It is extremely likely that you currently possess at least one device with a working Internet connection, which means that you have access to numerous online libraries and catalogs. Unfortunately, not all of them are well-organized and sometimes it is pretty hard to find the ebook you need there.
This website was designed to provide the best user experience and help you download by Wayne L. Here you can download by Wayne L. Winston Student Solutions Manual For Winston's Operations Research: Applications And Algorithms, 4th without having to wait or complete any advertising offers to gain access to the file you need.
You may say that by Wayne L. Well, we do our best to improve your experience with our service, and we make sure that you can download all files in various document formats.
There is no need for you to waste your time and Internet traffic on online file converters: we have already done that for you. We work on a daily basis to expand our database and make sure that we offer our users as many titles including some pretty rare handbooks and manuals as possible, which is also the reason why you are highly unlikely to find broken links on our website.
What is ebook? An eBook is an electronic version of a traditional print book that can be read by using a personal computer or by using an eBook reader.
An eBook reader can be a software application for use on a computer, such as Microsoft's free Reader application. These books are used by students of top universities, institutes and colleges. The goal of this Engineering Mechanics course is to expose students to problems in mechanics as applied to plausibly real-world scenarios. Problems of particular types are explored in detail in the hopes that students will gain an inductive understanding of the underlying principles at work; students should then be able to recognize problems of this sort in real-world situations and respond accordingly.
Problems of particular types are explored in detail in the hopes that students will gain an inductiv. Read more. June 11, John Finnemore and Joseph B. This book is well known and well respected in the civil engineering market and has a following among civil engineers.
Finally, perform a Type 2 ERO by replacing row 1 of 7. Translating 7. This explains why EROs do not change the set of solutions to a system of equations. Because 9 was obtained from 8 by a sequence of EROs, we know that 8 and 9 are equivalent linear systems. We now show how we can use EROs to transform a relatively complicated system such as 8 into a relatively simple system like 9. This is the essence of the Gauss—Jordan method.
We now use the Gauss—Jordan method to solve 8. We begin by using a Type 1 ERO to change the element of 8 in the first row and first column into a 1. Then we add multiples of row 1 to row 2 and then to row 3 these are Type 2 EROs.
The purpose of these Type 2 EROs is to put zeros in the rest of the first column. The following sequence of EROs will accomplish these goals. Then we use the resulting row 2 to perform the Type 2 EROs that are needed to put zeros in the rest of column 2.
Steps 4—6 accomplish these goals. Then we use Type 2 EROs to put zeros in the rest of column 3. Steps 7—9 accomplish these goals. The reader might be wondering why we defined Type 3 EROs interchanging of rows.
The following two examples illustrate how the Gauss—Jordan method can be used to recognize these cases. Thus, 12 has no solution. Because 12 was obtained from 11 by use of EROs, 11 also has no solution. Then Similarly, Of course, Thus, 14 has an infinite number of solutions one for each number k.
Because 14 was obtained from 13 via EROs, 13 also has an infinite number of solutions. A more formal characterization of linear systems that have an infinite number of solutions will be given after the following summary of the Gauss—Jordan method. At any stage, define a current row, current column, and current entry the entry in the current row and column. Begin with row 1 as the current row, column 1 as the current column, and a11 as the current entry.
Go to step 3. Then go to step 3. When finished, obtain the new current row, column, and entry. If this is impossible, then stop.
Otherwise, repeat step 3. If it is impossible, then stop. Suppose that Case 1 does not apply and NBV, the set of nonbasic variables, is empty. Suppose that Case 1 does not apply and NBV is nonempty. To obtain these, first assign each nonbasic variable an arbitrary value. Then solve for the value of each basic variable in terms of the nonbasic variables. From Because Our discussion of the Gauss—Jordan method is summarized in Figure 6.
We have devoted so much time to the Gauss—Jordan method because, in our study of linear programming, examples of Case 3 linear systems with an infinite number of solutions will occur repeatedly. Because the end result of the Gauss—Jordan method must always be one of Cases 1—3, we have shown that any linear system will have no solution, a unique solution, or an infinite number of solutions.
Indicate the solutions if any exist. These concepts will be useful in our study of matrix inverses. Before defining a linearly independent set of vectors, we need to define a linear combination of a set of vectors. The foregoing definition may also be applied to a set of column vectors. To determine whether V is a linearly independent set of vectors, we try to find a linear combination of the vectors in V that adds up to 0.
We may now define linearly independent and linearly dependent sets of vectors. Thus, there is a nontrivial linear combination of vectors in V that adds up to 0. The only linear combination of vectors in V that yields 0 is the trivial linear combination.
Therefore, V is a linearly independent set of vectors. Thus, V is a linearly dependent set of vectors. Intuitively, what does it mean for a set of vectors to be linearly dependent? To understand the concept of linear dependence, observe that a set of vectors V is linearly dependent as 2. For example, in two dimensions it can be shown that two vectors are linearly dependent if and only if they lie on the same line see Figure 7.
The Rank of a Matrix The Gauss—Jordan method can be used to determine whether a set of vectors is linearly independent or linearly dependent. Before describing how this is done, we define the concept of the rank of a matrix. This means that rank A cannot be 2. Thus, rank A must equal 1. From Example 9, we know that R is a linearly independent set of vectors. To find the rank of a given matrix A, simply apply the Gauss—Jordan method to the matrix A. It can be shown that performing a sequence of EROs on a matrix does not change the rank of the matrix.
Form the matrix A whose ith row is vi. A will have m rows. This shows that V is a linearly dependent set of vectors. This equation also shows that V is a linearly dependent set of vectors. Provide an argument showing that the collection must be linearly dependent.
We begin with some preliminary definitions. Thus, just as the number 1 serves as the unit element for multiplication of real numbers, Im serves as the unit element for multiplication of matrices. This motivates the following definition of the inverse of a matrix. Observe that See Figure 8 and file Minverse. Suppose we want to invert the matrix Minverse. What properties must be possessed by the columns of an orthogonal matrix? Find the inverse of the matrix AB.
Knowing how to compute the determinant of a square matrix will be useful in our study of nonlinear programming. We close our discussion of determinants by noting that they can be used to invert square matrices and to solve linear equation systems. Because we already have learned to use the Gauss—Jordan method to invert matrices and to solve linear equation systems, we will not discuss these uses of determinants.
This result is true for any upper triangular matrix. For the matrix A, we let aij represent the element of A in row i and column j. A matrix with only one row or one column may be thought of as a vector. Vectors appear in boldface type v. Suppose this is the case and A has m rows and B has n columns. Step 2 Begin with row 1 as the current row, column 1 as the current column, and a11 as the current entry. Obtain the new current row, column, and entry.
Transform the column using EROs and move to the next current entry. After the Gauss—Jordan method has been applied to any linear system, a variable that appears with a coefficient of 1 in a single equation and a coefficient of 0 in all other equations is called a basic variable. Any variable that is not a basic variable is called a nonbasic variable. Linear Independence, Linear Dependence, and the Rank of a Matrix A set V of m-dimensional vectors is linearly independent if the only linear combination of vectors in V that equals 0 is the trivial linear combination.
A set V of m-dimensional vectors is linearly dependent if there is a nontrivial linear combination of the vectors in V that adds to 0. The rank of A is the number of vectors in the largest linearly independent subset of R. To find the rank of a given matrix A, apply the Gauss—Jordan method to the matrix A.
Let Ut be the number of untenured S. Let Rt be the number of rural residents at the beginning of year t, and Ct be the number of city residents at the beginning of year t. Courses 1 and 2 are four-credit courses, and courses 3 and 4 are three-credit courses. Let GPAi be the semester grade point average for student i.
Determine a linear equation system to find the amounts paid in bonuses, state tax, and federal tax. What is the unique solution? During the coming year, Seriland wants to consume ds dollars of steel, dc dollars of cars, and dm dollars of machinery. This should give you the general idea. This will increase the value of the steel, cars, and machines that must be produced next year. To understand the theory of linear and nonlinear programming, master at least one of these books: Dantzig, G.
Linear Programming and Extensions. Princeton, N. Hadley, G. Linear Algebra. Reading, Mass. See references at end of chapter. Linear Algebra and Its Applications, 3d ed. Orlando, Fla. Leontief, W. Input—Output Economics. New York: Oxford University Press, Teichroew, D. New York: Wiley, A more extensive discussion of linear algebra than this chapter gives at a comparable level of difficulty. In , George Dantzig developed an efficient method, the simplex algorithm, for solving linear programming problems also called LP.
Since the development of the simplex algorithm, LP has been used to solve optimization problems in industries as diverse as banking, education, forestry, petroleum, and trucking.
In Section 3. In Sections 3. Solving these simple LPs will give us useful insights for solving more complex LPs. The remainder of the chapter explains how to formulate linear programming models of real-life situations. In this section, we introduce linear programming and define important terms that are used to describe linear programming problems.
The manufacture of wooden soldiers and trains requires two types of skilled labor: carpentry and finishing. A soldier requires 2 hours of finishing labor and 1 hour of carpentry labor.
A train requires 1 hour of finishing and 1 hour of carpentry labor. Each week, Giapetto can obtain all the needed raw material but only finishing hours and 80 carpentry hours. Demand for trains is unlimited, but at most 40 soldiers are bought each week. Solution In developing the Giapetto model, we explore characteristics shared by all linear programming problems. We begin by defining the relevant decision variables. In any linear programming model, the decision variables should completely describe the decisions to be made in this case, by Giapetto.
Clearly, Giapetto must decide how many soldiers and trains should be manufactured each week. The function to be maximized or minimized is called the objective function. For the Giapetto problem, we note that fixed costs such as rent and insurance do not depend on the values of x1 and x2.
It would be foolish for Giapetto to manufacture more soldiers than can be sold, so we assume that all toys produced will be sold. We use the variable z to denote the objective function value of any LP. The coefficient of a variable in the objective function is called the objective function coefficient of the variable. For example, the objective function coefficient for x1 is 3, and the objective function coefficient for x2 is 2.
This means that if Giapetto were free to choose any values for x1 and x2, the company could make an arbitrarily large profit by choosing x1 and x2 to be very large.
Unfortunately, the values of x1 and x2 are limited by the following three restrictions often called constraints : Constraints Constraint 1 Each week, no more than hours of finishing time may be used. Constraint 2 Each week, no more than 80 hours of carpentry time may be used. Constraint 3 Because of limited demand, at most 40 soldiers should be produced each week.
The amount of raw material available is assumed to be unlimited, so no restrictions have been placed on this. The next step in formulating a mathematical model of the Giapetto problem is to express Constraints 1—3 in terms of the decision variables x1 and x2. For a constraint to be reasonable, all terms in the constraint must have the same units. Finally, we express the fact that at most 40 soldiers per week can be sold by limiting the weekly production of soldiers to at most 40 soldiers.
The coefficients of the decision variables in the constraints are called technological coefficients. This is because the technological coefficients often reflect the technology used to produce different products. For example, the technological coefficient of x2 in 3 is 1, indicating that a soldier requires 1 carpentry hour.
The number on the right-hand side of each constraint is called 3. Often the rhs of a constraint represents the quantity of a resource that is available. To complete the formulation of a linear programming problem, the following question must be answered for each decision variable: Can the decision variable only assume nonnegative values, or is the decision variable allowed to assume both positive and negative values?
If a variable xi can assume both positive and negative or zero values, then we say that xi is unrestricted in sign often abbreviated urs. In other problems, however, some variables may be urs. In this case, it would be appropriate to classify xi as urs. Other uses of urs variables are discussed in Section 4.
Before formally defining a linear programming problem, we define the concepts of linear function and linear inequality. A function f x1, x2,. For any linear function f x1, x2,. The reason for this will become apparent when we study the simplex algorithm in Chapter 4. The function that is to be maximized or minimized is called the objective function. Each constraint must be a linear equation or linear inequality. The Proportionality and Additivity Assumptions The fact that the objective function for an LP must be a linear function of the decision variables has two implications.
The contribution of the objective function from each decision variable is proportional to the value of the decision variable. For example, no matter what the value of x2, the manufacture of x1 soldiers will always contribute 3x1 dollars to the objective function.
The contribution of each variable to the left-hand side of each constraint is proportional to the value of the variable. For example, no matter what the value of x1, the manufacture of x2 trains uses x2 finishing hours and x2 carpentry hours. Implication 2 of the first list implies that the value of the objective function is the sum of the contributions from individual variables, and implication 2 of the second list implies that the left-hand side of each constraint is the sum of the contributions from each variable.
For this reason, the second implication in each list is called the Additivity Assumption of Linear Programming. For an LP to be an appropriate representation of a real-life situation, the decision variables must satisfy both the Proportionality and Additivity Assumptions.
Two other assumptions must also be satisfied before an LP can appropriately represent a real situation: the Divisibility and Certainty Assumptions. For example, in the Giapetto problem, the Divisibility Assumption implies that it is acceptable to produce 1.
Because Giapetto cannot actually produce a fractional number of trains or soldiers, the Divisibility Assumption is not satisfied in the Giapetto problem. A linear programming problem in which some or all of the variables must be nonnegative integers is called an integer programming problem. The solution of integer programming problems is discussed in Chapter 9. In many situations where divisibility is not present, rounding off each variable in the optimal LP solution to an integer may yield a reasonable solution.
Suppose the optimal solution to an LP stated that an auto company should manufacture , In this case, you could tell the auto company to manufacture , or , compact cars and be fairly confident that this would reasonably approximate an optimal production plan.
On the other hand, if the number of missile sites that the United States should use were a variable in an LP and the optimal LP solution said that 0. In this situation, the integer programming methods of Chapter 9 would have to be used, because the number of missile sites is definitely not divisible.
The Certainty Assumption The Certainty Assumption is that each parameter objective function coefficient, righthand side, and technological coefficient is known with certainty. If we were unsure of the exact amount of carpentry and finishing hours required to build a train, the Certainty Assumption would be violated. Feasible Region and Optimal Solution Two of the most basic concepts associated with a linear programming problem are feasible region and optimal solution. For defining these concepts, we use the term point to mean a specification of the value for each decision variable.
The feasible region for the Giapetto problem is the set of possible production plans that Giapetto must consider in searching for the optimal production plan. Similarly, for a minimization problem, an optimal solution is a point in the feasible region with the smallest objective function value.
However, some LPs have no optimal solution, and some LPs have an infinite number of solutions these situations are discussed in Section 3. Giapetto can maximize profit by building 20 soldiers and 60 trains each week.
An acre of wheat yields 25 bushels of wheat and requires 10 hours of labor per week. An acre of corn yields 10 bushels of corn and requires 4 hours of labor per week. Seven acres of land and 40 hours per week of labor are available. Government regulations require that at least 30 bushels of corn be produced during the current year. Using these decision variables, formulate an LP whose solution will tell Farmer Jones how to maximize the total revenue from wheat and corn. Each truck must go through the painting shop and assembly shop.
If the painting shop were completely devoted to painting Type 1 trucks, then per day could be painted; if the painting shop were completely devoted to painting Type 2 trucks, then per day could be painted. If the assembly shop were completely devoted to assembling truck 1 engines, then 1, per day could be assembled; if the assembly shop were completely devoted to assembling truck 2 engines, then 1, per day could be assembled.
We always label the variables x1 and x2 and the coordinate axes the x1 and x2 axes. This set of points is indicated by darker shading in Figure 1. These points are shown by lighter shading in Figure 1. Determine whether P satisfies the inequality.
This agrees with Figure 1. Note that the only points satisfying 5 and 6 lie in the first quadrant of the x1—x2 plane. This is indicated in Figure 2 by the arrows pointing to the right from the x2 axis and upward from the x1 axis. Thus, any point that is outside the first quadrant cannot be in the feasible region.
This means that the feasible region will be the set of points in the first quadrant that satisfies 2 — 4. Our method for determining the set of points that satisfies a linear inequality will also identify those that meet 2 — 4. The side of a line that satisfies an inequality is indicated by the direction of the arrows in Figure 2.
From Figure 2, we see that the set of points in the first quadrant that satisfies 2 , 3 , and 4 is bounded by the five-sided polygon DGFEH. Any point on this polygon or in its interior is in the feasible region. Any other point fails to satisfy at least one of the inequalities 2 — 6.
Thus 40, 30 is infeasible, because it fails to satisfy 2. Note that all points above line AB in Figure 2 are infeasible, because they fail to satisfy 2. Similarly, all points above CD are infeasible, because they fail to satisfy 3. Also, all points to the right of the vertical line EF are infeasible, because they fail to satisfy 4. To find the optimal solution, we need to graph a line on which all points have the same z-value.
In a max problem, such a line is called an isoprofit line in a min problem, an isocost line. To draw an isoprofit line, choose any point in the feasible region and calculate its z-value. Let us choose 20, 0. This means that once we have drawn one isoprofit line, we can find all other isoprofit lines by moving parallel to the isoprofit line we have drawn.
It is now clear how to find the optimal solution to a two-variable LP. After you have drawn a single isoprofit line, generate other isoprofit lines by moving parallel to the drawn isoprofit line in a direction that increases z for a max problem.
After a point, the isoprofit lines will no longer intersect the feasible region. The last isoprofit line intersecting touching the feasible region defines the largest z-value of any point in the feasible region and indicates the optimal solution to the LP. From Figure 2, we see that the isoprofit line passing through point G is the last isoprofit line to intersect the feasible region.
Thus, G is the point in the feasible region with the largest z-value and is therefore the optimal solution to the Giapetto problem. The optimal value of z may be found by substituting these values of x1 and x2 into the objective function. Binding and Nonbinding Constraints Once the optimal solution to an LP has been found, it is useful see Chapters 5 and 6 to classify each constraint as being a binding constraint or a nonbinding constraint. In Figures 3a and 3b, each line segment joining two points in S contains only points in S.
Thus, in both these figures, S is convex. In Figures 3c and 3d, S is not convex. In our study of linear programming, a certain type of point in a convex set called an extreme point is of great interest.
Extreme points are sometimes called corner points, because if the set S is a polygon, the extreme points of S will be the vertices, or corners, of the polygon. The feasible region for the Giapetto problem is a convex set. This is no accident: It can be shown that the feasible region for any LP will be a convex set. It can be shown that the feasible region for any LP has only a finite number of extreme points. Also note that the optimal solution to the Giapetto problem point G is an extreme point of the feasible region.
It can be shown that any LP that has an optimal solution has an extreme point that is optimal. This result is very important, because it reduces the set of points that yield an optimal solution from the entire feasible region which generally contains an infinite number of points to the set of extreme points a finite set.
We note that z increases as we move isoprofit lines in a northeast direction, so the largest z-value in the feasible region must occur at some point P that has no points in the feasible region northeast of P. This means that the optimal solution must lie somewhere on the boundary of the feasible region DGFEH.
The LP must have an extreme point that is optimal, because for any line segment on the boundary of the feasible region, the largest z-value on that line segment must be assumed at one of the endpoints of the line segment. To see this, look at the line segment FG in Figure 2.
This means that moving along FG in a direction of decreasing x1 increases z. Thus, the value of z at point G must exceed the value of z at any other point on the line segment FG. A similar argument shows that for any objective function, the maximum value of z on a given line segment must occur at an endpoint of the line segment.
Therefore, for any LP, the largest z-value in the feasible region must be attained at an endpoint of one of the line segments forming the boundary of the feasible region. In short, one of the extreme points of the feasible region must be optimal. Our proof that an LP always has an optimal extreme point depended heavily on the fact that both the objective function and the constraints were linear functions.
In Chapter 11, we show that for an optimization problem in which the objective function or some of the constraints are not linear, the optimal solution to the optimization problem may not occur at an extreme point.
The company believes that its most likely customers are high-income women and men.
0コメント