Math II (Несколько текстов для зачёта), страница 3

One question that students frequently ask concerns the populations of the towns. I deliberately withhold this information initially, and I instruct students to devise possible solutions without knowing the populations.

Students should recognize that splitting the cost equally is an inferior solution for Bender. Students should devise two preferable kinds of solutions, either on the basis of the cost or on the basis of the savings involved. Splitting the savings equally between the two towns is an example of the latter. Since $5 million, that is, $24 million minus $19 million, represents the amount saved, each town should save $2.5 million, so that the $19 million cost would be divided in the ratio of 12.5 to 6.5, that is, ($15 - $2.5) to ($9 - $2.5).

A possible solution on the basis of cost is to allocate costs in proportion to opportunity, that is, stand-alone, costs. In this solution,

9/24 = 3/8

of the cost, or $7.125 million, should be borne by Bender; and

15/24 = 5/8,

or $11.875 million, by Amity. The same solution can be obtained by allocating savings in proportion to opportunity costs, so that the cost for Bender, for example, would be

9 - 9/24 x 5,

or $7.125 million. See table 1; where necessary, numbers in tables are rounded to three decimal places.

Students often rebel against first finding solutions without knowing the populations of the towns, and their concern is worthy of classroom discussion. But if the populations are cleverly constructed, the problem becomes more complex rather than easier. For example, if the population of Bender is 10 000 and the population of Amity is 40 000 and costs are allocated in proportion to population, then Amity should pay four-fifths of the cost, or $15.2 million. Such a solution is clearly not in Amity's best interest, just as splitting the cost equally is not in Bender's best interest. A question to ask students is, Under what circumstances does the ratio of the populations of the towns produce a solution that encourages each town to participate? However, if the savings are divided equally among the residents, then Amity pays $11 million, that is,

(15 - 4/5 x 5),

and Bender pays $8 million.

Three solutions appear to be in the best interests of both towns, as indicated in table 2:

Dividing the savings equally--A (Amity) pays $12.5 million, and B (Bender) pays $6.5 million

Dividing the savings equally among the residents--A pays $11 million and B pays $8 million on the basis of the given populations

Dividing the costs or the savings proportionally to opportunity costs or savings--A pays $11.875 million, and B pays $7.125 million

Which of the three solutions is the fairest? Young (1991) takes an interesting geometric approach to this question. Core is the term that game theorists and political economists give to the set of possible solutions in which neither player, or town, pays more than the opportunity costs. The colored segment in figure 1 represents the core. The x-axis represents Amity's payments; the y-axis, Bender's payments. The line segment joining the points (0, 19) and (19, 0) is the set of all possible allocations; the portion of that line segment between the horizontal at 9 and the vertical at 15 represents the core. Students can easily replicate this figure on a graphing calculator in a window that goes from 0 to 20 in each direction. The equation of the line segment in question is y = -x + 19, and the DRAW menu can be accessed from the home screen, as opposed to the graph, to obtain the desired horizontal and vertical segments. The previously discussed solutions, both those in the core and those outside it, are labeled in the figure.

A good case can be made for choosing the midpoint of the line segment representing the core as the solution to the problem. That point corresponds to equal savings for each town. In that solution, A pays $12.5 million and B pays $6.5 million. When students try to negotiate an equitable settlement in their groups, this solution is often the most appealing.

THE SEWAGE-TREATMENT-PLANT PROBLEM, PART 2

We next suppose that a third town, Cordial, is involved. The stand-alone cost for Cordial is $7 million, and the cost for a sewage-treatment plant that would service all three towns is $23 million.

Before students can break up into groups to decide how to solve this problem, costs for all possible coalitions must be assigned. One possible way follows:

• The cost for Amity and Bender together remains as before, $19 million.

• Were Amity and Cordial to participate together, the cost would be $17 million.

• Were Bender and Cordial to participate together, their cost would be $13 million.

If we use the method of proportional allocation, which gave us a solution in the core in the two-person game, then Amity contributes $15.862 million, a solution that is not in the core. Moreover, dividing savings equally among residents fails to fall within the core because Bender and Cordial can form a coalition that leaves Amity out and build the plant for roughly $2.5 million less than by joining with Amity and using that method. Table 3 summarizes results from the other methods used in the two-town game. For these results, we assume that the population of Cordial is 8000 and that the populations of the other towns are as stated initially. Students can investigate which of these methods fall within the core and which are outside it.

In the classroom, letting students play with the problem before analyzing it in this fashion is advisable; fascinating student interactions can result. If the class is divided into three groups, each representing one of the towns, students can caucus among themselves to determine a "strategy," or method that is equitable from their point of view, to divide costs. Pairs of students from each group are then randomly assigned to negotiate a settlement; in other words, two students from A (Amity), two from B (Bender), and two from C (Cordial) work as one group; another two from A, two from B, and two from C work in a second group; and so on.

Young (1991) presents a geometric analog to the line segment that denoted the core in the two-town game. We construct an equilateral triangle with its altitude numerically equal to the cost if all three towns cooperate. Each vertex of the triangle represents one town's payment of the full cost, and any point in the interior of the triangle represents the towns' splitting the $23 million in some fashion. The core in this game is the shaded area in figure 2.

A COMBINATORIC APPROACH

L. S. Shapley, a political economist at Princeton, developed a cost-allocation method (Shapley 1981) that is similar to his approach to power indices in voting games. We consider all possible permutations of the three towns. Each permutation is treated as if the towns join the coalition sequentially and make up the difference between what has already been contributed and the total cost for the coalition. For example, in the permutation ABC, A joins first and must contribute 15. When it joins the coalition, B must contribute 4, the difference between A's 15 and the cost for AB, which is 19. When C joins, C must also contribute 4, the difference between 23 and 19. The Shapley value is the average of all possible contributions for a town. The values for the problem are summarized in table 4.

THE GEOMETRIC SOLUTION FOR THREE PLAYERS

The Shapley solution obtained previously is within the core and is thus a valid solution to the problem, but we have no guarantee that the Shapley value will be in the core (Young 1991). Can we guarantee a solution that is in the core of a three-player game if a core exists? We can easily construct a situation in which the core does not exist. We consider the core in figure 2. We try to extend the midpoint solution of the two-player game, called the standard solution, to three players. The core here is a triangle, although we have no guarantee that the core will be a triangle. To visualize this result, we move the line designated "A and B pay 19" parallel to itself and away from vertex C. As that line moves, the core changes from a triangle to a quadrilateral to a pentagon. The upper vertex of the core triangle represents B's paying a share of 9. This amount is B's maximum payment within the core. B's minimum payment is represented by the line designating "A and C pay 17," or 6. We average those payments at 7.5 and construct through that point the horizontal segment with endpoints on the borders of the core. See figure 3. The left endpoint of the segment represents C's minimum cost, and therefore A's maximum cost, given that B will pay 7.5. The right endpoint represents A's minimum cost and C's maximum cost. If we simply average the maximum and minimum costs for A and C, we obtain the solution that A pays 10.75, B pays 7.5, and C pays 4.75.

A spreadsheet that neatly summarizes all these solutions in the three-town game can be constructed. Such a spreadsheet appears as table 5. Entries in the top half of the spreadsheet represent the costs to each town or coalition of towns for each possible solution. Entries in the bottom half of the spreadsheet represent the savings for each coalition. Any negative entry in the bottom half of the table indicates that the solution does not fall within the core of the game.

Students usually need help in arriving at either the geometric solution or the Shapley value. They do have quite a bit to say about these and the other solutions that they may generate on their own, and talking through the solutions in class has always been interesting and provocative.

Problems of cost allocation are inherently interesting to students and are rich in mathematical applications. Those that come to mind most readily include graphing straight lines, geometric constructions, parallelism, combinatorics, and proportions. The aspect that makes cost-allocation problems so valuable in the classroom, however, is that students are motivated to talk about mathematics with one another and to experience a real-life application of the mathematics that they know.

TABLE 1 Payments by Town on the Basis of Costs or Savings

B - Amity Share (Millions of $)

C - Bender Share (Millions of $)

A B C

Stand-alone costs 15 9

Split costs 9.5 9.5

Split savings 12.5 6.5

TABLE 2 Three Solutions in the Best Interests of Both Towns

Legend for Chart:

B - Amity Share (Millions of $)

C - Bender Share (Millions of $)

A B C

Dividing savings equally 12.5 6.5

Dividing savings equally

among residents 11 8

Dividing costs or savings in

proportion to opportunity

costs 11.875 7.125

TABLE 3 Payments by Town for the Three-Town Game

Legend for Chart:

B - Payments by Town Amity

C - Payments by Town Bender

D - Payments by Town Cordial

A B C

D

Stand-alone costs 15 9

7

Split costs 7.67 7.67

7.67

Split savings 12.33 6.33

4.33

Cost divided in proportion to stand-alone costs 11.129 6.677

5.194

Costs divided among residents 15.862 3.966

3.172

Savings divided among residents 9.483 7.621

5.897

TABLE 4 Allocation Using a Combinatoric Approach

Legend for Chart:

A - Coalition order

B - Individual Contributions A

C - Individual Contributions B

D - Individual Contributions C

A B C D

ABC 15 4 4

ACB 15 6 2

BAC 10 9 4

BCA 10 9 4

CAB 10 6 7

CBA 10 6 7

Total contribution 70 40 28

Shapley value 11.67 6.67 4.67

TABLE 5 Summary of All Solutions in the Three-Town Game

Legend for Chart:

B - Amity

C - Bender

D - Cordial

A B C D

Costs

Stand-alone costs 15 9 7

Split-cost solution 7.67 7.67 7.67

Split-savings solution 12.33 6.33 4.33

Costs prop. to oppty. 11.129 6.677 5.194

Prorated costs 15.862 3.966 3.172

Prorated savings 9.483 7.621 5.897

Geometric solution 10.75 7.5 4.75

Shapley solution 11.67 6.67 4.67

Savings

Split cost 7.33 1.33 -0.67

Split savings 2.67 2.67 2.67

Costs prop. to oppty. 3.871 2.323 1.806

Prorated costs -0.862 5.034 3.828

Prorated savings 5.517 1.379 1.103

Geometric 4.25 1.5 2.25

Shapley 3.33 2.33 3.33

DIAGRAM: Fig. 1; A diagram of possible solutions in a two-town game

DIAGRAM: Fig. 2; A geometric diagram of possible solutions in a three-town game

A MATH TWIST

Source: Christian Science Monitor, 09/12/2000, Vol. 92 Issue 203, p18, 0p, 13c

Author(s): Jacobsen, Pamela D.

How a simple paper loop became a major breakthrough - in theory and in practice

Everyone knows that a flat piece of paper has two sides - a back and a front. But there's a way to turn a two-sided strip of paper into a one-sided object. This is not a magic trick. It's called a Mobius (pronounced "MAY-bee-uss" or "MOH-bee-uss") strip.

Jeff Weeks is a geometrician (someone who studies geometry). He's an expert on the Mobius strip, and he also enjoys discussing his favorite math topic, "understanding the three-dimensional world we live in."

Making a Mobius strip from a sheet of paper, Dr. Weeks says, is a good way to start to understand one-sidedness. It illustrates the concept, he says, but it's not perfect. Paper has thickness. A perfect Mobius strip has no thickness.

Imagine a two-dimensional person, Weeks says - someone with height and width but no depth. Now imagine that person walking along a perfect Mobius strip. "They would go around and come back reversed," he says. When they returned to their starting point, "they would be their own mirror image." That's weird.

It gets weirder. Try making Mobius strips and cutting them lengthwise. (See diagrams at left.)

To check to see whether the figures that result are one-sided or two-sided, use a colored marker. Start coloring one side of the strip. When you meet up with where you began, look at the strip. If there's a side that is not colored, it's not a Mobius strip. If you don't see an uncolored side, it is one. Why? Because being one-sided means that what we think of as the "outside" and the "inside" are continuous - the same.

The Mobius strip belongs to a field of geometry called topology (toh-POHL-uh-jee). Those who study it are called topologists. They examine the mathematical properties of various surfaces. One way to describe what topology is all about is to think of an imaginary rubber doughnut.