In this section, we discuss two additional formulation examples. These examples are more sophisticated than the product-mix problem. Click on the titles below to view these examples (which are in the pdf format).

**Example 1: ****The Production-Planning Problem**

**Example 2: ****The Investment Problem**

For additional formulation examples, browse Section 3.4 of the text.

We now briefly discuss how to use the LINDO software.

Suppose you wish to solve the product-mix problem. Launch the LINDO package. We will use XR and XE to denote the decision variables. In the current window, enter:

MAX 5 XR + 7 XE ST 3 XR + 4 XE < 650 2 XR + 3 XE < 500 END

Note that

- In LINDO, the minute you use a variable in your model, it exists. You don't have to do anything other than enter it in a formula.
- LINDO interprets the "<" symbol as less-than-or-equal-to rather than strictly-less-than.
- The objective function should not contain any constant. For example, MAX 3 X1 + 4 is not allowed.
- All variables must appear on the left-hand side of the constraints, while numerical values must appear on the right-hand side of the constraints.
- All variables are assumed to be nonnegative. Therefore, do not enter the nonnegativity constraints.

Now, click "Solve" on the menu bar and then select "Solve". Click on "No" in the "Do Range (Sensitivity) Analysis?" dialog box. The solution will be displayed in a separate "Reports Window". The output is:

LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 1137.500 VARIABLE VALUE REDUCED COST XR 0.000000 0.250000 XE 162.500000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 1.750000 3) 12.500000 0.000000 NO. ITERATIONS= 2

Thus, the optimal solution for the problem is: *x _{r} = 0* and

For more-detailed instructions on using LINDO, click "Help". Appendix 4.1, p. 169, in the text also provides a discussion.

As another example, the investment problem can be entered as:

MAX 2 X4 + S5 ST X1 + S1 = 100 S1- 0.5 X1 - X2 - S2 = 0 2 X1 + S2 - 0.5 X2 - X3 - S3 = 0 2 X2 + S3 - 0.5 X3 - X4 - S4 = 0 2 X3 + S4 - 0.5 X4 - S5 = 0 END

This yields the output:

LP OPTIMUM FOUND AT STEP 6 OBJECTIVE FUNCTION VALUE 1) 208.1301 VARIABLE VALUE REDUCED COST X4 104.065041 0.000000 S5 0.000000 0.138211 X1 27.642277 0.000000 S1 72.357727 0.000000 X2 58.536587 0.000000 S2 0.000000 0.520325 X3 26.016260 0.000000 S3 0.000000 0.130081 S4 0.000000 0.292683 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 2.081301 3) 0.000000 -2.081301 4) 0.000000 -1.560976 5) 0.000000 -1.430894 6) 0.000000 -1.138211 NO. ITERATIONS= 6

It is interesting to note that the optimal objective-function value 208.1301 is approximately 17% higher than 177.7778, which is the corresponding value for the naive strategy of investing 66.6667 dollars on Monday. Moreover, an inspection of the listed values for the decision variables shows that the optimal strategy prescribes an initial investment of $27.6423 on Monday; and that this initial investment paves the way for three "perfectly-coupled" investment cycles that begin on Tuesday, Wednesday, and Thursday, respectively. A little bit of reflection should convince you that this strategy is, in fact, quite intuitive.

The product-mix problem and the investment problem can also be solved using Excel with the Premium Solver add-in. (The Premium Solver can be installed from the course CD.) Section 3.6 (pp. 67-72) in the text explains how to use the Solver. For a quick start, click on the following titles to view/download the Excel setups for these two problems: The Product-Mix Problem, The Investment Problem.

**Assignment No. 1**

Chapter 3: 3.4-13, 3.4-15, 3.4-16, 3.4-18, and 3.6-4.

Provide a brief explanation for each of your formulations.

For part (c) of 3.4-13, use Excel's Solver add-in. For part (b) of 3.4-15, 3.4-16, and 3.4-18, use LINDO.