optimconstr

Create empty optimization constraint array

Description

Use optimconstr to initialize a set of constraint expressions.

Tip

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

example

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

example

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

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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);
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 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",...
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

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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

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

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

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

Data Types: string | cell

Output Arguments

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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.