Multiobjective Programming

Abstract

As diverse as the problems in the previous chapters have been, they share one common feature: they all have one single objective function and the result is an optimal solution (or multiple optima, in case of dual degeneracy). However, the concept of optimality applies only in case of a single objective. If we state that something is “the best” or optimal, we always have an objective in mind: the fastest car, the most comfortable vehicle, the automobile that is cheapest to operate, and so forth. Whenever a second or even more objectives are included in a problem, the concept of optimality no longer applies. For instance, if the top speed of a vehicle and its gas mileage are relevant concerns, then the comparison between a car, whose speed may go up to 110 miles per hour and which gives 20 miles to the gallon (highway rating) and a vehicle that can go up to 90 miles per hour and which gives 25 miles to the gallon is no longer a simple one: the former car is faster at the expense of fuel efficiency. It will now depend on the decision maker which of the two criteria is considered more important. In other words, the decision maker will―sooner or later―have to specify a tradeoff between the criteria. This is the type of problems considered in this chapter.

Exercises

Problem 1 (improvement cone)

Consider the following two vector optimization problems (to simplify matters, the constraints have been ignored):

• (a) ${{\text{P}}_{{1}}}:{\text{Max }}{z_{{1}}} = {5}{x_{{1}}} + {2}{x_{{2}}},{\text{Max }}{z_{{2}}} = - {2}{x_{{1}}} - {2}{x_{{2}}},{\text{Max }}{z_{{3}}} = {3}{x_{{1}}},{\text{ and}}$

• (b) ${{\text{P}}_{{2}}}:{\text{Max }}{z_{{1}}} = {x_{{2}}},{\text{Max }}{z_{{2}}} = {4}{x_{{1}}} - {x_{{2}}},{\text{Min }}{z_{{3}}} = {x_{{1}}} + {x_{{2}}}.$

Plot each of these two problems individually and determine the improvement cone.

Solution

The improvement cone for (a) is shown in Fig. 3.8a, the improvement cone for (b) is empty as shown in Fig. 3.8b.

Fig. 3.8

Improvement cones

Problem 2 (nondominated frontier and composite objective, graphical)

Consider the following linear programming problem:

$$ \begin{array}{lll} {\text{P}}:{\text{ Max }}{z_{{1}}} = {x_{{1}}} + { 2}{x_{{2}}} \hfill \\{\text{s}}.{\text{t}}. - {2}{x_{{1}}} + {x_{{2}}} \le {2} \hfill \\{x_{{1}}} + {x_{{2}}} \le {5} \hfill \cr {x_{{1}}}\quad \quad \le {3} \hfill \cr {x_{{1}}}, {x_{{2}}} \ge 0. \end{array} $$

• (a) Plot the constraints and determine the feasible set.

• (b) Graph the gradient of the objective function and use the graphical solution technique to determine the optimal point. Compute the exact coordinates of the optimal point and its value of the objective function.

• (c) Consider a second objective function Max z ₂ = 2x ₁ − x ₂. Ignoring the first objective, what is the optimal point? Compute its exact coordinates and its value of the objective function.

• (d) Determine the nondominated frontier given the two objectives.

• (e) Use the two objectives above to construct the composite objective function with weights w ₁ = ¾ and w ₂ = ¼. What is the optimal solution with this objective?

Solution

• (a) The solutions are based on Fig. 3.9.

• (b) The exact coordinates of the optimal solution are ${\bar{\mathbf{x}}} = { }\left( {{1},{4}} \right)$ with value of the objective function ${\bar{z}_1} = 9.$

• (c) The optimal solution with the second objective is ${\bar{\mathbf{x}}} = { }\left( {{3},{ }0} \right)$ with ${\bar{z}_2} = 6.$

• (d) The efficient frontier is shown by the bold line in Fig. 3.10.

• (e) The composite objective function is ${\text{Max}}z = \tfrac{5}{4}{x_1} + \tfrac{5}{4}{x_2}$, it is shown as the gradient labeled “z” in Fig. 3.10.

Fig. 3.9

Graph for Problem 2

Fig. 3.10

Efficient frontier and composite objective function

Problem 3 (vector optimization, nondominated frontier)

Consider the following vector optimization problem:

• (a) Graph the constraints, clearly indicate the feasible set, and graph the directions of both objective functions.

• (b) Determine the improvement cone in a separate graph.

• (c) Plot the improvement cone at each extreme point of the feasible set and determine the efficient frontier. Clearly describe the efficient frontier.

Solution

Figure 3.11

Fig. 3.11

Solution for Problem 3(a) and 3(c)

Problem 4 (weighting method)

Consider again the vector optimization problem in Problem 3.

• (a) Use the weighting method with weight combinations w = (5, 1), (3, 1), (1, 1), (1, 3), and (1, 5) to determine nondominated solutions.

• (b) Use the constraint method by keeping the first objective and using the second objective as constraint with a variable right-hand side value b ₂.

• (c) Repeat (b) by keeping the second objective and using the first objective as a constraint with variable right-hand side b ₁.

Solution

• (a) The different weight combinations result in the solutions shown below.

  w	${\bar{\mathbf{x}}}$
  5, 1	0, 0
  3, 1	0, 0
  1, 1	3, 0
  1, 3	3, 2
  1, 5	3, 2

• (b) The method generates the noninferior solutions shown below for various values of b ₂ in the constraint ${2}{x_{{1}}} + {x_{{2}}} \ge{b_{{2}}}$.

  b 2	${\bar{\mathbf{x}}}$	${\bar{z}_1}$
  −10	0, 0	0
  0	0, 0	0
  5	2.5, 0	2.5
  6	3, 0	3
  7	3, 1	4
  8	3, 2	5
  9	no feasible solution

• (c) The method generates the noninferior solutions shown below for various values of b ₁ in the constraint ${x_{{1}}} + {x_{{2}}} \le {b_{{1}}}$.

  b 1	${\bar{\mathbf{x}}}$	${\bar{z}_2}$
  10	3, 2	8
  5	3, 2	8
  4	3, 1	7
  3	3, 0	6
  2	2, 0	4
  1	1, 0	2
  0	0, 0	0
  −1	No feasible solution

Problem 5 (goal programming formulation)

A product P is to be blended from three ingredients I ₁, I ₂, and I ₃. Firm requirements dictate that at least 100 lbs of P are to be blended, and that the average cost of the blend per pound do not exceed $2.80, given that one pound of the three ingredients costs $5, $3, and $2 for I ₁, I ₂, and I ₃, respectively. In addition, it would be desirable if 20 % of P were to be I ₁. Similarly, the decision maker would like to have P consist of no more than 50 % of the cheap ingredient I ₃, if possible. This desirable feature is about half as important as the former desirable feature. Formulate as a goal programming problem.

Solution

As there is only a single product, we only need variables with a single subscript. Define variables x ₁, x ₂, and x ₃ as the quantities of the three respective ingredients in the product. The constraints can then be written as

$$ {x_{{1}}} + {x_{{2}}} + {x_{{3}}} \ge {1}00 {\text{and}} $$

$$ {5}{x_{{1}}} + { 3}{x_{{2}}} + { 2}{x_{{3}}} \le { 2}.{8}\left( {{x_{{1}}} + {x_{{2}}} + {x_{{3}}}} \right). $$

The first of the two desirable properties, written as a constraint is

$$ {x_{{1}}} \ge { }0.{2}\left( {{x_{{1}}} + {x_{{2}}} + {x_{{3}}}} \right). $$

Rewriting as a goal constraint, we obtain

$$ {x_{{1}}} + d_1^{ - } - d_1^{ + } = { }.{2}\left( {{x_{{1}}} + {x_{{2}}} + {x_{{3}}}} \right) $$

with objective function contribution ${\text{Min }}d_1^{ - }$. The latter desirable property, written as a constraint, is

$$ {x_{{3}}} \le { }.{5}\left( {{x_{{1}}} + {x_{{2}}} + {x_{{3}}}} \right). $$

Rewritten as a goal constraint, we obtain

$$ {x_3} + d_2^{ - } - d_2^{ + } = { }.{5}\left( {{x_{{1}}} + {x_{{2}}} + {x_{{3}}}} \right). $$

with objective function contribution ${\text{Min }}d_2^{ + }$. The objective function is

$$ {\text{Min }}z = { 2}d_1^{ - } + d_2^{ + }. $$

Solving the problem results in 15, 35, and 50 lbs of the three ingredients being used, so that exactly 100 lbs are blended, whose price is exactly equal to the required value of 2.8. It is apparent that the product includes 15 % I ₁, 5 % short of the desired target. On the other hand, the upper limit of 50 % I ₃ is satisfied as equation.