### Question :

Consider the following (convex) optimization problem:

```
minimize 0.5 * y.T * y
s.t. A*x - b == y
```

where the optimization (vector) variables are `x`

and `y`

and `A`

, `b`

are a matrix and vector, respectively, of appropriate dimensions.

The code below finds a solution easily using the `SLSQP`

method from Scipy:

```
import numpy as np
from scipy.optimize import minimize
# problem dimensions:
n = 10 # arbitrary integer set by user
m = 2 * n
# generate parameters A, b:
np.random.seed(123) # for reproducibility of results
A = np.random.randn(m,n)
b = np.random.randn(m)
# objective function:
def obj(z):
vy = z[n:]
return 0.5 * vy.dot(vy)
# constraint function:
def cons(z):
vx = z[:n]
vy = z[n:]
return A.dot(vx) - b - vy
# constraints input for SLSQP:
cons = ({'type': 'eq','fun': cons})
# generate a random initial estimate:
z0 = np.random.randn(n+m)
sol = minimize(obj, x0 = z0, constraints = cons, method = 'SLSQP', options={'disp': True})
```

```
``````
Optimization terminated successfully. (Exit mode 0)
Current function value: 2.12236220865
Iterations: 6
Function evaluations: 192
Gradient evaluations: 6
```

Note that the constraint function is a convenient ‘array-output’ function.

Now, instead of an array-output function for the constraint, one could in principle use an equivalent set of ‘scalar-output’ constraint functions (actually, the scipy.optimize documentation discusses only this type of constraint functions as input to `minimize`

).

Here is the equivalent constraint set followed by the output of `minimize`

(same `A`

, `b`

, and initial value as the above listing):

```
# this is the i-th element of cons(z):
def cons_i(z, i):
vx = z[:n]
vy = z[n:]
return A[i].dot(vx) - b[i] - vy[i]
# listable of scalar-output constraints input for SLSQP:
cons_per_i = [{'type':'eq', 'fun': lambda z: cons_i(z, i)} for i in np.arange(m)]
sol2 = minimize(obj, x0 = z0, constraints = cons_per_i, method = 'SLSQP', options={'disp': True})
```

```
``````
Singular matrix C in LSQ subproblem (Exit mode 6)
Current function value: 6.87999270692
Iterations: 1
Function evaluations: 32
Gradient evaluations: 1
```

Evidently, the algorithm fails (the returning objective value is actually the objective value for the given initialization), which I find a bit weird. Note that running `[cons_per_i[i]['fun'](sol.x) for i in np.arange(m)]`

shows that `sol.x`

, obtained using the array-output constraint formulation, satisfies all scalar-output constraints of `cons_per_i`

as expected (within numerical tolerance).

I would appreciate if anyone has some explanation for this issue.

##
Answer #1:

You’ve run into the “late binding closures” *gotcha*. All the calls to `cons_i`

are being made with the second argument equal to 19.

A fix is to use the `args`

dictionary element in the dictionary that defines the constraints instead of the lambda function closures:

```
cons_per_i = [{'type':'eq', 'fun': cons_i, 'args': (i,)} for i in np.arange(m)]
```

With this, the minimization works:

```
In [417]: sol2 = minimize(obj, x0 = z0, constraints = cons_per_i, method = 'SLSQP', options={'disp': True})
Optimization terminated successfully. (Exit mode 0)
Current function value: 2.1223622086
Iterations: 6
Function evaluations: 192
Gradient evaluations: 6
```

You could also use the the suggestion made in the linked article, which is to use a lambda expression with a second argument that has the desired default value:

```
cons_per_i = [{'type':'eq', 'fun': lambda z, i=i: cons_i(z, i)} for i in np.arange(m)]
```