add conditional constrain in optimization
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if a=0 then q=0
if a=1 then q=24;
if a=2 then q=32;
if a=3 then q=44;
how to write this as constrain for optimization problem?
5 Kommentare
Dyuman Joshi
am 27 Feb. 2023
q=[0 24 32 44];
out = q(a+1)
Rik
am 27 Feb. 2023
Can you provide an example? This may require creating a custom function.
Torsten
am 27 Feb. 2023
Which optimizer do you use ?
Is a defined to be integer ?
Is q defined to be integer ?
Are a and q solution variables ?
Akzeptierte Antwort
Define x_it be the decision variable if pump i is active at time t. That means x_it can take values 0 and 1 and equals 0 of pump i is off at time t and equals 1 if pump i is on at time t.
Then your problem somehow looks like
min: sum_t (x1t + x2t + x3t)
24*x1t + 32*x2t + 44*x3t >= amount of water needed at time t (t=1,...,T)
0 <= x_it < = 1
x_it integer
You can use "intlinprog" to solve.
35 Kommentare
Fathima
am 28 Feb. 2023
Torsten
am 28 Feb. 2023
x = optimvar('x',3,n,'LowerBound',0,'UpperBound',1,'Type','integer');
Fathima
am 28 Feb. 2023
Please look this video to get accustomed to the problem-based approach in optimization.
Follow the steps in your problem setup.
Could you include the mathematical formulation of your optimization problem for comparison ?
Fathima
am 28 Feb. 2023
Torsten
am 28 Feb. 2023
As far as I understand your problem, this is the first step to solve your problem:
A mathematical formulation.
min: sum_{t=1}^{t=T} (x_1t + 2*x2t + 3*x3t)
under the constraints
A * (h_(t+1) - h_t) = deltaT * ( (x_1t*24 + x2t*32 + x3t*44) - Qd(t)) (t=1,...T)
x1t + x2t + x3t <= 1 (t=1,...,T)
0 <= h(t) <= Hmax (t=1,...,T)
0 <= x1t, x2t, x3t <= 1 (t=1,...,T)
x1t,x2t,x3t integer (t=1,...,T)
Now use "intlinprog" to solve with the help of the video.
Fathima
am 28 Feb. 2023
Note that in the above formulation,
x_1t = 0, x_2t = 0, x_3t = 0 means: no pump is active at time t
x_1t = 1, x_2t = 0, x_3t = 0 means: one pump is active at time t
x_1t = 0, x_2t = 1, x_3t = 0 means: two pumps are active at time t
x_1t = 0, x_2t = 0, x_3t = 1 means: three pumps are active at time t
This differs from my former definition of the x_it.
I think it will also be possible to formulate the problem with x_it = 0 or 1 depending on whether pump i is active at time t or not, but I did no longer follow this path of thought.
And are you sure you only want to minimize the sum of the pumps used and not the total energy consumption ? Because the three pumps seem to differ in power demand.
Thus a more flexible formulation would be
min: sum_{t=1}^{t=T} a_t
s.c.
a_t = sum_{i=1}^{i=I} P_i*x_it (t = 1,...,T) % total Power consumption at time t
Q_t = sum_{i=1}^{i=I} Q_i*x_it (t = 1,...,T) % total water supply at time t
A*(h(t+1)-h(t)) = deltaT * (Q_t - Q_d(t)) (t = 1,...,T) % variation of Tank height
0 <= h(t) <= Hmax (t = 1,...,T)
0 <= x_it <= 1 (i = 1,...,I ; t=1,...,T)
x_it integer (i = 1,...,I ; t=1,...,T) % = 1 if pump i is active at time t, = 0 else
Fathima
am 1 Mär. 2023
Fathima
am 1 Mär. 2023
Fathima
am 1 Mär. 2023
dh/dt means change in height with respect to time so dh/dt will be same as h[t+1]-h[t] right?
No. dh/dt has unit m/s, h[t+1]-h[t] has unit m.
as you have asked of objective , i wanted to minimize power consumption. but did not know how to connect power to decision variables. so i minimized number of pumps working at time t since its directly propotional to power consumption. if number of pumps working increases , power also will increase.. so thats the reason I took minimize number of pumps as objective.
The pumps you use seem to differ because a[t] does not increase linearily. Your approach to just add the pumps in action is only feasible if a[0] = 0, a[1] = 24, a[2] = 48 and a[3] = 72 and the power they consume is equal . So you have to take into account how much power each individual pump consumes and how much water it is able to supply.
x0t = 1; %no pump working
if h<6.8 && h>=4
x1t = 1; %one pump working
elseif h<4 && h>=2.3
x2t=1;
elseif h<2.3
x3t = 1;
end
how to add this as constraint?
If you add this as constraint, you do not need an optimizer any longer because you prescribe the variables "intlinprog" would solve for. So why do you want to add this constraint ? Since you used the constraints 0 <= h(t) <= Hmax, "intlinprog" will automatically adjust the number of pumps for each time such that water remains in the reservoir tank while at the same time minimum power is used to supply sufficient water. If there does not exist a control strategy for the liquid level such that 0 <= h(t) <= Hmax can be satisfied over the time period for your problem, "intlinprog" will give you a message that your problem is not feasible. But you don't need to include control mechanisms on your own as you try above - the solver will automatically find the best strategy.
Fathima
am 3 Mär. 2023
Torsten
am 3 Mär. 2023
But this is a completely different approach to the problem. You must decide which approach you want to use. If you know water demand for future periods in advance, the above approach using "intlinprog" is superior to the neural network approach. If water demand is uncertain, then it makes sense to first train your system with demand data from the past and then use the trained system to take decisions in similar situations in the future (at least this is how I understand the approach used in the link you supplied).
Fathima
am 3 Mär. 2023
Fathima
am 3 Mär. 2023
Fathima
am 4 Mär. 2023
If you include the Excel files with time and demand data, I'll take a look.
You did not set the initial liquid level h(1) in the tank - this is of course necessary to give the system an initial condition.
For this, you simply add a constraint
h(1) == H0;
where H0 is a numerical value defined somewhere in your code with 0 <= H0 <= Hmax.
Try this. It takes too long here to finish, so I cannot tell whether your output part makes sense.
rng("default")
num_days = 3;
[WaterDemand T_max] = generateWaterDemand(num_days);
time = WaterDemand.time;
Qd = WaterDemand.Data;
plot(time,Qd)
problem = optimproblem;
% Define the constants and parameters of the problem
T = length(time);
A = 40;
Hmax = 7;
H0 = 3.5;
% Define the decision variables
x1 = optimvar('x1', T-1,1, 'LowerBound', 0, 'UpperBound', 1, 'Type', 'integer');
x2 = optimvar('x2', T-1,1, 'LowerBound', 0, 'UpperBound', 1, 'Type', 'integer');
x3 = optimvar('x3', T-1,1, 'LowerBound', 0, 'UpperBound', 1, 'Type', 'integer');
h = optimvar('h', T,1, 'LowerBound', 0, 'UpperBound', Hmax);
% Define the objective function
problem.Objective = sum(x1+2*x2+3*x3);
% Define the linear constraints
constr1 = optimconstr(T-1);
constr2 = optimconstr(T-1);
constr1(1) = h(1) == H0;
for t = 2:T-1
constr1(t) = A*(h(t+1) - h(t)) == (( x1(t)*164 + x2(t)*279 + x3(t)*344) - Qd(t));
end
for t = 1:T-1
constr2(t) = x1(t)+x2(t)+x3(t)<=1;
end
problem.Constraints.constr1 = constr1;
problem.Constraints.constr2 = constr2;
show(problem)
[sol, fval, exitflag] = solve(problem);
figure(1)
plot(time,sol.h)
figure(2)
plot(time(1:end-1),sol.x1+2*sol.x2+3*sol.x3)
function [WaterDemand,T_max] = generateWaterDemand(num_days)
t = 0:(num_days*24)-1; % hr
T_max = t(end);
Demand_mean = [28, 28, 28, 45, 55, 110, 280, 450, 310, 170, 160, 145, 130, ...
150, 165, 155, 170, 265, 360, 240, 120, 83, 45, 28]'; % m^3/hr
Demand = repmat(Demand_mean,1,num_days);
Demand = Demand(:);
% Add noise to demand
a = -25; % m^3/hr
b = 25; % m^3/hr
Demand_noise = a + (b-a).*rand(numel(Demand),1);
WaterDemand = timeseries(Demand + Demand_noise,t);
WaterDemand.Name = "Water Demand";
WaterDemand.TimeInfo.Units = 'hours';
end
Fathima
am 4 Mär. 2023
I tried to run the code but not getting correct output. since for h=3.5 , I am getting 0 pumps working and at h=3.92 I am geting x1=1 that is one pump is working. which doesnot make sense since 3.5<3.9 and @3.5 no pump is working.
If at the time when h=3.5 there is a low water demand and when h=3.9 there is a large water demand, pumps won't work at h=3.5 and will work at h=3.9. So working or not working of pumps does not necessarily depend on the liquid level.
I'm almost certain the code above is correct.
% Load data from excel sheet
time= xlsread('mathlab_water_tank_pipe.xlsx', 'Sheet1', 'A1:A184');
Qd=xlsread('mathlab_water_tank_pipe.xlsx', 'Sheet1', 'B1:B184');
plot(time,Qd)
problem = optimproblem;
% Define the constants and parameters of the problem
T = length(time);
A = 40;
Hmax = 7;
H0 = 3.5;
% Define the decision variables
x1 = optimvar('x1', T-1,1, 'LowerBound', 0, 'UpperBound', 1, 'Type', 'integer');
x2 = optimvar('x2', T-1,1, 'LowerBound', 0, 'UpperBound', 1, 'Type', 'integer');
x3 = optimvar('x3', T-1,1, 'LowerBound', 0, 'UpperBound', 1, 'Type', 'integer');
h = optimvar('h', T,1, 'LowerBound', 0, 'UpperBound', Hmax);
% Define the objective function
problem.Objective = sum(x1+2*x2+3*x3);
% Define the linear constraints
constr1 = optimconstr(T-1);
constr2 = optimconstr(T-1);
constr1(1) = h(1) == H0;
for t = 2:T-1
constr1(t) = A*(h(t+1) - h(t)) == 0.1*(( x1(t)*164 + x2(t)*279 + x3(t)*344) - Qd(t));
end
for t = 1:T-1
constr2(t) = x1(t)+x2(t)+x3(t)<=1;
end
problem.Constraints.constr1 = constr1;
problem.Constraints.constr2 = constr2;
show(problem)
[sol, fval, exitflag] = solve(problem);
figure(1)
plot(time,sol.h)
figure(2)
plot(time(1:end-1),sol.x1+2*sol.x2+3*sol.x3)
Fathima
am 4 Mär. 2023
Fathima
am 4 Mär. 2023
Fathima
am 4 Mär. 2023
No feasible solution found is ur output... But in mathwork they are getting solution..
I got a solution to the optimization for a time horizon of 3 days when I used the function "generateWaterDemand" which generates the demand for the pump model from mathworks (see above).
In mathwork model , they have added function in simulink which is
function a = fcn(h)
break1 = 4.5;
break2 = 3.5;
a = 0;
if h<6.8 && h>=break1
a = 1;
elseif h<break1 && h>=break2
a = 2;
elseif h<break2
a = 3;
end
here a is number of pump
As I already wrote, this constraint does not make sense if you use "intlinprog" to optimize the usage of the pumps.
It already prescribes the solution of the optimization problem:
x1(t) = 1, x2(t) = 0, x3(t) = 0 if h < 3.5
x1(t) = 0, x2(t) = 1, x3(t) = 0 if 3.5 <= h < 4.5
x1(t) = 0, x2(t) = 0, x3(t) = 1 if h >= 4.5
Do you understand what I mean ?
If "intlinprog" tells you that there is no solution, then the water demand Qd cannot be satisfied with the supplied pumping power for at least one time instant t if the restriction 0 <= h <= hmax should hold.
Give me a "solution" from Reinforcement Learning for the demand curve you prescribe in your Excel sheet and I'm sure one of the constraints will fail for at least one time t.
As far as I understand the Mathworks example, the aim is - where possible - avoid violating constraints and at the same time activating the pumps efficiently. But this does not exclude possible failures. Optimization is more stringent: if the inputs are such that there is no strategy to avoid failure, the optimizer will return that there is no feasible solution.
Torsten
am 4 Mär. 2023
The code now also works for your Excel sheet. Seems you changed the time increment from 1 hour to 1/10 hour. This has to be reflected in the "deltat" used.
Why should "intlinprog" use the strategy to switch on three pumps if h < 3.5, use 2 pumps of 3.5 < = h < 4.5 and use one pump if h >= 4.5 ? The values 3.5 and 4.5 are absolutely empirical and not optimal.
"Intlinprog" is given the water demands over the time period it optimizes because you use the demands in your constraints. That's why "Intlinprog" can find a much better strategy than the one above.
But usually, the demands are uncertain and not exactly known in advance. That's what the Deep Learning Tool is for. It develops a strategy based on data in the past (learning phase) and uses this strategy to take present decisions (now without knowing what exactly the future will bring). Thus both methods try to optimize the usage of the pumps, but the circumstances under which they optimize, namely exact or uncertain knowledge about the future demands, make the mathematical methods used and the results completely different.
If the future demands are known, "intlinprog" will always give the "best" solution. If the future demands are uncertain, you will have to refer to neural networks, Deep Learning or related methods.
Fathima
am 15 Mär. 2023
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