Solution(s) of a problem
Solution(s) of a problem
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A mathematical problem usually is made up of a list of requirements that must be satisfied. Elements in the list can be translated into:
equations
inequalities
optimization
statement of a more complex requirement
etc.
Whatever the solution is, it needs to satisfy all the requirements. It could be exact or approximate solution. It could be temporary solution if the underlying situations are time dependent and they can change in the future. A solution could be a state described by a list of the values of some variables. For example, a solution to {x + y = 5; x > y; x, y are positive integers} is (x, y) = (4, 1). It is a state described by x = 4 and y = 1 which satisfies the requirements of the problem.
A solution could also be a procedure that performs the required tasks of the problem. It could be an identification of an object if the object is what is looked after in the problem. After translating the problem statements into equivalent and executable simplified forms, we can define the solution and design a way to get there. Once we have a solution, we could know more about it. If what we have found happens to be a non-solution, then we investigate and start again to find a correct solution.
It is also possible that a problem does not have a solution. That may mean that there is nothing we can imagine that can satisfy all requirements of the problem. A problem may also be not solvable due to our insufficient knowledge and computational capacity. In this case, the problem could be modified (often down graded) so that we are able to find a solution to the modified version.
Example (4a): Solve f(t) = 0, where f is a function that maps D to C, D and C are subsets of ℝ.
Solution to this problem is the set of all t-values (from D) that make f(t) equal to zero. There may be no such t, one or more than one in the solution set. If t represents a real variable in the world we live, then we understand that t can take one value at a time.
Example (4b): find all values of (x, y) ∈ A ⨯ B, where A and B are subsets of ℝ so that f(x) > 0, f(x) > g(y) and f(x) - g(y) is maximum.
Solutions to this problem are the values of the ordered pairs (x, y) which satisfies all the following requirements.
x is from A, y is from B.
A, B are subsets of ℝ.
f(x) > 0, f(x) > g(y), f(x) - g(y) is maximum.
One important thing is that for the value of f(x) - g(y) to exist, A ∩ B ≠ ϕ, empty set. This reduces the list of requirements to:
x, y are from A ∩ B.
A, B are subsets of ℝ.
f(x) > 0, f(x) > g(y).
f(x) - g(y) is maximum.
Equivalently, it can be written as,
x, y are from C, C = { t ∈ A ∩ B | f(t) > 0, f(t) > g(t)}.
A, B are subsets of ℝ.
f(x) - g(y) is maximum.
Problem (4c): Suppose we are attempting to build a formula that would estimate the construction cost of a single family two levels above ground houses in a region excluding the costs of land, various permits and licenses, and materials delivered to site. Our assumption is that the construction cost depends on four main factors: numbers of bedrooms and bathrooms, total floor area and grade of finishing. House design details, ceiling height and time interval of construction works are excluded. We are testing the following linear model.
C = aX + bY + cZ + dG,
C = estimated construction cost in $ amount.
X = number of bedrooms.
Y = number of baths.
Z = total floor area in square meters.
G = finishing grade (3 for grade-A, 2 for grade-B, 1 for grade-C).
Our work is to estimate the average values of the coefficients a, b, c and d in the formula for each region. The formula may last for only a period of time due to the fact that the costs of labour and equipment are changing. Therefore, the values of the coefficients need to be updated regularly.
The objective is to find a procedure to generate such a formula. We are allowed to use any high school level math, but not higher than that. For this kind of problem, we could only get a solution which is likely not the best.
Proposed procedure:
[1]: Obtain the data of past construction costs in the region for the last N years. The N value may be 10 years. Organize the data into a proper database.
[2]: Generate subsets of the houses so that all houses in each subset have the same X, Y, Z, and G values. One such subset shall have all houses that have 3 beds, 2 baths, 300 sq-m total floor area, and finishing grade A. Another such subset could be for all houses that have 4 beds, 3 baths, 350 sq-m total floor size, and finishing grade A. According to our model, the construction costs of the houses in one subset should not vary more than an allowed error. And, if S1 and S2 are two different subsets with the houses in S2 having one more bath than those in S1. Then, average construction cost of a houses in S2 should be greater than that in S1. These ideas will be used to check if our linear model works in next steps. However, before we can do that, one important thing is to ask if we have sufficient number of houses in each subset, and if we have sufficient number of different subsets. One subset must have at least five houses as a minimum. Assume that the number five is a recommendation found in a statistics text book. In addition, we must have at least two subsets where only X values are different, and two subsets where only Y values are different, and two subsets where only Z values are different, and finally two subsets where only G values are different. If these tests are not satisfied, then we stop and go back to gather more data. Otherwise, we continue.
[3]: First inspection of validity of the proposed model: First calculate the variation of costs in each subset. Define the error by three numbers: standard deviation of the costs, range = maximum cost - minimum cost, and skewness of the cost distribution (how far the distribution is from symmetry). The skewness may be not used if we believe it has little influence. Compare them with the pre-determined allowed error values. If the observed errores are smaller than allowed values for all sets, then this test is passed. Otherwise, we stop and gather more data or modify the model.
[4]: Second inspection of validity of the proposed model: We check each group of subsets where only one of four parameters are different. Suppose we have two subsets S2 and S3 corresponding to two and three bedrooms respectively. All other three parameters are the same. Then, average cost of S2 must be smaller than average cost of S3. Repeat this for other pairs of subsets with difference in Y only, Z only and G only. We may also have three subsets S2, S3, and S4 corresponding to two, three and four bedrooms respectively. In this case, we must have average cost of S2 < average cost of S3 < average cost of S4. If all these requirements are met, then we continue. Otherwise we stop.
[5]: Estimate of coefficient values: From the groups of subsets where only X-values are different, we estimate the coefficient a value for X. Let S1 and S2 be the subsets of the houses where only X values are different. Let C(S1) and C(S2) be average costs in respective subset. Then,
Estimated a = [ C(S2) - C(S1) ] ÷ [X(S2) - X(S1)].
If we have three subsets S1, S2, and S3 where only X values are different, we have three estimates of a. Then do average of these three to get the final estimate of a. Repeat these for estimates of b, c and d.
[6]: Final tests and reporting: AFter having the estimates of all four coefficients, we have the model formula:
C = aX + bY + cZ + dG,
Test this formula for each house in the databse and record the percent error to report together. Finally, all these steps should be repeated regularly. If we get enough new houses in the database each year, then the procedure should be performed once a year and publish the updated formula.
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