Solve the assessment in Pharo Smalltalk and please loop the whole code in two ways whileTrue: and do loops.
Here is what I tried but it gives me a wrong answer.
| x y z count elem sum |
x := #(84 45 54).
sum := 0.
z := 1.
y := x size.
elem := x at: z.
[z < y] whileTrue: [ elem even ifTrue: [sum := sum + elem.]. z := z + 1.].
Transcript clear.
Transcript show: sum.
Yes is wrong because "elem" is always the same.
You have to do like this:
| x y z count elem sum |
x := #(84 45 54 ).
sum := 0.
z := 1.
y := x size.
[z <= y] whileTrue: [
elem := x at: z.
elem even ifTrue: [sum := sum + elem.]. z := z + 1.
].
Transcript clear.
Transcript show: sum.
or more concise
| x |
x := #(84 45 54 ).
sum := 0.
x do:[:each | each even ifTrue: [sum := sum + each.].].
Transcript clear.
Transcript show: sum.
or
| x |
x := #(84 45 54 ).
sum := x inject: 0 into: [ :sum :each | each even ifTrue: [sum + each ] ifFalse: [sum + 0]].
Transcript clear.
Transcript show: sum.
Related
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.
Suppose I have
param m; #number of modes
param n; #number of individual
param a; #number of alternatives
param f; #number of household
set M, default{1..m}; #set of modes
set N, default{1..n}; #set of individuals
set A, default{1..a}; #set of alternatives
set F, default{1..f}; #set of family
set E, within F cross N
How I can initialize param X{E,M,A} ?
Suppose
a:=2 , m:=3 , n:= 4 f:=2;
and set E is defined:
set E:= 1 1 1 2 2 3 2 4 ;
You can declare the parameter just as you suggested:
param X{E,M,A};
Now, if you want to provide a default value (which I assume is what you are asking), you can do it in the usual way:
param X{E,M,A} default 0;
Then provide some non-default values in the .dat file, e.g.,:
param: X :=
1 1 1 2 5
2 3 2 1 6;
Note that AMPL doesn't fill the default values into the parameter until you call solve. From the AMPL book, p. 120:
The expression that gives the default value of a parameter is evaluated only when the parameter’s value is first needed, such as when an objective or constraint that uses the parameter is processed by a solve command.
So if you type display X; after you have issued the model and data commands but before you have issued the solve command, you'll only get the non-default values, e.g.:
X :=
1 1 1 2 5
2 3 2 1 6
;
But if you use display X; after you call solve, you'll get the full list:
X [1,*,*,1] (tr)
: 1 2 :=
1 0 0
2 0 0
3 0 0
[1,*,*,2] (tr)
: 1 2 :=
1 5 0
2 0 0
3 0 0
[2,*,*,1] (tr)
: 3 4 :=
1 0 0
2 6 0
3 0 0
[2,*,*,2] (tr)
: 3 4 :=
1 0 0
2 0 0
3 0 0
;
For completeness, here are the .mod and .dat files I used for this answer:
.mod:
param m; #number of modes
param n; #number of individual
param a; #number of alternatives
param f; #number of household
set M, default{1..m}; #set of modes
set N, default{1..n}; #set of individuals
set A, default{1..a}; #set of alternatives
set F, default{1..f}; #set of family
set E, within F cross N;
param X{E,M,A} default 0;
var myVar{E,M,A} >= 0;
minimize Obj: sum {(i,j) in E, mm in M, aa in A} X[i,j,mm,aa] * myVar[i,j,mm,aa];
.dat:
param a:=2;
param m:=3;
param n:= 4;
param f:=2;
set E:= 1 1 1 2 2 3 2 4 ;
param: X :=
1 1 1 2 5
2 3 2 1 6;
I think logically the following code is right but I get the wrong answer:
.mod file:
set R := {1,2};
set D1 := {1,2,4,5};
set P1 := {1,2,3,4,5};
var V{D1,R}, binary;
param Ud{D1,R} ;
param U{P1,R} ;
minimize obj{p in D1, r in R}: V[p,r] * (Ud[p,r]+ sum{j in P1: j!=p} U[j,r]);
s.t. a10{ r in R }: sum{p in D1} V[p,r]=2 ;
.dat file:
param Ud: 1 2:=
1 -10 -6
2 -20 -4
4 1 -10
5 -4 -4;
param U: 1 2 :=
1 -8.1 -3
2 -6.8 -8
3 -7.2 1
4 -16 -4
5 -6.8 -4;
Basically for each r and for two p , I want to minimize (Ud[p,r] + sum{j in P: j!=p} U[j,r])
But it always give me V[1,r]=v[5,r]=1 even if V[2,r] minimize obj function.
I except to get V[2,r]=1 because -20 + (-8.1-7.2 -16-6.8) is the most negative.
Your syntax for the objective function is incorrect; it should be
minimize obj: sum {p in D1, r in R} V[p,r] * (Ud[p,r]+ sum{j in P1: j != p} U[j,r]);
(Note the location of the colon (:), and the presence of the sum.) To be honest I'm not exactly sure what AMPL was doing in response to your objective function, but I would just treat the results as unpredictable.
With the revised objective function, the optimal solution is:
ampl: display V;
V :=
1 1 1
1 2 1
2 1 1
2 2 0
4 1 0
4 2 1
5 1 0
5 2 0
;
I am new in maxima, so I am really sorry if I ask simple question. I have a differential equation,
(%i1) -(x-x/2*sinh(x/2)+'diff(y,x))*(1/y+'diff(y,x)*x/y^2)+(x-x^2/sinh(x/2)+x^2*cosh(x/2)/(4*(sinh(x/2))^2)+'diff(y,x)*x+'diff(y,x,2)*x^2)/y+y^2-1-0.9*(x-x^2/(2*sinh(x/2)))=0;
2 x
2 2 x cosh(-)
2 d y dy x 2
x --- + x -- - ------- + ---------- + x
2 dx x 2 x
dx sinh(-) 4 sinh (-)
2 2
(%o1) ----------------------------------------
y
x dy
x sinh(-) x -- 2
dy 2 dx 1 2 x
+ (- -- + --------- - x) (---- + -) + y - 0.9 (x - ---------) - 1 = 0
dx 2 2 y x
y 2 sinh(-)
2
(%i2) ode2(%,y,x);
rat: replaced -0.9 by -9/10 = -0.9
(%o2) false
what should I do?
The equation you have is nonlinear. Maxima's ode2 can only solve a limited variety of differential equations, and it appears your equation doesn't fall into any of the categories it can handle.
I don't know if there is another symbolic diff eq solver in Maxima that you can try. If a numerical solution is enough, take a look at rk (a Runge-Kutta implementation).
Input:
G0894 x 1 x 3 x 1 k 1
C4458 x 1 k 5
C9057 x 7 x 4 x 4 x 3 x 5
Desired output:
G0894 x 1
G0894 x 3
G0894 x 1
G0894 k 1
C4458 x 1
C4458 k 5
C9057 x 7
C9057 x 4
C9057 x 4
C9057 x 3
C9057 x 5
This is what I came up with:
data want;
infile cards missover;
input id $ #;
do while (1);
input letter $ number #;
if letter EQ ' ' then leave;
output;
end;
cards;
G0894 x 1 x 3 x 1 k 1
C4458 x 1 k 5
C9057 x 7 x 4 x 4 x 3 x 5
;
run;
And it does work but since we've been talking about double trailing ## in class I think I'm supposed to use it. This was my other approach:
data want;
infile cards missover;
input id $ #;
input letter $ number ##;
cards;
G0894 x 1 x 3 x 1 k 1
C4458 x 1 k 5
C9057 x 7 x 4 x 4 x 3 x 5
;
run;
And it generates an error which says something about using missover and ## in an inconsistent manner. What am I doing wrong?
There is no way in your program for the data step to ever advance to the second row of input data. That is what the error message is telling you.
The ## tells SAS that it should keep the line pointer and column pointer the same when it starts the next data step iteration. The MISSOVER option tells SAS not to go to a new line when it cannot find data to meet the current input request. Hence there is no way for the line pointer to ever advance to line two.
The double trailing at sign (##) holds a record across multiple iterations of the DATA step until the end of the record is reached. However, the single trailing at sign (#) releases a record when control returns to the top of the DATA step.
Try this:
data want;
infile cards missover;
input id $ letter $ number #;
do while (letter ne '' or number ne .);
output;
input letter $ number #;
end;
cards;
G0894 x 1 x 3 x 1 k 1
C4458 x 1 k 5
C9057 x 7 x 4 x 4 x 3 x 5
;
run;