﻿﻿ Vector space. The linear dependence of vectors. Decomposition of the vector on the basis.
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Vector space. The linear dependence of vectors. Decomposition of the vector on the basis.

Lecture

Vector space. The linear dependence of vectors. Decomposition of the vector on the basis.

Vectors

Meaning, geometric representation, and types of vectors

Vector operations

Lecture

Matrices and operations on matrices. Determinants. Basic properties. Rank of the matrix.

Two of the importantproperties of scalar multiplicationof matrices are the following.

Lecture

System of linear algebraic equations and methods for their solutions (Formula Cramer, Gauss methods).

Lecture

The use of elements of linear algebra in the economy. Leontief model of a diversified economy.

The use of elements of linear algebra in the economy. Leontief model of a diversified economy.

Macroeconomics operation of a diversified economy requires a balance between the different branches. Each sector on the one hand, is prizvoditelem, and on the other - the consumer products produced by other industries. There is quite a difficult task of calculating the relationship between the branches through the production and consumption of products of different types. This problem was first formulated in the form of a mathematical model in 1936 in the writings of well-known American economist Leontief, who tried to analyze the causes of the economic depression of the United States 1929-1932. This model is based on matrix algebra and matrix analysis using the device.

Balance sheet ratio

For simplicity, we assume that the manufacturing economy is a sphere n branches, each of which produces a homogeneous product. To secure its production every industry needs from other sectors (industrial consumption). Typically the process of production is considered for a certain period of time; in some cases, such a unit is the year.

We introduce the following notation:

xi - total output of i-th branch (its gross output);

the volume of product i-th industry consumed j-th industry in the production volume

of products xj;

the volume of product i-th branch, intended for sale (consumption) in the non-manufacturing

sector, or the so-called end-use product. This includes personal consumption of the citizens,

satisfaction of social needs, the content of the state institutions, etc.

The balance principle of communication in various industries is that the gross output of the i-th sector should be equal to the sum of consumption volumes in the industrial and non-industrial sectors. In its simplest form (the hypothesis of linearity, or simple addition) balance relations are of the form

This equation (16.2) is called the balance of relations.

Since the products of different industries have different dimensions, we will in the future to keep in mind the value balance.

Lecture

Lecture

Lecture

Lecture

The first order.

General form

Definition

A second-order ordinary differential equation is an ordinary differential equation that may be written in the form

x"(t) = F(t, x(t), x'(t))

for some function F of three variables.

Equations of the form x"(t) = F(t, x'(t))

An equation of the form

x"(t) = F(t, x'(t)),

in which x(t) does not appear, can be reduced to a first-order equation by making the substitution z(t) =x'(t).

Example

Let u be a function of a single variable, with the interpretation that u(w) is the utility of an individual with wealth w. The function ρ defined by

ρ(w) = −wu"(w)/u'(w)

is known as the Arrow-Pratt measure of relative risk aversion. (If ρ(w) for the function u exceeds ρ(w) for the function v for all w then u reflects a greater degree of risk-aversion than does v.)

What utility functions have a degree of risk-aversion that is independent of the level of wealth? That is, for what utility functions u do we have

a = −wu"(w)/u'(w) for all w

for some number a?

This equation is a second-order differential equation in which the term u(w) does not appear. (The variable is w, rather than t.) Define z(w) = u'(w). Then we have

a = −wz'(w)/z(w)

or

z'(w) = −az(w)/w,

a separable first-order ordinary differential equation. Making the substitution y = z(w), so that dy =z'(w)dw, and integrating, the equation becomes

∫(1/y)dy = −a∫(1/w)dw,

so that

ln y = −aln w + C,

or

ln z(w) = −aln w + C,

or

z(w) = Cwa.

Now, z(w) = u'(w), so to get u we need to integrate:

u(w) =
 Cln w + B if a = 1 Cw1−a/(1 − a) + B if a ≠ 1.

We conclude that a utility function with a constant degree of relative risk-aversion equal to a takes this form.

Graphical Method

The graphical method for solving linear programming problems in two unknowns is as follows.

1. Graph the feasible region.
2. Compute the coordinates of the corner points.
3. 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.
4. 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 minimizing the objective function, and its coefficients are non-negative, then a solution exists, so this method yields the solution.

To determine if a solution exists in the general unbounded case:

1. 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.
2. Calculate the coordinates of the new corner points you obtain.
3. Find the corner point that gives the optimal value of the objective function.
4. 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.

Example

Minimize C = 3x + 4y subject to the constraints

3x - 4y ≤ 12,
x + 2y ≥ 4
x ≥ 1, y ≥ 0.

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 C = 3x + 4y, whose coefficients are non-negative, and so the method shown above left applies.

The following table shows the value of C at each corner point:

 Point C = 3x + 4y (1, 1.5) 3(1)+4(1.5) = 9 minimum (4, 0) 3(4)+4(0) = 12

Therefore, the solution is x = 1, y = 1.5, giving the minimum value C = 9.

14 lecture

Simplex method. The algorithm of the Simplex method.

15 lecture

The transportation problem. Finding the initial of the support solutions.

The Transportation Problem

There is a type of linear programming problem that may be solved using a simplified version of the simplex technique called transportation method. Because of its major application in solving problems involving several product sources and several destinations of products, this type of problem is frequently called the transportation problem. It gets its name from its application to problems involving transporting products from several sources to several destinations. Although the formation can be used to represent more general assignment and scheduling problems as well as transportation and distribution problems. The two common objectives of such problems are either (1) minimize the cost of shipping m units to n destinations or (2) maximize the profit of shipping m units to n destinations.

Let us assume there are m sources supplying n destinations. Source capacities, destinations requirements and costs of material shipping from each source to each destination are given constantly. The transportation problem can be described using following linear programming mathematical model and usually it appears in a transportation tableau.

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

Table 7.1 ”Data for Transportation Problem”

At first, it is necessary to prepare an initial feasible solution, which may be done in several different ways; the only requirement is that the destination needs be met within the constraints of source supply.

The Transportation Matrix

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 ”dummy” equals the difference between the row and column totals.

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.

Table 7.2 "Transportation Matrix for Chests Problem With an Additional Factory (Dummy)"

Initial Feasible Solution

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 northwest-corner assignment. (Cell A-E was assigned first, A-F second, B-F third, and so forth.) Total cost : 10*10 + 30*4 + 15*10 + 30*1 + 20*12 + 20*2 + 45*12 + 0*1 = 1220(\$).

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 least-cost assignment. (Cell Dummy-E was assigned first, C-E second, B-H third, A-H fourth, and so on.) Total cost : 30*3 + 25*6 + 15*5 +10*10 + 10*9 + 20*6 + 40*12 + 0*1= 1105 (\$).

Table 7.5 shows the VAM assignments. (Cell Dummy-G was assigned first, B-F second, C-E third, A-H fourth, and so on.) Note that this starting solution is very close to the optimal solution obtained after making all possible improvements (see next chapter) to the starting solution obtained using the northwest-comer method. (See Table 7.3.) Total cost: 15*14 + 15*10 + 10*10 + 20*4 + 20*1 + 40*5 + 35*7 + 0*1 = 1005 (\$).

Table 7.3 ”Northwest – Corner Assignment”

Table 7.4"Least - Cost Assignment"

Table 7.5 "VAM Assignment"

Develop Optimal Solution

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.

Table 7.6 "Optimal Matrix, With Minimum Transportation Cost of \$1,000."

Alternate Optimal Solutions

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.

Table 7.7 "Alternate Optimal Matrix for the Chest Transportation Problem, With Minimum Transportation Cost of \$1,000.

Degeneracy

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 either during the initial allocation when the first entry in a row or column satisfies both the row and column requirements or during the Stepping stone method application, when the added and subtracted values are equal. Degeneracy requires some adjustment in the matrix to evaluate the solution achieved. The form of this adjustment involves inserting some value in an empty cell so a closed path can be developed to evaluate other empty cells. This value may be thought of as an infinitely small amount, having no direct bearing on the cost of the solution.

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.

Table 7.8 "Degenerate Transportation Problem With Added. Number of filled cells = 4”

7.6 Transportation Problem with a Maximization as a Criterion

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 7.9 "The Euro Profit Per One Shipped Motor"

Table 7.10 shows a highest - profit assignment(Least Cost method modification). In contrast to the Least – Cost method it allocates as much as possible to the highest-cost cell. (Cell Hamburg - Budapest was assigned first, Munich - Warsaw second, Leipzig - Warsaw third, Leipzig – Budapest fourth, Dresden – Prague fifth and Leipzig – Prague sixth.) Total profit : 3 335 000 euro.

Table 7.10 "Highest - Profit Assignment"

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

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.
Let's consider the balanced transportation problem as an example.

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.

Picture 7.1 "Transshipment Example in the Form of a Network Model"

There are two possible conversions to a transportation model. In the first conversion, make each excess node a supply point and each shortage node a demand point. Then, find the cheapest method of shipping from surplus nodes to shortage nodes considering all transshipment possibilities. Let's perform the first conversion for the Picture 7.1 example. Because a transportation table Prague, Warsaw, and Vienna have excesses, they are the supply points. Because Krakow, Pilsen, Innsbruck, and Linz have shortages, they are the demand points. The cheapest cost from Warsaw to Innsbruck is 900 euro, transshipping through Pilsen. The cheapest cost from Prague to Innsbruck is 400 euro, transshipping through Pilsen too. The cheapest cost from all other supply points to demand points is obtained through direct shipment. Table 7.11 shows the balanced transportation table for this transshipment problem. For a simple transportation network, finding all of the cheapest routes from excess nodes to shortage nodes is easy. You can list all of the possible routes and select the cheapest. However, for a network with many nodes and arcs, listing all of the possible routes is difficult.

Table 7.11 "The Transshipment Problem After Conversion to a Transportation Model"

The secondconversion of a transshipment problem to a transportation model doesn't require finding all of the cheapest routes from excess nodes table to shortage nodes. The second conversion requires more supply and demand nodes than the first conversion, because the points where you can ship into and out of, occur in the converted transportation problem twice – first as a supply point and second as a demand point.

The Assignment Problem

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 n tasks to n people. Each person must do one and only one task, and each task must be done by only one person. You represent each person and each task by a node. Number the people 1 to n, and number the tasks 1to n. The assignment problem can be described simply using a verbal model or using a linear programming mathematical model .

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 ”Cost of Trainings According to the Groups of Users”

Picture 7.2 "Network Model for Assignment Problem"

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).

Conclusion

The transportation problem is only a special topic of the linear programming problems. It would be a rare instance when a linear programming problem would actually be solved by hand. There are too many computers around and too many LP software programs to justify spending time for manual solution.( There are also programs that assist in the construction of the LP or TP model itself. Probably the best known is GAMS—General Algebraic Modeling System (GAMS-General, San Francisco, CA). This provides a high-level language for easy representation of complex problems.)

Lecture

Vector space. The linear dependence of vectors. Decomposition of the vector on the basis.

Vectors

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