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Andiguys

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Fetching Results?...

Andiguys 85
July 28 2025
0 0

@janhardo I already have the results computed, but how can I fetch these values and automatically insert them into a table? What is the correct method or syntax to extract and organize the output data into a structured table format?

Ok...

Andiguys 85
July 28 2025
0 0

@sand15 Thankyou.

Need help!...

Andiguys 85
July 28 2025
0 0

@mmcdara 7891 can you help with the problem above?

Not working : Issue with Contour Plot Sy...

Andiguys 85
July 28 2025
0 0

@sand15This isn't working on my end. I'm getting an error saying 'ContoursWithLabels is not a command in the plots package' when I use your syntax. Also, the command gridelines = false,  doesn't seem to have any effect.

S1_vs_s2_contour_plot.mw

To export this table....

Andiguys 85
July 26 2025
0 0

@acer 

I would like to export this table in a document format (Word or PDF), but with the actual values substituted in place of the symbolic bounds (lower and upper).

Could you please guide me on the correct syntax or steps to perform this substitution and export?

wrong...

Andiguys 85
July 22 2025
0 0

@vv The constraint is perpendicular to x-axis, its horizontal?? something is wrong

 

constraint in plot?...

Andiguys 85
July 22 2025
0 0

@vv What i meant is constraint C1 and C2 are lines perpendicular to x-axis, How can we draw that in plot function??
I want syntax for that? it will be region right, so how to draw plot with constraints with feasible region.

what about the graph? Its not linear...

Andiguys 85
July 22 2025
0 0

@vv Can you show syntax for adding constraints in graph?

resolution above 300Dpi...

Andiguys 85
July 18 2025
0 0

@mmcdara is it possible to keep the image resolution above 300 dpi? If yes can you provide the syntax

label point where difference is zero?...

Andiguys 85
July 02 2025
0 0

@mmcdara Thankyou. Can you label the Cb point and give value where the difference between both is zero. pi(DR)-pi(DD)=0. Also in Cb axis can you provide 4.10^4 , 5 * 10^4. Or   Provide 10^4 in seperate bracket outside like Cb (10^4).

magnify it?...

Andiguys 85
July 01 2025
0 0

@dharr There may be some syntax to magnify it?

Error?...

Andiguys 85
June 14 2025
0 0

@mmcdara In all the cases, the solution for optimal Pc​ is showing as null. Can you please check if there's some error?

Q_Pc_KKT.mw
 

> 

``
> 

NULL
> 

NULL
> 

restart
> 

with(Optimization); with(plots); with(LinearAlgebra)
> 

_local(Pi)

Pi

 (1)
> 

q1 := -beta*p1+a
> 

q2 := -beta*p2+a
> 

r := proc (Pc) options operator, arrow; (1/3)*`ϕ`*q1-upsilon*(Pu-Pc) end proc
> 

``
> 

> 

`π_m` := proc (Pc) options operator, arrow; lambda*(w*r(Pc)+Ce*r(Pc)-Pc*r(Pc)-(1/2)*eta*Ce^2+U*r(Pc)^2) end proc

proc (Pc) options operator, arrow; lambda*(w*r(Pc)+Ce*r(Pc)-Pc*r(Pc)-(1/2)*eta*Ce^2+U*r(Pc)^2) end proc

 (2)
> 

C1 := Pc <= (`&varphi;`*q1-(1/3)*`&varphi;`*q1+upsilon*Pu)/upsilon

Pc <= ((2/3)*varphi*(-beta*p1+a)+upsilon*Pu)/upsilon

 (3)
> 

> 

C2 := Pc > -((1/3)*`&varphi;`*(-beta*p1+a)-upsilon*Pu)/upsilon

-((1/3)*varphi*(-beta*p1+a)-upsilon*Pu)/upsilon < Pc

 (4)
> 

NULL
> 

# No equality constraints
#
# Inequality constraints must be of the form f[i](p1) <= 0
#
# Thus
#
# (1) For C1
#     so
      f[1] := (Pc) -> Pc-(varphi*q1 - varphi*q1/3 + upsilon*Pu)/upsilon:
      

      
#
# (2) C2
#     so
      f[2] := (Pc) -> -(varphi*(-beta*p1 + a)/3 - upsilon*Pu)/upsilon-Pc:

 
> 

# Lagrangian (we want to maximize `&pi;_m` so to minimize -`&pi;_m`

L := -`&pi;_m`(Pc) + add(f[i](Pc)*mu[i], i=1..2);

-lambda*(w*((1/3)*varphi*(-beta*p1+a)-upsilon*(Pu-Pc))+Ce*((1/3)*varphi*(-beta*p1+a)-upsilon*(Pu-Pc))-Pc*((1/3)*varphi*(-beta*p1+a)-upsilon*(Pu-Pc))-(1/2)*eta*Ce^2+U*((1/3)*varphi*(-beta*p1+a)-upsilon*(Pu-Pc))^2)+(Pc-((2/3)*varphi*(-beta*p1+a)+upsilon*Pu)/upsilon)*mu[1]+(-((1/3)*varphi*(-beta*p1+a)-upsilon*Pu)/upsilon-Pc)*mu[2]

 (5)
> 

dLdPc := collect(diff(L, Pc), [Pc]);

-lambda*(2*U*upsilon^2-2*upsilon)*Pc-lambda*(w*upsilon+Ce*upsilon-(1/3)*varphi*(-beta*p1+a)+upsilon*Pu+2*U*((1/3)*varphi*(-beta*p1+a)-upsilon*Pu)*upsilon)+mu[1]-mu[2]

 (6)
> 

KKT_conditions := [
                    seq(mu[i] >= 0, i=1..2),         # Dual feasibility conditions
                    dLdPc = 0,                       # Stationarity condition
                    seq(``(f[i](Pc)) <= 0, i=1..2),  # Primal feasibility conditions
                    add(mu[i]*f[i](Pc) = 0, i=1..2)  # Complementary slackness
                  ]:

print~(KKT_conditions);

0 <= mu[1]

  

0 <= mu[2]

  

-lambda*(2*U*upsilon^2-2*upsilon)*Pc-lambda*(w*upsilon+Ce*upsilon-(1/3)*varphi*(-beta*p1+a)+upsilon*Pu+2*U*((1/3)*varphi*(-beta*p1+a)-upsilon*Pu)*upsilon)+mu[1]-mu[2] = 0

  

``(Pc-((2/3)*varphi*(-beta*p1+a)+upsilon*Pu)/upsilon) <= 0

  

``(-((1/3)*varphi*(-beta*p1+a)-upsilon*Pu)/upsilon-Pc) <= 0

  

(Pc-((2/3)*varphi*(-beta*p1+a)+upsilon*Pu)/upsilon)*mu[1]+(-((1/3)*varphi*(-beta*p1+a)-upsilon*Pu)/upsilon-Pc)*mu[2] = 0

  

[]

 (7)

 

 

> 

# dLdP1 is linear wrt p1

with(LargeExpressions):

DLDPc := collect(dLdPc, Pc, Veil[K]);

-2*K[1]*Pc-(1/3)*K[2]

 (8)
> 

# Thus dLdp1 = 0 verifies

isolate((8), Pc)
 

Pc = -(1/6)*K[2]/K[1]

 (9)
> 

i := 'i'; KS := [seq(K[i] = Unveil[K](K[i]), i = 1 .. LastUsed[K])]

[K[1] = lambda*upsilon*(U*upsilon-1), K[2] = -2*U*beta*lambda*p1*upsilon*varphi-6*Pu*U*lambda*upsilon^2+2*U*a*lambda*upsilon*varphi+beta*lambda*p1*varphi+3*Ce*lambda*upsilon+3*Pu*lambda*upsilon-a*lambda*varphi+3*lambda*upsilon*w-3*mu[1]+3*mu[2]]

 (10)
> 

beta

beta

 (11)
> 

Cs := {seq(beta[i] = subs(p1 = 0, f[i](Pc)), i = 1 .. 2)}

{beta[1] = Pc-((2/3)*a*varphi+upsilon*Pu)/upsilon, beta[2] = -((1/3)*a*varphi-upsilon*Pu)/upsilon-Pc}

 (12)
> 

infolevel[solve] := 2; sols := solve({seq(mu[i] >= 0, i = 1 .. 2), seq(Pc-beta[i] = 0, i = 1 .. 2), dLdPc = 0}, {Pc, seq(mu[i], i = 1 .. 2)})

Main: Entering solver with 5 equations in 3 variables
Main: attempting to solve as a linear system
Main: system cannot be directly solved as a linear system
Main: solving successful - now forming solutions
Main: Exiting solver returning 0 solutions
solve: Warning: no solutions found

  

(13)
> 

Cmin_value := -((1/3)*`&varphi;`*(-beta*p1+a)-upsilon*Pu)/upsilon; Cmax_value := (`&varphi;`*q1-(1/3)*`&varphi;`*q1+upsilon*Pu)/upsilon; Cmin_name := beta[2]; Cmax_name := beta[1]; Cmin_name := LowerBound; Cmax_name := UpperBound; Complementary_Slackness := mu[1]*(Cmin_name-Pc), mu[2]*(Pc-Cmax_name); L2 := -`&pi;_m`(Pc)+add(Complementary_Slackness); dL2dp1 := diff(L2, Pc); dL2dp1 := map(simplify, %, size)

LowerBound

  

mu[1]*(LowerBound-Pc), mu[2]*(Pc-UpperBound)

  

-(2/3)*lambda*(3*U*(Pc-Pu)*upsilon^2+(U*(-beta*p1+a)*varphi+(3/2)*w+(3/2)*Ce-3*Pc+(3/2)*Pu)*upsilon-(1/2)*varphi*(-beta*p1+a))-mu[1]+mu[2]

 (14)
> 

SC_CS_sols := solve({op(`~`[`=`]([Complementary_Slackness], 0)), dL2dPc = 0}, {Pc, seq(mu[i], i = 1 .. 2)})

false

  

Main: Entering solver with 3 equations in 3 variables
Main: attempting to solve as a linear system
Main: attempting to solve as a polynomial system

Main: Polynomial solver successful. Exiting solver returning 1 solution
solve: Warning: no solutions found

  
> 

"numelems([SC_CS_sols]); #` And those solutions are charecterized by` map(s -> if eval(lambda[1], s) = 0 then if eval(lambda[2], s) = 0 then "Pc belongs to interval (LowerBound, UpperBound)" else "Pc is equal to the UpperBound" end if else "Pc is equal to the LowerBound" end if , [SC_CS_sols]); "

0

  

[]

 (15)
> 

SC_CS_sols
> 

"LowerBound:=: UpperBound:= :"
> 

NULL
> 

NULL
> 

SC_CS_sols[1][1]; simplify(SC_CS_sols[1][3]) > 0

SC_CS_sols[1][1]

  

0 < SC_CS_sols[1][3]

 (16)
> 

NULL
> 

SC_CS_sols[2][1]; simplify(SC_CS_sols[2][2]) > 0

SC_CS_sols[2][1]

  

0 < SC_CS_sols[2][2]

 (17)
> 

NULL
> 

simplify(SC_CS_sols[3][1])

SC_CS_sols[3][1]

 (18)
> 

NULL
> 

NULL
> 

NULL
> 

``


 

Download Q_Pc_KKT.mw

 

Error...

Andiguys 85
June 11 2025
0 0

@acer 
Showing error "Error, (in assuming) when calling 'solve'. Received: 'cannot solve for an unknown function with other operations in its arguments'
"

Attaching sheet:

> 

restart
> 

kernelopts(version)

`Maple 2019.1, X86 64 WINDOWS, May 21 2019, Build ID 1399874`

 (1)
> 

B := w+t+((3*t+3*Pc-3*Pu)*upsilon+((-t-p1)*beta+a)*`&varphi;`)/(3*eta); A := (1/3)*`&varphi;`*(-beta*(p1+t)+a)-upsilon*Pu
> 

K1 := (A(-2*U*upsilon+1)-2*U*t*upsilon^2+(-B+2*t)*upsilon)/(2*upsilon*(U*upsilon-1))

(1/2)*((1/3)*varphi(-2*U*upsilon+1)*(-beta(-2*U*upsilon+1)*(p1(-2*U*upsilon+1)+t(-2*U*upsilon+1))+a(-2*U*upsilon+1))-upsilon(-2*U*upsilon+1)*Pu(-2*U*upsilon+1)-2*U*t*upsilon^2+(-w+t-(1/3)*((3*t+3*Pc-3*Pu)*upsilon+((-t-p1)*beta+a)*varphi)/eta)*upsilon)/(upsilon*(U*upsilon-1))

 (2)
> 

K2 := simplify(solve(K1 = Pc, Pc))

(((-t(-2*U*upsilon+1)-p1(-2*U*upsilon+1))*beta(-2*U*upsilon+1)+a(-2*U*upsilon+1))*eta*varphi(-2*U*upsilon+1)-3*upsilon(-2*U*upsilon+1)*Pu(-2*U*upsilon+1)*eta-6*((U*t*eta+(1/2)*t-(1/2)*Pu)*upsilon+(-(1/2)*t+(1/2)*w)*eta+(1/6)*varphi*(-beta*p1-beta*t+a))*upsilon)/((6*U*eta+3)*upsilon^2-6*upsilon*eta)

 (3)
> 

``
> 

ineq := -A/upsilon-t <= K2 and K2 <= (`&varphi;`*(-beta*p1+a)-A)/upsilon-t

 

> 

You gave us the (new, updated) information that all variables
are positive. Literally, that means p1 as well (unless you now exclude it).

extra := indets(ineq,And(name,Not(constant))) >~ 0;

{0 < Pu, 0 < U, 0 < a, 0 < beta, 0 < eta, 0 < p1, 0 < t, 0 < upsilon, 0 < varphi, 0 < w}

 (4)

 

> 

ineq1 := op(1,ineq):
> 

R1 := op(solve({ineq1},p1)) assuming extra[];

Error, (in assuming) when calling 'solve'. Received: 'cannot solve for an unknown function with other operations in its arguments'

  
> 

ineq2 := op(2,ineq):
> 

R2 := op(solve({ineq2},p1) assuming extra[]);

Error, (in assuming) when calling 'solve'. Received: 'cannot solve for an unknown function with other operations in its arguments'

  
> 

Maybe this is all you are after...

`and`(R1, R2, p1>0);

R1 and R2 and 0 < p1

 (5)
> 

Or maybe something like this...

p1>0 and p1 <= min(rhs(R1),rhs(R2));

Error, invalid input: rhs received R1, which is not valid for its 1st argument, expr

  
> 

Or maybe something like this...
simplify(solve(`and`(R1, R2, p1>0), p1)) assuming extra[];

Error, (in assuming) when calling 'simplify'. Received: 'invalid input: simplify uses a 1st argument, s, which is missing'

  
 

``

Download Q_isolate_acc_2_me.mw

Reminder!...

Andiguys 85
June 10 2025
0 0

@mmcdara Hi...if possible can you verify that the substitutions i made and the interpretation on optimal p1 which i have taken in above sheet is right or not? Thankyou

Clarification on p₁ Solution Type and Co...

Andiguys 85
June 08 2025
0 0

@mmcdara 

I have two questions:

  1. Is my approach for finding the optimal value of p1—based on whether the solution lies at the lower bound, upper bound, or within the interior—correct as per the syntax you provided?

  2. Are the lower and upper bounds that I substituted at the end of the sheet accurate? If not, what should they be?

Question_Lower_upper.mw

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