Biomathematics PDF

Biomathematics Dr. Mohamed Bondok Biomedical Engineer Second year Biomathematics Biomathematics is the application of mathematical, statistical, and computational methods in biology. Traditionally this was mainly in developing mathematical models of biological systems, from the molecular and...

Biomathematics Dr. Mohamed Bondok Biomedical Engineer Second year Biomathematics Biomathematics is the application of mathematical, statistical, and computational methods in biology. Traditionally this was mainly in developing mathematical models of biological systems, from the molecular and cellular levels through to whole organism models, and population and ecological studies. A mathematical model is a description of the behavior of a system using mathematical language. Objectives of this course It is an introduction to the application of mathematical models in the study of biological systems. The main objectives of this course is to introduce (i) Some basic techniques for developing mathematical models of biological processes, (ii) Some basic mathematical knowledge and skills required in the analysis of the models. Many examples of the models introduced in this course are taken from population biology. Why should you also study mathematics and statistics if you are interested in biology or medicine? Mathematics as a language is often far more suitable for precise description of the characteristics of various biological processes. 2 INDICES AND LOGARITHMS INDICES Indices or powers show how many times a number is multiplied by itself. Properties of indices 1) am an = am+n 5) (a/b)m = am/bm 2) am/an = am-n 6) a0 = 1 3) (am)n = amn 7) a-m = 1/(am) 4) (ab)m = ambm 8) am/n = n√am Exponential functions Are functions where f(x) = ax + B where a is any real constant and B is any expression. For example, f(x) = e-x - 1 is an exponential function. The exponential function exp(t) is given by exp(t) = et, where e = 2.71828.... Logarithmic functions Are the inverse of exponential functions. For example, the inverse of y = ax is y = logax, which is the same as x = ay. (Logarithms written without a base are understood to be base 10.) Ex. Convert to logarithmic form: 8 = 2x Solution: Remember that the logarithm is the exponent. x = log2 8 3 Logarithms and Solving Equations The logarithm of x to the base a is written "loga x" and is defined as follows. loga x = b if and only if ab = x where a > 0 (a≠ 1), and x >0. Note that the base must be positive and different from 1, and the expression that you are taking the logarithm of must also be positive. Some logarithms can be evaluated easily. For example, log4 64 = 3 since 43 = 64. Special Bases If the log base is 10, then the log is called the common logarithm and we write "log" for log10. If the log base is the number "e", then the logarithm is called the natural logarithm and we write "ln" for loge. The "ln" key on a scientific calculator gives values for the natural logarithm. 4 Properties of Logarithms Here, a > 0 and a≠1. 1. loga1 = 0 2. loga a = 1 3. loga(ax) = x 4. a(logax) = x, if x > 0 5. logax + loga y = loga(xy) if x, y > 0 6. logax - loga y = loga (x/y) if x, y > 0 7. loga (xr) = r loga x if x > 0 Example 1 : Find parameters a and b so that f(0) = 0 and f(1) = 2, where f is a logarithmic function given by f(x) = a*Log2(x + b) Solution to Example 1: Use the fact that f(0) = 0 to obtain 0 = a*Log2 (0 + b) Divide both sides by a to obtain Log2b = 0 Solve for b b = 20 =1 Function f can be written as a*Log2(x + 1) Use the fact that f(1) = 2 to obtain 2 = a*Log2 (1 + 1) Simplify, Log22 = 1 a=2 Function f is given by f(x) = 2*Log2(x + 1) 5 Problem: Solve for x: 3x = 8. Solution: Take the logarithm of both sides. log 3x = log 8 Use theorem 7 to simplify the equation. x * log 3 = log 8 Solve for x by dividing each side by log 3. x = (log 8/log 3) A decimal approximation may be found if desired - x = 1.8929. Problem: Solve log3 (5x + 7) = 2 for x. Solution: Write an equivalent exponential expression. 5x + 7 = 32 5x + 7 = 9 Solve for x. 5x = 2 x = (2/5) Problem: Solve log2 x = -3. Solution: Convert the logarithm to exponential form. 2-3 = x x = (1/8) abc² Example Apply the laws of logarithms to log. d3 Answer. According to the 5,6 laws, abc² log = log (abc²) − log d 3 d3 = log a + log b + log c² − log d 3 = log a + log b + 2 log c − 3 log d, 6 Logarithms and Modeling Many phenomena in nature seem to follow the law that an amount A varies with time according to the formula A=A0 ekt Where A0 is the original amount (the amount at time t=0) and k is a non-zero constant. If k is positive, then the amount A gets larger or grows with time, and the amount is said to be experiencing exponential growth. If k is negative, then the amount A gets smaller or diminishes with time, and the amount is said to be experiencing exponential decay. When solving exponential growth/decay problems, two situations often occur: Time is given; you are to find the amount at that given time. This usually just involves evaluating the amount function. The amount is given; you are to determine at what time this amount occurs. This usually involves solving an exponential equation, which means logarithms will be needed. Modeling Exponential Decay - Using Logarithms A common example of exponential decay is radioactive decay. Radioactive materials, and some other substances, decompose according to a formula for exponential decay. That is, the amount of radioactive material A present at time t is given by the formula A=A0ekt where k < 0. A radioactive substance is often described in terms of its 7 half-life, which is the time required for half the material to decompose. Problem After 500 years, a sample of radium-226 has decayed to 80.4% of its original mass. Find the half-life of radium-226. Solution Let A= the mass of radium present at time t (t=0 corresponds to 500 years ago). We want to know for what time t is A = (1/2)A0. However, we do not even know what k is yet. Once we know what k is, we can set A in the formula for exponential decay to be equal to (1/2)A0, and then solve for t. First we must determine k. We are given that after 500 years, the amount present is 80.4% of its original mass. That is, when t=500, A=0.804 A0. Substituting these values into the formula for exponential decay, we obtain: 0.804 A0=A0ek(500). Dividing through by A0 gives us 0.804 = e500k Which is an exponential equation. To solve this equation, we take natural logs (ie. ln) of both sides. (Common logs could be used as well.) ln ( 0.804) = ln (e500k) We know that ln (e500k) = 500k by the cancellation properties of ln and e. So the equation becomes ln ( 0.804) = 500k and k= (ln 0.804)/500. This is the exact solution; evaluate the natural log with a calculator to get the decimal approximation k = -0.000436. 8 Since we now know k, we can write the formula (function) for the amount of radium present at time t as A=A0 e-0.000436 t. Now, we can finally find the half-life. We set A=1/2 A0 and solve for t. (1/2)A0=A0 e-0.000436 t Dividing through by A0 again, we get: 1/2 = e-0.000436 t. To solve for t, take natural logs: ln(1/2) = ln[e-0.000436 t]. Then applying the cancellation property for logarithms yields ln (1/2) = -0.000436 t So t= ln(1/2) /(-0.000436) or t = 1590. The half-life is approximately 1590 years. 9 Modeling Exponential Growth - Using Logarithms A common example of exponential growth is bacteria growth Example The count in a bacteria culture was 400 after 2 hours and 25,600 after 6 hours. (Assume exponential growth.) a. What was the initial size of the culture? b. Find an expression for the population after t hours. c. In what period of time does the population double? Solution Let N be the number of bacteria at time t. Then an exponential growth model says that the formula for N in terms of t is: N= N0 ekt And we have that when t=2, N=400 and when t=6, N= 25,600. Substituting each pair of t and N into the formula we get: 400= N0 ek(2) and 25600 = N0 ek(6) If we divide the second equation by the first, we get: 64 = e6k / e2k because N0 cancels. And subtracting the exponent of e gives: 64= e4k 10  taking the logarithm for each side of the equation 64= e4k ln (64) = ln e4k = 4k Solving for k: k = ln(64) /4 = 1.0397 (approx.). 64= e4k To find N0, the original size of the culture, substitute into the first equation we established: 400= N0 e2(1.0397) You should get 400 = 8N0, so N0= 50. N= N0 ekt = 50e1.0397t. For doubling time, N=100, t=? Set N=100 in the formula: 100=50e1.0397t Solving for t gives t= ln(2) / 1.0397 = 0.67. Or the doubling time is approximately 40 minutes. Example The number of bacteria in a dish has tripled every month. If there were 2500 bacteria at first, how many bacteria were there 10 months later. Using A= prt and by creating an equation. A(t) = Ao(3)t A(10) = 2500(3)10 A(10) = 2500 x 59049 A(10) = 147,633,500 11 EXAMPLE : A colony of bacteria double every 6 days. If there were 3000 bacteria at the start of an experiment how many bacteria will there be in 15 days? Solution A=? A=A°Mt/p A°=3000 =3000(2)15/6 M=2 =16970.5628 t=15 ≈16971 bacteria p=6 Example Two populations of bacteria are growing at different rates. Their populations at time t are given by 5t+2 and e2t respectively. At what time are the populations the same? This problem requires us to solve the equation: 5t+2 = e2t We need to use loge because of the base e on the right hand side. ln (5t+2) = ln (e2t) (t + 2) ln 5 = 2t ln e Now, ln e = 1, and we need to collect t terms together: t ln 5 + 2 ln 5 = 2t t (ln 5 - 2) = - 2 ln 5 So is the required time. 12 Example Assume that a population P is growing exponentially, so P = aebt, where t is measured in years. If P = 15000 in 2006, and P has grown to 17000 in 2009, find the formula for P. Let t be the number of years since 2006. Then a = 15000, the value of P when t = 0. Note that t could have been chosen differently. For instance, we could let t be the number of years since 2006. But then we would not know the value of P when t = 0, so we would not know the value of a immediately. We still need to find b. All we know about b is that it is positive, since the population is growing. Using the value we have found for a, we have P = 15000 ebt. The year 2009 corresponds to t = 3, so we substitute P = 17000 and t = 3 in the equation above and solve for b. 17000 = 15000 e3b We have solved equations of this form several times. The first step is to isolate the exponential term. Then take the natural logarithm of both sides. 17/15 = e3b ln (17/15) = ln e3b ln (17/15) = 3b b = (ln(17/15))/3 = 0.0417 (approximately) Therefore P = 15000 e0.0417 t Based on this model, when will the population reach 20000? 13 Set P equal to 20000 and solve for t. 20000 = 15000 e0.0417 t 4/3 = e0.0417 t ln (4/3) = ln e0.0417 t ln (4/3) = 0.0417 t t = (ln (4/3))/0.0417 = 6.9 years So, we expect the population to reach 20000 toward the end of 2012. Exercise : Find a model of the type P = aebt, where t is the number of years since 2002, if P = 30000 in 2002 and P = 36000 in 2009. Use this model to predict the value of P in 2012. Solution P = 30000 e 0.026 t In 2012, P was approximately 38,900 Example A lab technologist places a bacterial cell into a vial at 5 am. The cells divide in such a way that the number of cells doubles every 4 minutes. The vial is full 1 hour later. a) How long does it take for the cells to divide to 4095? b) At what time is the vial 1/16 full? Ptotal = Po ekt Atotal = Ao Ct/p e = is the irrational number 2.71828... One way to do this problem using the first or third equations: When t= 0, ek0= e0 so Ptotal= Po= 1 since there was one bacterium when t= 0. After 4 minutes, that bacterium has doubled to 2 bacteria: as long as t is measured in minutes, Ptotal= 2= 1(ek*4) so e4k= 2. Taking ln of both sides K =(1/4)ln(2). Now you know that Ptotal= e(ln 2)(t/4). 14 To answer (a), solve the equation e(ln 2)t = 4095 for t. Another way, using the second equation above: Since the number of bacteria double every 4 minutes, we could just multiply by t every 4 minutes: if t is measured in minutes, that means we double every t/4 so Atotal= Ao (2)t/4. Again, if t= 0, Atotal= 1= Ao so the formula is just Atotal= 2t/4. Solve 2t/4= 4095 to t. You can get a good approximation to that by recognizing that 4096 to our calculator is 212. 212 = 2t/4 t/4= 12 t = 48 b) At what time is the vial 1/16 full? Work backwards from one hour. Since the number of bacteria double every 4 minutes, 4 minutes before the hour (56 minutes) the vial must have been 1/2 full. 4 minutes before that (52 minutes) the vial must have been 1/4 full, etc. 15 linear programming Introduction. In "real life", linear programming is part of a very important area of mathematics called "optimization techniques". This field of study is used every day in the organization and allocation of resources. These "real life" systems can have dozens or hundreds of variables, or more. In algebra, though, you'll only work with the simple (and graphable) two-variable linear case. The mathematical models which tell to optimize (minimize or maximize) the objective function Z subject to certain condition on the variables is called a Linear programming problem (LPP). The general process for solving linear-programming exercises is to graph the inequalities (called the "constraints") to form a walled-off area on the x,y-plane (called the "feasibility region"). Then you figure out the coordinates of the corners of this feasibility region (that is, you find the intersection points of the various pairs of lines), and test these corner points in the formula (called the "optimization equation") for which you're trying to find the highest or lowest value. The standard form of the linear programming problem is used to develop the procedure for solving a general programming problem. A general LPP is of the form Max (or min) Z = c1x1 + c2x2 + … +cnxn x1, x2,....xn are called decision variable. 54 Basic Concept of Linear Programming Problem Objective Function: The Objective Function is a linear function of variables which is to be optimised i.e., maximised or minimised. e.g., profit function, cost function etc. The objective function may be expressed as a linear expression. 2x + 4y Constraints: A linear equation represents a straight line. Limited time, labour etc. may be expressed as linear inequations or equations and are called constraints. 2x + y > 8 Optimisation: A decision which is considered the best one, taking into consideration all the circumstances is called an optimal decision. The process of getting the best possible outcome is called optimisation. Minimize Maximize Solution of a LPP: A set of values of the variables x1, x2,….xn which satisfy all the constraints is called the solution of the LPP.. Feasible Solution: A set of values of the variables x1, x2, x3,….,xn which satisfy all the constraints and also the non- negativity conditions is called the feasible solution of the LPP. The feasible solutions must lie within a certain well-defined region of the graph. Optimal Solution: The feasible solution, which optimises (i.e., maximizes or minimizes as the case may be) the objective function is called the optimal solution. 55 Mathematical Formulation of Linear Programming Problems There are mainly four steps in the mathematical formulation of linear programming problem as a mathematical model. We will discuss formulation of those problems which involve only two variables. 1. Identify the decision variables and assign symbols x and y to them. These decision variables are those quantities whose values we wish to determine. 2. Identify the set of constraints and express them as linear equations/inequations in terms of the decision variables. These constraints are the given conditions. 3. Identify the objective function and express it as a linear function of decision variables. It might take the form of maximizing profit or production or minimizing cost. 4. Add the non-negativity restrictions on the decision variables, as in the physical problems, negative values of decision variables have no valid interpretation. 56 Graphical Method of Solution of a Linear Programming Problem The graphical method is applicable to solve the LPP involving two decision variables x1, and x2, we usually take these decision variables as x, y instead of x1, x2. To solve an LPP, the graphical method includes two major steps. a) The determination of the solution space that defines the feasible solution (Note that the set of values of the variable x1, x2, x3,....xn which satisfy all the constraints and also the non- negative conditions is called the feasible solution of the LPP). b) The determination of the optimal solution from the feasible region. A linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to linear constraints. The constraints may be equalities or inequalities. Here is a simple example. Find numbers x1 and x2 that maximize the sum x1 + x2 subject to the constraints x1 ،> 0, x2 >.0, and x1 + 2x2 < 4 4x1 + 2x2 < 12 -x1 + x2< 1 In this problem there are two unknowns, and five constraints. All the constraints are inequalities and they are all linear in the sense that each involves an inequality in some linear function of the variables. The first two constraints, x1 > 0 and x2 > 0, are special. These are called nonnegativity constraints and are often found in linear programming problems. The other constraints are then called the main constraints. The function to be maximized (or minimized) is called the objective function. Here, the objective function is x1 + x2. 57 The largest or smallest value of the objective function is called the optimal value Since there are only two variables, we can solve this problem by graphing the set of points in the plane that satisfies all the constraints (called the constraint set) and then finding which point of this set maximizes the value of the objective function. Each inequality constraint is satisfied by a half-plane of points, and the constraint set is the intersection of all the half-planes. In the present example, the constraint set is the five sided figure shaded in Figure 1. We seek the point (x1, x2), that achieves the maximum of x1 +x2 as (x1, x2) ranges over this constraint set. The function x1 + x2 is constant on lines with slope -1, for example the line -x1 + x2 = 1, and as we move this line further from the origin up and to the right, the value of x1 + x2 increases. Therefore, we seek the line of slope -1 that is farthest from the origin and still touches the constraint set. This occurs at the intersection of the lines x1 +2x2 = 4 and 4x1 +2x2 = 12, namely, (x1, x2) = (8/3, 2/3). The value of the objective function there is (8/3) + (2/3) = 10/3. 58  Sketching the Solution Set of a Linear Inequality To sketch the region represented by a linear inequality in two variables: A. Sketch the straight line obtained by replacing the inequality with an equality. B. Choose a test point not on the line ((0,0) is a good choice if the line does not pass through the origin, and if the line does pass through the origin a point on one of the axes would be a good choice). C. If the test point satisfies the inequality, then the set of solutions is the entire region on the same side of the line as the test point. Otherwise it is the region on the other side of the line. In either case, shade out the side that does not contain the solutions, leaving the solution region showing. Feasible Region The feasible region determined by a collection of linear inequalities is the collection of points that satisfy all of the inequalities. To sketch the feasible region determined by a collection of linear inequalities in two variables: Sketch the regions represented by each inequality on the same graph, remembering to shade the parts of the plane that you do not want. What is unshaded when you are done is the feasible region. 59 Graphical Method The graphical method for solving linear programming problems in two unknowns is as follows. A. Graph the feasible region. B. Compute the coordinates of the corner points. C. Substitute the coordinates of the corner points into the objective function to see which gives the optimal value. Example To sketch the linear inequality 3x - 4y ≤ 12, first sketch the line 3x - 4y = 12. y= - (12-3x)/4 = -3+ ¾ x Next, choose the origin (0, 0) as the test point (since it is not on the line). Substituting x=0, y=0 in the inequality gives 3(0) - 4(0) ≤ 12. Since this is a true statement, (0, 0) is in the solution set, so the solution set consists of all points on the same side as (0, 0). This region is left unshaded, while the (grey) shaded region is blocked out. 60 Example The feasible region for the following collection of inequalities is the unshaded region shown below (including its boundary). 3x - 4y ≤ 12, x + 2y ≥ 4 x≥1 y ≥ 0. Example Minimize C = 3x + 4y subject to the constraints 3x - 4y ≤ 12, x + 2y ≥ 4 x ≥ 1, y ≥ 0. Solution The feasible region for this set of constraints was shown above. Here it is again with the corner points shown. The following table shows the value of C at each corner point: Therefore, the solution is x = 1, y = 1.5, giving the minimum value C = 9. Point C = 3x + 4y (1, 1.5) 3(1)+4(1.5) = 9 minimum (4, 0) 3(4)+4(0) = 12 61 Example Find the maximal and minimal value of z = 3x + 4y subject to the following constraints: Solution The three inequalities in the curly braces are the constraints. The area of the plane that they mark off will be the feasibility region. The formula "z = 3x + 4y" is the optimization equation. We need to find the (x, y) corner points of the feasibility region that return the largest and smallest values of z. The first step is to solve each inequality for the more-easily graphed equivalent forms: It's easy to graph the system: 62  To find the corner points -- which aren't always clear from the graph. We'll pair the lines (thus forming a system of linear equations) and solve: y = –( 1/2 )x + 7 y = –( 1/2 )x + 7 y = 3x y = 3x y=x–2 y=x–2 –( 1/2 )x + 7 = –( 1/2 )x + 7 = x 3x = x – 2 3x –2 2x = –2 –x + 14 = 6x –x + 14 = 2x – 4 x = –1 14 = 7x 18 = 3x 2=x 6=x y = 3(–1) = –3 y = 3(2) = 6 y = (6) – 2 = 4 corner point at corner point corner pt. at (2, 6) at (6, 4) (–1, –3) So the corner points are (2, 6), (6, 4), and (–1, –3). Somebody really smart proved that, for linear systems like this, the maximum and minimum values of the optimization equation will always be on the corners of the feasibility region. So, to find the solution to this exercise, we only need to plug these three points into "z = 3x + 4y". (2, 6): z = 3(2) + 4(6) = 6 + 24 = 30 (6, 4): z = 3(6) + 4(4) = 18 + 16 = 34 (–1, –3): z = 3(–1) + 4(–3) = –3 – 12 = –15 Then the maximum of z = 34 occurs at (6, 4), and the minimum of z = –15 occurs at (–1, –3). 63 Example The objective is to maximize 2x + 3y, constraints are The implied constraints tell us that the graph will be in the upper right quadrant of the plane. When we graph the first constraint, the x and y intercepts are both 3 Since this is a "greater than" constraint the side of the line where the x's and y's get bigger will be the correct side. This region goes on forever. In the next constraint, the x intercept is 80/20 = 4, and the y intercept is 80/10 = 8. 64 Since this is a "less than" constraint, the points on the side where x and y get small should be shaded. The set of points which satisfy both of theses constraints are shaded above. Finally In this type of equation, the x and y intercept will both be at (0, 0), so we will need to plot at least one other point. If we let x = 4 then if y = 2x we will have y = 8. Plot the point (4, 8) and connect it with (0, 0). To tell which side of this line is which, notice that the x's are on the large side of the inequality and the y's are on the small side. So we shade the side of the line where the x's are big and the y's are small. We label the lines with the equations from which they come so that we can 65  Find the feasible corner points. Where we see a y we substitute 2x, and the first equation becomes x + 2x = 3 3x = 3 x=1 If x = 1 then y = 2. Hence A = (1, 2) Again, where we see a y we substitute 2x, and the first equation becomes 20x + 10(2x) = 80 20x + 20x = 80 40x = 80 x=2 If x = 2 and y = 2x then y = 4. Hence B = (2, 4) 66 C is the x intercept for x + y = 3, so C = (3, 0) D is the x intercept for 20x + 10y = 80 so D = (4, 0) Evaluate the objective function. At (1, 2) we get 2(1) + 3(2) = 8.. At (2, 4) we get 2(2) + 3(4) = 16. At (3, 0) we get 2(3) + 3(0) = 6. At (4, 0) we get 2(4) + 3(0) = 8 The maximum yield 16 by processing 2 x and 4 y 67 Example - designing a diet A dietitian wants to design a breakfast menu for certain hospital patients. The menu is to include two items A and B. Suppose that each ounce of A provides 2 units of vitamin C and 2 units of iron and each ounce of B provides 1 unit of vitamin C and 2 units of iron. Suppose the cost of A is 4¢/ounce and the cost of B is 3¢/ounce. If the breakfast menu must provide at least 8 units of vitamin C and 10 units of iron, how many ounces of each item should be provided in order to meet the iron and vitamin C requirements for the least cost? What will this breakfast cost? Solution Let x = #oz. of A Let y = #oz. of B vit. C: 2x + 1y > 8 iron: 2x + 2y > 10 x > 0, y >0 Cost = C = 4x+3y 68 Example 69 The diet problem can be easily stated as follows: Minimize the cost of food eaten during one day Subject to the requirements that the diet satisfy a person's nutritional requirements and that not too much of any one food be eaten. The first step is to assign variables for the amounts of food that we are solving for and parameters for the data that we know. An Example: The Diet Problem. A student is trying to decide on lowest cost diet that provides sufficient amount of protein, with two choices: – steak: 2 units of protein/pound, $3/pound – peanut butter: 1 unit of protein/pound, $2/pound. In proper diet, need 4 units protein/day. Let x = # pounds peanut butter/day in the diet. Let y = # pounds steak/day in the diet. Goal: minimize 2x + 3y (total cost) subject to constraints: x + 2y ≥ 4 x ≥ 0, y ≥ 0 This is an LP- f Assume for simplicity that there are only two foods, which for definiteness we shall assume are cereal and steak. Further, we assume that there are only two products people need to stay alive, iron and protein; each day a person must consume at least 60 units of iron and at least 70 units of protein to stay alive. Let us assume that one unit of cereal costs $20 and contains 30 unit of iron and 5 units of protein, and that one unit of steak costs $2 and contains 15 units of iron and 10 units of protein. The goal is to find the cheapest diet which will satisfy the minimum daily requirements Let x1 represent the number of units of cereal that the person consumes a day, and x2 the number of units of iron consumed. For the diet to meet the minimum requirements, we must have 30x1 + 5x2 ≥ 60, 15x1 + 10x2 ≥ 70, x1 ≥ 0, x2 ≥ 0, Minimize cost(x1, x2) = 20x1 + 2x2, 70