specified nutritional profile. Each dimension of this profile gives rise to a separate con-
straint. An example of such a constraint states, in words, that the number of grams of
vitamins provided in the package must be greater than or equal to the number of grams
required by the specified profile. Laying out those words in the format of an inequality,
we can write
Grams of vitamins provided ≥ Grams of vitamins required
where we chose to place “grams required” on the RHS because it is represented by a
parameter of the model (16 grams, in this case). Converting the inequality to symbols,
we can then write
Vitamin content = 10S + 20R + 10F + 30P + 20W ≥ 16 (Vitamin floor)
Similar constraints must hold for mineral protein, and calorie content. The entire
model, stated in algebraic terms, reads as follows.
Minimize z = 4S + 5R + 3F + 7P + 6W
subject to
10S + 20R + 10F + 30P + 20W ≥ 16
5S + 7R + 4F + 9P + 2W ≥ 10
1S + 4R + 10F + 2P + 1W ≥ 15
500S
+ 450R + 160F + 300P + 500W ≥ 600
In this basic scenario, no other constraints arise, although we could imagine that there
could also be limited quantities of the ingredients available, expressed as LT con-
straints, or a weight requirement for the package, expressed as an EQ constraint.
A spreadsheet model for the basic scenario appears in Figure 2.10. Again, we see
the three modules: a highlighted row for decision variables, a highlighted single cell
for the objective function value, and a set of constraint relationships with highlighted
RHS’s. If we were to displ ay the formulas for this model, we would again see that the
only formula in the worksheet is the SUMPRODUCT formula.
Once we have persuaded ourselves that the model is valid, we proceed to the
Model tab in the task pane and enter the following information
Objective:
Variables:
Constraints:
G8 (minimize)
B5:F5
G11:G14 ≥ I11:I14
As in the allocation model, we move to the Engine tab, specify the linear solver, and
invoke the option for nonnegative variables.
After contemplating some hypotheses about the problem (e.g., will the solution
require all five ingredients?) we run Solver and find the result message in the solution
log. The optim al solution is reproduced in Figure 2.11. It calls for 0.48 lb of seeds,
0.33 lb of raisins, and 1.32 lb of flakes, with no nuts at all. Evidently, nuts are prohi-
bitively expensive, given the nature of the required nutritional profile and the other
40
Chapter 2 Linear Programming: Allocation, Covering, and Blending Models