I'm trying to implemenet an assignment problem. I have the following problem when trying to multiply two variables in linear programming (using glpk gusek) in my goal function:
minimize PATH_COST: sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z])*y[k,z]; #y is a binary variable; Koszty_Suma is total cost for ordez z and car type k
The following error is arising: "model.mod:47: multiplication of linear forms not allowed".
Code (.dat file):
data;
set numery_Zlecen := 1, 2, 3; #order numbers
set Miasta := '*some data: *' #cities.
#order numer (from city to city)
set Zlecenie[1] := Warszawa Paris;
set Zlecenie[2] := Berlin Praha;
set Zlecenie[3] := Praha Amsterdam;
#number of packages for transport for a particular order
param Ilosc_Wyrobow :=
1 10
2 50
3 110;
param Godziny_Pracy := 9; #number of working hours during the day
param Pojemnosc_Samochodu := 35; #capacity of the car (how many packages it can take)
param Srednia_Predkosc := 80; #average car speed
param Spalenie_Paliwa := 0.25; #fuel combustion
param Wynagrodzenie_za_Godzine := 20; #salary for one working hour
param Cena_Noclegu := 100; #price of accommodation
param Dystans: '*some data: *' #km between cities.
param Koszt_Paliwa : '*some data: *' #fuel consumption depends on country.
end;
Code (.mod file):
#INDEXY
#=====================================================================
set Miasta; #i,j
set numery_Zlecen; #z
set Zlecenie{numery_Zlecen} dimen 2; #p,q
set Rodzaj_Transportu; #k
#PARAMETRY
#=====================================================================
param Dystans {Miasta,Miasta};
param Ilosc_Wyrobow{numery_Zlecen};
param Godziny_Pracy >= 0;
param Pojemnosc_Samochodu {Rodzaj_Transportu}>= 0;
param Srednia_Predkosc >=0;
param Spalenie_Paliwa >=0;
param Koszt_Paliwa {Miasta,Miasta};
param Wynagrodzenie_za_Godzine >= 0;
param Cena_Noclegu >= 0;
#ZMIENE
#=====================================================================
var x{Miasta,Miasta,numery_Zlecen} <= 1, >= 0; #variable x equal 1 when we're going the path from city A to city B; otherwise it equals 0
var y{Rodzaj_Transportu,numery_Zlecen} binary <=1, >=0; #variable that shows what types of car/s we are using for order (can be 0 or 1)
var Koszty_Suma{Rodzaj_Transportu,numery_Zlecen}; #total costs
var Koszty_Transportu{numery_Zlecen}; #transport costs
var Koszty_Odpoczynku{numery_Zlecen}; #rest costs
var Koszty_Wynagrodzenia{numery_Zlecen}; #salary costs
#FUNKCjA CELU
#=====================================================================
minimize PATH_COST: sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z])*y[k,z];
#OGRANICZENIA (constraints)
#=====================================================================
s.t. SOURCE{z in numery_Zlecen, (p,q) in Zlecenie[z], i in Miasta: i = p && p != q}:
sum {j in Miasta} (x[i ,j ,z ]) - sum {j in Miasta}( x[j ,i ,z ]) = 1;
s.t. INTERNAL {z in numery_Zlecen, (p,q) in Zlecenie[z],i in Miasta: i != p && i != q && p != q }:
sum {j in Miasta} (x[i ,j ,z ]) - sum {j in Miasta}( x[j ,i ,z ]) = 0;
s.t. OGR_KM_DZIEN{z in numery_Zlecen,(p,q) in Zlecenie[z], j in Miasta, i in Miasta: i != q}:
if (Dystans[i,j] > (Godziny_Pracy*Srednia_Predkosc)) and i != q then x[i,j,z] = 0;
s.t. OGR_KOSZTY_SUMA{z in numery_Zlecen, k in Rodzaj_Transportu}:
Koszty_Suma[k,z] = (Koszty_Transportu[z] + Koszty_Odpoczynku[z] + Koszty_Wynagrodzenia[z])*ceil(Ilosc_Wyrobow[z]/Pojemnosc_Samochodu[k]);
s.t. OGR_KOSZTY_TRANSPORTU{z in numery_Zlecen}:
Koszty_Transportu[z] = (sum{i in Miasta} (sum{j in Miasta} ( Dystans[i,j]*x[i,j, z]*Koszt_Paliwa[i,j] ) ))*Spalenie_Paliwa;
s.t. OGR_KOSZTY_ODPOCZYNKU{z in numery_Zlecen}:
Koszty_Odpoczynku[z] =
(sum{i in Miasta} (sum{j in Miasta} ( Dystans[i,j]*x[i,j, z] ) ))/(Godziny_Pracy*Srednia_Predkosc) * Cena_Noclegu;
s.t. OGR_KOSZTY_WYNAGRODZENIA{z in numery_Zlecen}:
Koszty_Wynagrodzenia[z] =
((sum{i in Miasta} (sum{j in Miasta} ( Dystans[i,j]*x[i,j, z] ) ))/(Srednia_Predkosc)) * Wynagrodzenie_za_Godzine;
s.t. OGR_Y_JEDEN{z in numery_Zlecen}:
sum{k in Rodzaj_Transportu}(y[k,z]) = 1;
solve;
How is it possible to get rid of this error? Any hints how to solve this kind of problem are welcome.
First I think the parentheses are incorrect (note that y[k,z] depends on z). The expression
sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z])*y[k,z];
is not mathematically correct. So, I assume what you meant is:
sum{k in Rodzaj_Transportu}(sum{z in numery_Zlecen}Koszty_Suma[k,z]*y[k,z]);
Let me restate the problem a little bit. I assume we can write this as:
sum((i,j), x[i,j]*y[i,j])
with y a binary variable and x a continuous variable. I also assume 0 <= x[i,j] <= U[i,j]. (U is an upper bound).
Here is a way to linearize this quadratic term. We can introduce a variable z[i,j]=x[i,j]*y[i,j] using the following inequalities:
z[i,j] <= U[i,j]*y[i,j]
z[i,j] <= x[i,j]
z[i,j] >= x[i,j]-U[i,j]*(1-y[i,j])
0 <= z[i,j] <= U[i,j]
Now you just can minimize sum((i,j),z[i,j]). For a similar linearization see link.
Related
I need help with looping in Mata. I have to write a code for Beta coefficients for OLS in Mata using a loop. I am not sure how to call for the variables and create the code. Here is what I have so far.
foreach j of local X {
if { //for X'X
matrix XX = [mata:XX = cross(X,1 , X,1)]
XX
}
else {
mata:Xy = cross(X,1 , y,0)
Xy
}
I am getting an error message "invalid syntax".
I'm not sure what you need the loop for. Perhaps you can provide more information about that. However the following example may help you implement OLS in mata.
Load example data from bcuse:
ssc install bcuse
clear
bcuse bwght
mata
x = st_data(., ("male", "parity","lfaminc","packs"))
cons = J(rows(x), 1, 1)
X = (x, cons)
y = st_data(., ("lbwght"))
beta_hat = (invsym(X'*X))*(X'*y)
e_hat = y - X * beta_hat
s2 = (1 / (rows(X) - cols(X))) * (e_hat' * e_hat)
B = J(cols(X), cols(X), 0)
n = rows(X)
for (i=1; i<=n; i++) {
B =B+(e_hat[i,1]*X[i,.])'*(e_hat[i,1]*X[i,.])
}
V_robust = (n/(n-cols(X)))*invsym(X'*X)*B*invsym(X'*X)
se_robust = sqrt(diagonal(V_robust))
V_ols = s2 * invsym(X'*X)
se_ols = sqrt(diagonal(V_ols))
beta_hat
se_robust
end
This is far from the only way to implement OLS using mata. See the Stata Blog for another example using quadcross, I like my example because it preserves a little more of the matrix algebra in the code.
The following is a program for self avoiding random walk. The program works fine but I need to make a minor modification but I do not know how.
Currently the program receives n and ns as inputs and then calculates a distance (dis). I want the program to receive more than one n and calculate the distance for each n.
Example of current output
n = 100 ns = 100 dis = 10.8
I want the program to output
n = 100 ns = 100 dis = 10.8
n = 200 ns = 100 dis = 11.6
and go on for all input vales of n.
This can be done by running the program every time with different n but I need to do it with one run.
PROGRAM Two_dimensional_Self_Avoiding__Random_Walks
implicit none
integer, dimension(:,:), allocatable :: lattice
integer :: i,x,y,xt,yt,id,step,xx, ns,n
real :: r,dis,dis2,square,d,d2
Logical :: terminate,newsite
print*, ' Enter ns and n '
read(*,*) ns,n
allocate(lattice(-n:n,-n:n))
CALL RANDOM_SEED()
dis = 0.0; dis2 = 0.0 ! intial values for end to end distance
CALL walks() ! self avoiding walks
dis = dis/float(ns); dis2 = dis2/float(ns)
print*,ns,n,dis,dis2
CONTAINS
SUBROUTINE walks
DO i = 1,ns
lattice = 0; x = 0; y = 0
step = 0; terminate = .FALSE.
!do ii = 1, n
DO WHILE ((.NOT. terminate) .AND. (step <= n))
xt = x; yt = y
xx = lattice(x+1,y)+lattice(x-1,y) &
+lattice(x,y+1)+lattice(x,y-1)
IF (xx == 4) THEN
terminate = .TRUE.
ELSE
newsite = .FALSE.
DO WHILE (.NOT. newsite)
CALL RANDOM_NUMBER(r)
id = INT(r*4.0)
IF (id == 0) THEN
x = xt + 1; y = yt
ELSEIF (id == 1) THEN
x = xt - 1; y = yt
ELSEIF (id == 2) THEN
x = xt; y = yt + 1
ELSEIF (id == 3) THEN
x = xt; y = yt - 1
ENDIF
IF (lattice(x,y) == 0) newsite = .TRUE.
ENDDO
step = step + 1; lattice(x,y) = 1
ENDIF
write(10,*),step
!print*, x,y
write(7,*) x,y
ENDDO
square = float(x**2+y**2)
dis = dis + sqrt(square); dis2 = dis2 + square
d = dis/ns; d2=dis2/ns
write(8,*) step, d, d2
!enddo
ENDDO
END SUBROUTINE walks
END PROGRAM Two_dimensional_Self_Avoiding__Random_Walks
I wanted to convert this C++ code to ocaml, but i'm getting syntax error
C++ code
int** matrix(int n,int **a,int**b)
{
t=n/2;
a11=new int*[t];
for(i=0;i<t;i++)
a11[i]=new int [t];
for(i=0;i<t;i++)
for(j=0;j<t;j++)
a11[i][j]=a[i][j];
a12=new int*[t];
for(i=0;i<t;i++)
a12[i]=new int [t];
for(i=0;i<t;i++)
for(j=0;j<t;j++)
a12[i][j]=a[i][j+t];
a21=new int*[t];
for(i=0;i<t;i++)
a21[i]=new int [t];
for(i=0;i<t;i++)
for(j=0;j<t;j++)
a21[i][j]=a[i+t][j];
}
Ocaml code
let matrix n x y =
let t = n/2 in
let a11 = Array.make_matrix t t 0 in
for i = 0 to t-1 do
for j = 0 to t-1 do
a11.(i).(j) <- x.(i).(j)
done
done
and
a12 = Array.make_matrix t t 0 in
for i = 0 to t-1 do
for j = 0 to t-1 do
a12.(i).(j) <- x.(i).(j+t)
done
done
and
a21 = Array.make_matrix t t 0 in
for i = 0 to t-1 do
for j = 0 to t-1 do
a21.(i).(j) <- x.(i+t).(j)
done
done
;;
The problem is that the value of t is not getting passed inside the a12 and a21 arrays and it is getting unbounded.
The and for a let can't appear in the body of the let. Instead, use multiple lets:
let matrix n x y =
let t = n/2 in
let a11 = Array.make_matrix t t 0 in
for i = 0 to t-1 do
for j = 0 to t-1 do
a11.(i).(j) <- x.(i).(j)
done
done;
let a12 = Array.make_matrix t t 0 in
for i = 0 to t-1 do
for j = 0 to t-1 do
a12.(i).(j) <- x.(i).(j+t)
done
done;
let a21 = Array.make_matrix t t 0 in
for i = 0 to t-1 do
for j = 0 to t-1 do
a21.(i).(j) <- x.(i+t).(j)
done
done
Just to clarify, the following has the same problem as your original code:
# let x = 4 in () and y = 7 in ();;
Error: Syntax error
Since let defines an expression, this amounts to writing () and (), which indeed is not legal OCaml.
The correct use of let ... and looks like this:
# let x = 4 and y = 7 in ();;
Warning 26: unused variable x.
Warning 26: unused variable y.
- : unit = ()
There are several let ... and let ... but just one in.
How to implement a piecewise linear regression model in PHREG procedure of SAS?
For example with one knot at X=T:
Y = β_10 + β_11 . X if X ≤ T
Y = β_20 + β_21 . X if X >T
Given the model with the constraint of continuity:
Y = β_10 + β_11 . X if X ≤ T
Y = β_10 + (β_11 - β_21) T + β_21 . X if X >T
i.e :
Y= β_0 + β_1 . X + S_1
where
S_1 = ( β_11 - β_21 ) T if X >T and 0 otherwise.
Finally i would like to include it in a Cox model:
Proc PHREG
Model time * cas (censure) = X S_1 ;
Run ;
But the problem is S_1 has unknown beta coefficients in it.
Thanks for your help!
I have this SML code. I don't know why I cannot compile this :
fun score =
let
val sum = 3; (* error at this line : SYNTAX ERROR : inserting LPAREN *)
if sum div 2 > 0
then sum = 0
else sum = 1
(*some other code*)
in
sum (* I want to return sum after some steps of calculation *)
end
There are more issues with your code, than jacobm points out.
You are also missing a function argument. Functions in SML always takes one argument. For example
fun score () =
let val sum = 3
val sum = if sum div 2 > 0
then sum = 0
else sum = 1
in
sum
end
However this still doesn't make much sense. since the expressions sum = 0 and sum = 1 evaluates to a Boolean.
A let-expression is used to make some local declarations which are only visible inside the in ... end part. Thus the calculations you wan't to do with sum, should probably be done inside the in ... end part, unless you wan't to express it as a means of a function.
One such example is
fun score () =
let val sum = 3
in
if sum div 2 > 0
then ...
else ...
end
If we look at the syntax of a let-expression, it probably makes more sense
let
<declaration>
in
<expr> ; ... ; <expr>
end
Since if-then-else is an expression, it can't be in the "declarations part" by itself.
That syntax just isn't legal -- in between let and in all you're allowed to have is a series of val name = expr fragments. You can do this, though:
fun score =
let val sum = 3
val sum = if sum div 2 > 0
then sum = 0
else sum = 1
in
sum
end
I would consider it a bit of a weird style to use sum for both variable names, but it's legal.