Moe is deciding how much Regular Duff beer and how much Duff...

Key Concepts

Feasible Region

The feasible region is the set of all possible solutions that satisfy all the constraints of the linear programming problem. It is typically a convex polygon (or polyhedron in higher dimensions) and identifies the space in which the optimal solution must lie.

Graphical (Geometric) Solution

The graphical solution method is a technique used to solve linear programming problems with two decision variables by plotting the constraints on a graph, identifying the feasible region, and then determining the optimal value of the objective function at the vertices (corner points) of that region.

Corner Point Principle

The corner point principle states that if an optimal solution to a linear programming problem exists, it will be found at one of the vertices of the feasible region. This concept forms the basis for many solution methods, including the graphical method for low-dimensional problems and the simplex method in higher dimensions.

Objective Function

The objective function in a linear programming problem represents the goal of the problem, typically expressed as a linear combination of the decision variables. In many practical problems, such as profit maximization or cost minimization, the objective function quantifies the measure one wishes to optimize.

Decision Variables

Decision variables represent the quantities that can be controlled in the optimization problem. They are used to model the decisions, such as the number of pints or units to purchase or produce, and are the unknowns that the linear program seeks to determine for optimal results.

Linear Programming

Linear programming is an optimization method for a problem in which a linear objective function is maximized or minimized subject to a set of linear inequality or equality constraints. It provides a systematic procedure to determine the best outcome in a mathematical model whose requirements are represented by linear relationships.

Constraints

Constraints are the linear inequalities or equations that represent the limitations or requirements of the problem. They model the available resources, restrictions, or relationships between the decision variables, ensuring that the solution is feasible within the defined limits.

Transcript

00:01 Okay, for this problem, we are trying to help a brewery figure out how many units of different kinds of beer to produce.

00:07 And if you read through all of it in section a, you'll see that this is a minimization problem.

00:14 We are trying to minimize production costs.

00:22 So what is the cost of producing this beer? well, two kinds of beer we can produce.

00:28 Regular and a light beer.

00:30 Regular beer costs $32 ,000 per unit.

00:34 So $32 ,000 for every unit of regular beer.

00:40 And i'm just going to mark that this is my regular y -sub 1.

00:44 It's always a good idea when you're using these variables in word problems.

00:49 Mark what they are.

00:50 So when you get to the very end, you remember what you let each of your variables be to make sure you're reading your answer correctly at the end.

00:57 Okay, so regular beer costs $32 ,000 per unit.

01:01 Light beer costs $50 ,000 per unit.

01:06 So those are our total production costs.

01:11 What are our constraints? well, i need to first look at how many units of each of these beers i need to do.

01:20 My regular beer, i need to produce at least 10 units.

01:25 That's what we're going through right now with our regular customers.

01:30 We're also going through 15 units of light beer monthly.

01:33 So i have to have at least 15 units of light beer.

01:37 And when i go to do more beer here, if you read through the, problem, i can do at least 20 additional units of beer.

01:45 So if i add these, y sub 1 plus y sub 2, 25 units being produced already, 10 of 1, 15 of another, i can do at least 20 more.

01:55 So y sub 1 plus y sub 2 has to be greater than or equal to 45, what i'm already producing plus that minimum 20 unit increase.

02:08 And i also want to make a certain amount of money.

02:10 Well, i make i make a hundred twenty thousand in revenue for my regular beer and i make 300 ,000 in revenue for my, for my light beer.

02:30 And altogether, i want to make at least $9 million.

02:37 Okay.

02:38 So those are my constraints.

02:40 This is a minimization problem.

02:43 I can write this as a maximization problem and solve it if i write the dual problem.

02:47 So let's make a matrix with the coefficients and the constants.

02:54 And let's start with our constraints first.

02:58 Y sub 1 is 1.

03:00 No y sub 2s.

03:02 That's 10.

03:03 Second one is 0, 1, and 15.

03:07 Third one, 1 ,1, and 45.

03:11 And third, and i'm going to have to write kind of small.

03:13 In fact, i'm going to abbreviate k for 1 ,000, 120 ,000, 300 ,000, 9 million.

03:23 I'll expand these back out again when we're using them, but for now it just makes my matrix a little bit easier to handle.

03:31 Okay.

03:32 And let's finish with our minimization, the objective function.

03:36 That is 32 ,000, 50 ,000.

03:40 And there is no constant.

03:43 Okay, so let's transpose this matrix.

03:46 That means my first row is now going to be my first column and so on.

03:50 So my first row, one, 10, that's now column 1.

03:54 Row 2, 0 -1 -15.

03:58 Row 3, 1145.

04:02 Row 4, 120 ,000, 300 ,000, and 9 million.

04:12 And the last row becomes the last column.

04:16 32 ,000, 50 ,000, and zero.

04:22 Okay, from here we can write our dual problem, the maximization problem.

04:29 So i want to maximize.

04:34 Now remember, minimization, we typically use w's and subscripted y's.

04:38 Maximization problems, we typically use z and subscripted x's.

04:42 It's just a reminder that these are different problems, though they're related, the variables are different.

04:48 So in order to find my new objective function, i look at that bottom.

04:53 So z equals 10x sub 1 plus 15 x sub 2 plus 45 x sub 3 plus 9 million x sub 4.

05:07 Okay.

05:09 What are our constraints? well, we're going to find those from our other rows.

05:14 X sub 1 plus x sub 3 plus 120 ,000 x sub 4 is going to be less than or equal to 32 ,000.

05:29 And our second one, we have x sub 2 plus x sub 3 plus 300 ,000 x sub 4 is less than or equal to 50 ,000.

05:42 Now, when we go to use our simplex method, we need to have some slack variables.

05:46 So i'm just going to make a quick change.

05:48 Instead of making these inequalities, i'm going to write them as our equalities with our slack variables added, s sub 1 and s sub 2.

05:59 Okay, let's move up.

06:00 We've got a little bit of room, and let's set this up.

06:04 I have x, 4 x's, x1, 2, 3, 4, 2 slack variables, and z.

06:14 And i'm trying to make this spread out with some room.

06:18 If there's not enough room, we'll just write small...

Please give Ace some feedback