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import numpy 
from numpy import asarray

Initial = numpy.asarray [2.0, 4.0, 5.0, 3.0, 5.0, 6.0]       # Initial values to start with


bounds = [(1, 5000), (1, 6000), (2, 100000), (1, 50000), (1.0, 5000), (2, 1000000)] 

# actual passed bounds

b1 = lambda x: numpy.asarray([1.4*x[0] - x[0]])  
b2 = lambda x: numpy.asarray([1.4*x[1] - x[1]])  
b3 = lambda x: numpy.asarray([x[2] - x[3]])     
constraints = numpy.asarray([b1, b2, b3])

opt= optimize.fmin_slsqp(func,Initial,ieqcons=constraints,bounds=bounds, full_output=True,iter=200,iprint=2, acc=0.01)

Problem: I want to pass in inequality constraints. Consider that I have 6 parameters

[ a, b, c, d, e, f]

in the Initial values, and my constraints are:

a<=e<=1.4*a   ('e' varies from a to 1.4*a)
b<=f<=1.4*b   ('f' varies from b to 1.4*b)
c>d           ('c' must always be greater than d)

But this is not working properly. I don't know what the mistake is. Is there any better way to pass my constraints as a function? Please help me.

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  • I don't know numpy, but are a and b negative? otherwise, I can't seee how any values of e and f can satisfy 1.4*a <= e <= a and 1.4*b <= f <= b. Commented Oct 2, 2009 at 19:45
  • It would help if you state clearly what it is exactly that you are doing, what is it you want to happen, and what actually happens, instead of just pasting a code fragment. Commented Oct 2, 2009 at 19:53
  • @pear, I've tried to answer your question, but as hughdbrown says, the constraints in your code above don't seem to work for positive numbers. Maybe the signs are backwards on the first two? Commented Oct 2, 2009 at 20:27
  • Sorry, all are positive values greater than 0. I have changed, i hope its correct. Commented Oct 5, 2009 at 9:41

1 Answer 1

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Based on the comment from Robert Kern, I have removed my previous answer. Here are the constraints as continuous functions:

b1 = lambda x: x[4]-x[0] if x[4]<1.2*x[0] else 1.4*x[0]-x[4]
b2 = lambda x: x[5]-x[1] if x[5]<1.2*x[1] else 1.4*x[1]-x[5]
b3 = lambda x: x[2]-x[3]

Note: Python 2.5 or greater is required for this syntax.1

To get the constraint a<=e<=1.4*a, note that 1.2*a is the halfway point between a and 1.4*a.

Below this point, that is, all e<1.2*a, we use the continuous function e-a. Thus the overall constraint function is negative when e<a, handling the lower out-of-bounds condition, zero on the lower boundary e==a, and then positive for e>a up to the halfway point.

Above the halfway point, that is, all e>1.2*a, we use instead the continuous function 1.4*a-e. This means the overall constraint function is is negative when e>1.4*a, handling the upper out-of-bounds condition, zero on the upper boundary e==1.4*a, and then positive when e<1.4*a, down to the halfway point.

At the halfway point, where e==1.2*a, both functions have the same value. This means that the overall function is continuous.

Reference: documentation for ieqcons.

1 - Here is pre-Python 2.5 syntax: b1 = lambda x: (1.4*x[0]-x[4], x[4]-x[0])[x[4]<1.2*x[0]]

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5 Comments

No, this will not work. The constraint functions should be continuous as much as possible, >0 when they are satisfied, <0 when they are not, and ==0 when they are precisely on the boundary.
system Pause, as Robert says function has to continuous. so i have made lambda x: ([1.4*x[0] - x[0]]), ie.,lambda x:(max-min). But what am i made is correct or not i don't know.
@pear, I don't think that can be correct, as it does not use e (aka x[4]) at all in the expression. That expression is equivalent to 1.4*a - a which is always going to have the value 0.4a.
@system Pause, Thank you, it seems b1 and b2 are working good, but b3 = lambda x: x[2]-x[3] not working as expected.For example some times func() passes [150, 192, 1.8487, 2.07364, 194, 216], here b1 and b2 are well defined, but b3 is violated. Like this it happens couple of times.what to do? any better way of defining 'b2'?
@pear, Sorry, I don't know why b3 is violated. Subtracting c-d (aka x[2]-x[3]) is a continuous function that is be positive when c>d and zero/negative when c<=d. But clearly 1.8487 < 2.07364, so I am stumped.

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