A section of a chemical plant makes two specialty products (E, F) from two raw materials (A, B) that are in limited
supply. Each product is formed in a separate process as shown in Fig. 19.3. Raw materials A and B do not have to
be totally consumed. The reactions involving A and B are as follows:

Question

Process 1: A + B → E
Process 2: A + 2B → F

The processing cost includes the costs of utilities and supplies. Labor and other costs are $200/day for process 1 and
$350/day for process 2. These costs occur even if the production of E or F is zero. Formulate the objective function
as the total operating profit per day. List the equality and inequality constraints (Steps 1, 2, and 3).

Cost (¢/lb)	Maximum Available (lb/day)	Raw Material
15	40,000	A
20	30,000	B

Maximum Production Level (lb/day)	Selling Price of Product	Processing Cost	Reactant Requirements (lb) per lb Product	Product	Process
30,000	40 ¢/lb E	15 ¢/lb E	2/3 A, 1/3 B	E	1
30,000	33 ¢/lb F	5 ¢/lb F	1/2 A, 1/2 B	F	2

Accepted Answer

The optimization problem is formulated using the first three steps delineated above.

Step 1. The relevant process variables are the mass flow rates of reactants and products (see Fig. 19.3):

[latex]x_{1}= lb∕day A[/latex] consumed

[latex]x_{2} = lb∕day B[/latex] consumed

[latex]x_{3} = lb∕day E[/latex] produced

[latex]x_{4} = lb∕day F[/latex] produced

Step 2. In order to use Eq. 19-1 to compute the operating product per day, we need to specify product sales income, feedstock costs, and operating costs:

[latex]P=\sum\limits_{s}{F_{s} V_{s} } - \sum\limits_{r}{F_{r} C_{r} } -OC[/latex] (19-1)

Sales income ($∕day) = [latex]\sum\limits_{s}{F_{s} V_{s} } =0.4x_{3}+0.33x_{4}[/latex] (19-2)

Feedstock costs ($∕day) =[latex]\sum\limits_{r}{F_{r} C_{r} } =0.15x_{1}+0.2x_{2}[/latex] (19-3)

Operating costs ($∕day) =[latex]OC = 0.15x_{3}+0.05x_{4}+350+200[/latex] (19-4)

Substituting into Eq. 19-1 yields the daily profit:

[latex]P= 0.4x_{3}+0.33 x_{4}-0.15x_{1}-0.2x_{2}-0.15x_{3}-0.05x_{4}-350-200[/latex]

[latex]=0.25x_{3}+0.28x_{4}-0.15x_{1}-0.2x_{2}-550[/latex] (19-5)

Step 3. Not all variables in this problem are unconstrained. First consider the material balance equations, obtained from the reactant requirements, which in this case comprise the process operating model:
[latex]x_{1} = 0.667x_{3} + 0.5x_{4} [/latex] (19-6a)

[latex]x_{2} = 0.333x_{3} + 0.5x_{4} [/latex] (19-6b)

The limits on the feedstocks and production levels are
[latex]0 ≤ x_{1} ≤ 40,000[/latex] (19-7a)

[latex]0 ≤ x_{2} ≤ 30,000[/latex] (19-7b)

[latex]0 ≤ x_{3} ≤ 30,000 [/latex] (19-7c)

[latex]0 ≤ x_{4} ≤ 30,000 [/latex] (19-7d)

Equations 19-5 through 19-7 constitute the optimization problem to be solved. Because the variables appear linearly in both the objective function and constraints, this formulation is referred to as a linear programming problem, which is discussed in Section 19.4.

Question 19.1: A section of a chemical plant makes two specialty products (...

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