I have a time dependent variable that changes with some predefined timesteps. I would like get the sum of the variable over the course of several timesteps. How can I write the sum of the variable?
.mod file:
set Time default {}; #number of timesteps
param top{Time}; #number of hours in each timestep
var E{Time} >= 0.001; #the timedependent variable
subject to annualElectricity{t in Time}:
Ea = sum{E[t]*top[t]};
.dat file:
set Time := 1 2 3 4 5 6;
param top := #operating time (hours)
1 609
2 321
3 532
4 425
5 351
6 7
;
I'm guessing that the annualElectricity constraint collection that you have declared is your attempt to do what you want, but I'm not sure you want to use a constraint for that.
If you want to sum all variables in the collection E, the sum expression will be sum{t in Time} E[t]. If you want the sum of all products E[t]*top[t], the sum expression will be sum{t in Time} E[t]*top[t].
If you won't need this value to solve the problem, you can declare a param Esum and then, after you have solved the problem, you can call let Esum := sum{t in Time} E[t]. But if you need this value in your problem, I believe you can define a variable var Esum = sum{t in Time} E[t] and use it in your constraints/objective function.
I can see you are new to AMPL, so you may benefit from reading some basic AMPL introduction, such as this https://pifop.com/help/ampl_programs.html that I've written.
Related
So if I was given a sorted list/array i.e. [1,6,8,15,40], the size of the array, and the requested number..
How would you find the minimum number of values required from that list to sum to the requested number?
For example given the array [1,6,8,15,40], I requested the number 23, it would take 2 values from the list (8 and 15) to equal 23. The function would then return 2 (# of values). Furthermore, there are an unlimited number of 1s in the array (so you the function will always return a value)
Any help is appreciated
The NP-complete subset-sum problem trivially reduces to your problem: given a set S of integers and a target value s, we construct set S' having values (n+1) xk for each xk in S and set the target equal to (n+1) s. If there's a subset of the original set S summing to s, then there will be a subset of size at most n in the new set summing to (n+1) s, and such a set cannot involve extra 1s. If there is no such subset, then the subset produced as an answer must contain at least n+1 elements since it needs enough 1s to get to a multiple of n+1.
So, the problem will not admit any polynomial-time solution without a revolution in computing. With that disclaimer out of the way, you can consider some pseudopolynomial-time solutions to the problem which work well in practice if the maximum size of the set is small.
Here's a Python algorithm that will do this:
import functools
S = [1, 6, 8, 15, 40] # must contain only positive integers
#functools.lru_cache(maxsize=None) # memoizing decorator
def min_subset(k, s):
# returns the minimum size of a subset of S[:k] summing to s, including any extra 1s needed to get there
best = s # use all ones
for i, j in enumerate(S[:k]):
if j <= s:
sz = min_subset(i, s-j)+1
if sz < best: best = sz
return best
print min_subset(len(S), 23) # prints 2
This is tractable even for fairly large lists (I tested a random list of n=50 elements), provided their values are bounded. With S = [random.randint(1, 500) for _ in xrange(50)], min_subset(len(S), 8489) takes less than 10 seconds to run.
There may be a simpler solution, but if your lists are sufficiently short, you can just try every set of values, i.e.:
1 --> Not 23
6 --> Not 23
...
1 + 6 = 7 --> Not 23
1 + 8 = 9 --> Not 23
...
1 + 40 = 41 --> Not 23
6 + 8 = 14 --> Not 23
...
8 + 15 = 23 --> Oh look, it's 23, and we added 2 values
If you know your list is sorted, you can skip some tests, since if 6 + 20 > 23, then there's no need to test 6 + 40.
I have to check how many hundreds are there in a number and translate that number to letters. For example the number 700. I have done the following code:
DATA(lv_dmbtr) = ZDS_FG-DMBTR. //Declared local variable of type DMBTR, thus DMBTR=700.
lv_dmbtr = ZDS_FG-DMBTR MOD 100. //Finding how many times 700 is in 100 via MOD and putting the value in lv_dmbtr.
IF lv_dmbtr LE 9. //The value is less or equal than 9(if larger means that the DMBTR is larger than hundreds,
e.g. 8000)
lv_hundred = lv_dmbtr / 100. // Divide the 700 with 100, taking the number 7.
lv_hundred_check = lv_hundred MOD 1. // Then taking the value of 7 into the new variable, done in case the
lv_hundred is a decimal value, e.g. 7.32.
IF lv_hundred_check > 0.
CALL FUNCTION 'SPELL_AMOUNT'
EXPORTING
amount = lv_hundred_check
* CURRENCY = ' '
* FILLER = ' '
LANGUAGE = SY-LANGU
IMPORTING
in_words = lv_hundred_string // the value is put in the new string
EXCEPTIONS
not_found = 1
too_large = 2
OTHERS = 3.
ENDIF.
Now when I debugg the code, all the variables have the value 0. Thus, lv_dmbtr, lv_hundred, lv_hundred_check all have the value 0.
May anyone of you know where the problem may be?
Thank you in advance!
Sorry for writing a lot in the code, just wanted to clarify as much as I could what I had done.
yes so I want to display the value of a specific number 700-> seven, 1400-> four.
So the basic formula to get the hundred in a number is the following: Find out how many times 100 fits completely into your number with integer division.
99 / 100 = 0
700 / 100 = 7
701 / 100 = 7
1400 / 100 = 14
1401 / 100 = 14
Now you can simply take this number MOD 10 to get the the individual hundreds.
0 MOD 10 = 0
7 MOD 10 = 7
14 MOD 10 = 4
Keep in mind that ABAP, in contrast to many other programming languages, rounds automatically. So in code this would be:
CONSTANTS lc_hundred TYPE f VALUE '100.0'.
DATA(lv_number) = 1403.
DATA(lv_hundred_count) = CONV i( floor( ( abs( lv_number ) / lc_hundred ) ) MOD 10 ).
I am trying to formulate a flowshop scheduling problem in Pyomo. This is an Abstract model
Problem description
There are 3 jobs (chest, door and chair) and 3 machines (cutting, welding, packing in that order). Objective is to minimise the makespan. The python code and the data are as follows.
## flowshop.py ##
from pyomo.environ import *
flowshop = AbstractModel()
flowshop.jobs = Set()
flowshop.machines = Set()
flowshop.machinesN = Param()
flowshop.jobsN = Param()
flowshop.proc_T = Param(flowshop.jobs,
flowshop.machines,
within=NonNegativeReals)
flowshop.start_T = Var(flowshop.jobs,
flowshop.machines,
within=NonNegativeReals)
flowshop.makespan = Var(within=NonNegativeReals)
def makespan_rule(flowshop,i,j):
return flowshop.makespan >= flowshop.start_T[i,j]+flowshop.proc_T[i,j]
flowshop.makespan_cons = Constraint(flowshop.jobs,
flowshop.machines,
rule=makespan_rule)
def objective_rule(flowshop):
return flowshop.makespan
flowshop.objc = Objective(rule=objective_rule,sense=minimize)
## data.dat ##
set jobs := chest door chair ;
set machines := cutting welding packing ;
param: machinesN := 3 ;
param: jobsN := 3 ;
param proc_T:
cutting welding packing :=
chest 10 40 45
door 30 20 25
chair 05 30 15
;
I havent added all the constraints yet, I plan to add them after this issue gets fixed. In the code (flowhop.py) above, for the makespan_rule, I want the makespan to be more that the completion time of only the last machine.
Currently, it is set to be more than completion times of all the machines.
For that, I believe, I have to get the last index of the machines set.
For that, I tried flowshop.machines[-1], but it gives an error saying:
Cannot index unordered set machines
How do I solve this issue?
Thanks for the help.
PS - I am also struggling to model the binary variables used to define the precedence of a job. If you have any ideas regarding that, that would also be helpful.
As the error says Cannot index unordered sets, the set flowshop.machines is not ordered. One needs to provide ordered=True argument in the while declaring the set -
flowshop.machines = Set(ordered=True)
After this, one can access any element by normal indexing - flowshop.machines[i]
For the binary variables, one can declare them as -
c = flowshop.jobsN*(flowshop.jobsN-1)/2
flowshop.prec = Var(RangeSet(1,c),within=Binary)
Then, this variable can be used to decide the precedence between 2 jobs and to formulate the assignment constraints. The precedence variable corresponding to a pair of jobs can be found out using the indices of the jobs (for which the flowshop.jobs has to be an ordered set - flowshop.jobs = Set(ordered=True))
Is there a way to give while loop a condition which makes it give an an output every ten times' executions, however it continues running after this output?
I hope I made myself clear...
Thanks!
Amy
Modulo is useful for this.
As an example; In swift to do modulo you use the % symbol. Essentially modulo outputs the remainder of the given terms.
So;
Value 1 MODULO Value 2 outputs Remainder.
Furthermore;
6 % 2 = 0
6 % 5 = 1
6 % 4.5 = 1.5
Essentially you want every nth element to output a value, with n being the rate. You need to track how many loops of the while you have gone through already.
The code below will run through the while 1000 times, and print out every 10 times ( for a total of 100 prints of output. )
var execution : Int = 0
while ( execution != 1000 ) {
if ( execution % 10 == 0 ) {
print("output")
}
execution = execution + 1
}
Here is the same answer as given by adam but then in Labview.
I need to set up all my variables to accept only values from two intervals, e.g. for var C1:
C1 = 0 or
100 <= C1 <= 1000
and so on for each variable.
I can specify either first (=0) or second ([100;1000]) constraints for every variable but I need lpsolve model to to be able to assign either 0 ot any value from given interval.
Any advice appreciated.