optimconstr

Create empty optimization constraint array

Syntax

constr = optimconstr(N)

constr = optimconstr(cstr)

constr = optimconstr(cstr1,N2,...,cstrk)

constr = optimconstr({cstr1,cstr2,...,cstrk})

constr = optimconstr([N1,N2,...,Nk])

Description

Use optimconstr to initialize a set of constraint expressions.

Tip

For the full workflow, see Problem-Based Optimization Workflow.

constr = optimconstr(N) creates an N-by-1 array of empty optimization constraints. Use constr to initialize a loop that creates constraint expressions.

constr = optimconstr(cstr) creates an array of empty optimization constraints that are indexed by cstr, a cell array of character vectors or string vectors.

If cstr is 1-by-ncstr, where ncstr is the number of elements of cstr, then constr is also 1-by-ncstr. Otherwise, constr is ncstr-by-1.

constr = optimconstr(cstr1,N2,...,cstrk) or constr = optimconstr({cstr1,cstr2,...,cstrk}) or constr = optimconstr([N1,N2,...,Nk]), for any combination of cstr and N arguments, creates an ncstr1-by-N2-by-...-by-ncstrk array of empty optimization constraints, where ncstr is the number of elements in cstr.

Examples

Create Constraints in Loop

Create constraints for an inventory model. The stock of goods at the start of each period is equal to the stock at the end of the previous period. During each period, the stock increases by buy and decreases by sell. The variable stock is the stock at the end of the period.

N = 12;
stock = optimvar('stock',N,1,'Type','integer','LowerBound',0);
buy = optimvar('buy',N,1,'Type','integer','LowerBound',0);
sell = optimvar('sell',N,1,'Type','integer','LowerBound',0);
initialstock = 100;

stockbalance = optimconstr(N,1);

for t = 1:N
    if t == 1
        enterstock = initialstock;
    else
        enterstock = stock(t-1);
    end
    stockbalance(t) = stock(t) == enterstock + buy(t) - sell(t);
end

show(stockbalance)

(1, 1)

  -buy(1) + sell(1) + stock(1) == 100

(2, 1)

  -buy(2) + sell(2) - stock(1) + stock(2) == 0

(3, 1)

  -buy(3) + sell(3) - stock(2) + stock(3) == 0

(4, 1)

  -buy(4) + sell(4) - stock(3) + stock(4) == 0

(5, 1)

  -buy(5) + sell(5) - stock(4) + stock(5) == 0

(6, 1)

  -buy(6) + sell(6) - stock(5) + stock(6) == 0

(7, 1)

  -buy(7) + sell(7) - stock(6) + stock(7) == 0

(8, 1)

  -buy(8) + sell(8) - stock(7) + stock(8) == 0

(9, 1)

  -buy(9) + sell(9) - stock(8) + stock(9) == 0

(10, 1)

  -buy(10) + sell(10) - stock(9) + stock(10) == 0

(11, 1)

  -buy(11) + sell(11) - stock(10) + stock(11) == 0

(12, 1)

  -buy(12) + sell(12) - stock(11) + stock(12) == 0

Include the constraints in a problem.

prob = optimproblem;
prob.Constraints.stockbalance = stockbalance;

Instead of using a loop, you can create the same constraints by using matrix operations on the variables.

tt = ones(N-1,1);
d = diag(tt,-1); % shift index by -1
stockbalance2 = stock == d*stock + buy - sell;
stockbalance2(1) = stock(1) == initialstock + buy(1) - sell(1);

Show the new constraints to verify that they are the same as the constraints in stockbalance.

show(stockbalance2)

(1, 1)

  -buy(1) + sell(1) + stock(1) == 100

(2, 1)

  -buy(2) + sell(2) - stock(1) + stock(2) == 0

(3, 1)

  -buy(3) + sell(3) - stock(2) + stock(3) == 0

(4, 1)

  -buy(4) + sell(4) - stock(3) + stock(4) == 0

(5, 1)

  -buy(5) + sell(5) - stock(4) + stock(5) == 0

(6, 1)

  -buy(6) + sell(6) - stock(5) + stock(6) == 0

(7, 1)

  -buy(7) + sell(7) - stock(6) + stock(7) == 0

(8, 1)

  -buy(8) + sell(8) - stock(7) + stock(8) == 0

(9, 1)

  -buy(9) + sell(9) - stock(8) + stock(9) == 0

(10, 1)

  -buy(10) + sell(10) - stock(9) + stock(10) == 0

(11, 1)

  -buy(11) + sell(11) - stock(10) + stock(11) == 0

(12, 1)

  -buy(12) + sell(12) - stock(11) + stock(12) == 0

Creating constraints in a loop can be more time-consuming than creating constraints by matrix operations. However, you are less likely to create an erroneous constraint by using loops.

Create Indexed Constraints in Loop

Create indexed constraints and variables to represent the calories consumed in a diet. Each meal has a different calorie limit.

meals = ["breakfast","lunch","dinner"];
constr = optimconstr(meals);
foods = ["cereal","oatmeal","yogurt","peanut butter sandwich","pizza","hamburger",...
    "salad","steak","casserole","ice cream"];
diet = optimvar('diet',foods,meals,'LowerBound',0);
calories = [200,175,150,450,350,800,150,650,350,300]';
for i = 1:3
    constr(i) = diet(:,i)'*calories <= 250*i;
end

Check the constraint for dinner.

show(constr("dinner"))

  200*diet('cereal', 'dinner') + 175*diet('oatmeal', 'dinner') + 150*diet('yogurt', 'dinner') + 450*diet('peanut butter sandwich', 'dinner') + 350*diet('pizza', 'dinner') + 800*diet('hamburger', 'dinner') + 150*diet('salad', 'dinner') + 650*diet('steak', 'dinner') + 350*diet('casserole', 'dinner') + 300*diet('ice cream', 'dinner') <= 750

Input Arguments

N — Size of constraint dimension

positive integer

Size of the constraint dimension, specified as a positive integer.

• The size of constr = optimconstr(N) is N-by-1.

• The size of constr = optimconstr(N1,N2) is N1-by-N2.

• The size of constr = optimconstr(N1,N2,...,Nk) is N1-by-N2-by-...-by-Nk.

Example: 5

Data Types: double

cstr — Names for indexing

cell array of character vectors | string vector

Names for indexing, specified as a cell array of character vectors or a string vector.

Note

cstr cannot be a string scalar such as "Tp", but must be a vector such as ["Tp" "ul"]. To specify a single name, use {'Tp'} or the equivalent cellstr("Tp").

Example: {'red','orange','green','blue'}

Example: ["red";"orange";"green";"blue"]

Data Types: string | cell

Output Arguments

constr — Constraints

empty OptimizationConstraint array

Constraints, returned as an empty OptimizationConstraint array. Use constr to initialize a loop that creates constraint expressions.

For example:

x = optimvar('x',8);
constr = optimconstr(4);
for k = 1:4
    constr(k) = 5*k*(x(2*k) - x(2*k-1)) <= 10 - 2*k;
end

Limitations

• Each constraint expression in a problem must use the same comparison. For example, the following code leads to an error, because cons1 uses the <= comparison, cons2 uses the >= comparison, and cons1 and cons2 are in the same expression.

  prob = optimproblem;
  x = optimvar('x',2,'LowerBound',0);
  cons1 = x(1) + x(2) <= 10;
  cons2 = 3*x(1) + 4*x(2) >= 2;
  prob.Constraints = [cons1;cons2]; % This line throws an error

  You can avoid this error by using separate expressions for the constraints.

  prob.Constraints.cons1 = cons1;
  prob.Constraints.cons2 = cons2;

Tips

• It is generally more efficient to create constraints by vectorized expressions rather than loops. See Create Efficient Optimization Problems.

• You can use optimineq instead of optimconstr to create inequality expressions. Similarly, you can use optimeq instead of optimconstr to create equality expressions.

Version History

Introduced in R2017b