Notebook

Example¶

find a point in the intersection of two lines

$\begin{align*} 2x + y = 4\\ -x + 5y = 0 \end{align*}$

In [5]:

from cvxpy import *

x = Variable()
y = Variable()

obj = Minimize(0)
constr = [2*x + y == 4,
          -x + 5*y == 0]

Problem(obj, constr).solve()

print x.value, y.value

1.81818181818 0.363636363636

Diet Problem¶

a classic problem in optimization is to meet the nutritional requirements

of an army via various foods (with different nutritional benefits and prices) under cost constraints

one soldier requires 1, 2.1, and 1.7 units of meat, vegetables, and grain, respectively, per day ( $r = (1, 2.1, 1.7)$ )
one unit of hamburgers has nutritional value $h = (.8, .4, .5)$ and costs $1
one unit of cheerios has nutritional vlaue $c = (0, .3, 2.0)$ and costs $0.25
prices $p = (1, 0.25)$
you have a budget of $130 to buy hamburgers and cheerios for one day
can you meet the dietary needs of 50 soldiers?

Formulation¶

write as optimization problem

$\begin{array}{ll} \mbox{minimize} & 0 \\ \mbox{subject to} & p^T x \leq 130\\ &x_1 h + x_2 c \geq 50 r\\ &x \geq 0 \end{array}$

with $x$ giving units of hamburgers and cheerios

or, with $A = [h, c]$ ,

$\begin{array}{ll} \mbox{minimize} & 0 \\ \mbox{subject to} & p^T x \leq 130\\ &Ax \geq 50 r\\ &x \geq 0 \end{array}$

In [6]:

from cvxpy import *
import numpy as np
h = np.array([.8, .4, .5])
c = np.array([0, .3, 2.0])
r = np.array([1, 2.1, 1.7])
p = np.array([1, .25])

solve the feasibility problem

In [9]:

x = Variable(2) # amounts of hamburgers and cheerios

obj = Minimize(0)
constr = [x.T*p <= 130,
          h*x[0] + c*x[1] >= 50*r,
          x >= 0]

prob = Problem(obj, constr)
# use the SCS solver because the default, ECOS, has a bug with 0 objectives
prob.solve(solver='SCS')
print x.value

[[  62.82945211]
 [ 266.57466712]]

solve the minimization problem to find the cheapest acceptable diet

In [11]:

x = Variable(2) # amounts of hamburgers and cheerios

obj = Minimize(x.T*p)
constr = [h*x[0] + c*x[1] >= 50*r,
          x >= 0]

prob = Problem(obj, constr)
prob.solve(solver='SCS')
print prob.value
print x.value

129.139602104
[[  62.44796859]
 [ 266.76653406]]

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