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Andiguys

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Apologies and Updated Worksheet with Obj...

Andiguys 100
October 01 2025
0 0

@sand15 

Thank you for pointing that out. I understand your concern and I apologize for mixing different problems together earlier.
Below is a clean worksheet with only the objective function and constraints so it’s clearer.

Also, I’m sorry if my way of following up seemed inconsiderate. I truly appreciate the time and effort you put into answering my queries, and I’ll make sure to acknowledge the solutions properly before moving on to new questions.

The last syntax you shared worked well and its correct.

> 

restart
> 

_local(Pi)

Pi

 (1)
> 

``
> 

NULL
> 

Pi := proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*alpha*rho0*(1-(Pn-Pr)/(1-delta))-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc
> 

C1 := (2*(-(1-(Pn-Pr)/(1-delta))/(2*tau)-Cr*(1-(Pn-Pr)/(1-delta))^2/(8*tau^2)))*((2*rho0-1)*tau+i2)+(w/(2*tau)-1/2+i2/(2*tau))*(1-(Pn-Pr)/(1-delta))-Cr*(1/2-i2/(2*tau))*(1-(Pn-Pr)/(1-delta))^2/(2*tau) > 0
 

 

Download Proposal_follow_up.mw

Solving optimization with three decision...

Andiguys 100
October 01 2025
0 0

@sand15 Thankyou for the syntax. Just one more question.

If I have three decision variables and one constraint, can I still solve it using the same syntax? How would that work?

> 

restart
> 

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

_local(Pi)

Pi

 (1)
> 

NULL
> 

NULL
> 

> 

SC_CS_sols := solve(
  {
    diff(?????) - mu[1] = 0,
    mu[1]*(-2*(-(1 - (Pn - Pr)/(1 - delta))/(2*tau) - Cr*(1 - (Pn - Pr)/(1 - delta))^2/(8*tau^2))*((2*rho0 - 1)*tau + i2) + (w/(2*tau) - 1/2 + i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta)) - Cr*(1/2 - i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta))^2/(2*tau)) = 0
   
  }
  ,
  {i1, mu[1]}
  , 'parametric'='full', 'parameters'={P, Q, A, B}
)

SC_CS_sols := piecewise(P = 0, piecewise(Q <> 0, [], Q = 0, [{i1 = i1, mu[1] = 0}]), P <> 0, [{i1 = -Q/(2*P), mu[1] = 0}])

 (2)
> 

pi__m := (Pn - Cn)*(1 - (Pn - Pr)/(1 - delta)) + (Pr - w - Crm)*alpha*(1/2 + (i1 - i2)/(2*tau))*(1 - (Pn - Pr)/(1 - delta)) - Ce*rho0*(1 - (Pn - Pr)/(1 - delta)):

collect(pi__m, [Pn,Pr,w]):
PQR := [R, Q, P] =~ [coeffs(%, i1)]:

print~(PQR):

R = w*(1/2-(1/2)*i2/tau)*(1-(Pn-Pr)/(1-delta))-C0-(1/2)*Cr*(1/2-(1/2)*i2/tau)^2*(1-(Pn-Pr)/(1-delta))^2+Ce*rho0*(1-(Pn-Pr)/(1-delta))

  

Q = ((1/2)*w/tau-1/2+(1/2)*i2/tau)*(1-(Pn-Pr)/(1-delta))-(1/2)*Cr*(1/2-(1/2)*i2/tau)*(1-(Pn-Pr)/(1-delta))^2/tau

  

P = -(1/2)*(1-(Pn-Pr)/(1-delta))/tau-(1/8)*Cr*(1-(Pn-Pr)/(1-delta))^2/tau^2

 (3)
> 

AB := [A = `2*`(-(1 - (Pn - Pr)/(1 - delta))/(2*tau) - Cr*(1 - (Pn - Pr)/(1 - delta))^2/(8*tau^2))*((2*rho0 - 1)*tau + i2) + (w/(2*tau) - 1/2 + i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta)) - Cr*(1/2 - i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta))^2/(2*tau) ]

[A = (2*rho0-1)*tau+i2, B = tau+i2]

 (4)
> 

ind  := indets(rhs~(PQR)) union indets(rhs~(AB));

{C0, Ce, Cr, Pn, Pr, delta, i2, rho0, tau, w}

 (5)
> 

randomize();
data:=[C0 = 0.02, Ce = 0.05, Cr = 0.05, Pn = 0.603882321, Pr = 0.45, delta = 0.7, i2 = 0.1, rho0 = 0.5, tau = 0.05, w = 0.243374];

PQRAB := { eval(PQR, data)[], eval(AB, data)[] };

all_solutions := allvalues~(eval(SC_CS_sols, PQRAB)):
all_pi        := map(s -> eval(eval(pi__m, data), s), all_solutions):
here          := ListTools:-Search(min(all_pi), all_pi):
SOLUTION      := [all_pi[here], all_solutions[here]];


plot(eval(pi__m, data), i1=eval(eval(A..B, AB), data), gridlines=true)

Download Proposal_follow_up.mw

Got it. Thankyou...

Andiguys 100
September 28 2025
0 0

@sand15 

Thank you very much for your detailed and helpful comments

You’re right , I’ll substitute the solved expression for K2 (the w solution) into {K3,K4} and solve it to simplify Maple’s task.

Point well taken, I will carefully check feasibility and not assume any solution is correct without verifying it myself.

Thanks again for the clear, constructive feedback.

Query on Selecting the Correct Solution ...

Andiguys 100
September 27 2025
0 0

 rcorless 750,  @sand15 and @dharr Thankyou for the reply.

I have modified my objective function and, as before, I am simultaneously solving three equations with three variables and multiple parameters. This process is giving me three different solution sets, and I am unsure which one is correct or should be considered.

Note: All parameters are positive and greater than zero. All decision variable Pn, Pr and w should be positive equal to or greater than zero.

Download Q_solve.mw

How to Simultaneously solve?...

Andiguys 100
September 22 2025
0 0

@acer I am optimizing a profit function with respect to three decision variables Pn,Pr, and w. When I solve the problem, I obtain conditions such as Pn(Pr,w), Pr​(Pn​,w), and w(Pn,Pr), which makes them appear dependent on each other. How can I simultaneuosly solve K2,K3 and K4 so that I can obtain the optimal values of Pn, Pr​, and w simultaneously as independent decision variables?

 

> 

restart
> 

``
> 

Pi1 := proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Cr)*((Pn-Pr)/(1-delta)-(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta)-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Cr)*((Pn-Pr)/(1-delta)-(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta)-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

 (1)
> 

K2 := diff(Pi1(Pn, Pr, w), w) = 0

-(Pn-Pr)/(1-delta)+(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta+(1/2)*(Pr-w-Cr)*beta*upsilon/delta = 0

 (2)
> 

diff(Pi1(Pn, Pr, w), `$`(w, 2))

-beta*upsilon/delta

 (3)
> 

solve(K2, w)

-(1/2)*(Cr*beta*delta*upsilon-Pr*beta*delta*upsilon+beta*delta*i2*upsilon-beta*delta*tau*upsilon-Cr*beta*upsilon+Pr*beta*upsilon-beta*i2*upsilon+beta*tau*upsilon-2*Pn*delta+2*Pr)/((-1+delta)*beta*upsilon)

 (4)
> 

K3 := diff(Pi1(Pn, Pr, w), Pn) = 0

1-(Pn-Pr)/(1-delta)-(Pn-Cn)/(1-delta)+(Pr-w-Cr)/(1-delta)+Ce*rho0/(1-delta) = 0

 (5)
> 

diff(Pi1(Pn, Pr, w), `$`(Pn, 2))

-2/(1-delta)

 (6)
> 

solve(K3, Pn)

(1/2)*Ce*rho0+(1/2)*Cn-(1/2)*Cr+Pr-(1/2)*delta-(1/2)*w+1/2

 (7)
> 

K4 := diff(Pi1(Pn, Pr, w), Pr) = 0

(Pn-Cn)/(1-delta)+(Pn-Pr)/(1-delta)-(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta+(Pr-w-Cr)*(-1/(1-delta)-1/delta)-Ce*rho0/(1-delta) = 0

 (8)
> 

diff(Pi1(Pn, Pr, w), `$`(Pr, 2))

-2/(1-delta)-2/delta

 (9)
> 

solve(K4, Pr)

-(1/4)*beta*delta*i2*upsilon+(1/4)*beta*delta*tau*upsilon-(1/4)*beta*delta*upsilon*w-(1/2)*Ce*rho0*delta+(1/4)*beta*upsilon*i2-(1/4)*beta*upsilon*tau+(1/4)*beta*upsilon*w-(1/2)*delta*Cn+delta*Pn+(1/2)*Cr+(1/2)*w

 (10)
> 

``
> 

``
> 

``
> 

> 

``


 

Download Simultaneously_solve.mw

Follow-up...

Andiguys 100
August 12 2025
0 0

@acer
I had shared the complete details of the substitutions and their explanations in my message last week. Could you please confirm if you were able to review that? I can resend the details if needed. Thankyou. 

Got it...

Andiguys 100
August 11 2025
0 0

@sand15 Thankyou so much.

Still showing the same...

Andiguys 100
August 11 2025
0 0

@sand15 The z-axis values are still shown only up to zero—how can I extend it to display positive values? Also, how can I display the z-value at point P?

Thankyou...

Andiguys 100
August 08 2025
0 0

@sand15 @acer Thankyou 

Correction...

Andiguys 100
August 08 2025
0 0

@Carl Love Correction: only s1, s2​, and d are greater than or equal to zero. All other parameters can take positive or negative values

> 

restart
> 

kernelopts(version)

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

 (1)
> 

X3 := 0 >= A*s1^2+B*s1+C

A*s1^2+B*s1+C <= 0

 (2)
> 

X4 := s1 <= (-4*A*d*s2-F)/(4*d*A)

s1 <= (1/4)*(-4*A*d*s2-F)/(d*A)

 (3)
 
> 

`assuming`([solve({X3, X4}, {s1})], [s1 > 0, s2 > 0, d > 0])

Download Q_s.mw

Clarification on Substitution and KKT Ca...

Andiguys 100
August 08 2025
0 0

@acer 
 

Substitution
In the sheet attached below, I’ve substituted Cmin⁡=β1and Cmax⁡=β2​.
Is this substitution correct, or should it instead be Cmin⁡=−β1 and Cmax⁡=−β2​ (or the other way around)?
This is my first point of confusion.

Regarding the table
From the solution set named SC_CS_sols, I have values for:

  • Pc∗​ (optimal Pc​)

  • μ1​

  • μ2​

I want these results presented in a table where:

  • Column 1 → optimal Pc value

  • Cloumn 2 → μ1​ value

  • Column 3 → μ2​ value

As you know, in the KKT conditions there are three cases, which the above table signifies:

  1. μ1=0, μ2>0

  2. μ2=0, μ1>0

  3. μ1=μ2=0

The table should show the optimal Pc​ value for each of these cases.I just want to tabulate it and export as doc or pdf format.

Questions are highlighted in yellow in sheet.

KKT_Table_Q-22.mw

unable to solve for s1...

Andiguys 100
August 08 2025
0 0

can you help me in  solving the values of s1 that satisfy both inequality constraints simultaneously — in other words, find a condition or range for s1​, such as s1=… , given ...... > …, under which both inequalities are true.

> 

restart
> 

kernelopts(version)

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

 (1)
> 

X3 := 0 >= A*s1^2+B*s1+C

A*s1^2+B*s1+C <= 0

 (2)
> 

X4 := s1 <= (-4*A*d*s2-F)/(4*d*A)

s1 <= (1/4)*(-4*A*d*s2-F)/(d*A)

 (3)
> 

``
> 

extra1 := `~`[`>`](indets(X3, And(name, Not(constant))), 0)

{0 < A, 0 < B, 0 < C, 0 < s1}

 (4)
> 

extra2 := `~`[`>`](indets(X4, And(name, Not(constant))), 0)

{0 < A, 0 < F, 0 < d, 0 < s1, 0 < s2}

 (5)
> 

`assuming`([solve({X3, X4}, {s1})], [extra1[], extra2[]])

Download Q_s.mw

Need help....

Andiguys 100
August 07 2025
0 0

@acer Hi can you help me to tabulate the result? and verify whether the substititution i made are right. Thankyou

how to take out delta3?...

Andiguys 100
August 07 2025
0 0

@sand15 Thankyou.

Can in anyway possible to take out  delta3  from the equaltion so that i get condition like delta > or < (delta3) * L expression.

Basically to show relationship between delta and delta3.

Got it...

Andiguys 100
August 05 2025
0 0

@sand15 Thankyou

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