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Module 4 – Home

LINEAR PROGRAMMING

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Modular Learning Outcomes

Upon successful completion of this module, the student will be able to satisfy the following outcomes:

• Case
• Write the profit and loss factors pertaining to a production decision in algebraic form.
• Given the relevant profit and loss factors, write the profit function for a production decision.
• Write the constraints pertaining to a production decision, using algebraic inequalities.
• Given the profit function and constraints of a production decision, use an online application to solve the optimization problem.
• SLP
• Conclude a Delphi decision-making process.
• Summarize and present the results of a Delphi decision-making process.
• Discussion
• Review of the course.
• Discuss the limitations of linear programming when applied to actual allocation problems.

Module Overview

Life is about constraints. We could make up millions of examples. When a couple earning a total of \$5,000 per month sit down to do the household budget, they’re faced by one immediate constraint; the sum of all their monthly expenditures must be less than or equal to \$5,000. A roofer is constrained by the size of his crew, the number of hours they’ll work per day, and the weather forecast. Students are constrained by the number of hours available for study, the courses they’re taking, and their academic schedules; that is, which assignments are due, and when.

Most of the time, we don’t handle constraints rationally. We don’t usually sit down and make plans for maximizing our productivity (or profitability, or whatever) while staying within our constraints. That’s what this module is about; how to make those plans.

To get a taste of the problem, let’s consider an example (Staple, 2014a).

A calculator company produces a scientific calculator and a graphing calculator. The market demands at least 100 scientific and 80 graphing calculators per day. Because of various limitations, no more than 200 scientific and 170 graphing calculators can be produced per day. To satisfy a contract with UPS, at least 200 calculators must be shipped each day.

Each scientific calculator sells at a \$2 loss, but the company makes them anyway, to maintain market position. Each graphing calculator sells at a \$5 profit. How many of each type should be made daily, to maximize profits?

How on Earth do you even begin to solve a problem like that? The short answer is, you don’t. A computer solves it. The application uses a procedure called linear programming. But how, you may ask, does linear programming work? The answer: It works just great!

NOTE: SECTIONS IN ITALICS ARE BACKGROUND. YOU CAN SKIP THEM AT FIRST READING (BUT SHOULD STUDY THEM LATER).

Even today, most textbooks teach various ways to solve linear programming problems using pencil and paper. That’s fine, from a pedagogical point of view, but let’s be honest – if faced with a real problem, involving real money, with real jobs on the line, would you attempt to find a solution using pencil and paper? Of course not. You’d either use a computer, or tell HR to hire a consultant, who would then use a computer.

(End of digression.)

Not all problems have solutions, including linear programming problems, and the computer will definitely tell you if the problem you’re trying to solve doesn’t have one. If it doesn’t, then there are two possibilities: either the problem itself is bogus, or you made a mistake when you entered the problem into the computer. The problems in this module aren’t bogus – they have solutions. And we’re going to spend a lot of time explaining and practicing the proper way of entering problems.

What’s Going On.

It’s worthwhile to stop and explain, in rough qualitative terms, exactly what linear programming does. This is background. No need to panic, since you won’t have to do anything like this in the assignments. In fact, you may want to skip ahead to the next session, entitled Setting Up Constraints. I strongly recommend, however, that you return to this section later. Sooner or later, you’ll need to know what’s going on “inside the box.”

Let’s take a closer look at the example. Begin with the constraints. I’ve numbered them in the “story” below, and then listed them separately.

A calculator company produces a scientific calculator and a graphing calculator. The market demands at least 100 scientific (1) and 80 graphing (2) calculators per day. Because of various limitations, no more than 200 scientific (3) and 170 graphing (4) calculators can be produced per day. To satisfy a contract with UPS, at least 200 calculators (5) must be shipped each day.

Just to make things simpler, let’s define two abbreviations: “SCI” = number of scientific calculators manufactured and shipped per day, “GPH” = number of graphing calculators manufactured and shipped per day. (We can use any abbreviations we like. These just happen to be easy to remember.) Here are the constraints.

1. SCI must be at least 100.
2. GPH must be at least 80.
3. SCI cannot exceed 200.
4. GPH cannot exceed 170.
5. SCI plus GPH must be at least 200.
 You may remember (if not, it’s OK) that relationships between two numbers can be shown as a plot on an X-Y plane. Here, we’re using SCI and GPH instead of X and Y. Every point on the plot represents one value of SCI and one value of GPH. For example, the black dot represents 50 scientific and 75 graphing calculators. (To review these ideas, see Staple, 2014b.) The heavy vertical line shows the equality SCI = 100. The shaded area to the right, which includes the heavy line, consists of all combinations of SCI and GPH for which SCI is at least 100; or to put it another way, greater than or equal to 100. This is abbreviated SCI >= 100.

Notice that the shaded area is bounded on the bottom by the horizontal axis, which represents GPH=0. This ought to make perfect sense, because you can’t make a negative quantity of anything. The minimum number of calculators of any type that you can manufacture is zero.

Here are the areas of the plot that correspond to the other constraints.

 Each of the plots above represents a separate constraint. When we’re planning calculator production, however, we can’t consider the constraints one at a time; we have to consider all of them at once. To visualize them all at once, we plot them together, as shown on the right. The numbers indicate the constraints, and the arrows show where the permitted points are located with respect to each equality line. If this isn’t clear, please compare (1) ð with the first plot above. The region where all the permitted points overlap is called the Feasibility Region. For future use, we’ve labeled the corners of the feasibility region with letters; A, B, C, D and E.

The points in the feasibility region, which include the points on the boundary lines, represent all the possible combinations of scientific and graphing calculators that the company could possibly manufacture, given these constraints. Examples: The green dot, representing 150 scientific calculators and 100 graphing calculators, is a permitted combination; the red dot, representing 50 calculators of each type, is not.

We still haven’t found the optimum combination of calculators, and we’ll return to that problem in a moment. But first let’s take a detour, and consider a combination of constraints that looks like a linear programming problem, but isn’t. It is, to use an earlier, non-technical term, a “bogus problem.”

 It’s not hard to construct a bogus problem. We’ll simply take the problem we’ve been looking at, and remove one of the constraints. Let’s remove constraint (4), which specifies that the number of graphing calculators cannot exceed 170.   With that constraint removed, we no longer have a problem that can be solved using linear programming. That’s because there are an infinite number of points in the feasibility region. The boundaries on the left and right are the SCI constraints: between 100 and 200, inclusive. There are bottom boundaries to the region, but no top boundary. The company could, in theory, make an infinite number of graphical calculators! This is bogus. And the computer app would tell us it’s bogus.

Having looked at something that isn’t a liner programming problem, let’s return to the solution of one that is.

 On the right, we’ve expanded the labels on the corners of the feasibility region to include their coordinates; these are the number of scientific and graphics calculators associated with each. For point A, that would be 100 scientific and 170 graphics calculators. Let’s list them in a table.                    Number of Calculators   Point    Scientific   Graphic                A         100        170                       B         200        170                            C         200         80                              D         80         120                           E         100        100     The coordinates can be determined by inspection (that is, simply looking), or from the equations of the constraints, by using various algebraic techniques that we won’t discuss here.

The points A, B, C, D, and E are called the extrema (sing. extremum). It’s a useful fact that the optimum mix of scientific and graphics calculators is associated with one of the extrema; that is, one of the coordinate pairs listed above will give the company their maximum profit, given these constraints.

To determine which coordinate pair that would be, let’s go all the way back to “the story problem.”

Each scientific calculator sells at a \$2 loss, but the company makes them anyway, to maintain market position. Each graphing calculator sells at a \$5 profit.

From this information, we can write the so-called profit equation. Let p = profit, SCI = the number of scientific calculators, GPH = the number of graphing calculators by GPH. Then the equation is,

p = -2(SCI) + 5(GPH)

Example: Imagine a really terrible day, during which the company sells exactly one calculator of each type. They make \$5 on the graphing calculator and lose \$2 on the scientific calculator, for a total profit of \$3. We can do that much in our heads. But let’s use the equation.

p = -2(SCI) + 5(GPH)

Since SCI = 1 and GPH = 1,

p= (-2)(1) + (5)(1) = -2 + 5 = 3.

(This is always a good way to make sure we’re using a correct equation. Try it out, using very simple variables. If it works as expected, then it’s probably OK.)

We now calculate the profit for each of the extrema.

 Point SCI GPH p = -2(SCI) + 5(GPH) A 100 170 (-2)(100) + (5)(170) = -200 + 850 = 650 B 200 170 (-2)(200) + (5)(170) = -400 + 850 = 450 C 200 80 (-2)(200) + (5)(80) = -400 + 400 = 0 D 80 120 (-2)(80) + (5)(120) = -160 + 600 = 440 E 100 100 (-2)(100) + (5)(100) = -200 + 500 = 300

Point A yields the greatest profit, \$650. The optimum mix of scientific calculators and graphing calculators that the company should product, per day, is 100 and 170 respectively.

To summarize, this is how linear programming works:

1. Specify the constraints in the form of valid equations.
2. Define a feasibility region.
3. Find the extrema; which are, the corners of the feasibility region.
4. Test the coordinates of each extremum against the profit equation.
5. The coordinates of the extremum that yields the maximum profit are the solution of the problem.

We’ve worked through all five steps. The good news is, the only important step is the first step. If we can do that, then a computer can do the rest.

(Digression.)

One final, non-mandatory bit of information, and we’ll move on. The procedure works with any number of variables, corresponding to any number of dimensions. The example we’ve been beating to death has two variables, SCI and GPH. The feasibility region is an area on a two-dimensional plane. If we had three variables, the feasibility region would be a volume in a three-dimensional space.

 For example: Suppose that instead of manufacturing only two types of calculators, the company made three; scientific, graphing, and accounting. Further suppose there was a set of constraints, similar to those above, but involving three variables: SCI, GPH, and ACC. The feasibility region, if one existed, would be the points on and inside a three-dimensional object. For reasons relating to the linear nature of linear programming, which we didn’t go into, the object would be a prism, bounded by flat surfaces. The corners where the surfaces came together would be the extrema, and the optimum solution of the profit equation would consist of the coordinates of one of those extrema.

It’s not uncommon to encounter a linear programming problem with even more variables; for example, we could imagine having to optimize the production of eight different types of calculators. In that case, the feasibility region would be a prism in eight-dimensional space. Such a region is, of course, impossible to draw, or even to imagine; but the algorithm doesn’t care. If a solution exists, the computer will find it.

(End of digression.)

If you’re feeling a bit panicky by now, take a deep breath and get over it. Everything you’ve just read in this section is background. The only thing you’ll need to know is what follows: that is, how to set up a linear programming problem in a form the computer can understand. We’re moving on to that right now.

Setting Up Constraints

If there’s one thing sixth graders detest, it’s “word problems,” or “story problems.” Adding, subtracting, multiplying and dividing numbers is one thing. Reading about a real-world situation, and then deciding which numbers are involved and what to do with them, is something else entirely. The sixth graders get no sympathy from their teachers. The teachers know that business, science, engineering and everyday life never hand us pages of numbers to crunch. They hand us situations. We have to think though the situations, decide which numbers are relevant, and perform the necessary calculations.

Let’s take it from the top. Here, once again, is the example we’ve been working to death:

A calculator company produces a scientific calculator and a graphing calculator. The market demands at least 100 scientific and 80 graphing calculators per day. Because of various limitations, no more than 200 scientific and 170 graphing calculators can be produced per day. To satisfy a contract with UPS, at least 200 calculators must be shipped each day.

Each scientific calculator sells at a \$2 loss, but the company makes them anyway, to maintain market position. Each graphing calculator sells at a \$5 profit. How many of each type should be made daily, to maximize profits?

The first step is to identify the variables; these are the things for which we need to find numbers. Don’t be confused by the fact that there are lots of numbers in the problem, such as the minimum number of each type needed daily; these aren’t variables, they’re constants. They’re given; we don’t need to find them. The numbers we need to find are the following.

• Scientific calculators produced, per day
• Graphing calculators produced, per day
• Daily profit (to be maximized)

Because we’re going to be writing equations, the next step is to come up with good labels, or abbreviations, for the variables. Theory doesn’t tell us what sort of labels to use, and almost any label would work. Some computer programs even permit variable labels with several words; the only requirement is they be enclosed with quotation remarks; as an example, “Scientific calculators produced, per day.” We won’t go that route, for two reasons. First: it takes a lot of space to type out such long labels, and it produces some confusing clutter. Second: a simple error, such as misspelling one of the labels or forgetting to close a set of quotation marks, would cause an error that could be difficult to find. So we’ll use labels that are both easy to remember, and short.

Returning to the example: because it’s the same for all the variables, we can take the time dimension “daily” or “per day,” as a given, and drop it.  We can also take “produced” as a given; it’s a calculator company, so it’s producing calculators, and it’s also trying to “produce” a profit. Because calculators are all we’re interested in at the moment, we can also drop the word “calculators” as being redundant. So the variables boil down to

• scientific
• graphing
• profit (maximized)

Some programs don’t permit any flexibility with respect the variable that’s being either maximized or minimized. It has to be represented by one letter. In this case, we’ll let “Profit” be simply “p,” and the final list becomes

• scientific
• graphing
• p

In many textbooks and online presentations (Staple, 2014b), the authors refer the variables to the traditional Cartesian coordinate axes; for example, “scientific” = x, “graphing” = y. I see no point in doing that. If you accidentally swap X and Y when writing your equations, then the program won’t work; or even worse, it may work, but produce incorrect results.

Let’s stop and summarize. The first steps of solving a linear programming problem are:

2. Identify the variables.
3. Define simple labels that unambiguously define the variables.

Let’s make steps 2 and 3 more systematic by putting the full variables and corresponding labels into a table. This is overkill for such a simple problem, but it may be useful for more complicated ones; further, it forces you to concentrate on the narrative, and be sure you’re not overlooking anything.

 VARIABLE LABEL Scientific calculators produced, per day scientific Graphing calculators produced, per day graphic Daily profit (to be maximized) p

Once we’ve defined the variables, there’s a lot more to do. We still have to write the equations that represent the constraints, and enter them into a computer program. But before we go on to do that, let’s practice defining variable labels.

Example 2 (After Staple, 2014b)

You need to buy some filing cabinets. The model made by Sauder costs \$100 per unit, needs 6 square feet of floor space, and holds 8 cubic feet of files. The unit by Steelcase costs \$200 per unit, needs 8 square feet of space, and holds 12 cubic feet. You have been given \$1400 for the purchase, but don’t have to spend all of it. Your office has enough floor space for 72 square feet of cabinets. How many of each model should you buy, to maximize the storage volume?

 VARIABLE LABEL Number of Sauder cabinets required sauder Number of Steelcase cabinets required steelcase Storage volume (to be maximized) v

Example 3 (after Staple, 2014b)

At a certain refinery, the refining process requires the production of at least two gallons of gasoline for each gallon of fuel oil. With winter coming, at least three million gallons of fuel oil will be required per day. On the other hand, winter sees a decrease in the requirement for gasoline; no more than 6.4 million will be required, per day. If gasoline sells for \$3.90 per gallon and fuel oil sells for \$2.50 per gallon, how many gallons of each should be produced, per day, to maximize revenue?

 VARIABLE LABEL Millions of gallons of gasoline produced, per day gas Millions of gallons of fuel oil produced, per day oil Revenue per day, dollars (to be maximized) r

Example 4 (after Staple, 2015)

A lab rabbit need a daily diet containing at least 24 grams (g) of fat, 36 g of carbs, and 4 g of protein. It should not be fed more than 5 ounces (oz) of food daily.  The lab tech decides to blend two commercially available feeds, Bunny Chow and Hop-To-It. An oz of Bunny Chow contains 8 g of fat, 12 g of carbs, and 2 g of protein, and costs \$0.20. An oz of Hop-To-It contains 12 g of fat, 12 g of carbs, and 1 g of protein, and it costs \$0.30. What is the optimum blend of the two feeds; that is, the blend that meets a rabbit’s requirements, at minimum cost?

 VARIABLE LABEL Oz of Bunny Chow in the blend, per rabbit per day bunny Oz of Hop-To-It in the blend, per rabbit per day hop Feed cost per rabbit per day (to be minimized) c

Once we have variables, the next task is to express their relationships in a form a computer can understand.

Writing Equations for Inequalities

The “=” sign, meaning “is equal to,” is something we learn how to use in the second grade (or thereabouts). The entities on either side of the “=” sign can be numbers, in which case we have an identity. Probably the most familiar identity, again familiar from early childhood, is 2+2=4.

Another use of the “=” sign is in equations, where one or more unknowns are indicated by letters. Here are some examples.

• X = 2+2 (Solution: X=4)
• F = (9/5)C + 32 (Temperature conversion: F = degrees Fahrenheit, C = degrees Celsius)
• R = P(1+T) (R = retail price, P = list price, T = sales tax)
• USD = R(EUR) (Currency conversion: USD = US dollars, R = exchange rate, EUR = Eurodollars)

We’re interested in a particular type of equation called the constraint, in which a variable is specified to be either one number, or within a range of numbers. Here are some examples.

• JAN = 31 (in which “JAN” is defined as, “The number of days in January.”)
• CTS = 100 (“CTS” = “The number of cents in a dollar.”)

These specify the value of a variable as one number. To specify a range of numbers, we need signs other than “=.” Here they are, along with their meanings. (For a complete treatment of this topic, please see Khan,2015.)

• < “Is less than”
• <= “Is less than or equal to”
• > “Is greater than”
• >= “Is greater than or equal to”

Here are some trivial numerical examples.

• 2 < 3
• 3 <= 3 (It’s not less than, but it’s definitely equal!)
• 4 > 3
• 4 >= 3 (It’s not equal, but it’s definitely greater than!)

Here are some less trivial examples.

• Vote >= 18 (where “Vote” = legal voting age)
• Drink >= 21 (where “Drink” = legal drinking age)
• Feb <= 29 (where “Feb” = number of days in the month of February)
• Prez <= 8 (where “Prez” = number of whole years a president has been in office)

Now let’s combine all of the above, and write out constraints that can be used in a linear programming application. It’s a four-step process;

• Identify the variables
• Create variable labels
• Translate the constraints into “less/equal/greater” language
• Write the constraints in algebraic form.

Here are some examples.

1. Joe’s Sports Bar is hiring a Chief of Security (aka bouncer). Joe will only interview karate black belts* 21 or older, at least 6 feet four inches tall, and weighing at least 220 pounds.
 Variables Labels “Less/equal/greater” Algebraic form Age in years Age Age 21 or older (greater than or equal to 21) Age >=21 Height in feet and inches Height Height at least 6’ 4” (greater than or equal to 76”) Height >= 76 Weight in pounds Weight Weight at least 220 (greater than or equal to 220) Weight >= 220

*Note: Karate qualification isn’t a variable, because it doesn’t vary. No black belt, no interview!

1. Joe’s Sports Bar is hiring table servers. Only the following candidates will be interviewed: women* between the ages of 21 and 35 years of age inclusive, not more than 5 feet 8 inches tall, weighing not more than 140 pounds. (Joe may be in trouble with the law concerning age and sex discrimination, but that’s a problem for another course.)
 Variables Labels “Less/equal/greater” Algebraic form Age in years Age Age 21 or older (greater than or equal to 21) Age 35 or younger (less than or equal to 35) Age >=21 Age <=35 Height in feet and inches Height Height less than or equal to 68 (inches) Height <= 68 Weight in pounds Weight Weight less than or equal to 140 Weight <= 140

*Note: Sex isn’t a variable, because men won’t be interviewed.

1. Peggy Potter makes coffee cups and vases. She’s able to make a maximum of 100 pieces per day.
 Variables Labels “Less/equal/greater” Algebraic form Number of cups per day Cups (no constraint) Vases per day Vases (no constraint) Cups plus vases is not more than 100 (less than or equal to 100) Cups + Vases <= 100
1. Peggy Potter makes coffee cups and vases. She gets a special rate from UPS if she ships at least 50 pieces per day, so she had adopted that as a business constraint.
 Variables Labels “Less/equal/greater” Algebraic form Number of cups per day Cups (no constraint) Vases per day Vases (no constraint) Cups plus vases is at least 50 (greater than or equal to 50) Cups + Vases >= 50
1. Peggy Potter makes coffee cups and vases. Because of the clay needed for each item, and the expected demand for each, Peggy decides she should make, at most, two cups for each vase.
 Variables Labels “Less/equal/greater” Algebraic form Number of cups per day Cups (no constraint) Vases per day Vases (no constraint) Number of cups not more than twice the number of vases (cups less than or equal to two times vases) Cups <= 2(vases)
1. Eye Full Optics makes binoculars and telescopes. Both instruments ship with the same eyepieces; each telescope needs one eyepiece, each pair of binoculars needs two. EFO’s supplier can send them no more than 300 eyepieces per day.
 Variables Labels “Less/equal/greater” Algebraic form Number of binoculars per day binocs Each pair requires two eyepieces (300 eyepieces max) Number of scopes per day scopes Each pair requires one eyepiece (300 eyepieces max) Number of eyepieces required for both binocs and scopes is not more than 300 (less than or equal to 300) Scopes + 2(binocs) <= 300
1. A calculator company produces a scientific calculator and a graphing calculator. The market demands at least 100 scientific and 80 graphing calculators per day. Because of supply limitations, no more than 200 scientific and 170 graphing calculators can be produced per day. A shipping contract requires at least 200 calculators be shipped per day.
 Variables Labels “Less/equal/greater” Algebraic form Number of scientific calculators produced per day SCI (Reminder: A label can be almost anything) At least 100 (greater than or equal to 100) No more than 200 (less than or equal to 200) SCI >= 100 SCI <= 200 Number of graphing calculators produced per day GPH At least 80 (greater than or equal to 80) No more than 170 (less than or equal to 170) GPH >= 80 GPH <= 170 Total calculators produced (SCI plus GPH) is at least 200 (greater than or equal to 200) SCI + GPH >= 200

The Profit Function

There’s one last thing we need to discuss; the profit function. The ultimate goal of linear programming is to maximize profit, by finding the particular values of the variables that give us the largest value of profit. (The actual entity may be something other than profit, and the goal may be to minimize rather than maximize. For example, we may be interesting in minimizing workspace, maximizing production volume, or minimizing waste. The same principles apply.)

Profit is a function of the variables, given the constraints. Suppose Peggy Potter earns \$5 for every coffee cup she makes. Then her profit (p) for the day, in dollars, is simply the number of cups she makes, multiplied by 5, which in algebraic form is p = 5(cups). But what’s the constraint? Well, in this case, there isn’t one. Peggy could, in theory, make a million cups in a day, and earn \$5M. But that’s not realistic. A realistic constraint would be, not more than 50 cups per day. Then Peggy’s profit is p = 5(cups), subject to cups <=50. If she puts in a long day at the wheel and makes 50 cups, she makes \$250. If she takes the day off and doesn’t make anything at all, then she earns nothing.

Suppose Peggy makes both cups and vases, earning \$5 for each cup and \$7 for each vase. Then her profit function would be

P = 5(cups) + 7(vases)

…where “p” is the profit for the day’s work, and the variables “cups” and “vases” simply refer to the number of each, produced that day.

Going back to the calculator company: If the company loses \$2 on each scientific calculator and earns \$5 on each graphing calculator, then their daily profit is

p = -2(scientific) + 5(graphing)

…where, as before “p” is the profit for the day’s work, and the variables “scientific” and “graphing” refer to the number of scientific and graphing calculators produced each day. (If \$5 doesn’t seem like much, remember that’s the profit; which is,the selling price minus the costs of production, shipping, advertising, and other expenses.)

Putting It All Together

We’ve come a long way. We’ve learned how to analyze a situation, define variables, set up constraints, and write the profit function. What we’ve skipped over (or rather relegated to an optional section) is the math that’s required to make use of all that information. We’ve skipped over it because a computer app is going to do it for us.

There are lots of linear programming apps on the Web. We’ll use Stefan Waner’s (2010).

(Digression.)

The example shown above illustrates something important. An app, such as Waner’s, allows us to evaluate a profit function at the extrema of a feasibility region that cannot be sketched, or even visualized. Let’s expand on that.

Up until now, we’ve been limited to two variables, x and y (cups and vases, etc.) The example above has four variables, labeled x, y, z and w. Just as x and y are perpendicular in two-dimensional space (the X-Y plane), the variables x, y, z and w are mutually perpendicular in four-dimensional space. The feasibility region is a volume in that four-dimensional space, with three-dimensional faces and corners defined by the intersections of those faces. Obviously, we can’t visualize them — but mathematical objects like that exist anyway, even if we can’t “see” them in the usual sense. And computer apps have no trouble working with them. You’ll find some three- and higher-dimensional problems in the Case exercises. You’ll solve them using an app.

(End of digression.)

Now, let’s finally do the entire scientific vs. graphing calculator problem.

A calculator company produces a scientific calculator and a graphing calculator. The market demands at least 100 scientific and 80 graphing calculators per day. Because of various limitations, no more than 200 scientific and 170 graphing calculators can be produced per day. To satisfy a contract with UPS, at least 200 calculators must be shipped each day.

Each scientific calculator sells at a \$2 loss, but the company makes them anyway, to maintain market position. Each graphing calculator sells at a \$5 profit. How many of each type should be made daily, to maximize profits?

As we’ve seen, the constraints are

• “at least 100 scientific and 80 graphing calculators per day”
• scientific >=100
• graphing >= 80
• “no more than 200 scientific and 170 graphing calculators produced per day”
• scientific <=200
• graphing >= 170
• “at least 200 calculators must be shipped each day”
• Scientific + graphing >= 200

The profit function is

• “scientific calculator sells at a \$2 loss, ….graphing calculator sells at a \$5 profit”
• p = -2(scientific) + 5(graphing)

The profit function goes on the top line. The format is prescribed; the line must read, “Maximize (the profit function) subject to.” The constraints go next, with each constraint on its own line.

Let’s check and make sure those values of the variables do, in fact, give that optimal solution.

p = -2(scientific) + 5(graphing)

= -2(100) + 5(170)

= -200 + 850

= 650.

We’re not expecting the computer to make a mistake, but it’s nice to check, if only to remind ourselves what the solution means.

So ends our discussion of linear programming. Be sure you understand it before moving on the problems; if you don’t, then you’re setting yourself up for major frustration. In addition, be sure you understand the information here, in this Module, before surfing the Web and looking for more. There’s a lot out there, but a lot of it confusing. (Why confusing? Because it includes lots of technical details that we don’t think are particularly important. You don’t need a degree in mechanical engineering to drive a car, and you don’t need a degree in math to do linear programming.)

What’s the takeaway? That is, what do we want you to remember, ten years from now? Well, we want you to remember that there’s a procedure called linear programming, and it’s highly effective for solving certain types of problems. When you encounter a problem of that type, then you’ll either go online and brush up on linear programming, or (more likely) tell HR to go out and hire a stats consultant. Either way, you’ll be a step ahead of your technically illiterate competition.

Module 4 – SLP

LINEAR PROGRAMMING

Homework help_ http://customwritings-us.com/orders.php

Complete the wrapup of a three-round Delphi decision-making exercise, following the detailed example cited in the Home Page discussion(SEE UPLOADED WORD FILE).  As before, you may copy and / or adapt verbiage from the example without citing it.

SLP Assignment Expectations

The SLP writeup should consist of:

• The Letters to the Participants, which include
• Thanks for their participation
• A summary of their third-round responses
• A short narrative discussing the evolution of the decision-making process, how opinions shifted, what relevant factors the group identified, and what consensus (if any) the group arrived at.
• Follow the instructions in the BSBA Writing Style Guide (July 2014 edition), available online at
https://mytlc.trident.edu/files/Writing-Guide_Trident_2014.pdf.
• There are no guidelines concerning length.  Write what you need to write – neither more, nor less.
• In the SLP ONLY, references and citations are NOT required.  However:  If you state a fact, express an opinion, or use a turn of phrase that isn’t your own, then you should credit the source, just like you would in everyday conversation.  (Example:  “In the words of Monty Python, ‘And now for something completely different.’ “)

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The linear equations corresponding to the constraints are:

1. 2y = 3x
2. 2x + 3y = 15
3. 3y = x

Here’s the plot, with the lines, extrema, and the region labeled.  It was created with Relplot, and the labels were added using the Snagit graphics editor.  Using Relplot, it’s possible to create the sketch without knowing the coordinates of the extrema.  That’s because the app takes the line equations as input.

Here’s another version.  It’s less elaborate, but still perfectly acceptable.  If you want to upload a hand sketch, however, you’ll have to do the calculations first, so you’ll know where to put the extrema.

Here’s how to find the coordinates of the extrema:

A:  The only values of x and y that satisfy the equation 1 (that is, 2y=3x) is (0,0) .  Ditto for equation 3.  So the coordinates for A are

A(0,0)

B: This point is the simultaneous solution of equations 1 and 2;  that is, of

2y=3x

2x + 3y = 15.

We’ll use the Webmath solver (Discovery, 2014) to find the values of x and y that satisfy both equations.  There are many such apps on the Web;  look for them using Google, or your favorite search engine.

Here’s what the setup looks like:

Proceed in the same way to find the coordinates of point C, which is simultaneous solution of equations 2 and 3;  that is,

2x + 3y = 15

3y=x

A(0,0)

B(2,31, 3.46)

C(5, 1.67)

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