t, t, argmax[t], t, 0, 1, 
 PlotRange -> 0, 1, 0, 1, Frame -> True]

enter image description here

edited 1 hour ago

answered 2 hours ago

kglr

167k8188390

are these the optimal (x,y) combinations for 0<=c<=1?
â€“Â Ovunc
5 secs ago

add a commentÂ |Â

up vote
3
down vote

Until someone comes up with a method that yields an analytical solution, you can use NArgMax:

argmax[c_?NumericQ] := NArgMax[Dbuy*x + Dwait*c*y, 
 0 <= c <= 1, 0 <= x <=1, 0<= y < 1, x <= y, x, y]
ParametricPlot[Evaluate@t, t, argmax[t], t, 0, 1, 
 PlotRange -> 0, 1, 0, 1, Frame -> True]

enter image description here

edited 1 hour ago

answered 2 hours ago

kglr

167k8188390

Until someone comes up with a method that yields an analytical solution, you can use NArgMax:

argmax[c_?NumericQ] := NArgMax[Dbuy*x + Dwait*c*y, 
 0 <= c <= 1, 0 <= x <=1, 0<= y < 1, x <= y, x, y]
ParametricPlot[Evaluate@t, t, argmax[t], t, 0, 1, 
 PlotRange -> 0, 1, 0, 1, Frame -> True]

enter image description here

edited 1 hour ago

answered 2 hours ago

kglr

167k8188390

edited 1 hour ago

answered 2 hours ago

kglr

167k8188390

answered 2 hours ago

kglr

167k8188390

answered 2 hours ago

kglr

167k8188390

are these the optimal (x,y) combinations for 0<=c<=1?
â€“Â Ovunc
5 secs ago

add a commentÂ |Â

are these the optimal (x,y) combinations for 0<=c<=1?
â€“Â Ovunc
5 secs ago

are these the optimal (x,y) combinations for 0<=c<=1?
â€“Â Ovunc
5 secs ago

add a commentÂ |Â

up vote
2
down vote

Here's a way to get a sense for the kinds of things for want. We'll solve this (exactly!) for a large number of values for CCC, plot them, and extract meaning out of that.

Here's the basic code I'll fux with:

blurg[CCC_] :=
 Block[
 
 ccc = Rationalize[CCC],
 Dbuy,
 Dwait,
 revx,
 revy,
 x, y
 ,
 Dbuy = 1 - (x/y) - x*Log[y/x];
 Dwait = (x/y) - x;
 revx = D[(Dbuy*x + Dwait*ccc*y), x]; 
 revy = D[(Dbuy*x + Dwait*ccc*y), y];
 Solve[revx == 0, revy == 0, x, y, Reals]
 ]

solns =
 blurg /@ Rest@Subdivide[0, 1, 100];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

You can see you always have two solutions and they converge down to those end-points at CCC=1 and apparently diverge as CCC->0

You can also look at the symbolic solutions returned but they will be nasty. Lots of complicated Root objects

Here's another fun example:

solns =
 blurg /@ Rest@Subdivide[0, .01, 25];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

This shows the divergence over that side of the solutions.

edited 1 hour ago

answered 1 hour ago

b3m2a1

25k254147

add a commentÂ |Â

up vote
2
down vote

Here's a way to get a sense for the kinds of things for want. We'll solve this (exactly!) for a large number of values for CCC, plot them, and extract meaning out of that.

Here's the basic code I'll fux with:

blurg[CCC_] :=
 Block[
 
 ccc = Rationalize[CCC],
 Dbuy,
 Dwait,
 revx,
 revy,
 x, y
 ,
 Dbuy = 1 - (x/y) - x*Log[y/x];
 Dwait = (x/y) - x;
 revx = D[(Dbuy*x + Dwait*ccc*y), x]; 
 revy = D[(Dbuy*x + Dwait*ccc*y), y];
 Solve[revx == 0, revy == 0, x, y, Reals]
 ]

solns =
 blurg /@ Rest@Subdivide[0, 1, 100];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

You can see you always have two solutions and they converge down to those end-points at CCC=1 and apparently diverge as CCC->0

You can also look at the symbolic solutions returned but they will be nasty. Lots of complicated Root objects

Here's another fun example:

solns =
 blurg /@ Rest@Subdivide[0, .01, 25];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

This shows the divergence over that side of the solutions.

edited 1 hour ago

answered 1 hour ago

b3m2a1

25k254147

add a commentÂ |Â

up vote
2
down vote

Here's a way to get a sense for the kinds of things for want. We'll solve this (exactly!) for a large number of values for CCC, plot them, and extract meaning out of that.

Here's the basic code I'll fux with:

blurg[CCC_] :=
 Block[
 
 ccc = Rationalize[CCC],
 Dbuy,
 Dwait,
 revx,
 revy,
 x, y
 ,
 Dbuy = 1 - (x/y) - x*Log[y/x];
 Dwait = (x/y) - x;
 revx = D[(Dbuy*x + Dwait*ccc*y), x]; 
 revy = D[(Dbuy*x + Dwait*ccc*y), y];
 Solve[revx == 0, revy == 0, x, y, Reals]
 ]

solns =
 blurg /@ Rest@Subdivide[0, 1, 100];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

You can see you always have two solutions and they converge down to those end-points at CCC=1 and apparently diverge as CCC->0

You can also look at the symbolic solutions returned but they will be nasty. Lots of complicated Root objects

Here's another fun example:

solns =
 blurg /@ Rest@Subdivide[0, .01, 25];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

This shows the divergence over that side of the solutions.

edited 1 hour ago

answered 1 hour ago

b3m2a1

25k254147

Here's a way to get a sense for the kinds of things for want. We'll solve this (exactly!) for a large number of values for CCC, plot them, and extract meaning out of that.

Here's the basic code I'll fux with:

blurg[CCC_] :=
 Block[
 
 ccc = Rationalize[CCC],
 Dbuy,
 Dwait,
 revx,
 revy,
 x, y
 ,
 Dbuy = 1 - (x/y) - x*Log[y/x];
 Dwait = (x/y) - x;
 revx = D[(Dbuy*x + Dwait*ccc*y), x]; 
 revy = D[(Dbuy*x + Dwait*ccc*y), y];
 Solve[revx == 0, revy == 0, x, y, Reals]
 ]

solns =
 blurg /@ Rest@Subdivide[0, 1, 100];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

You can see you always have two solutions and they converge down to those end-points at CCC=1 and apparently diverge as CCC->0

You can also look at the symbolic solutions returned but they will be nasty. Lots of complicated Root objects

Here's another fun example:

solns =
 blurg /@ Rest@Subdivide[0, .01, 25];

N[x, y /. #, 50] & /@ solns // 
 ListPlot[#, PlotStyle -> Map[Hue, Range[0, 1, 1/Length@solns]]] &

enter image description here

This shows the divergence over that side of the solutions.

edited 1 hour ago

answered 1 hour ago

b3m2a1

25k254147

edited 1 hour ago

answered 1 hour ago

b3m2a1

25k254147

answered 1 hour ago

b3m2a1

25k254147

answered 1 hour ago

b3m2a1

25k254147

add a commentÂ |Â

Ovunc is a new contributor. Be nice, and check out our Code of Conduct.

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