A distillery produces custom-blended whiskey. A particular blend consists of rye

Question

A distillery produces custom-blended whiskey. A particular blend consists of rye and bourbon whiskies. The company has received an order for a minimum of 400 gallons of this custom-blended whiskey. The customer has specified that the order must contain at least 40% rye and not more than 250 gallons of bourbon. The customer has also specified that the blend should be mixed in the ratio of two parts rye to one part bourbon. The company can produce 500 gallons per week regardless of the blend, and it wants to complete this order in one week. The blend is sold for $5 per gallon. The brewing company’s cost per gallon is $2 for rye and $1 for bourbon. The company wants to determine the blend mix that will meet customers requirements and maximize profits. Formulate a LP model for this problem

Explanation / Answer

Hi Below i have explained a similar example solution in LP model to understand and get better with any problem solving technique with LP model,

Lets go with this example model in LP, A division of a plastics company manufactures three basic products: sporks, packets, and school packs. A spork is a plastic utensil which purports to be a combination spoon, fork, and knife. The packets consist of a spork, a napkin, and a straw wrapped in cellophane. The school packs are boxes of 100 packets with an additional 10 loose sporks included. Production of 1000 sporks requires 0.8 standard hours of molding machine capacity, 0.2 standard hours of supervisory time, and $2.50 in direct costs. Production of 1000 packets, including 1 spork, 1 napkin, and 1 straw, requires 1.5 standard hours of the packaging-area capacity, 0.5 standard hours of supervisory time, and $4.00 in direct costs. There is an unlimited supply of napkins and straws. Production of 1000 school packs requires 2.5 standard hours of packaging-area capacity, 0.5 standard hours of supervisory time, 10 sporks, 100 packets, and $8.00 in direct costs. Any of the three products may be sold in unlimited quantities at prices of $5.00, $15.00, and $300.00 per thousand, respectively. If there are 200 hours of production time in the coming month, what products, and how much of each, should be manufactured to yield the most profit?

The first decision one has to make in formulating a linear programming model is the selection of the proper variables to represent the problem under consideration. In this example there are at least two different sets of variables that can be chosen as decision variables. Let us represent them by x’s and y’s and define them as follows:

x1 = Total number of sporks produced in thousands,

x2 = Total number of packets produced in thousands,

x3 = Total number of school packs produced in thousands,

and

y1 = Total number of sporks sold as sporks in thousands,

y2 = Total number of packets sold as packets in thousands,

y3 = Total number of school packs sold as school packs in thousands.

We can determine the relationship between these two groups of variables. Since each packet needs one spork, and each school pack needs ten sporks, the total number of sporks sold as sporks is given by:

y1 = x1 x2 10x3 -> (1)

Similarly, for the total number of packets sold as packets we have:

y2 = x2 100x3 -> (2)

Finally, since all the school packs are sold as such, we have:

y3 = x3 -> (3)

From Eqs. (1), (2), and (3) it is easy to express the x’s in terms of the y’s, obtaining:

x1 = y1 + y2 + 110y3, -> (4)

x2 = y2 + 100y3, -> (5)

x3 = y3. -> (6)

As a matter of exercise, let us formulate the linear program corresponding to the present example in two forms: first, using the x’s as decision variables, and second, using the y’s as decision variables. The objective function is easily determined in terms of both y- and x-variables by using the information provided in the statement of the problem with respect to the selling prices and the direct costs of each of the units produced. The total profit is given by:

Total profit = 5y1 + 15y2 + 300y3 2.5x1 4x2 8x3. -> (7)

Equations (1), (2), and (3) allow us to express this total profit in terms of the x-variables alone. After performing the transformation, we get:

Total profit = 2.5x1 + 6x2 1258x3. -> (8)

Now we have to set up the restrictions imposed by the maximum availability of 200 hours of production time. Since the sporks are the only items requiring time in the injection-molding area, and they consume 0.8 standard hours of production per 1000 sporks, we have:

0.8x1 200 -> (9)

For packaging-area capacity we have:

1.5x2 + 2.5x3 200 -> (10)

while supervisory time requires that

0.2x1 + 0.5x2 + 0.5x3 200 -> (11)

In addition to these constraints, we have to make sure that the number of sporks, packets, and school packs sold as such (i.e., the y-variables) are nonnegative. Therefore, we have to add the following constraints:

x1 x2 10x3 0, -> (12)

x2 100x3 0, -> (13)

x3 0. -> (14)

Finally, we have the trivial nonnegativity conditions

x1 0, x2 0, x3 0. -> (15)

Note that, besides the nonnegativity of all the variables, which is a condition always implicit in the methods of solution in linear programming, this form of stating the problem has generated six constraints, (9) to (14). If we expressed the problem in terms of the y-variables, however, conditions (12) to (14) correspond merely to the nonnegativity of the y-variables, and these constraints are automatically guaranteed because the y’s are nonnegative and the x’s expressed by (4), (5), and (6), in terms of the y’s, are the sum of nonnegative variables, and therefore the x’s are always nonnegative.

By performing the proper changes of variables, it is easy to see that the linear-programming formulation in terms of the y-variables is given in tableau form as follows:

y1 y2 y3 Relation Limit

Modeling 0.8 0.8 88 <= 200

Packaging 1.5 152.5 <= 200

Supervisory 0.2 0.7 72.5 <= 200

Objective 2.5 8.5 -383 = z(max)

(Optimal Solution) 116.7 133.5 0 1425

Since the computation time required to solve a linear programming problem increases roughly with the cube of the number of rows of the problem, in this example the y’s constitute better decision variables than the x’s. The values of the x’s are easily determined from Eqs. (4), (5), and (6).

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