Maximum-Minimum Problems

Listing 5.6. Application of partial derivative to maximum, minimum problem. 

The problem involved in the application which is the subject of this chapter is the optimisation of a property of a mixture of compounds (a common situation in pharmaceutical practice, where for example tablet formulations have to be optimised). This property has to be optimised with respect to a certain goal (maximum, minimum or target value) and also with respect to the robustness or ruggedness of the mixture property. This means that despite any variation in the response or the independent variables (mixture variables in our case) due to unknown variation the response values have to be as close as possible to a desired value (target value). 

Kerkhof and Vissers (1978) combined the minimum time and the maximum distillate problems into an economic profit function P to be maximised. 

The minimum time problem (shown as a negative of a maximum time problem) can be written as ... 

The Hamiltonian, the adjoint equations and the optimal reflux ratio correlation will be same as those in Equations P.10-P.13 (Diwekar, 1992). However, note that the final conditions (stopping criteria) for the minimum time and the maximum distillate problems are different. The stopping criterion for the minimum time problem is when (D, xq) is achieved, while the stopping criterion for the maximum distillate problem is when t, xo) is achieved. See Coward (1967) for an example problem. 

The operation strategies and the arguments presented above decomposes MDO problem (OP) into a series of three independent SDO problems as (1) Maximum productivity problem for Task 1 and (2) Minimum time problems for Tasks 2 and 3. 

Mujtaba (1997) used the maximum distillate problem to compare the performances of the two types of distillation columns (CBD and continuous). With the amount of initial charge and the feed flow rate fixed in a continuous column, the operation time (pass time) also becomes fixed. The performance measure using maximum distillate problem allows fixing of the operation time. Other types of optimisation problems such as minimum time or maximum profit problems (presented in the previous chapters) are not suitable for the purpose of comparing the performances of... 

If we have a constrained maximum or minimum problem with more than two variables, the direct method of substituting the constraint relation into the function is usually not practical. Lagrange s method finds a constrained maximum or minimum without substituting the constraint relation into the function. If the constraint is written in the form g(x, y) = 0, the method for finding the constrained maximum or minimum in f x, y) is as follows ... 

Example.—Find the curve of quickest descent from one given curve to another. Ex. (2), page 572, has taught us that the curve of quickest descent is a cycloid. The problem now before us is to find the relation between the cycloid and the two given curves. We see from (15) and (43) that the maximum-minimum condition is... 

In research, development, and above all in production, increased variability of a product implies reduced quality. Due to this fact a product s variability is itself a response. Therefore, the problem of choosing an optimum formulation, or conditions for manufacturing it, is not only one of obtaining the best values of the responses (whether maximum, minimum or target values), but also that of finding conditions where those characteristics vary as little as possible. 

A negative multiplier, on the other hand, leads to the maximum principle, which stipulates that the Hamiltonian be maximized in a minimum problem. 

As indirectly mentioned in Sect. 11.3, you are probably not yet familiar with the concept of derivatives, which is essential in handling maximum and minimum problems. Although we will present and solve interesting and applicable maximum and minimum engineering problems, it is not our intention to familiarize you with the concept of derivatives. Soon in your career you will be introduced by experts to this important and key mathematical concept for aU engineers. You are most probably familiar with the use of spreadsheets. In this section we will show you how to solve maximum and minimum problems using spreadsheets. 

In this section we present and develop three maximum and minimum problems and two OR problems. In both cases the emphasis will be on problem formulatiOTi because in the warm-up examples we already detailed the solution procedure with the help of the Excel spreadsheet. 

Statement 5. The maximum and minimum problems, providing the maximum and minimum values of the variation of amount of material in the intermediate storage in any failure cycle, are reduced to finding maxima and minima of finite number of functions. 

Example 16.1 The process stream data for a heat recovery network problem are given in Table 16.1. A problem table analysis on these data reveals that the minimum hot utility requirement for the process is 15 MW and the minimum cold utility requirement is 26 MW for a minimum allowable temperature diflFerence of 20°C. The analysis also reveals that the pinch is located at a temperature of 120°C for hot streams and 100°C for cold streams. Design a heat exchanger network for maximum energy recovery in the minimum number of units. 

For a long time the official specifications for diesel fuel set only a mciximum viscosity of 9.5 mm /s at 20°C. Henceforth, a range of 2.5 mm /s minimum to 4.5 mm /s maximum has been set no longer for 20°C but at 40°C which seems to be more representative of injection pump operation. Except for special cases such as very low temperature very fluid diesel fuel and very heavy products, meeting the viscosity standards is not a major problem in refining. 

In so doing, we obtain the condition of maximum probability (or, more properly, minimum probable prediction error) for the entire distribution of events, that is, the most probable distribution. The minimization condition [condition (3-4)] requires that the sum of squares of the differences between p and all of the values xi be simultaneously as small as possible. We cannot change the xi, which are experimental measurements, so the problem becomes one of selecting the value of p that best satisfies condition (3-4). It is reasonable to suppose that p, subject to the minimization condition, will be the arithmetic mean, x = )/ > provided that... 

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