Chapter 5

SINGLE VARIABLE SEARCH TECHNIQUES

Problems 5.1 to 5.5 | Problems 5.6 to 5.10 | Problems 5.11 to 5.14 

problem5.1 | problem5.2 | problem5.3 | problem5.4 | problem5.5

 

5-1. In Figure 5-10, the profit function is given for a sulfuric acid alkylation reactor as a function of feed rate and catalyst concentration. Plot the profit function as a function of feed rate for a constant catalyst concentration of 95%. Place six golden section experiments on the interval giving their location, the corresponding value of the profit function and the length and location of the final interval of uncertainty.

Fig5-10.jpg (26299 bytes)

Solution:

The function to be maximized is shown on p. 5-2. Equation (5-53) is used to compute the final interval I6, and equations (5-51) and (5-52) are used to compute the location of the first two points.

Equ5-53.jpg (3064 bytes)

n = 6,      t = 1.618       and Io = 15

Soln5-1.jpg (3356 bytes)

x1 = 0.382 Io                                      equation (5-52)

x1 = 5 + 0.382(15) = 10.729

x2 = 0.618 Io                                      equation (5-51)

x2 = 5 + 0.618(15) = 14.271

The location of x1 and x2 are shown on the following page.

The search proceeds as follows and is shown on the diagram seen here.



x1 = 10.729                                                    y(x1) = 147

x2 = 14.271                                                    y(x2) = 100, eliminate > 14.27

x3 = 14.271 - (10.279 - 5) = 8.542                 y(x3) = 125, eliminate < 8.542

x4 = 8.542 + (14.271 - 10.729) = 12.084       y(x4) = 130, eliminate > 12.084

x5 = 12.084 - (10.729 - 8.542) = 9.897          y(x5) = 149, eliminate > 10.729

x6 = 10.729 - (9.897 - 8.542) = 9.374            y(x6) = 147, eliminate < 9.374

y(x5) = 149, is the maximum value which is on the interval 9.374 < x* < 10.729.



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5-2. Show that equation (5-43) for x2 can be put in the following form in terms of = Io .

x2 = [1 + Î(An-1 - An-2An+1/An)] (An/An+1) Io

Solution:

From equation (5-43)

x2 = I2 = AnIn - ÎAn-2 Io


Using equation (5-41)

In = [(1 + ÎAn-1)/An+1] Io

x2 = An[(1 + ÎAn-1)/An+1] Io - ÎAn-2 Io

     = [An/An+1 (1 + ÎAn-1) Io - ÎAn-2 Io]

     = [1 + ÎAn-1 - ÎAn-2 (An+1/An)] (An/An+1) Io

and x2 = [1 + Î(An-1 - An-2An+1/An)] (An/An+1) Io

 

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5-3. Derive equation (5-65) from equation (5-64).

Obtain equation (5-65)

Soln5-3a.jpg (8808 bytes)

 

from equation (5-64)

Soln5-3b.jpg (7038 bytes)

Solution:

The matrix of coefficients are known values from experiments placed at y(x1) = y1, y(x2) = y2 and y(x3) = y3. Thus, the values of a and b can be determined by Cramer's rule for use in the following equation for the optimum (stationary point) x* of a single variable quadratic equation (y = ax2 + bx + c).

x*= - b/2a


Cramer's rule is:

a = Da/D and b = Db/D

  Soln5-3c.jpg (7747 bytes)

Soln5-3d.jpg (7947 bytes)

 Soln5-3e.jpg (7927 bytes)

x* = -b/2a = -(Db/D)/(2Da/D) = -Db/2Da

Substituting for Da and Db gives:

Soln5-3f.jpg (8766 bytes)

which is equation (5-65).

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5-4. Compare the final intervals of uncertainty (initial being 1.0) for simultaneous elimination, Fibonacci search and golden section search using 2, 5 and 10 experiments. Assume perfect resolution. Give conclusions from this conclusion.

Solution:

Compare simultaneous elimination, Fibonacci search and golden section search for n = 2, 5, 10.

For simultaneous elimination, equation (5-29) is applicable for an even number of experiments, and for perfect resolution this equation is:

Soln5-4a.jpg (2387 bytes)

This equation is also applicable for an odd number of experiments. (See equation (5-30).) Substituting for values of n = 2, 5 and 10 with Io = 1 gives:

n = 2        p = 1        I*2 = 1/2

n = 5        p = 2        I*5 = 1/3

n = 10      p = 5        I*10 = 1/6

For Fibonacci search equation (5-40) is applicable, and for perfect resolution d = 0 to give:

In = Io/An+1

Substituting for values of n with Io = 1 gives:

n = 2        A3 = 2          I2 = 1/2 = 0.5

n = 5        A6 = 8          I5 = 1/8 = 0.125

n = 10      A11 = 89       I10 = 1/89 = 0.0122

For Golden Section search equation (5-53) is applicable and is:

In = Io/tn+1

Substituting for values of n with Io = 1 gives:

n = 2        I2 = 1/1.618 = 0.618

n = 5        I5 = 1/(1.618)4 = 0.1455

n = 10      I10 = 1/(1.618)9 = 0.0135

For comparison, the above results are tabulated below.

Soln5-4b.jpg (19924 bytes)

 

As seen in the above table, simultaneous search and Fibonacci search have the same final interval for two experiments, but golden section search's interval is larger. Both Fibonacci search and golden section search are more efficient than simultaneous search for five and ten experiments, and Fibonacci search is only 8% better than golden section search for ten experiments.

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5-5. Compare the anticipated performance of Fibonacci search and the quadratic method for the following types of unimodal, single variable functions: quadratic, continous, arbitrary and discrete.

Solution:

Function Fibonacci Search Quadratic Method
1. Quadratic

Soln5-5a.jpg (2382 bytes)

 Moves to optimum rapidly eliminating section that do not contain the optimum. In 11 experiments, the optimum is located in a line segment that is 1/144 the length of the original interval. Locates the optimum in one applicationof the method, if the starting point is sufficiently close to the optimum.
2. Continous

Soln5-5b.jpg (2319 bytes)

 Same as above. Proceeds to the optimum at some rate determined by the form of the function if not confounded by a discontinous first derivative.
3. Arbitrary

 Soln5-5c.jpg (2673 bytes)

 Same as above. Quadratic Search is not applicable if the function is discontinous which gives discontinous first derivatives required by the method.
4. Discrete

  Soln5-5d.jpg (2746 bytes)

 Latticesearch version of Fibonnacci search may require adding fictitious points which will reduce search effectiveness. Quadratic search is not applicable if the function is not continous.  However, it may be modified but with difficulty.
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