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## Linear second-order equations with constant coefficients
Definition A
for a function Suppose that
which we refer to subsequently as the “original equation”. Define
Conversely, let Thus we have the following result. Proposition Let
Then
The practical importance of this result is that the set of all solutions of the original equation may be found by finding Procedure for finding general solution of linear second-order ordinary differential equation with constant coefficients The general solution of the differential equation
may be found as follows. 1. Find the general solution of the associated homogeneous equation 2. Find a single solution of the original equation 3. Add together the solutions found in steps 1 and 2. I now explain how to perform the first two steps and illustrate step 3. 1. Find the general solution of a homogeneous equation You might guess, based on the solutions we found for first-order equations, that the homogeneous equation has a solution of the form x(t) = Ae then ^{rt}x'(t) = rAe and ^{rt}x"(t) = r^{2}Ae, so that^{rt}
Thus for
This equation is knows as the Suppose that x(t) = Be, for any values of ^{st}A and B, are solutions of the equation. Thus also x(t) = Ae + ^{rt}Be is a solution. It can be shown, in fact, that ^{st}every solution of the equation takes this form.The cases in which the characteristic equation has a single real root ( Proposition SOURCE Consider the homogeneous linear second-order ordinary differential equation with constant coefficients
The general solution of this equation depends on the character of the roots of the characteristic equation Distinct real roots If
Be.^{st}Single real root If ( where Complex roots If ( where α = −
t + ω),where the relationships between the constants You may be asking yourself how it is possible that trigonometric functions (sin and cos) appear in the solutions. The answer lies in fact that
x + isin x for all x,where Example Consider the differential equation
The characteristic equation is
so the roots are 1 and −2. That is, the roots are real and distinct. Thus the general solution of the differential equation is
Be^{−2t}.Example Consider the differential equation
The characteristic equation has a single real root, −3. Thus the general solution of the differential equation is
Example Consider the equation
The characteristic roots are complex. We have [ 2. Find a solution of a nonhomogeneous equation Consider the nonhomogeneous equation
If B, then you should try to find a value of Asuch that Ae is a solution.^{Bt}Example Consider the differential equation
A linear combination of the function on the right-hand side and its first and second derivatives is 2 For this condition to be satisfied (note that the equation must be satisfied
The unique solution of these equations is
is a solution of the differential equation. 3. Add together the solutions found in steps 1 and 2 This step is quite trivial! Example Consider the equation from the previous example, namely
We saw above that the general solution of the associated homogeneous equation is
Be^{−2t}.and that
is a solution of the original equation. Thus the general solution of the original equation is
Be^{−2t} − 3/4 − t/2 − t^{2}/2.Second-order initial value problems A first-order initial value problem consists of a first-order ordinary differential equation Definition A
and
where Example Consider the second-order initial value problem in which the differential equation is the one in theprevious example, namely
and the initial conditions are
From the previous example, the general solution of the equation is
Be^{−2t} − 3/4 − t/2 − t^{2}/2.We have
so that for the initial conditions to be satisfied we need
The unique solution of these equations is
e^{−2t} − 3/4 − t/2 − t^{2}/2.Equilibrium and stability The notions of equilibrium and stability for a second-order initial value problem are closely related to the corresponding notion for a first-order initial value problem. Definition If for some initial conditions a second-order initial value problem has a solution that is a constant, the value of the constant is an Consider the homogeneous linear second-order equation
If Characteristic equation has two real roots If the characteristic equation has two real roots, Be. This function converges to 0 for all values of ^{st}A and B, so that the equilibrium is globally stable, if and only if r < 0 and s < 0.Characteristic equation has a single real root If the characteristic equation has a single real root, A and B, so that the equilibrium is globally stable, if and only if r < 0. (If r < 0 then for any value of k, t converges to 0 as ^{k}e^{rt}t increases without bound.)Characteristic equation has complex roots If the characteristic equation has complex roots, the general solution of the equation is ( a/2, β = √(b − a^{2}/4). This function converges to 0 for all values of A and B, so that the equilibrium is globally stable, if and only if α < 0, or equivalently a > 0. (Remember that cos θ and sin θ lie between +1 and −1 for all values of θ.)Now, the roots of the characteristic equation
are, by the quadratic formula, (− Thus if these roots are complex, then the If Thus an equilibrium of the second-order linear homogeneous equation
is globally stable if and only if the real parts of each root of the characteristic equation
is negative. If Thus we have the following result. Proposition An equilibrium of the homogeneous linear second-order ordinary differential equation with constant coefficients Now consider the (nonhomogeneous) linear second-order equation
where
Thus the equilibrium (0) of this homogeneous equation is globally stable if and only if the equilibrium of the original equation is globally stable. That is, we have the following result. Proposition For any value of Example Consider the following macroeconomic model. Denote by
Denote by
where π'( for some One way to answer this question is to reduce the system to a single second-order differential equation by differentiating the equation for
Given In particular, if
11 lecture
Random events and their properties. The full group of random events. Classical statistical and geometric probability. Addition theorems of probability of joint and incompatible events. Independent random events.
Geometric Probability Random events that take place in Some problems, like , or the problem of the Bertrand's Paradox do, by their nature, arise in a geometric setting. The latter also admits multiple reformulations which require comparison of the areas of geometric figures. In general, we may think of Stick Broken Into Three Piecesgeometric probabilities as non-negative quantities (not exceeding 1) being assigned to subregions of a given domain subject to certain rules. If function μ is an expression of this assignment defined on a domain D, then, for example, we require0 ≤ μ(A) ≤ 1, A ⊂ D and The function μ is usually not defined for all A ⊂ D. Those subsets of D for which μ is defined are the Problem 1
In a Cartesian system of coordinates (s, t), a square of side 20 (minutes) represents all the possibilities of the morning arrivals of the two friends at the metro station. The gray area A is bounded by two straight lines, t = s + 5 and t = s - 5, so that inside A, |s - t| ≤ 5. It follows that the two friends will meet only provided their arrivals s and t fall into region A. The probability of this happening is given by the ratio of the area of A to the are of the square: [400 - (15×15/2 + 15×15/2)] / 400 = 175/400 = 7/16. Problem 2 ([
Fix point C. The positions of points A and B are then defined by arcs α and β extending from C in two directions. A priori we know that 0 < α + β < 2π. The favorable for our problem values of α and β (as subtending acute angles satisfy) 0 < α < π and 0 < β < π. Their sum could not be less than π as this would make angle C obtuse, therefore,α + β > π. The situation is presented in the following diagram where the square has the side 2π. Region D is the intersection of three half-planes: 0 < α, 0 < β, and α + β < 2π. This is the big triangle in the above diagram. The favorable events belong to the shaded triangle which is the intersection of the half-planes α < π, β < π,and α + β > π. The ratio of the areas of the two is obviously 1/4. Now observe, that unless the random triangle is acute it can be thought of as obtuse since the probability of two of the three points A, B, C forming a diameter is 0. (For BC to be a diameter, one should have α + β = π which is a straight line, with zero as the only possible assignment of area.) Thus we can say that the probability of ΔABC being obtuse is 3/4. For an obtuse triangle, the circle can be divided into two halves with the triangle lying entirely in one of the halves. It follows that 3/4 is the answer to the following question:
12 lecture
Conditional probability. Theorem of multiplication of probabilities. The formula of total probability. Bayes' formula.
13 lecture
Elements of linear programming. Graphical method.
The - Graph the feasible region.
- Compute the coordinates of the corner points.
- Substitute the coordinates of the corner points into the objective function to see which gives the optimal value. This point gives the solution to the linear programming problem.
- If the feasible region is not bounded, this method can be misleading: optimal solutions always exist when the feasible region is bounded, but may or may not exist when the feasible region is unbounded.
If the feasible region is unbounded, we are To determine if a solution exists in the general unbounded case: - Bound the feasible region by adding a vertical line to the right of the rightmost corner point, and a horizontal line above the highest corner point.
- Calculate the coordinates of the new corner points you obtain.
- Find the corner point that gives the optimal value of the objective function.
- If this optimal value occurs at a point of the
*original*(unbounded) region, then the LP problem has a solution at that point. If not, then the LP problem has no optimal solution.
Minimize 3 The feasible region for this set of constraints was shown above. Here it is again with the corner points shown. Although the feasible region is unbonded, we are minimizing The following table shows the value of
Therefore, the solution is
14 lecture
Simplex method. The algorithm of the Simplex method.
15 lecture
The transportation problem. Finding the initial of the support solutions.
There is a type of linear programming problem that may be solved using a simplified version of the simplex technique called Let us assume there are There are three general steps in solving transportation problems. We will now discuss each one in the context of a simple example. Suppose one company has four factories supplying four warehouses and its management wants to determine the minimum-cost shipping schedule for its weekly output of chests. Factory supply, warehouse demands, and shipping costs per one chest (unit) are shown in Table 7.1
At first, it is necessary to prepare an
The transportation matrix for this example appears in Table 7.2, where supply availability at each factory is shown in the far right column and the warehouse demands are shown in the bottom row. The unit shipping costs are shown in the small boxes within the cells (see transportation tableau – at the initiation of solving all cells are empty). It is important at this step to make sure that the total supply availabilities and total demand requirements are equal. Often there is an excess supply or demand. In such situations, for the transportation method to work, a dummy warehouse or factory must be added. Procedurally, this involves inserting an extra row (for an additional factory) or an extra column (for an ad warehouse). The amount of supply or demand required by the ” In this case there is: Total factory supply … 51 Total warehouse requirements … 52 This involves inserting an extra row - an additional factory. The amount of supply by the dummy equals the difference between the row and column totals. In this case there is 52 – 51 = 1. The cost figures in each cell of the dummy row would be set at zero so any units sent there would not incur a transportation cost. Theoretically, this adjustment is equivalent to the simplex procedure of inserting a slack variable in a constraint inequality to convert it to an equation, and, as in the simplex, the cost of the dummy would be zero in the objective function.
Initial allocation entails assigning numbers to cells to satisfy supply and demand constraints. Next we will discuss several methods for doing this: the Northwest-Corner method, Least-Cost method, and Vogel's approximation method (VAM). Table 7.3 shows a Inspection of Table 7.3 indicates some high-cost cells were assigned and some low-cost cells bypassed by using the northwest-comer method. Indeed, this is to be expected since this method ignores costs in favor of following an easily programmable allocation algorithm. Table 7.4 shows a Table 7.5 shows the
To develop an optimal solution in a transportation problem involves evaluating each unused cell to determine whether a shift into it is advantageous from a total-cost stand point. If it is, the shift is made, and the process is repeated. When all cells have been evaluated and appropriate shifts made, the problem is solved. One approach to making this evaluation is the Stepping stone method. The term stepping stone appeared in early descriptions of the method, in which unused cells were referred to as "water" and used cells as "stones"— from the analogy of walking on a path of stones half-submerged in water. The stepping stone method was applied to the VAM initial solution, as shown in Table 7.5 Table 7.6 shows the optimal solutions reached by the Stepping stone method. Such solution is very close to the solution found using VAM method.
When the evaluation of any empty cell yields the same cost as the existing allocation, an alternate optimal solution exists (see Stepping Stone Method – alternate solutions). Assume that all other cells are optimally assigned. In such cases, management has additional flexibility and can invoke nontransportation cost factors in deciding on a final shipping schedule.
Degeneracy exists in a transportation problem when the number of filled cells is less than the number of rows plus the number of columns minus one (m + n - 1). Degeneracy may be observed Procedurally, the value (often denoted by the Greek letter epsilon, - ) is used in exactly the same manner as a real number except that it may initially be placed in any empty cell, even though row and column requirements have been met by real numbers. A degenerate transportation problem showing a Northwest Corner initial allocation is presented in Table 7.8, where we can see that if were not assigned to the matrix, it would be impossible to evaluate several cells. Once a has been inserted into the solution, it remains there until it is removed by subtraction or until a final solution is reached. While the choice of where to put an is arbitrary, it saves time if it is placed where it may be used to evaluate as many cells as possible without being shifted.
A fictive corporation A has a contract to supply motors for all tractors produced by a fictive corporation B. Corporation B manufactures the tractors at four locations around Central Europe: Prague, Warsaw, Budapest and Vienna. Plans call for the following numbers of tractors to be produced at each location: Prague 9 000 Warsaw 12 000 Budapest 9 000 Corporation A has three plants that can produce the motors. The plants and production capacities are Hamburg 8 000 Munich 7 000 Leipzig 10 000 Dresden 5 000 Due to varying production and transportation costs, the profit earns on each motor depends on where they were produced and where they were shipped. The following transportation table (Table 7.9) gives the accounting department estimates of the euro profit per unit (motor).
Table 7.10 shows a
Applying the Stepping Stone method (modified for maximization purposes) to the initial solution we can see that no other transportation schedule can increase the profit and so the Highest – Profit initial allocation is also an optimal solution of this transportation problem.
The transshipment problem is similar to the transportation problem except that in the transshipment problem it is possible to both ship into and out of the same node (point). For the transportation problem, you can ship only from supply points to demand points. For the transshipment problem, you can ship from one supply point to another or from one demand point to another. Actually, designating nodes as supply points or demand points becomes confusing when you can ship both into and out of a node. You can make the designations clearer if you classify nodes by their net stock position-excess (+), shortage (-), or 0. One reason to consider transshipping is that units can sometimes be shipped into one city at a very low cost and then transshipped to other cities. In some situations, this can be less expensive than direct shipment. Picture 7.1 shows the net stock positions for the three warehouses and four customers. Say that it is possible to transship through Pilsen to both Innsbruck and Linz. The transportation cost from Pilsen to Innsbruck is 300 euro per unit, so it costs less to ship from Warsaw to Innsbruck by going through Pilsen. The direct cost is 950 euro, and the transshipping cost is 600 + 300 = 900 euro. Because the transportation cost is 300 euro from Pilsen to Innsbruck, the cost of transshipping from Prague through Pilsen to Innsbruck is 400 euro per unit. It is cheaper to transship from Prague through Pilsen than to ship directly from Prague to Innsbruck.
There are two possible conversions to a transportation model. In the
The
Another transportation problem is the assignment problem. You can use this problem to assign tasks to people or jobs to machines. You can also use it to award contracts to bidders. Let's describe the assignment problem as assigning For example, say that five groups of computer users must be trained for five new types of software. Because the users have different computer skill levels, the total cost of trainings depends on the assignments.
Table 7.12 shows the cost of training for each assignment of a user group (A through E) to a software type (S1 through S5). Picture 7.2 is a network model of this problem. A balanced assignment problem has the same number of people and tasks. For a balanced assignment problem, the relationships are all equal. Each person must do a task. For an unbalanced assignment problem with more people than tasks, some people don't have to do a task and the first class of constraints is of the type. In general, the simplex method does not guarantee that the optimal values of the decision variables are integers. Fortunately, for the assignment model, all of the corner point solutions have integer values for all of the variables. Therefore, when the simplex method determines the optimal corner point, all of the variable values are integers and the constraints require that the integers be either 1 or 0 (Boolean). |
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