Mathematical Modeling, Lecture 1

Transcription

1 Mathematical Modeling, Lecture 1 Gudrun Gudmundsdottir January

2 Some practical issues A lecture each wednesday , with some exceptions Text book: Meerschaert We go through the text and also do exercises Two projects during the course Written exam on May 26, based on the exercises The slides give an overview of the course, they will be on the home page Two main themes: Optimization and Dynamical Systems

3 The modeling process Mathematical modeling is to take a problem from the real world, state it in mathematical terms, then solve it and relate the solution to the original problem. It is useful to have a system for how to proceed, for example: 1. Identify the problem. Collect information, formulate a question, state the objective. 2. Formulate the problem mathematically, select the modeling approach. 3. Solve the problem analytically as far as possible. 4. Solve the problem numerically, or do a simulation. 5. Interprete the answer, present it, relate to the original question. Each step may require going back and reconsider earlier steps.

4 Optimization Optimus= best Find the maximum of a function f (x) for x in a specified set S. The set S may be specified by a number of functions, g j (x) b j, j = 1... m. Different branches Linear optimization, with linear functions Non-linear optimization Discrete optimization Dynamical optimization Combinatorial optimization

5 A simple example, one variable optimization A farmer has a field with potatoes weighing 1000 kg. The potatoes will grow on if they are left in the ground, but the prices of potatoes are declining. When should the farmer harvest and sell? Assume: 1. Weight on day t = w(t) = t kg 2. Price per kg on day t = p(t) = t kr 3. Cost for keeping the field (water, fertilizer..) 100 kr per day We want to maximize the profit.

6 Mathematical formulation and solution The profit is f (t) = ( t)(35 0.5t) 100t = 12.5t t for 0 t t max. The derivative is zero when t = 11. So if t max 11, the farmer should harvest after 11 days.

7 Sensitivity analysis In our example, we assumed that The rate by which the price is falling is r = 0.5 The rate of growth is g = 25. The cost per day is c = 100 The value of these parameters is uncertain. Let us find out how the result t = 11 depends on them.

8 Dependence on r We have f (t) = ( t)(35 rt) 100t = 25rt 2 +t( r) and f (t) = 0 when t = t r = 15.5/r 20. To see how sensitive t r is for changes in r we can differentiate, or look at the graph. We have t r < 0 if r > 15.5/20 = In that case, the farmer should sell immediately, for the profit is decreasing for t greater than the maximum point t r.

9 Dependence on g We have f (t) = (1000+gt)(35 0.5t) 100t = 0.5gt 2 +t(35g 600) and f (t) = 0 when t = t g = /g. To see how sensitive t g is for changes in g we can differentiate, or look at the graph. We have t g < 0 if g < In that case, the farmer should sell immediately, for the profit is decreasing for t greater than the maximum point t g.

10 MATM20 Lecture 1 Sensitivity as relative change If a quantity x changes by x then the relative change is x/x or 100 x/x%. A change by r in the rate r of price decline results in a change t in the optimal time t to sell. Consider the ratio between the relative changes and let r 0. Then t/t r/r dt r dr t = S(t, r). This quantity is called the sensitivity of t to r.

11 Sensitivity of t to r and to g For the potato problem we have and dt dr = 15.5r 2, S(t, r) = 15.5, S(11, 0.5) = 2.8. rt dt dg = 600g 2, S(t, g) = 600, S(11, 25) = 2.2. gt This means that if r increases by 10% then the optimal time until harvest is shortened by 28%. If g increases by 10% then the optimal time until harvest will be about 22% longer.

12 Robustness A mathematical model will never be perfect. The conclusions a mathematical model leads to should preferably be correct even though the model is not completely accurate. Such a model is called robust. Robustness is not a precisely defined property. In the potato problem we assumed that the price p and weight w were linear functions of the time. Even without this assumption we have f (t) = p(t)w(t) 100t The derivative is zero when p w + pw = 100.

13 Conclusions The term p w + pw is the increase in value of the potatoes. The model tells us not to harvest as long as the increase in value is greater than the cost of keeping the field. But there is no way of getting a precise formula for p(t) and w(t). We can still follow the increase in value of the whole crop and harvest when it is less than 100 per day. If we want to predict how many days there will be until this happens, we can take the current values of p and w and assume that these will be constant for the next few days and do the calculation above.

14 MATM20 Lecture 1 Conclusions The sensitivity analysis tells us that if the growth rate (g = w ) is within 10% of 25 then the optimal number of days to wait will be within 22% of 11, or something between 9 and 13. The price derivative will probable be less negative in the future. So the advice is to wait about 9 days, or a week, and then check again.

15 Optimization in several variables A continuous function f on a compact set S attains its maximum at a point which is either 1. An inner point of S where all the derivatives f / x j = 0 2. A boundary point of S 3. or a point where one or more of the partial derivatives is undefined

16 A simple example in two variables A company is going to start manufacturing two kinds of a product, product A and product B. The startup cost is and one unit of A costs kr. 195 to make and one unit of B costs 225 to make. The company will make x units of A and y units of B per year. We assume that all the units made can be sold, but the average price for a unit will depend on how many are made per year in the following way: p A = x 0.003y p B = x 0.01y

17 Maximize the profit The profit will be f (x, y) = xp A + yp B 195x 225y = 0.01x y xy + 144x + 174y , This is a negative definite quadratic function, and therefore has a global maximum where the derivatives are zero.

18 The Solution We solve the system and get f / x = 0.02x 0.007y = 0 f / y = 0.02y 0.07x = 0 The profit will be x = , y =

19 MATM20 Lecture 1 Computational tools PYTHON from