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2.4 BIM production for worst case

2.4 BIM production for worst case#

Preamble: Install Pyomo and a solver#

The following cell sets and verifies a global SOLVER for the notebook. If run on Google Colab, the cell installs Pyomo and the HiGHS solver, while, if run elsewhere, it assumes Pyomo and HiGHS have been previously installed. It then sets to use HiGHS as solver via the appsi module and a test is performed to verify that it is available. The solver interface is stored in a global object SOLVER for later use.

import sys
 
if 'google.colab' in sys.modules:
    %pip install pyomo >/dev/null 2>/dev/null
    %pip install highspy >/dev/null 2>/dev/null
 
solver = 'appsi_highs'
 
import pyomo.environ as pyo
SOLVER = pyo.SolverFactory(solver)

assert SOLVER.available(), f"Solver {solver} is not available."

Minmax objective function#

Another class of seemingly complicated objective functions that can be easily rewritten as an LO problem are those stated as maxima over several linear functions. Given a finite set of indices \(K\) and a collection of vectors \(\{c_k\}_{k \in K}\), the minimax problem given by

\[ \begin{align} \min \; \max_{k \in K} \; c^\top_{k} x \end{align} \]

General expressions like the latter can be linearized by introducing an auxiliary variable \(z\) and setting

\[\begin{split} \begin{align*} \min \quad & z \\ \text{s.t.} \quad & c^\top_{k} x \leq z \qquad \forall\, k \in K. \end{align*} \end{split}\]

This technique works because if all the quantities corresponding to different indices \(k \in K\) are below the auxiliary variable \(z\), then we are guaranteed that also their maximum is also below \(z\) and vice versa. Note that the absolute value function can be rewritten \(|x_i|= \max\{x_i, -x_i\}\), hence the linearization of the optimization problem involving absolute values in the objective functions is a special case of this.

BIM problem variant: Maximizing the lowest possible profit#

In the same way we can minimize a maximum like above, we can also maximize the minimum. Let us consider the BIM microchip production problem, but suppose that there is uncertainty regarding the selling prices of the microchips. Instead of just the nominal prices 12\( and 9\), BIM estimates that prices may more generally take values \(P=\{ (12,9), (11,10), (8, 11) \}\). The optimization problem for a production plan that achieves the maximum among the lowest possible profits can be formulated using the trick mentioned above and can be implemented in Pyomo as follows.

def BIM_maxmin(costs):
    m = pyo.ConcreteModel("BIM production planning with maxmin objective")

    m.x1 = pyo.Var(domain=pyo.NonNegativeReals)
    m.x2 = pyo.Var(domain=pyo.NonNegativeReals)
    m.z = pyo.Var()

    m.profit = pyo.Objective(sense=pyo.maximize, expr=m.z)

    m.maxmin = pyo.ConstraintList()
    for c1, c2 in costs:
        m.maxmin.add(expr=m.z <= c1 * m.x1 + c2 * m.x2)

    m.silicon = pyo.Constraint(expr=m.x1 <= 1000)
    m.germanium = pyo.Constraint(expr=m.x2 <= 1500)
    m.plastic = pyo.Constraint(expr=m.x1 + m.x2 <= 1750)
    m.copper = pyo.Constraint(expr=4 * m.x1 + 2 * m.x2 <= 4800)

    return m


m = BIM_maxmin([[12, 9], [11, 10], [8, 11]])
SOLVER.solve(m)

print(f"x = ({pyo.value(m.x1):.1f}, {pyo.value(m.x2):.1f})")
print(f"revenue = {pyo.value(m.profit):.2f}")

x = (583.3, 1166.7)
revenue = 17500.00
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