Consider a simple version of a refinery blending and production problem. This example is more illustrative of a scheduling application (Level 5 in Fig. 19.1) that has been used extensively since the 1960s in the chemical process industries. Figure 19.7 is a schematic diagram of feedstocks and produ

Question

Consider a simple version of a refinery blending and production problem. This example is more illustrative of a scheduling application (Level 5 in Fig. 19.1) that has been used extensively since the 1960s in the chemical process industries. Figure 19.7 is a schematic diagram of feedstocks and products for the refinery (costs and selling prices are given in parentheses). Table 19.3 lists the information pertaining to the expected yields of the two types of crude oils when processed by the refinery. Note that the product distribution from the refinery is quite different for the two crude oils. Table 19.3 also lists the limitations on the stablished markets for the various products in terms of the allowed maximum daily production. In addition, processing costs are given.

To set up the linear programming problem, formulate an objective function and constraints for the refinery operation. From Fig. 19.7, six variables are involved, namely, the flow rates of the two raw materials and the four products. Solve the LP using the Excel Solver.

Table 19.3 Data for the Refinery Feeds and Products

	Volume percent yield		Maximum allowable production(bbl/day)
	Crude #1	Crude #2	Maximum allowable production(bbl/day)
Gasoline	80	44	24,000
Kerosene	5	10	2000
Fuel oil	10	36	6000
Processing	0.50	1.00
cost ($/bbl)

Accepted Answer

Let the variables be

[latex]x_{1}[/latex] = bbl∕day of crude #1
[latex]x_{2}[/latex] = bbl∕day of crude #2
[latex]x_{3}[/latex] = bbl∕day of gasoline
[latex]x_{4}[/latex] = bbl∕day of kerosene
[latex]x_{5}[/latex] = bbl∕day of fuel oil
[latex]x_{6}[/latex] = bbl∕day of residual

The linear objective function [latex]f[/latex] (to be maximized) is the profit, the difference between income and costs:

f = income − raw material cost − processing cost

where the following items are expressed as dollars per day:

[latex]Income =72x_{3}+48x_{4} +42x_{5}+20x_{6}[/latex]
[latex]Raw material cost = 48x_{1} + 30x_{2}[/latex] (19-22)
[latex]Processing cost = 1x_{1} + 2x_{2}[/latex]

[latex] f = 72x_{3} + 48x_{4} + 42x_{5} + 20x_{6} − 49x_{1} − 32x_{2}[/latex] (19-23)

The yield data provide four linear equality constraints (material balances) relating [latex]x_{1} [/latex]through [latex]x_{6}[/latex]:

Gasoline: [latex]x_{3} = 0.80x_{1} + 0.44x_{2}[/latex] (19-24)

Kerosene: [latex]x_{4} = 0.05x_{1} + 0.10x_{2}[/latex] (19-25)

Fuel oil: [latex]x_{5} = 0.10x_{1} + 0.36x_{2} [/latex] (19-26)

Residual: [latex]x_{6} = 0.05x_{1} + 0.10x_{2}[/latex] (19-27)

Other constraints that exist or are implied in this problem are given in Table 19.3, which lists certain restrictions on the [latex]{x_{i}}[/latex] in terms of production limits. These can be formulated as inequality constraints:

Gasoline: [latex]x_{3} ≤ 24,000[/latex] (19-28)
Kerosene: [latex]x_{4} ≤ 2,000[/latex] (19-29)
Fuel oil: [latex]x_{5} ≤ 6,000[/latex] (19-30)

One other set of constraints, although not explicitly stated in the formulation of the problem, is composed of the nonnegativity restrictions, namely, [latex]{x_{i}}[/latex] ≥ 0. All process variables must be zero or positive, because it is meaningless to have negative production rates.

The formal statement of the linear programming problem is now complete, consisting of Eqs. (19-23 to 19-30). We can now proceed to solve the LP problem using the Excel Solver option. The problem statement can be introduced into the spreadsheet as illustrated in the Solver Parameter dialog box in Fig. 19.8. There are four equality constraints and three inequality constraints; the first three equality constraints are shown in the dialog box in Fig. 19.8. The objective function is in the target cell A10, and the six variable cells are in cells A4–F4.

In the refinery blending problem, the optimum x obtained by Excel occurs at the intersection of the gasoline and kerosene constraints. For these active constraints, the optimum is therefore

[latex]x_{1} = 26,207[/latex]
[latex]x_{2} = 6897[/latex]
[latex]x_{3} = 24,000[/latex] (gasoline constraint)
[latex]x_{3} = 24,000[/latex] (gasoline constraint)
[latex]x_{5} = 5103[/latex]
[latex]x_{6} = 2000[/latex]
[latex]f[/latex]= $573,516 ∕ day

Question 19.3: Consider a simple version of a refinery blending and product...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions