The General Optimization Problem

Abstract

The problems studied in the previous two chapters have all been special cases of a more general optimization problem. In most practical problems, there is no close relationship between individual design variables and constraints, as was the case in the fully stressed design of a simple truss structure. A box beam is used to illustrate the complex relations between design variables and constraints in a more representative optimization problem. Since we generally cannot know in advance which constraints will prove to be active at the optimum, the task of numerical optimization is both to select the active constraints and to locate the optimum on those constraints. The general form of the optimization problem is defined, and a distinction drawn between an intersection optimum and a mathematical optimum. The classical method of Lagrange multipliers might be considered the mathematical basis of optimization, while applying only to equality constrained problems. For inequality constrained problems, Lagrange multipliers are used to identify those constraints that have been correctly selected as active at the optimum. The Kuhn–Tucker conditions are the necessary conditions for an inequality constrained optimum. A spreadsheet program is presented for the optimization of an eccentrically loaded column, taking into account the effect of yielding of the material at higher stresses and extending the scope of the spreadsheet in the previous chapter.

Exercises

• 3.1 Use the Lagrange multiplier method to find the minimum of the function:

  $$ f\left( {\mathbf{x}} \right) = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} $$

  subject to the constraint:

  $$ h({\mathbf{x}}) = x_{1} + 2x_{2} + 3x_{3} - 7 = 0. $$

  Follow the method of Example 3.1.

• 3.2 Find the minimum of the function:

  $$ f({\mathbf{x}}) = 2x_{2} + x_{1} $$

  subject to inequality constraints:

  $$ \begin{aligned} g_{1} ({\mathbf{x}}) & = 2x_{2} - x_{1} \ge 0, \\ g_{2} ({\mathbf{x}}) & = x_{2} - 2x_{1} + 4 \ge 0, \\ g_{3} ({\mathbf{x}}) & = x_{2} + x_{1} - 3 \ge 0, \\ g_{4} ({\mathbf{x}}) & = x_{1} - 1 \ge 0. \\ \end{aligned} $$

  Only two of the above constraints are active at the minimum. Draw the design space to find the active constraints. With the active constraints known, treat these as equalities and solve the problem analytically by the Lagrange multiplier method. Notice that the Lagrange multipliers are positive, confirming correctly chosen active constraints.

• 3.3 A thin circular tube of radius R and length L has flat, closed ends. Use the Lagrange multiplier method to find an expression for the minimum total surface area A of the material of the tube and the two ends, if it has a required internal volume $V_{0}$.

  Derive expressions for the total surface area and for the internal volume to set up the Lagrangian function.

• 3.4 A rectangular container is to be made of materials that cost €20/m² for the bottom, €30/m² for the sides and €10/m² for the top. The volume of the container has to be 4 m³. Use the Lagrange multiplier method to calculate the minimum cost of the container and its corresponding dimensions.

  Use the value of the Lagrange multiplier found above to estimate the minimum cost if the volume of the container is increased to 5 m³.

  The Lagrange multiplier gives the rate of change of the objective function (cost) with the value of the constraint (volume).

• 3.5 Verify the principle of simultaneous buckling modes for a circular tube loaded in compression by deriving expressions for the Lagrange multipliers for this problem.

  Use the Lagrange multiplier method to optimize the circular tube, with only flexural and local buckling constraints (use the formulae in Sect. 2.1 ). Derive expressions for the optimum radius and thickness and for the Lagrange multipliers. Observe that the Lagrange multipliers are positive for all values of P, L and E.

• 3.6 Set up a spreadsheet for the problem in Exercise 3.2, and use Solver to find the minimum of the function.

  Compare the values of the Lagrange multipliers with those found in the exercise. These are found by selecting the Sensitivity Report in the Solver result box after optimization (see the Appendix). Notice that the Lagrange multipliers are zero for the inactive constraints.

• 3.7 Repeat Exercise 3.4, using Solver to find the minimum cost of the 4 m³ container and again after increase in volume to 5 m³.

  Use the formulae derived in Exercise 3.4 to set up a spreadsheet for this problem. The cost has to be minimized, with variables the dimensions of the container and its volume as the single constraint. Compare the value of the Lagrange multiplier in the Sensitivity Report with the value found in Exercise 3.4. By repeating the optimization for a volume of 5 m ³ , the increase in cost can be compared with the estimated increase based on the Lagrange multiplier in Exercise 3.4.

• 3.8 Use the spreadsheet ‘Eccentrically Loaded Column’ to show the effect of eccentric load on the minimum cross-sectional area of a column of effective length 1000 mm, under a compressive load of 10,000 N.

  Take a range of eccentricity from 0 to 50 mm, with $K/K_{0} = 1.0$ for no local imperfection. Use the material data already present in the spreadsheet. Plot a graph of cross-sectional area, diameter and thickness after optimization against eccentricity. Observe the different values of stress at each eccentricity.

• 3.9 Use the spreadsheet ‘Eccentrically Loaded Column’ to show the effect of yielding on the minimum cross-sectional area of a column of effective length 1000 mm, over a range of compressive load from 5000 to 50,000 N.

  Take $e = 0$ for a perfectly straight column and $K/K_{0} = 1.0$ for no local imperfection. Use the material data already present in the spreadsheet. Note the cross-sectional area and the average stress at each load, and observe the reduction in modulus with increasing load. Plot a graph of stress after optimization against structural index $P/L^{2}$ on a log-log basis. Add a line to this graph for the same column with no yielding (efficiency $\upeta = 0.780$ ).

• 3.10 Modify the spreadsheet ‘Eccentrically Loaded Column’ to optimize a square-section tube under eccentric load.

  Use the formulae in Sect. 2.1.1 for the Euler and local buckling stresses of a square-section tube.