Practicing Optimization—Larger Examples - USING OPTIMIZATION—PRACTICAL ESSENTIALS - Optimization in Practice with MATLAB for Engineering Students and Professionals (2015)

Optimization in Practice with MATLAB for Engineering Students and Professionals (2015)

PART 3


USING OPTIMIZATION—PRACTICAL ESSENTIALS

10


Practicing Optimization—Larger Examples

10.1 Overview

This chapter covers the practice of optimization in diverse technical areas. What has been learned thus far is the practical knowledge that is needed to apply optimization to different types of designs or systems. We will explore optimizing systems/designs in the following disciplines: chemistry, mechanics, aerospace, automotive, mathematics (data fitting), nuclear, electrical, portfolio management, and business (Ref. [1]). Our ability to optimize these various designs provides us with the skills to apply optimization beyond the confines of this book or of the class you might be taking, and successfully venture into the real world!

10.2 Mechanical Engineering Example

10.2.1 Structural Example

Figure 10.1 represents a ten-bar truss. The members are connected to each other at six nodes, numbered 1 to 6. The truss is assumed to be planar, and every node has two degrees of freedom (i.e., it is free to move both in the x and y directions). The left edge of the truss is fixed to the wall. The displacements u1 to u8 are specified at the specific non-fixed nodes. F1 to F8 represent loads applied to these nodes. The variables by x1 to x10 denote the cross sectional areas of the truss members. The Young’s modulus of the material is E = 1 × 106N/mm2. The maximum allowable stress in each bar is σult = ±100N/mm2 (i.e., tension/compression), and the maximum allowable deflection is δmax =±2 mm. We note that the unit of length in the figure and in this problem is mm, and that of force is N. While the current example is commonly used in the literature, the reader is also encouraged to review the topology optimizationsurvey by Rozvany in Ref. [2].

Figure 10.1. Ten-Bar Truss

Imagine you are hired as the optimization expert for the ten-bar truss project. The structural engineer working on the project provides you with a program blackbox-10bar.m, which allows you to enter the loads at the nodes and the area of each member, and obtain the corresponding nodal displacements and stresses in the members. Figure 10.2 illustrates this task. The ten-bar truss is subjected to 10 different loading conditions given in Table 10.1.

Figure 10.2. Black Box

Table 10.1. Loading Conditions

Your task is to generate the most optimal design for the ten-bar truss by minimizing the mass of the configuration along with the maximum stress for a given set of conditions. The following steps will help you accomplish this optimization process.

1.Generate the row vector F that contains all the 8 forces:

F = [61,540 5,790 1,530 83,810 19,340 49,660 72,710 79,480]

2.Generate the row vector x that contains all the 10 areas:

x = [16.49 4.67 13.34 0.40 0.10 0.10 17.57 11.36 4.06 16.19]

3.Use these values in the program blackbox-10bar.m to determine (1) stresses and (2) deflections.

4.Double the value of the forces and use these new values in the program blackbox-10bar.m to obtain the stresses and deflections.

5.Halve the areas and repeat the process to obtain the corresponding stresses and deflections.

6.Minimize the total material volume of the truss for the loading condition specified in Part 1 and the initial areas specified in Part 2. Note that to obtain the corresponding minimum mass, we would simply multiply by the density. (Hint: In forming the constraint function, use the black box program to determine your stresses and displacements, which should not exceed the maximum design conditions).

7.Minimize the total material volume of the truss for all ten loading conditions individually. (Hint: Change the set of loading conditions in each minimization subroutine. The different set of loading conditions are given inTable 10.1. In addition to the area vector provided in Part 2., randomly generate 9 row sets of area values to obtain the initial area matrix).

8.Minimize the maximum stress along with the total mass of the truss for the loading condition specified in Part 1 and the initial areas specified in Part 2. Plot the Pareto frontier for 11 solutions. (Hint: This is a bi-objective problem where the first objective is the combined mass of the truss, and the second objective is the maximum stress).

9.In this case, you are not so blessed as to have the Black Box simply provided to you. You started your job, and since you graduated in Mechanical Engineering (or Civil or Aerospace), your team simply expects that you will develop the Black Box yourself. Your task is to develop this Black Box. (i) Provide clear documentation and explain your assumptions in developing the analysis for this Black Box. Comment on how truss or frame structures assumptions apply here, and your pertinent decisions. (ii) Code your development in your own Black Box. (iii) Solve Item 8 above using your own Black Box; then using the one provided. Comment on the relative accuraciesand computation times between the two Black Boxes, yours and the one provided. (This part of the problem (9.) is geared towards graduate students in mechanical, aerospace, or civil engineering).

10.2.2 Tolerance Allocation Problem

Consider the example of a two-part assembly shown in Fig. 10.3. The dimensions of the two parts, A and B, are given by x1 through x12. Parts A and B are mass produced by the same manufacturing company. To understand the concept of tolerance, assume that the design requirement for the dimension x1 is 50 in. Consider a batch of 10,000 parts of A. If we measure the dimension x1 of each part in the batch, it will not be exactly equal to 50 in. There is some variability in the measurement of x1 in all the 10,000 parts. This variability can result from measurement errors or the manufacturing process itself. Each measurement of the dimension x1 for the batch falls within a range 50 ±tol inch, where tol is termed as the tolerance of the dimension.

Figure 10.3. Tolerance Design Example

It is desirable to have the tolerances be as close to zero as possible to ensure a high quality product. However, smaller tolerances require precise manufacturing processes, which, in turn, result in higher manufacturing costs. The range of 50 ± tol inch mentioned above specifies the acceptable range of the dimension x1. Parts that do not conform to the above specification are rejected, which are either scrapped or reworked, thus adding to the manufacturingcosts. Setting a “good" tolerance value for a dimension involves understanding the tradeoff between cost and quality. Multiobjective optimization can be effectively applied in these problems.

To understand how to mathematically represent the variability in the part dimensions in the optimization formulation, the concept of a random variable and probability theory are used. A detailed description of probability theorycan be found in probability textbooks (Ref. [3, 4]). In this problem, we present a brief summary of the basics of probability, normal distribution in particular.

Basic Concepts in Probability Theory

An event that can result in different outcomes, even though it is repeated in the same manner every time, is known as a random event. In the above example, each manufacturing event for the 10,000 parts in the batch results in a different length x1. In probability theory, a variable referred to as a random variable is associated with the outcome of the random event. For the above example, the length of the dimension x1 in each part of the batch is a random variable. Random variables are denoted by upper case letters. The dimensions of Parts A and B, x1 through x12, can be thought of as random variables, denoted X1 through X12.

A function that assigns probability values to a random variable is known as the probability density function (PDF). There are several random variables with standard PDF definitions that are typically used in engineeringapplications, such as uniform, Gaussian, and exponential distributions. In this example, we use Gaussian (or normal) distribution to represent the probabilities of the random variables X1 through X12. Figure 16.3 shows the PDF of a normal distribution.

The quantities μ and σ in Fig. 10.4 are referred to as the mean and the standard deviation of the normal distribution, respectively. The PDF curve provides information about the probabilities of random variables. Consider twopoints on the x-axis, such as μ + σ and μσ in Fig. 10.4. The probability that the random variable, X, lies between μ + σ and μσ is given by the area under the PDF curve, denoted by the shaded region in Fig. 10.4. The probabilities that X falls within the intervals μ ± σ, μ ± 2σ, and μ ± 3σ (see Fig. 10.4) for a normal distribution are 0.6827, 0.9545, and 0.9973, respectively [4]. The area under the entire PDF curve is always equal to one.

Figure 10.4. Normal Distribution

The two quantities, mean and standard deviation, define the bell-shape of the normal distribution illustrated in Fig. 10.4. The normal distribution is symmetric about its mean. The smaller the value of σ, the narrower the bell-shaped PDF. In the manufacturing example, the closer the tolerance on the dimension X1, the lower its variability. Thus, achieving closer tolerances requires reducing the standard deviation of the random variable X1.

Finding Optimal Tolerances

To formulate the multiobjective problem that models the tradeoff between the tolerances and the manufacturing costs, the tolerances of the variables X1 through X12 are assumed to be six times their standard deviations. The nominal or the mean values of X1 through X12 are given as

The quality of the assembly requires that the angles θ1 and θ2 in Fig. 10.3 be as close to each other as possible. Satisfaction of this requirement imposes tight tolerances on the parts dimensions, which increases manufacturing costs.

The design variables for the problem are the standard deviations of the dimensions. The first objective, J1, is to minimize cost, which is given as a function of the standard deviations of the design variables as follows.

(10.1)

where b = [50 50 50 50 50 50 50 50 50 50 50 50], and σXi is the standard deviation of the i-th dimension. Note that the above cost function increases as the standard deviations decrease.

The second objective, J2, ensures that the angles θ1 and θ2 are as close to each other as possible. Note that the tangent of the angles θ1 and θ2 can be expressed in terms of the dimensions X6, X5, X2, X3 and X9, X10, X7, X8, respectively. Since the above dimensions are random variables, the tangent of the angles θ1 and θ2 are also random. To ensure that θ1 and θ2 are as close to each other as possible, consider the random variable θ1θ2 and minimize its variation. One possible measure of the variation of θ1θ2 is its standard deviation, J2, given as follows.

(10.2)





(10.3)

Where Xn(i) is the i-th element of the mean vector of X (given as Xn); V is the variance of θ1θ2; and the square root of the variance yields the standard deviation.

The multiobjective optimization problem formed to obtain the optimal tolerances from a cost perspective is presented below.

(10.4)

subject to

0.0001 ≤ σX ≤ 1

(10.5)

J1 ≤ 100

(10.6)

J2 ≤ 1

(10.7)

-(Xn(6) - Xn(5)) + (Xn(8) - Xn(7))

(10.8)

-(Xn(3) - Xn(4)) + (Xn(11) - Xn(10))

(10.9)

Equation 10.5 represents the maximum and minimum allowable standard deviations for X1 through X12. Equations 10.8 and 10.9 ensure that the clearances between Parts A and B are positive, as shown in Fig. 10.3.

The Pareto frontier for the above bi-objective problem is illustrated in Fig. 10.5. The weighted sum method is used to obtain the Pareto frontier. Based on the relative preferences, any of the Pareto solutions may be chosen as thefinal design. From the design variable values at the chosen Pareto solution, the tolerances can be computed.

Figure 10.5. Pareto Frontier for the Tolerance Allocation Problem

10.3 Aerospace Engineering Example

When designing an aircraft landing gear, the wheel track (the length between the left-most and the right-most wheels when looking at a front view) has several requirements that must be considered. These include (i) the ground lateral control, (ii) the ground lateral stability, and (iii) the structural integrity. The minimum allowable value for the wheel track must satisfy the ground lateral control and the ground lateral stability requirements. The maximum allowable value for the wheel track must satisfy the structural integrity requirements (Ref. [5]).

10.3.1 Ground Controllability

The wheel track must be sufficiently wide so that the aircraft does not roll over while taxiing on the ground. The centrifugal force (Fc) during a turn, which may cause the aircraft to roll, is given by

(10.10)

where m represents the aircraft mass, V denotes the aircraft ground speed, and R is the radius of the turn (see Fig. 10.6(b)). The sum of the two contributing moments (the moment of the centrifugal force (Fc) and the moment due to the aircraft weight, W = mg) about the gear is given by

Mo = 0 mgT∕2 + FcHcg

(10.11)

Figure 10.6. An Aircraft in a Ground Turn

where the parameter Hcg is the distance of the aircraft’s center of gravity, cg, from the ground, and T is the length of the wheel track. The minimum constraint for the wheel track (T) is given by

(10.12)

10.3.2 Ground Stability

While taxiing on the ground, wind affects the stability of the aircraft and, as such, must be considered during the design process. A cross wind (i.e., perpendicular to the ground path or fuselage centerline of the aircraft) is the most important wind force when designing for ground stability.

The cross wind force (FW) on an aircraft can be modeled as a drag force given by

(10.13)

where ρ is the air density, V W denotes the wind speed, and AS is the aircraft side area (hatched area in Fig. 10.7). The parameter CDs is the side drag coefficient of the aircraft, and it varies from 0.3 to 0.8. To prevent an aircraft from overturning due to the cross wind, the wheel track (T) satisfies

T > 2

(10.14)

where Hc is the distance of the centroid from the ground.

Figure 10.7. Aircraft Side Area and Its Centroid

10.3.3 Structural Integrity

When seen from the front, the aircraft structure can be modeled as a beam with a few simple supports (see Fig. 10.8). At the main gear station, the wing is the beam and the two main wheels are the simple supports. The wheel track is then the distance between these two supports.

Figure 10.8. Front View of the Aircraft Structure Modeled as a Beam with Two Simple Supports

The maximum deflection (ymax) in a beam (wing) is given by

ymax = −

(10.15)

where Fmmax is the maximum load on the main gear, E is the modulus of elasticity, and I is the beam area moment of inertia. The maximum static load, which is carried by the main gear, can also be estimated as

Fmmax = mg

(10.16)

where B is the wheel base (the distance between the nose gear and the main gear along the x-axis), and Bnmax is the maximum length between the aircraft cg and the nose gear along the x-axis. Substituting Eq. 10.16 into Eq. 10.15,the wheel track in terms of maximum allowable deflection is given by

T =

(10.17)

For a twin engine jet transport aircraft, the maximum allowable wing deflection is 0.03 m and the take-off mass is 50,000 kg.

The question at hand: Your job is to determine the range of wheel tracks that will satisfy all of the requirements discussed in this Aerospace Engineering Example. Explain the practical meaning of the nature of your answer, including an examination of which constraints are active of inactive.

Constants are given as

(10.18)

(10.19)

(10.20)

(10.21)

(10.22)

(10.23)

(10.24)

and constraints as

(10.25)

(10.26)

(10.27)

(10.28)

(10.29)

(10.30)

(10.31)

10.4 Mathematical Example

10.4.1 Data Fitting

Data fitting is the process of constructing a mathematical function to fit a series of data points. Data points can be experiment results, simulation results, or observations of n natural phenomena. There are many different methods for data fitting. This section will use the least squares approach, which requires the use of optimization.

Let us look at the three points marked as circles in Fig. 10.9. The coordinates of the three points are (2, 9), (6, 1), and (9, 16).

Figure 10.9. Data Fitting Through Three Points

We will use a quadratic function, expressed by y = ax2 + bx + c, to fit the data. To fit a quadratic function that passes through all three points, the following system of equations must be solved. In the following system of equations,a, b, and c are unknowns and their values can be obtained exactly by solving the equations. The equations are satisfied by a = 1, b = −10, and c = 25.

(10.32)

(10.33)

(10.34)

The fitted curve of the above example passes through all three points. If a quadratic function is required to fit more than three points, its curve can pass through all points only in special cases. The method of least squares is a popular and easy-to-implement approach to approximate the solution of overdetermined systems (sets of equations in which there are more equations than unknowns). The overall solution obtained by the least squares data fitting minimizes the sum of the squares of the errors when solving every single equation. It is used in Sec. 10.4.2 to fit a quadratic function to four points.

10.4.2 Least Squares Data Fitting

The least squares data fitting minimizes the sum of the squared residuals. The residuals are the differences between the recorded values and the fitted (or approximate) values provided by a model (Ref. [6]). In this example, there is one more point (7.5, 6.5) in addition to the three points in Sec. 10.4.1. This new point is not on the curve given by y = x2 − 10x + 25. A fitted quadratic function cannot pass through all the four points. Instead, the least squares data fitting is used to fit a quadratic function that minimizes the sum of the errors between the fours points and the corresponding points on its curve. Let represent the fitted quadratic function. The sum of the errors is expressed as

(10.35)

Perform an unconstrained minimization using Eq. 10.35 as the objective function. After the minimization problem is solved using MATLAB, the corresponding values of the three parameters are a = 0.9937, b = −9.9198, and c = 24.8524. The sum of the squared residuals is 0.0383. The four points, as well as the curve given by the fitted quadratic function, are plotted in Fig. 10.10. Note that the points are very close to, but not exactly on, the fitted curve.

Figure 10.10. Least Squares Data Fitting Through Four Points

10.5 Civil Engineering Example

The company you work for is planning on building a new facility to account for recent growth. You have been chosen to design the exterior walls of the building. The indoor temperature is to remain at Ti = 25°C for a comfortableindoor environment. During the winter months, the outdoor temperature is consistently at To = 0°C. Your objective is to minimize the heat loss during the winter months, while also being cost conscious.

In the building and construction industry, insulating materials are commonly described by their thermal resistance (R-value). The R-value being discussed is the unit thermal resistance. This is used for a unit value of any particularmaterial. It is expressed as the thickness of the material divided by the thermal conductivity. For the thermal resistance of an entire section of material, the unit thermal resistance is divided by the thickness of the material. The largerthe R-value, the better the building insulation’s effectiveness (Ref. [7]).

Heat transfer through an insulating layer is analogous to electrical resistance. The heat transfer in a simple system can be solved by using electrical resistance in series with a fixed potential. The differences between electrical and thermal are: (i) the resistances of thermal systems are thermal resistances, and (ii) the potential is the difference in temperature from one side of the material to the other. The resistance of each material to heat transfer depends on the specific thermal resistance [R-value]/[unit thickness].

Assuming 1-D steady heat transfer, given by Eq. 10.36.

= 0

(10.36)

The boundary conditions are given by

k = h0(T0T(0))

(10.37)

and

k = hi(T(L) − Ti)

(10.38)

where k is the thermal conduction coefficient, and h is the convection heat transfer coefficient. The convective heat transfer coefficients are hi = 2 [] and h0 = 22 [] for the inner and outer regions, respectively.

Your boss said that the wall must cost no more than $100 and be no thicker than 1m. Instead of solving the differential equation by hand, you are provided with the file steadyeqn.m to calculate the inner and outer wall temperatures. Using the above equations along with the material properties in Table 10.2, minimize the heat transfer rate and minimize the cost per unit wall area.

Table 10.2. Material Properties



Material

k (W/mK)

Cost ($/cm)


Concrete block

0.688

1

R-15 insulation board

0.04

15

Wood

0.212

3

Brick

1.24

0.50



1.Plot the Pareto frontier for each of the four materials on the same graph.

2.Plot the Pareto frontier for the entire range of materials and thicknesses.

10.6 Electrical Engineering Example

10.6.1 Introduction to Thermoelectric Window Design

The Thermoelectric (TE) window design presented in this example uses TE units to actively transfer heat to maintain the desired indoor temperatures. TE units are solid state devices that actively transfer heat in designated directionswhen supplied with electric power (see Refs. [8, 9, 10, 11]). They are sufficiently small to be integrated into a window. The simplified schematic of the TE window depicted in Fig. 10.11 indicates that the TE units are installedwithin the frame of the window, which, for practical purposes, also performs as a heat enhancer (heat sink). TE units in the window facilitate heat flow (solid lines) in the direction opposite that of the passive heat flow through the window panes (broken lines), as illustrated in Fig. 10.11(a). Though not shown in Fig. 10.11(a), the frame of the TE window will be equipped with fins to maximize heat transfer. With this design, it is possible to integrate TE units in any glazing system once the frame is modified to accommodate the TE units.

Figure 10.11. Schematic of TE Window

To achieve a high heat transfer rate through the TE units installed on the window frame, the connection of the TE units in their electric network should be optimized. Suppose the TE units are connected in Np parallel circuits. Ineach of the parallel circuits, there are Ns TE units connected in series. The electric network is shown in Fig. 10.12.

Figure 10.12. Electric Network

The TE units shown in Fig. 10.12 are integrated into the window frames to transfer heat from inside to the outside in summer. Under this condition, the function of the TE units is to cool the air inside the room. The higher the rateof heat transfer through the TE units, the better the performance. The objective of this optimization is to maximize the total heat transfer rate through the TE units to the outside, expressed as cold. The numbers of TE units in series and in parallel, the total current, and the temperature difference across the TE units are the four parameters to be varied in the optimization. In this problem, the TE units are modeled based on their configuration and properties. The trust region method is used to generate an approximation for the optimization problem.

10.6.2 Brief Introduction to the Trust Region Method

The trust region method generates a series of intermediate steps with the help of a quadratic model of the objective function. This method defines a region around the current iterate within which it trusts the model to be an adequaterepresentation of the objective function. It then defines the step leading to an approximate minimum of the model in this region. A detailed discussion of the theory and implementation of the method can be found in (Refs. [12, 13, 14, 15]).

The trust region method chooses the direction and length of the step simultaneously. If a step is not acceptable, it reduces the size of the region and finds a new minimum. In general, the direction of the step changes whenever the size of the trust region is altered. The size of the trust region is critical to the effectiveness of each step. If the region is too small, the algorithm misses an opportunity to take a substantial step that will move it much closer to the minimum of the objective function. If it is too large, the minimum of the model may be far from that of the objective function in the region, and the size of the region will need to be reduced. In practical algorithms, the size of the region is chosen according to the performance of the algorithm during previous iterations. If the model is consistently reliable, producing good steps and accurately predicting the behavior of the objective function along these steps, the size of the trust region may be increased to allow longer, more ambitious, steps to be taken. A failed step is an indication that our model is an inadequate representation of the objective function over the current trust region. Aftersuch a step, the size of the region is reduced, and the step is run again. [12] The detailed implementation of the trust region method for this optimization is explained in Sec. 10.6.4.

10.6.3 Modeling TE Units

The ATI windows use TE units to control the heat transferred through the inner panes (Ref. [9]). Each TE unit consists of thermocouples which, when supplied with electric current, induce heat flow in the direction of the current. This is known as the Peltier effect. Because of the thermocouples’ electrical resistance, heat is generated. This is known as the Joules effect. As a result of the two conflicting effects, heat is absorbed on the cold side and released from the hot side. A temperature difference is created across the TE units. On the cold side of the TE units, the heat rate is predicted as

cold = 2NteN

(10.39)

where Nte is the number of TE units; N is the number of thermocouples in each TE unit; α is the Seebeck coefficient; Ite is the electric current; Tc is the cold side temperature; ρ is the resistivity; G is the geometry factor, which represents the area to thickness ratio of the thermocouple; κ is the thermal conductivity; and ΔTte is the temperature difference across the thermocouple. For a given TE unit, α, ρ, and κ are temperature dependent properties, N and Gare constants, and are all provided by the manufacturer. The remaining four variables, Ite, Tte, Tc, and Nte, are design variables for the problem presented in Sec. 10.6.4.

TE units are connected in an electrical network, and power is supplied to every TE unit. The TE units used for the TE window design are divided into several groups. The number of the groups is Np. There is the same number of TE units in each group, which is Ns. Within each group, the TE units are connected in series. Then, all the groups are connected in parallel. The electric voltage supplied to each group is the same. The connection of the TE units is shown in Fig. 10.12. The total number of TE units is

Nte = NsNp

(10.40)

The voltage drop across the TE unit is given by

V te = 2N

(10.41)

The heat released from the hot side of the TE unit is the combination of the heat absorbed by the TE and the heat generated by electric current, which is given by

hot = NteIteV te + cold

(10.42)

The maximum allowable applied current, Imax, is given by

(10.43)

where Z is the figure-of-merit provided by the manufacturer, Th is the hot side temperature, and the other variables are defined above.

The maximum allowable temperature difference, ΔTmax, is given by

ΔTmax = Th

(10.44)

10.6.4 Solving Optimization Problem

The optimization of the connection of the TE units can be solved as follows:

1.Objective of Optimization

The TE units are integrated into the window frames to transfer the heat from the inside of a room to the outside in summer. In this condition, the function of the TE units is to cool the air inside the room. The objective of optimization is to maximize the overall heat transfer rate through the TE units to the outside, which is cold. The number of the TE units in series, Ns, the number of the TE units in parallel, Np, the total electric current, I, and the temperature difference across the TE units, δT, are the four variables for optimization.

2.Physical Constraints

The TE units are fixed on the frame of the window. The total area of all the TE units should not exceed the area of the frame, Aframe. The total area of the TE units is expressed as

Aall = NsNpAte

(10.45)

The electric current in each set of TE units should not exceed the maximum allowable current.

IteImax

(10.46)

According to energy conservation, the outside temperature is less than the sum of the inside temperature and the temperature difference across the TE units.

ToutTcold + NsΔTte

(10.47)

Bounds on the total current are used to avoid excessively large power requirements or unreasonably small power consumption.

0.01 ≤ NpIte ≤ 100

(10.48)

3.Modeling the Optimization Problem

Considering the objective of the optimization and the constraints, the optimization problem is as follows.

(10.49)

subject to

(10.50)

(10.51)

(10.52)

(10.53)

(10.54)

(10.55)

4.Optimization Procedure

The trust region method is used for optimization. At the k iteration, the model of the TE units is approximated by a quadratic Taylor series expansion.

coldapprox(p) = cold(xk) + GkTp + pTH kp

(10.56)

Where coldapprox(p) is the approximated objective function; p is the step length; cold(xk) is the optimal result from the last iteration and also the initial value of the objective function in the current iteration; GkT is the gradient at the starting point, and Hk is the Hessian at the starting point.

In each iteration, a ratio is defined to evaluate the agreement between the approximate objective function and the actual objective function, as given by

ρk =

(10.57)

The approximate objective in Eq. 10.56 is optimized using MATLAB. In each iteration, the approximate model is optimized as follows.

Given a small number > 0, γ = 0.3, Δ0 = γ, α = 1, and μ = ;
for k=1, 2, ...

Solve optimization problem for the approximate model of Eq. 10.56 in MATLAB.

Check the improvement that Δcold has on the value of the actual objective function.

if Δcoldα
Stop optimization. The actual value of the objective
function cold is the final optimal value.

else continue.
Evaluate ρk in Eq. 10.57
if ρk
Δk+1 = μΔk
else
if ρk ≥ and pk = Δk
Δk+1 = min(2Δk,)
else Δk+1 = Δk
if ρkμ
xk+1 = xk + pk
else
xk+1 = xk
end (for).

10.6.5 Results

The optimization of the problem in Sec. 10.6.4 is performed by MATLAB. The maximum value of the objective function is Qcold = 1924. The optimal values of the design variables obtained from the optimization are ΔT = 6, I = 23.36, Ns = 16.8, and Np = 6.6. Since the numbers of the TE units in series and in parallel are integers, the values of Ns and Np are rounded to the closest integers. The final optimal results are ΔT = 6, I = 23.36, Ns = 17, and Np = 7. The optimization stops only after four iterations to evaluate the approximate model.

10.7 Business Example

You are put in charge of rearranging the dining room at the restaurant you work at. The owner has told you that your salary will now be measured as a percentage of all food sold. Therefore, you have chosen to use optimization to maximize your paycheck.

The restaurant is a square with each side measuring 40 ft. You are free to select the size, position, and number of circular tables within the dining hall. In order for the kitchen staff to operate properly and the patrons to feelcomfortable, the owner has decided that there must be a minimum distance of 3 ft between any two surfaces. This includes between tables, as well as away from the walls. In order to please the wait staff (whose salary is based on tips), you have chosen a minimum of two seats per table. The minimum radius of these two person tables is 1 ft.

For every 2.5 ft. of additional circumference, you may add a place setting. If you assume that your salary is directly proportional to the number of seats you can fit (neglecting atmosphere, quality of food, sanitary requirements,fire regulations, etc.), formulate and solve a problem to maximize your paycheck.

10.8 Summary

In order to fully appreciate the imperative role of optimization in designing systems - from simple everyday products and complex engineering systems, to non-engineering systems - it is important to begin the application of optimization to practical problems. With this perspective in mind, this chapter provided the opportunity to apply optimization to diverse practical problems. These included (i) designing a truss structure, (ii) designing an aircraft landing gear, (iii) data fitting, (iv) designing a thermoelectric window, and (v) arranging the seating pattern in a restaurant. Pertinent mathematical models were also provided for each problem, such that additional references were not required to formulate and solve each optimization problem.

10.9 Problems

Warm-up Problems

10.1Prove the results of Sec. 10.4.1. Briefly discuss the issues as you see them.

Intermediate Problems

10.2A structural engineering example is presented in Sec. 10.2.1, solve Items 1. through 5. Explain your work throughout.

10.3A structural engineering example is presented in Sec. 10.2.1, solve Items 1. through 8. Explain your work throughout.

10.4Generate the Pareto frontier for the tolerance allocation problem shown in Fig. 10.5 with 20 points. Prepare a five-page PowerPoint presentation for a customer who wants to understand the tradeoffs involved in the problem. Also submit a concise report that explains how you developed your presentation results. The customer will give your report to her technical group leader who will pass judgement on the credibility of your findings.

10.5In Section 10.3, an aerospace engineering example is provided. You are to determine the range of wheel track values that will satisfy all of the requirements discussed in this Aerospace Engineering Example. Explain the practical meaning of the nature of your answer, including an examination of which constraints are active or inactive, and any pertinent implications. Discover any tradeoffs that result from the range provided in your answer. Prepare a five-page PowerPoint presentation of your findings, and a concise report that supports your results.

10.6Solve the problem in Section 10.4.2, using computational optimization.

10.7Solve the problem in Section 10.4.2, using analytical optimization. That is, find the necessary conditions, and solve them using linear algebra.

10.8(i) Solve the problem in Section 10.5. (ii) Prepare a five-page PowerPoint presentation of your findings. (iii) Submit a brief report that supports your analytical developments.

10.9HEAT EXCHANGER: In this problem, we are interested in designing a three-stage heat exchanger network (Refs. [16, 17]). A schematic of the heat exchanger network is shown in Fig. 10.13. A cold stream of fluid of a given flow rate, W, and specific heat, Cp, is heated from temperature T0 oF to T˙3 oF to T3 oF using three heat exchangers arranged in three stages. At each stage, the cold fluid is heated by a hot fluid of the same flow rate, W, and specific heat, Cp, as for the cold fluid. The temperatures of the hot fluid entering Stages 1, 2, and 3 are t11, t12, and t13, respectively. The overall heat transfer coefficients U1, U2, and U3 of the heat exchangers are known.

Figure 10.13. Heat Exchanger Network Design Problem

Our focus is to minimize the sum of the areas of the individual heat exchangers, while maximizing the final temperature of the cold fluid, T3. We assume that the initial temperature of the cold fluid, T0, is given.

A total of six inequality constraints are imposed in this design. The first three arise from the fact that the rate of heat transferred to the cold fluid is less than or equal to the rate of heat lost by the hot fluid. For the i-th heatexchanger, this constraint can be expressed as

WCp(TiTi-1) ≤ WCp(ti1ti2) i = {1, 2, 3}

(10.58)

The remaining inequality constraints reflect that the heat gained by the cold fluid is less than or equal to the heat lost by the heat exchanger. For the i-th heat exchanger, this constraint can be expressed in a simplified form as

(10.59)

The areas and temperatures are given by the following upper and lower bounds.

(10.60)

(10.61)

(10.62)

The given design parameters are T0 = 100oF, WC p = 105 BTU/Hr-oF, ti1 = {300, 400, 600}oF, and Ui = {120, 80, 40} BTU/hr-sq.ft-oF, where i = {1, 2, 3}. Given the above information, solve the following problems.

(a)Assume that the initial temperature of the cold fluid is 100 oF, and the desired final temperature is 500 oF. Minimize the total area of the heat exchanger network that accomplishes the desired heating of the cold fluid, subject to the constraints discussed above.

(b)Assume that the initial temperature of the cold fluid is still 100oF. Find a design that maximizes the final temperature of the cold fluid, subject to the given constraints. (c) We now wish to simultaneously minimize the total area, and maximize the final temperature of the cold fluid, subject to the above constraints. Formulate the multiobjective problem, and solve it using the weighted sum method. With supporting information, comment on how easy/difficult it was to obtain the Pareto frontier using the weighted sum method.

(d)Now that we have obtained the Pareto frontier, use it as a design tool to choose the “optimum” design for the heat exchanger. You are given the ranges of desirability for each of the objectives (see Table 10.3). On thePareto frontier plot, mark each of these ranges for the two objectives. A grid of different possibilities in the design space (in terms of ranges of differing desirabilities) is obtained. As an example, a region in the design space could have total area as “unacceptable” and the final temperature as “desirable.”

Table 10.3. Ranges of Desirability for the Objectives



Total area, AT (×104 ft2)

Final Temperature (oF )




Ideal

AT ≤ 0.5

≥ 600

Desirable

0.5 ≤ AT ≤ 1

600 ≤ T3 ≤ 550

Tolerable

1 ≤ AT ≤ 1.5

550 ≤ T3 ≤ 500

Undesirable

1.5 ≤ AT ≤ 2

500 ≤ T3 ≤ 450

Unacceptable

AT ≥ 2

T3 ≤ 450



Is it possible for both of the objectives to be in the “ideal" or “desirable" range of desirability? Explain.

(e)Our task now is to choose a final design that achieves the “best of both worlds" from the above generated grid of possibilites. Make sure that your final design does not fall into“undesirable" or “unacceptable” ranges for either objectives. Discuss your thought process behind your choice of final design (note that there is generally no single correct answer for such problems).

Advanced Problems

10.10UNIVERSAL MOTOR: Refer to the universal motor problem given in Sec. 7.7. Use the information in that example to complete the following tasks.

1.Use the following values for the design variables, Na = 1236.8,Nf = 53.23,Awf = 0.2596×10-6,A wa = 0.2601×10-6,I = 6,r o = 0.025448,t = 0.007184, and L = 0.024894. Input the information as a row vector and use thefunction umotor.m to determine the mass, power, efficiency, torque, and magnetizing intensity of the motor from the design variables.

2.Minimize the mass of the motor using the values in Part 1 as initial guesses.

3.Minimize the mass and maximize the efficiency of the motor, for both objective functions bearing the same weight (i.e., w = 0.5 in f = w(μ1) + (1 − w)(μ2)). This is a bi-objective problem where μ1 is the mass and μ2 is efficiency. What is the result? Discuss possible reasons for your results.

4.Implement the scaling of the design variables discussed in the chapter. Minimize the mass and maximize the efficiency of the motor for both objective functions bearing the same weight. (Remember to specify appropriate upper and lower bounds.)

5.Plot the Pareto frontier for the bi-objective problem. Maximize efficiency and minimize mass. Obtain 100 Pareto points.

6.Design the motor for a mass target of 0.5kg and an efficiency target of 0.7.

10.11WIND Energy: You have installed a small wind turbine at Syracuse University. In the next 36 hours, it can be operated for only 4 hours 30 minutes (4.5 hours). The turbine can be switched on and off twice over the concerned36 hours (i.e., it can be operated only twice). It cannot be operated when the chance of precipitation (C) is 50% or more. The power generated (P) by the turbine in watts depends on the wind speed (U), as given by

P = 3.0 × U3

(10.63)

The energy generated (E) over a period of T minutes can be estimated as:

(10.64)

where Pi is the power generated by the turbine in the ith minute.

1.Model the variations of wind speed (U) and chance of precipitation (C) with time (T) over the concerned 36 hours. You may use quadratic or cubic polynomial functions for this purpose.

2.Formulate, model, and solve an optimization problem to maximize the total energy produced (ET) by the turbine over the next 36 hours, starting from a point of data recording. The design variables are the two starting times, TT1 and TT2, and the two stopping times, TP1 and TP2. All four design variables are to be expressed in minutes.

The hourly (and 15-min interval) weather data (i.e., wind speed (U) and chance of precipitation (C)) can be obtained from the website: www.weather.com.

10.12Solve the business problem presented in Sec. 10.7. Provide brief and helpful pertinent discussions.

Graduate Level Problems

10.13A structural engineering example is presented in Sec. 10.2.1. Solve Items 1. through 9. (This problem is geared towards graduate students in mechanical, civil, or aerospace engineering). This problem may require someindependent research for some students.

10.14We are given the means, μX, and standard deviations, σX, of a set of independent random variables, X = {X1,X2,...Xnx}. The mean and standard deviation of a function of X, say g(X), can be found using a Taylor series approximation, given as follows [18].

(10.65)

(10.66)

(10.67)

where μg, V g, and σg are the mean, variance, and standard deviation of g, respectively. Using the above equation, derive the expression for the variance and standard deviation of θ1θ2 (Eqs. 10.2 and 10.3) for the tolerance allocation problem.

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