I am building linear programming model of machines manufacturing some parts in AMPL using CPLEX solver.
My problem is how to model 10% efficiency step drop of machine after 100 hours of work. I know that proper approach in my case is to use binary variable. Unfortunately I have no idea how to manage this.
Could you show an example how to model described above behavior in linear programming?
EDIT: attaching current AMPL code
Data:
data;
param M := 3; # machines count
param N := 5; # parts count
param efficiency: # efficiency [pc / h]
1 2 3 4 5 :=
1 0.85 1.30 0.65 1.50 0.40
2 0.65 0.80 0.55 1.50 0.70
3 1.20 0.95 0.35 1.70 0.40;
param R := 1 45 # machines costs [usd / h]
2 35
3 40;
param C := 1 60 # minimal manufactured counts of particulat parts
2 60
3 60
4 120
5 120;
Model:
model;
param M; # machines count
param N; # parts count
param efficiency {1..M, 1..N}; # efficiency [pc / h]
param R {1..M}; # machines costs [usd / h]
param C {1..N}; # minimal manufactured counts of particulat parts
var t {1..M, 1..N}; # time[machine,part]
var d {n in 1..N} = sum {m in 1..M} t[m,n] * efficiency[m,n];
minimize cost: sum {m in 1..M, n in 1..N} R[m] * t[m,n];
subject to c1 {n in 1..N}:
d[n] >= C[n];
subject to t1 {m in 1..M}:
sum {n in 1..N} t[m,n] <= 180; # machine time must be less than 180
subject to t2 {m in 1..M, n in 1..N}:
t[m,n] >= 0; # times must non-negative
Related
Winbugs trap error
model
{
for (i in 1:5323) {
Y[i] ~ dpois(mu[i]) # NB model as a Poisson-gamma mixture
mu[i] ~ dgamma(b[i], a[i]) # NB model as a poisson-gamma mixture
a[i] <- b[i] / Emu[i]
b[i] <- B * X[i]
Emu[i] <- beta0 * pow(X[i], beta1) # model equation
}
# Priors
beta0 ~ dunif(0,10) # parameter
beta1 ~ dunif(0,10) # parameter
B ~ dunif(0,10) # over-dispersion parameter
}
X[] Y[]
1.5 0
2.9 0
1.49 0
0.39 0
3.89 0
2.03 0
0.91 0
0.89 0
0.97 0
2.16 0
0.04 0
1.12 1s
2.26 0
3.6 1
1.94 0
0.41 1
2 0
0.9 0
0.9 0
0.9 0
0.1 0
0.88 1
0.91 0
6.84 2
3.14 3
End ```
This is just a sample of the data, the model question is coming from Ezra Hauer 8.3.2, the art of regression of road safety, the model is providing an **error undefined real result. **
The aim of model is to fully Bayesian and a one step model and not use empirical bayes.
The results should be similar to MLE where beta0 is 1.65, beta1 0.871, overdispersion is 0.531
X is the only variable and y is actual collision,
So X cannot be zero or negative, while y cannot be lower than zero, if the model in solved as Poisson gamma mixture using maximum likelihood then it can be created
How can I make this model work
Solving an error in winbugs?
the data is in excel, the model worked fine when I selected the biggest 1000 observations only.
We make phones. We have selling price, production cost, profit.
The goal is to maximize profits.
The following components are required to assemble each phone.
Maximum quantity of components .
Orders (so many phones were ordered from us, we sold them) :
Here is my mod file:
set PHONE;
set COMPONENTS;
param price {PHONE} >= 0;
param cost {PHONE} >= 0;
param maxComponents {COMPONENTS} >= 0;
param ordered {PHONE} >= 0;
param matrix {COMPONENTS, PHONE}; #The amount of components needed to make a particular phone.
var x {PHONE} >= 0; # Number of manufactured telephones.
maximize profit: sum {i in PHONE} ( ordered[i] * price[i] - x[i] * cost[i] );
subject to min_manufacture {i in PHONE}:
x[i] >= ordered[i]; # We must produce a minimum of what is ordered
subject to component {i in COMPONENTS}:
sum {j in PHONE} matrix[i,j] * x[j] <= maxComponents[i]; # The number of components used must not exceed the maximum.
subject to min_quantity {i in COMPONENTS, l in PHONE}:
sum {j in PHONE} matrix[i,j] * x[j] >= matrix[i,l]; # Minimum quantity used per component if we manufacture at least one telephone. For example, a triple phone requires at least 2 of the five components.
and dat file:
set PHONE := 1 2 3 4 5;
set COMPONENTS:= 1 2 3 4 5 6 7;
param price :=
1 450
2 120
3 500
4 390
5 100;
param cost :=
1 370
2 90
3 400
4 320
5 70;
param maxComponents :=
1 28
2 20
3 8
4 30
5 47
6 27
7 15;
param ordered :=
1 3
2 5
3 5
4 0
5 10;
param matrix: 1 2 3 4 5 :=
1 1 1 0 0 0
2 1 1 0 0 0
3 1 0 0 0 0
4 1 0 1 1 0
5 0 0 2 1 1
6 0 0 2 1 0
7 0 0 1 1 0;
The problem is that if, for example, the maximum amount of sixth components is three, the maximum amount of seventh components is two , then 1.5 is produced from the triple phone which cannot be . And quantity used of the fourth, fifth, sixth, seventh components for the triple phone 1,5 3 3 1,5 which also cannot be.
How do I do it to just get a integer solution?
Because if I write to the variable x that it's an integer, I get zero for everything.
My run file:
model phone.mod;
data phone.dat;
option presolve 0;
option solver cplex;
solve;
display profit, x;
display {i in COMPONENTS, j in PHONE} matrix[i,j] * x[j];
You need to declare the relevant variables as integer, like so:
var x {PHONE} >= 0 integer;
Some solvers are not able to deal with integer constraints and may ignore that constraint (with a warning message) but CPLEX should be fine.
I am trying to solve below network problem in GAMS Cplex. I am unable to get the desired output as the GAMS is giving the arcs which doesnt exist as output. Could you please help me in correcting this.
Program:
Set
i supply nodes /1,2,3,4/;
Alias(i,j);
Set
arc(i,j) arcs from node i to j
/1 .2
2 .3
3 .4
1 .4
2 .4/;
Parameter b(i) number of units available or required at node i
/ 1 5
2 2
3 -4
4 -3/ ;
Table c(i,j) cost of shipping from node i to node j
1 2 3 4
1 0 3 0 1
2 0 0 6 5
3 0 0 0 0
4 0 0 2 0 ;
Positive variables
x(i,j) number of units shipped along arc from i to j;
Variable z;
Equations obj, cons(i);
obj.. z =E= sum(arc(i,j),c(arc)*x(arc));
cons(i).. sum(j,x(i,j)) - sum(j,x(j,i)) =E= b(i);
I think you need to use the set "arc" in your constraints "cons" like this:
cons(i).. sum(arc(i,j),x(i,j)) - sum(arc(j,i),x(j,i)) =E= b(i);
Hope that helps,
Lutz
I have the next LP problem
Maximize
1000 x1 + 500 x2 - 500 x5 - 250 x6
Subject To
c1: x1 + x2 - x3 - x4 = 0
c2: - x3 + x5 = 0
c3: - x4 + x6 = 0
With these Bounds
0 <= x1 <= 10
0 <= x2 <= 15
0 <= x5 <= 15
0 <= x6 <= 5
By solving this problem with Cplex dual algorithm I get an optimal solution of 6250. But checking the reduced costs of the variables I get the next results
Variable value reduced cost
1 10.0 500.0
1 0.0 -0.0
2 5.0 -0.0
3 5.0 -0.0
4 5.0 -0.0
5 5.0 250.0
Is it possible to have a positive reduced cost on a positive valued variable? Because the reduced cost value indicates how much the objective function coefficient on the corresponding variable must be improved before the value of the variable will be positive in the optimal solution, what does a positive reduced cost means on a postive valued variable?
Variable 1 is listed twice in the solution?
Note that you need to distinguish between nonbasic at lower bound and nonbasic at upper bound. The reduced cost indicates how much the objective can change when the corresponding bound changes by one unit.
Note also that most textbooks focus on the special case x >= 0 while practical solvers support both lower and upper bounds: L <= x <= U.
I am relatively new to proc optmodel and have been struggling with syntax/structure. I was able to get help once before and am stuck again.
Here is my dataset:
data have;
input NAME $ TEAM $ LEAD GRADE XXX MIN MAX YYY RATE;
cards;
HAL A 1 1 50 45 55 100 1.1
SAL A 0 2 55 0 9999 200 1
KIM A 0 3 70 0 9999 50 1.4
JIM B 1 2 100 90 110 300 .95
GIO B 0 3 120 0 9999 50 1
CAL B 0 4 130 0 9999 20 .9
TOM C 1 1 2 1 5 20 .7
SUE C 0 3 5 0 9999 10 .5
VAL D 1 7 20 15 25 100 .6
WHO D 0 4 10 0 9999 10 .9
;
run;
Here are the specifics:
1. Only the "team lead" has any meaningful constraints.
2. However, the other members of the team will be adjusted accordingly. The value of XXX will be ten percent lower or higher relative to the difference in grade from the team lead. So, if HAL's NEW_XXX is 50 (stays same), then SAL will be 10% higher than HAL's (2 is 1 unit greater than 1) which is 55. KIM's NEW_XXX is 60, since this is twenty percent higher than HAL (3 is 2 units greater than 1. SImilarly, WHO's NEW_XXX will be 30% lower than VAL's.
Does that make sense?
Below is what I have so far, which is the skeleton from a similar project.
proc optmodel;
*set variables and inputs;
set<string>NAME;
string TEAM{NAME};
number LEAD{NAME};
number GRADE{NAME};
number XXX{NAME};
number MIN{NAME};
number MAX{NAME};
number YYY{NAME};
number RATE{NAME};
set TEAMS = setof{i in NAME} TEAM[i];
set NAMEperTEAM{gi in TEAMS} = {i in NAME: TEAM[i] = gi};
var NEW_XXX{i in NAME}>=MIN[i]<=MAX[i];
*read data into procedure;
read data have into
NAME=[NAME]
TEAM
LEAD
GRADE
XXX
MIN
MAX
YYY
RATE;
*state function to optimize;
max metric=sum{gi in TEAMS}
sum{i in NAMEperTEAM[gi]}
(NEW_XXX[i])*(1-(NEW_XXX[i]-XXX[i])*RATE[i]/XXX[i])*YYY[i];
expand;
solve;
*write output dataset;
create data results
from [NAME]={NAME}
TEAM
LEAD
GRADE
XXX
NEW_XXX
MIN
MAX
RATE
YYY;
*write results to window;
print NEW_XXX metric;
quit;
If I understand this correctly, you need set the non-team leads NEW_XXX variable in an equality constraint. That leaves only the team lead NEW_XXX variables free for the optimization.
Let me know if this is what you are trying to accomplish.
Here's how I did it:
proc optmodel;
*set variables and inputs;
set<string> NAME;
string TEAM{NAME};
number LEAD{NAME};
number GRADE{NAME};
number XXX{NAME};
number MIN{NAME};
number MAX{NAME};
number YYY{NAME};
number RATE{NAME};
*read data into procedure;
read data have into
NAME=[NAME]
TEAM
LEAD
GRADE
XXX
MIN
MAX
YYY
RATE;
set TEAMS = setof{i in NAME} TEAM[i];
set NAMEperTEAM{gi in TEAMS} = {i in NAME: TEAM[i] = gi};
/*Helper array that gives me the team leader for each team*/
str LEADS{TEAMS};
for {i in NAME: LEAD[i] = 1} do;
LEADS[TEAM[i]] = i;
end;
var NEW_XXX{i in NAME} init XXX[i] >=MIN[i]<=MAX[i];
*state function to optimize;
max metric=sum{gi in TEAMS}(
sum{i in NAMEperTEAM[gi]} (
(NEW_XXX[i])*(1-(NEW_XXX[i]-XXX[i])*RATE[i]/XXX[i])*YYY[i]
)
);
/*Constrain the non-lead members*/
con NonLeads{i in NAME: LEAD[i] = 0}: NEW_XXX[i] = (1 + (GRADE[i] - GRADE[LEADS[TEAM[i]]]) * 0.1) * NEW_XXX[LEADS[TEAM[i]]] ;
expand;
solve;
*write output dataset;
create data results
from [NAME]={NAME}
TEAM
LEAD
GRADE
XXX
NEW_XXX
MIN
MAX
RATE
YYY;
*write results to window;
print new_xxx metric;
quit;