Model Building: Selected Case Studies

Chapter 2 Model Building: Selected Case Studies The goal of Chapter 2 is to illustrate the basic process in a variety of selfcontained situations where the process of model building can be well illustrated using mathematical ideas generally involving only basic concepts from calculus and probability. The topics have been selected in part because in most cases there are natural alternatives to the assumptions used in the text. This provides an opportunity to discuss the validity of the assumptions made and the limitations of the models. Section 2. is a model for classical Mendelian genetics. This approach to the topic is of great historical interest, and continues to be useful in certain applications. Section 2.2 includes some models for single species population dynamics and illustrates the cycle of model building. Additional models for the same situation are included in Sections 2.5 and 2.6 and these can be used to illustrate how a model changes when the basic assumptions are changed. The models for Social Choice (Section 2.3) illustrate models of a very different sort, models involving only logic and basic counting principles. This section also illustrates how a negative result (a conclusion that something cannot be true) is sometimes a very useful outcome of the modeling process. Sections 2.4 and 2.7 provide models that are common and extremely useful to help understand situations arising in economics and business. Finally, Section 2.8 introduces some of the issues to be considered in model calibration and validation. In a one semester course we would typically discuss 3 or 4 of these sections over a period of about 3 weeks. 3

4 CHAPTER 2 Model Building: Selected Case Studies Exercises and Solutions for Chapter 2 2. Mendelian Genetics. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by selfing. At the nth observation, these frequencies are [.3.2.5 ]. (a) What will the frequencies be at the (n + ) observation? (b) What were the frequencies at the (n ) observation? (c) What will the frequencies be at the (n + 2) observation? (d) What were the frequencies at the (n 2) observation? Solution. Let x 0 denote the initial and x i the ith genotypic distribution. By Theorem 2.2, x = x 0 M. Using x as our initial genotypic distribution, we get x 2 = x M = x 0 M 2. Continuing in this manner, we conclude that x n = x 0 M n. (a) [.35..55 ]. (b) Solve the equation xm = [.3.2.5 ] for x and find x = [.2.4.4 ]. (c) We need to find x n+2 : x n+2 = x n M 2 = [.375.05.575 ]. (d) The goal is to find a probability vector x such that (xm)m = [.3.2.5 ]. It will be helpful to use the inverse M of M. We have 0 0 M = 2 2 2 0 0 Using M we find x = [.3.2.5 ] (M )(M ) = [ 0.8.2 ]. This vector x has the desired properties. 2. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by selfing. Use mathematical induction to show that if the initial genotypic distribution is [ x y z ], then the genotypic distribution in the nth generation is [ x + y 2 n 2 y n+ 2 y 2n n 2 + z ]. n+ Solution. The population reproduces by selfing, and using Theorem 2.2, the nth genotypic distribution is given by [ x y z ] M n. We need to calculate M n. To this end, let us suppose that 0 0 M n = 2 n 2 n. 2 n+ 2 n 2 n+ 0 0

2. Mendelian Genetics 5 Then M M n = = = 0 0 4 + 2 n 2 n 2 2 n+ 2 2 n 2 2 + n+ 4 0 0 0 0 2 n 2 + 2n 2 n n+2 2 n+2 2 n+ 2 + 2n n+2 2 n+2 0 0 0 0 2 n+. 2 n+ 2 n+2 0 0 2 n+ 2 n+2 Hence, M n+ also has this form. Clearly, M = M has this form, so by the principle of mathematical induction, 0 0 M n = 2 n 2 n 2 n+ 2 n 2 n+ 0 0 for any n. It follows that the nth genotypic distribution is [ x + y(2 n )/2 n+ y/2 n y(2 n )/2 n+ + z ]. 3. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by random mating. If the initial genotypic distribution is [.5.5 0 ], find the genotypic distribution in the second filial generation and show that this distribution does not change in future generations. Solution. There are only three possible matings: AA AA, AA Aa, and Aa Aa. The frequency of each mating (as a proportion of all matings) and the resulting distributions are: Mating Pair AA AA AA Aa Aa Aa Frequency 4 Distribution of Offspring [ 0 0 ] [ 2 2 2 0 ] [ 4 We calculate the genotypic distribution of the offspring through vector addition: [ ] [ 0 0 + 4 2 2 = [ 4 + 4 + 6 2 0 ] + 4 4 + 8 6 [ 4 2 ] = [ 9 6 ] 4 3 8 6]. So the genotypic distribution in the first filial generation is [ 9 6 6 6 6]. Although we show in Exercise 4 that the genotypic distribution does not change after the first filial generation in a population which reproduces by 4 2 ] 4

6 CHAPTER 2 Model Building: Selected Case Studies random mating, we will do the explicit calculation here for the second filial generation. Recall that a pairing of mixed genomes can occur in two ways, while a pairing of the same genomes occurs in just one. That is, AA Aa occurs when the male parent is genotype AA or when the female parent is genotype AA. This accounts for the multiplication by 2 in the table below. We have the following possible matings and offspring: Pairing AA AA AA Aa Aa Aa Aa aa aa aa AA aa 8 Freq. [ ] [ Dist. 0 0 2 256 2 54 256 2 0 ] [ 4 36 256 2 6 256 ] [ 2 4 0 2 2 Again we use vector arithmetic: [ ] [ 8 256 0 0 + 2 54 256 2 2 0 ] + 36 [ ] + 2 6 256 0 2 2 + 256 = [ 8+54+9 256 54+8+6+8 256 [ 256 4 2 256 2 9 256 ] [ 0 0 ] [ 0 0 ] 4 ] [ 0 0 ] + 2 9 256 9+6+ 256 ] = [ 9 6 6 6 [ 0 0 ] ] 6 This is the same genotypic distribution that we started with, so the genotypic distribution will not change after the first filial generation. 4. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by random mating. Show that the genotypic frequencies are constant after the first filial generation. The result is the Hardy- Weinberg law. Solution. Assume the genotypic distribution is initially given by the probability vector [x y z]. We have the following possible matings and offspring: Pairing AA AA AA Aa Aa Aa Aa aa aa aa AA aa Freq. [ x 2 ] [ 2xy y 2 2yz z 2 2xz Dist. 0 0 2 2 0 ] [ ] [ ] [ ] [ ] 4 2 4 0 2 2 0 0 0 0 The genotypic distribution of the first filial generation is given by [ x 2 + 2xy 2 + y2 4 2xy 2 + y2 2 + 2yz 2 + 2xz y2 4 + 2yz 2 + z2] = [ (x + y 2 )2 xy + y 2 2 + yz + 2xz ( y 2 + z)2] so the genotypic distribution of the first filial generation is v = [ (x + y 2 )2 2(x + y 2 )( y 2 + z) ( y 2 + z)2]. We could do exactly the same procedure using v as our initial distribution to get a distribution v 2 for the second filial generation. Reducing the equations in v 2 by using the relation x + y + z = will give us v = v 2. However, we can do this step directly, by making an appropriate change of variables.

2. Mendelian Genetics 7 Setting a = x + y 2 and b = z + y 2 we see that v has the form [a 2 2ab b 2 ] where a + b =. We apply the same procedure as above to find v 2 and we have v 2 = [ (a 2 + ab) 2 2a 3 b + a 2 b 2 + 2ab 3 + 2a 2 b 2 (ab + b 2 ) 2] = [ a 2 (a + b) 2 2ab(a 2 + ab + b 2 + ab) b 2 (a + b) 2] = [ a 2 2ab(a + b) 2 b 2] = [ a 2 2ab b 2]. Therefore v 2 = v as required. Note that in the proof above, we established the following result. Corollary. If a population has genotypic distribution [ x y z ], then under random mating, the offspring population will have genotypic distribution [ a 2 2ab b 2], where a = x + y/2 and b = z + y/2. This will be useful in other exercises. 5. Consider a situation in which the transmission of two characteristics is tracked from one generation to the next in the case where Assumption 2.4 holds. If an individual with genotype AaBb reproduces by selfing, show that the genotypic distribution of the next generation is [ 6 8 6 8 4 8 6 8 6 ] AABB AABb AAbb AaBB AaBb Aabb aabb aabb aabb where the genotypes are written below the respective coordinate of the distribution vector. Solution. In this situation, that is, when 2 characteristics are tracked, the genotype of an individual is denoted by XY where X is a genotype for texture and Y is a genotype for color. Let p(x) and p(y ) denote the probabilities that an individual has the genotypes X and Y, respectively. Under Assumption 2.4, the probability that an individual has genotipe XY in the first filial generation is p(x)p(y ). We have p(aa) = p(aa) = p(bb) = p(bb) = 4 and p(aa) = p(bb) = 2, and consequently [ 4 4 4 2 4 4 2 4 2 2 2 4 4 4 4 2 4 4 ] AABB AABb AAbb AaBB AaBb Aabb aabb aabb aabb. 6. Assume that an initial population has a genotypic distribution [.5.2.3 ] with respect to a single characteristic. Find the genotypic distribution (a) in the third filial generation under selfing, and (b) in the third filial generation under random mating.

8 CHAPTER 2 Model Building: Selected Case Studies Solution. (a) Using Theeorem 2.2 we know that if a population has the genotypic distribution x, then under selfing the genotypic distribution in the next generation is given by xm, where M is the matrix shown in Theorem 2.2. Therefore we have the genotypic distribution in first filial generation = [.5.2.3 ] M = [.55..35 ] in second filial generation = [.55..35 ] M = [.575.05.375 ] in third filial generation = [.575.05.375 ] M = [.5825.025.3875 ] (b) By Exercise 4, we need only calculate the first filial generation since subsequent genotypic distributions do not change under random mating. Using the formula established in Exercise 4, we have a =.5 + 2.2 =.6 and b =.3 + 2.2 =.4, so the genotypic distribution of the first (and third) filial generation is [.36.48.6 ]. 7. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by random mating and in which the initial genotypic distribution is [ 2 2 0 ]. Assume selection is adverse with respect to the recessive form of the characteristic in that all individuals with genotype aa die prior to reproduction. Find the genotypic distribution in the first two filial generations immediately after reproduction. Solution. The genotypic distribution in the first filial generation immediately after reproduction is [ 9 6 6 6 6], and the genotypic distribution in the second filial generation immediately after reproduction is [ 44 225 72 225 9 225]. 8. Genotypic frequencies associated with a single characteristic are tracked in a population whose initial genotypic distribution is [.4.6 0 ]. Suppose that in each generation, ten percent of the population reproduces by selfing and ninety percent reproduces by random mating (which includes, of course, crossing of individuals with the same genotype). Find the genotypic distribution in the first filial generation. Solution. Under selfing, a population with the genotype [.4.6 0 ] has genotypic distribution [.55.3.5 ] in the first filial generation, and under random mating the genotypic distribution in the first filial generation is [.49.42.09 ]. Thus, with the reproduction described in the exercise, the genotypic distribution in the first filial generation is. [.55.3.5 ] +.9 [.49.42.09 ] = [.496.408.096 ]. 9. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by random mating and in which the initial genotypic distribution immediately prior to reproduction is [.4.4.2 ]. Assume that selection is adverse with respect to the hybrid form of the characteristic in

2. Mendelian Genetics 9 that twenty-five percent of the individuals with genotype Aa die in the interval between successive generations. Find the genotypic distribution in the first two filial generations immediately after reproduction. Solution. With the reproduction and adverse selection described in the exercise, the genotypic distributions evolve as follows (entries rounded to 4 decimal places): initial population [.4.4.2 ] random mating first filial immediately after reproduction [.36.48.6 ] first filial immediately after reproduction [.36.48.6 ] adverse selection reproducing first filial [.409.409.88 ] reproducing first filial [.409.409.88 ] random mating second filial immediately after reproduction [.3766.4742.493 ] 0. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by random mating and in which the initial genotypic distribution immediately prior to reproduction is [.5.3.2 ]. Assume selection is adverse with respect to the pure recessive form of the characteristic. (a) In particular, assume that only sixty percent of the individual with genotype aa are capable of reproduction and that one hundred percent of the individuals in genotypes AA and Aa are capable of reproduction. Find the genotypic distribution in the first two filial generations immediately after reproduction. (b) Suppose that a fraction f of the individuals with genotype aa are capable of reproduction and that one hundred percent of the individuals with genotypes AA and Aa are capable of reproduction. If the genotypic distribution of the reproducing members of the first filial generation has about five percent of genotype aa, estimate the fraction f. Solution. (a) With the reproduction and adverse selection described in the exercise, the genotypic distributions evolve as follows (entries rounded to 4

20 CHAPTER 2 Model Building: Selected Case Studies decimal places): initial population [.5.3.2 ] adverse selection reproducing initial population [.5435.326.304 ] reproducing initial population [.5435.326.304 ] random mating first filial immediately after reproduction [.4992.447.086 ] first filial immediately after reproduction [.4992.447.086 ] adverse selection reproducing first filial generation [.570.4295.0535 ] reproducing first filial generation [.570.4295.0535 ] random mating second filial immediately after reproduction [.5354.3926.0720 ] (b) Each of the above calculations depends on the reproducing fraction of the aa genotype. To solve part (b), replace the given value of.6 by a parameter f and solve for f so that the third coordinate of the genotypic distribution of the first filial generation immediately after reproduction is.05. Using Excel we find f = 0.5749, a value very close to the original reproducing percentage of 60%.. Genotypic frequencies associated with a single characteristic are tracked in a population that reproduces by random mating and in which the initial genotypic distribution immediately prior to reproduction is [.6.4.0 ]. Assume that in the period between successive reproductions any recessive allele mutates into the dominant form of the allele with probability.0, and the mutations of any two recessive alleles are independent. (a) Find the genotypic distribution in the first filial generation immediately after the first reproduction. (b) Find the genotypic distribution in the first filial generation immediately prior to the second reproduction. Solution. (a) [.64.32.04 ] (b) [.6432.376.0392 ] (to 4 decimal place accuracy)

2. Mendelian Genetics 2 2. The genotypic frequencies associated with a single characteristic are tracked in a population where reproduction is defined as in Assumption 2.3 except that the table in Definition 2. is replaced by Table 2.5. (a) Find the reproduction function, that is, find the genotypic distribution resulting from each of the six possible matings. (b) An initial population contains only individuals with genotype Aa. If reproduction is by random mating, find the genotypic distribution in the first two filial generations. (c) An initial population has genotypic distribution [ x y z ]. If reproduction is by selfing, find the genotypic distribution in the first two filial generations. Solution. (a) u v r(u, v) AA AA [ 0 0 ] [ AA Aa 2 3 3 0 ] [ ] AA aa 0 0 [ Aa Aa 4 ] 4 [ 9 9 9] Aa aa 0 2 [ 3 3] aa aa 0 0 (b) The genotypic distribution in the first filial generation is [ 4 9 second filial generation it is [.5487.384.0672 ]. 4 9 9] ; in (c) Using the same approach as in Theorem 2.2 with the matrix M replaced by the following 0 0 4 4 9 9 9 0 0 we find that the genotype in the first filial generation is [ x + 4 9 y 4 9 y 9 y + z], and in the second filial generation it is [ x +.6420y.975y.605y + z ]. 3. Consider a population in which two characteristics are monitored, and suppose they are texture and color. A plant that is pure dominant with respect to both characteristics is crossed with a plant that is pure recessive, and the resulting plant (hybrid with respect to both characteristics, a dihybrid) reproduces by selfing. Suppose that each form of the color gene is equally likely to be contributed from the parent to an offspring. Also suppose that the genes for texture and color are linked in the following way: If the dominant form of the color gene is contributed, then the dominant form of the texture gene is three times as likely to be contributed as the recessive form. If the recessive form of the color gene is contributed, then the dominant and recessive forms of the texture gene are equally likely to be contributed.

22 CHAPTER 2 Model Building: Selected Case Studies (a) Find the genotypic distribution of the second filial generation the result of the dihybrid cross. (b) Find the phenotypic distribution of the second filial generation the result of the dihybrid cross. (c) How does this phenotypic distribution compare qualitatively with the 9:3:3: ratio predicted when the genes are not linked? Solution. (a) (b) [ 9 64 6 64 64 3 6 4 6 6 2 6 6 ] AABB AABb AAbb AaBB AaBb Aabb aabb aabb aabb Phenotype: dominant color and texture 43 64 dominant color and recessive texture 5 64 recessive color and dominant texture 3 6 recessive color and texture 6 (c) 43:5:2:4 compared to 9:3:3: or.679:.078:.875:.0625 compared to.5625:.875:.875:.0625 4. Consider a two-characteristic genetics model where the characteristics are texture (alleles S and W) and color (alleles G and Y ); S and G are dominant. Suppose that during reproduction each form of the texture gene is equally likely to be selected if both are available. If the texture allele S is selected, then the color allele G is twice as likely to be selected as Y, assuming both are available, and if the texture allele W is selected, then each color allele is equally likely to be selected. A pure-line SG is crossed with a dihybrid. (a) Find the phenotypic distribution of the resulting population. (b) Find the genotypic distribution of the resulting population. Solution. (a) (b) Phenotype frequency S texture, G color S texture, Y color 0 W texture, G color 0 W texture, Y color 0 2 [ 6 6 0 4 4 0 0 0 0 ] SSGG SSGY SSY Y SWGG SWGY SWY Y WWGG WWGY WWY Y

2.2 Models for Growth Processes 2.2 Models for Growth Processes 23. Assume that Equation (2.) holds and give a mathematical induction proof that Equation (2.3) holds for k =, 2, 3,.... Solution. Assume x k+ x k = A, k = 0,, 2,... Show x k = x 0 + ka, k =, 2,... Proof. By assumption, x x 0 = A, or x = x 0 + A, so (2.3) holds for k =. Suppose x k = x 0 + ka holds for k =, 2,..., N, and show x N+ = x 0 + (N + )A. By the induction hypothesis, x N = x 0 + NA. Also x N+ x N = A or x N+ = x N + A, so x N+ = (x 0 + NA) + A = x 0 + (N + )A. Induction proof is complete. 2. Consider the discrete time model for growth in which the amount added in each time interval is proportional to the population size at the beginning of the interval: Model 2 and Equation (2.5). For growth rates r between 0.025 and 0.25, graph the number n = n(r) of time periods necessary for the initial population to double in size. Also graph the function 72/r for the same range of values of r. For what values of r is the relative error between n(r) and 72/r less than 2%? Solution. Equation (2.5) in the text is x k = ( + r) k x 0. The number of time periods n for the population to double (not necessarily an integer) is the solution n of 2x 0 = ( + r) n x 0 or n ln( + r) = ln(2). Therefore, n = ln(2)/ ln( + r). The graph of the function n(r) = ln(2)/ ln( + r) for 0.025 < r < 0.25 is shown in Figure S 2.. In that figure the graph of the function g(r) = 72/r is also shown. The relative error between n(r) and g(r) = 72/r (that is, the ratio (n(r) g(r))/g(r)) is less than 2 percent for 0.037 < r < 0.22. 3. Verify that the function x defined by Equation (2.0) is a solution of the Problem (2.9) Solution. With x defined by x(t) = we have x 0 M x 0 + (M x 0 )e kt dx dt = kx 0M(M x 0 )e kt [x 0 + (M x 0 )e kt ] 2 and dx x dt = k(m x 0)e kt x 0 + (M x 0 )e kt.

24 CHAPTER 2 Model Building: Selected Case Studies 30 25 20 n 5 0 5 n(r) g(r) 0 0.05 0. 0.5 0.2 0.25 r Figure S 2. Also, x M = x 0 x o + [M x 0 ]e kt = (M x 0)e kt x 0 + (M x 0 )e kt Therefore, dx ( x dt = k x ) for t > 0 M x 0 M Finally, x(0) = x 0 + (M x 0 ) = x 0, and (2.9) is satisfied. 4. Show that with an appropriate change of variables involving x n, k, and M Equation (2.) can be transformed into Equation (2.2). ( ) Solution. Equation (2.) is x n+ x n = kx n x n M. Set xn = (k+) k Mp n, n + 0,, 2,.... Then (2.) becomes ( ) ( ) ( ( ) ) k + k + k + Mp n+ = Mp n + M(k + )p n p n k k k ( ) k + = Mp n [ + k (k + )p n ]. k So p n+ = (k + )p n ( p n ) = bp n ( p n ), where b = k +.

2.2 Models for Growth Processes 25 5. Let g be a continuous function defined on an interval [0, T]. Show that no positive solution of a problem of the form dx dt = g(x), x(0) = x 0 can oscillate on that interval. Solution. Oscillation means that there is an interval within which the function x increases, reaches a local maximum, and then decreases (or, an interval within which the function decreases, reaches a local minimum, then increases). Suppose the former, and that t 0 is such that x has a local maximum at t 0 Also, suppose that t and t 2 are such that t < t 0 < t 2, x(t ) = x(t 2 ) < x(t 0 ), and dx dt (t ) > 0, dx dt (t 2) < 0. Then since x(t ) = x(t 2 ), we have g(x(t )) = g(x(t 2 )) and the equation dx dt (t) = g(x(t)) cannot hold for both t and t 2. 6. Consider the discrete model described by x n+ = bx n ( x 2 n), x 0 given. Discuss the behavior of {x n }, including results similar to those developed in this section. That is, find a range of parameter values b for which the sequence {x n } generated by this recurrence relation can be used as a population model. Discuss the convergence of the sequence {x n } for various values of b. In particular, identify the ranges of values of b for which the sequence converges to 0, converges monotonically to a nonzero limit, and converges with oscillations. Solution. Set f(x) = bx( x 2 ), b > 0. Then, the recurrence relation is x n+ = f(x n ). (a) The sequence {x n } can be used as a population model if x n 0 for all n (given x 0 ). This requires x n for all n, which means that the maximum value of f must be. This holds if b.5(3.5 ) or b 2.598. (b) The sequence {x n } converges to 0 (for any x 0, x 0 ) if f (0). Since f (x) = b( 3x 2 ), f (0) = b and therefore the condition is b. (c) If {x n } converges to x (where x = f(x ) 0) then b must be and ( ).5. x = The sequence converges monotonically (except possibly for b b the first step) if f (x ) 0. This holds for < b.5. (d) The sequence {x n } converges to x, but oscillates, if < f (x ) < 0. This holds if.5 < b < 2. 7. For the model of Exercise 6, find a value of the parameter b for which the subsequences of odd and even iterates, {x 2n+ } and {x 2n }, each converges but to different limits. Find the limits. Solution. For b = 2. the sequence of even iterates converges and the sequence of odd iterates converges, but the two limits are different. The limits are approximately 0.5895 and 0.8077.

26 CHAPTER 2 Model Building: Selected Case Studies 8. Consider the discrete model described by x n+ = bx n exp( 5x n ), x 0 given. Discuss the behavior of {x n } as in Exercise 6 for this situation. Solution. Set f(x) = b x exp( 5x). Then the given recurrence relation is x n+ = f(x n ), x 0 given. (a) For any x 0 > 0, the sequence {x n } can be used as a population model for any value of b, b > 0. (b) The sequence {x n } converges to 0 for b. (c) The sequence {x n } converges monotonically to x, x = f(x ), for values of b for which f(x ) > 0, that is, < b e. (d) The sequence converges, but not monotonically, for e < b e 2. 9. Consider the discrete model described by x n+ = b(x n )/( + x n ) 2, x 0 given. Discuss the behavior of {x n } as in Exercise 6 for this situation. bx Solution. Set f(x) =, b > 0. Then the given recurrence relation is ( + x) 2 x n+ = f(x n ), x 0 given. (a) The sequence {x n } can be used as a population model for any value of b, b > 0. (b) The sequence {x n } converges to 0 for b. (c) The sequence {x n } converges monotonically to x, x = f(x ), for < b 4. (d) The sequence {x n } converges, but not monotonically, for b > 4. 0. In the discrete model, x n+ = bx n ( x n ), and in the models of Exercises 6 and 8, the parameter b can be viewed as a reproduction factor for small population sizes. In each of these cases the population dies out, that is, x n approaches 0 as n becomes large, if the reproduction factor is small, but does not die out if the reproduction factor is large enough. This conclusion holds for any initial population size. There are populations which have the characteristic that for all values of the reproduction factor, or its equivalent, the population dies out when the initial population size is small enough. Examples include animal populations distributed over a wide geographic area. The model described by the equation below has this characteristic: x n+ = bx 2 n exp( 5x n ), x 0 > 0 given.

2.2 Models for Growth Processes 27 (a) Find a value of b for which there is some initial value x 0 for which the population does not die out. (b) For this value of b, show that if x 0 is small enough then the population does die out. Solution. (a) Any value of b > 5e will do, for example, b = 4. (b) If x is the smallest solution of x = 4(x ) 2 exp( 5x ), then with x 0 < x, the sequence {x n } converges to 0. Indeed, if x satisfies 4x exp( 5x ) = and if x 0 < x, then 4x 0 exp( 5x 0 ) = k <. It follows that x = 4x 2 0 exp( 5x 0 ) = kx 0 x 2 = 4x 2 exp( 5x ) = kx 0 (4x exp( 5x )) < k 2 x 0 since the function 4xexp( 5x) is increasing on (0, x ). Similarly, x 3 < k 3 x 0 and, in general, x n < k n x 0. Since k < the proof that {x n } converges to 0 is complete.. For the model of Exercise 0, find the smallest value of b, call it b, for which the population does not die out for some choice of the initial population size. (a) For b = b find an initial population size for which {x n } does not converge to 0. (b) For b =.2b there is a range of initial population sizes, that is, a set of values of x 0 for which the population dies out. Find this range. Solution. The desired value of b is the smallest value of b for which there is a non-zero solution of x = b(x ) 2 exp( 5x ). This value is 5e. (a) For b = 5e, the population {x n } with x 0 =.2 does not converge to zero. Indeed, x n =.2 for all n. (b) For b =.2b = 6e the sequence {x n } converges to 0 for 0 x 0 <.02234 and for x 0 >.0239623. 2. Consider the continuous model described by ( dx dt = kx ( x M ) 2 ), for t > 0, x(0) = x 0. Find an explicit solution of this equation and describe the geometry of the solutions. Compare these solutions with those of the standard logistic model for selected parameter values.

28 CHAPTER 2 Model Building: Selected Case Studies Solution. Multiplying both sides of the differential equation by M y = x M, we have the equation and setting dy dt = ky[ y2 ] = ky( + y)( y), for t > 0, y(0) = y 0 = x 0 M for y 0, y, y, we have Integrating, we have dy/dt y + dy/dt y( + y)( y) = k ( ) ( ) dy/dt dy/dt 2 + y + 2 y = k ln(y) 2 ln( + y) ln( y) = kt + c [ 2 ] y ln = kt + c [( + y)( y)] /2 [ y 2 ] ln [ y 2 = 2kt + c ] y 2 [ y 2 ] = c 2e 2kt y 2 = c 2e 2kt + c 2 e 2kt = + c 3 e 2kt and finally y = + c3 e, 2kt where the positive square root is chosen because the solution of the differential equation provides a population model. Finally, expressing this result in terms of the original variables, we have x(t) = Mx 0, t > 0. x 2 0 + (M 2 x 2 0 )e 2kt Graphs of the function x for M = 0, k =.5 and four different initial population sizes (x 0 =, x 0 = 3, x 0 = 7, and x 0 = 5) are shown in Figure S 2.2 and may be compared with the graphs of the solution of the standard logistic model shown in Figure 2.3 in the text. 2.3 Social Choice. There are five individuals and three alternatives with the preference rankings shown below.

2.3 Social Choice 29 4 2 0 8 6 4 2 0 2 4 6 8 0 t Figure S 2.2 Preference Ranking Individual A B C D E X X Y Z Y Y Z X Y Z Z Y Z X X (a) Using pairwise comparisons and a simple-majority vote, determine (if possible) a group preference ranking and the most preferred alternative. (b) Use the sequential-voting scheme, as in Figure 2.0a, to find the most preferred alternative. Solution. (a) Since A, B, and C prefer X to Z, the group will prefer X to Z. Since A, C, and E prefer Y to Z, the group will prefer Y to Z. Finally, since C, D, and E prefer Y to X, the group will prefer Y to X. These preferences are transitive, Y is most preferred by the group, and the group preference ranking is Y is preferred to X is preferred to Z. (b) First a vote is taken between X and Y resulting in Y. Then a vote is taken between Y and Z, resulting again in Y. So the process of sequential voting will yield Y as the most preferred alternative. 2. There are five individuals with the preference rankings for four alternatives shown below.

30 CHAPTER 2 Model Building: Selected Case Studies Preference Ranking Individual A B C D E X W Y Z Y Y Z W W Z Z Y X X W W X Z Y X (a) Use the sequential-voting method of Figure 2.a to determine the preferred alternative for the group. (b) Suppose alternative W is eliminated. Use the sequential-voting method of Figure 2.0b to determine the preferred alternative for the group. Solution. (a) When X and Y are compared, Y is preferred. Y is then compared to W, and Y is again chosen. Finally, Y is compared to Z, and Y is once again chosen. Hence the group most prefers Y when we use the sequential voting method of Figure 2.a. (b) First X is compared to Z. Since Z is preferred over X, Z is selected to be compared to Y. But Y is preferred over Z, so the group most prefers Y among X, Y, and Z under the sequential voting of Figure 2.0b. 3. Use the preference schedule of Exercise and the Borda count decision process of Example 2.2 to determine the preference ranking for the group. Solution. By the Borda count decision process of Example 2.2, we have So the group prefers Y to X to Z. Alternative Points X 0 Y Z 9 4. Use the preference schedule of Exercise 2 and the Borda count decision process of Example 2.2 to determine the preference ranking for the group. Solution. As above, we have a table: Alternative Points X 0 Y 4 Z 3 W 3 So the Borda count decision process does not yield a group preference ranking. However, we do see that this process produces a most preferred alternative, Y.

2.3 Social Choice 3 5. Suppose there are five individuals with the preference rankings for four alternatives shown in Exercise 2. Find a method of changing the preferences for voter E such that sequential voting can be used to obtain different winners by starting the voting with different pairs. Solution. If the preference ranking for individual E is replaced by X preferred to Y preferred to Z preferred to W, then the two sequential voting methods shown in Figure S 2.3 have different results. X W X Y Y W Y Y W Z Z X Z X Figure S 2.3 6. There are twenty-one individuals and three alternatives. The numbers of individuals with various preference rankings are as shown below. Preference Ranking Number of Individuals 5 3 3 6 4 X X Z Z Y Y Z X Y X Z Y Y X Z (a) Using pairwise comparisons and a simple-majority vote, determine (if possible) a group preference ranking and the most preferred alternative. (b) Use sequential voting as in Figure 2.0a to find the most preferred alternative. Solution. (a) We have that X is preferred to Y by 5 + 3 + 3 = individuals, Z is preferred to Y by 3 + 3 + 6 = 2 individuals, and X is preferred to Z by 5 + 3 + 4 = 2 individuals. These preferences determine the group preferences since the group s preference for one alternative over the other is determined by simple majority, and or more individuals hold each of the above preferences. Hence the group will prefer X to Z to Y.

32 CHAPTER 2 Model Building: Selected Case Studies (b) A vote between X and Y yields X since individuals prefer X to Y. The next vote between X and Z yields X since 2 individuals prefer X to Z. Therefore, X is the most preferred alternative. 7. The plurality method of making a group decision is to select, as the most preferable of the alternatives, the one that is the first choice of the largest number of individuals. (a) Using the data of Exercise 6, use the plurality method to determine the most preferred alternative for the group. (b) If alternative Y is eliminated, what happens to the group preference? Solution. (a) In Exercise 6, Z is most preferred by 9 people, X is most preferred by 8 people, and Y is most preferred by 4 people. So the plurality method produces Z as the most preferred by the group. (b) If alternative Y is eliminated, then X will be most preferred by those who had previously most preferred Y. Hence X will be most preferred by 2 people, and so will be selected by the plurality method. 8. There are five individuals with the preference rankings shown below for four alternatives. Preference Ranking Individual A B C D E X W Y Z Y Y Z W W Z Z X X X W W Y Z Y X Does a pairwise comparison and simple-majority decision process result in a group preference ranking? Solution. No. Since A, B, and D prefer X to Y, the group prefers X to Y. Since A, C, and E prefer Y to W, the group prefers Y to W. Since B, C, D, and E prefer W to X, the group prefers W to X. So the group prefers X to Y to W to X. In other words, the group has an intransitive preference ranking. 9. Suppose there are three individuals, there are three alternatives, and the decision process is to determine pairwise group preferences using a simple-majority vote. (a) Show that there are 68 different preference schedules for which this method yields a group preference ranking and a majority vote winner. (b) Show that there are 2 different preference schedules that result in intransitivity.

2.3 Social Choice 33 (c) Show that there are 36 different preference schedules that result in neither a majority winner with a group preference ranking nor intransitivity. Solution. (a) There are 3 2 3 = 24 group preference rankings in which each individual has the same most preferred alternative There are 3 2 2 3 2 2 = 44 group preference rankings in which exactly 2 individuals have the same most preferred alternative. Therefore, there are 24+44=68 group preference rankings with a winner using majority vote. Once a winner is determined, one of the two remaining alternative is preferred to the other by two individuals (or more), and therefore there is a group preference ranking. (b) There are 6 group preference schedules in which the 3 individual preference rankings are X, Y, and Z Y, Z, X Z, X, Y There are 6 group preference schedules in which the 3 individual preference rankings are X, Z, and Y Z, Y, X Y, X, Z Each of these schedules results in intransitivity. (c) There are 6 2 3 = 36 group preference schedules with 3 different individual preference rankings, no 2 of them with a common most preferred, and 2 of them with a common least preferred alternative. Each such schedule results in neither a majority winner or intransitivity. 0. In the situation of Exercise 9, use a computer to generate group preference schedules at random, and then use these data to estimate the probability that there is no majority vote winner when individual preferences ranking are selected at random. Solution. Five repetitions of a simulation consisting of 000 random selections of individual preference rankings give the data in the following table: # of selections (of 000) with no majority winner 225 The average is 226.4, and the associated estimate of the probability d, 208 224 no majority winner, is.2264. This 244 compares with the theoretical result 288 (problem 9) of.2222.

34 CHAPTER 2 Model Building: Selected Case Studies. Suppose there are five individuals, there are three alternatives, and the decision process is to determine pairwise group preferences using simple-majority vote. (a) Find the number of different group preference schedules for which this method yields a group preference ranking and a majority vote winner. (b) If individual preference rankings are selected at random, find the probability that a group preference ranking cannot be determined using this method. Solution. Recall the discussion of the case of 3 individuals and 3 alternatives on pages 5 53. In that discussion it was shown that with pairwise majority voting there are group preference schedules that do not yield group preference rankings because of intransitivity. The situation is the same with 5 voters and 3 alternatives: a group preference schedule and pairwise simple majority voting results in a group preference ranking if and only if there are no situations such as X is preferred to Y is preferred to Z is preferred to X. The latter situation will be called a 3-cycle and written as X Y Z X. That is, a 3-cycle consists of 3 alternatives and an order: for example, the alternatives X, Y, and Z and the order X Y Z X. (a) In this situation, for there to be a majority vote winner there must be an alternative which is the top choice of 3, 4, or 5 of the individuals. Then one of the remaining alternatives must be preferred to the other, and there can be no 3-cycles. There must be a group preference ranking. A group preference schedule can have the same top choice for all individuals in C(3, ) 2 5 = 96 different ways, the same top choice for exactly 4 individuals in C(3, ) C(5, 4) C(2, ) 2 5 = 960 different ways, the same top choice for exactly 3 individuals in C(3, ) C(5, 3) C(2, ) C(2, ) 2 5 = 3840 different ways. Consequently, there are 4896 different group preference schedules which result in a majority vote winner, and therefore a group preference schedule. (b) There are a total of 6 5 = 7776 different group preference schedules, and if one is selected at random, the probability it will result in a majority vote winner is 4896 7776 =.6296. Consequently, the probability that a random selected group preference schedule will not yield a group preference schedule using this method is.3704. 2. In the situation of Exercise, find the probability that if individuals select their preference rankings at random, then the group preferences will be intransitive.

2.3 Social Choice 35 Solution. There are six individual preference rankings, and using the representation introduced on page 52 we partition this set of six into two groups of 3: group I X Y Z Y Z X Z X Y and group II X Y Z Z X Y Y Z X We introduce these groups to facilitate the discussion of the solution. There are two possible 3-cycles: X Y Z X and X Z Y X, and the individual preference rankings in group I are associated with the first and those in group II are associated with the second. For a group preference schedule to have a 3-cycle X Y Z X, it must consist of only individual preference rankings from group I, or 4 individual preference rankings from group I and one from group II. Moreover, it cannot have more than two individual preference rankings with the same top choice or more than two with the same bottom choice. Using these facts, the only possible sets of individual preference rankings which result in intransitivity are the following: X X Y Y Z Y Y Z Z X Z Z X X Y X X Y Z Z Y Y Z X Y Z Z X Y X X X Y Y Z Y Z Z Z X Z Y X X Y X Y Y Z Z Y Z Z X X Z X X Y Y X X Y Z Z Y Y Z X X Z Z X Y Y X Y Y Z Z Y X Z X X Z Z X Y Y Considering the sets in the first row, there are 20 4 = 30 different group preference schedules with the individual preference rankings in the first set, 20 2 = 60 different group preference schedules with the individual preference rankings in the second set, and 20 4 = 30 different group preference schedules with the individual preference rankings in the third set. The corresponding numbers for the second row are 60, 30, and 60. Consequently, there are a total of 270 different group preference schedules with a 3-cycle of the form X Y Z X. A similar argument shows that there are 270 different group preference schedules with a 3-cycle of the form X Z Y X. Consequently, there are 540 different group preference schedules resulting in intransitivity, and if individual preference rankings are selected independently and at random, the probability of intransitivity is 540 6 = 540 5 7776 =.06944. 3. Suppose there are three individuals, there are four alternatives, and the decision process is to determine pairwise group preferences using simple-majority vote. Find the number of different group preference schedules for which this method yields a group preference ranking. Solution. We recall the discussion of the case of 3 individuals and 3 alternatives on pages 5 53 of the text. In that discussion it was shown that there are group

36 CHAPTER 2 Model Building: Selected Case Studies preference schedules that with pairwise majority voting do not yield group preference rankings because of intransitivity. In general, pairwise majority voting does yield a group preference ranking as long as there is no intransitivity. We omit further discussion of the general case, and we consider only the situation (3 individuals and 4 alternatives) of the exercise. Suppose the 3 individuals are labeled A, B, and C, and the 4 alternatives are labeled X, Y, Z, and W. The essential fact that we use to solve this exercise can be stated as follows: a group preference schedule and pairwise simple majority voting results in a group preference ranking if and only if there are no situations such as X is preferred to Y is preferred to Z is preferred to X. The latter situation will be called a 3-cycle. That is, a 3-cycle consists of 3 alternatives and an order: for example, the alternatives X, Y, and Z and the order X Y Z X, which should be read X is preferred to Y is preferred to Z is preferred to X. The validity of this fact for the case of 3 individuals and either 3 or 4 alternatives can be verified directly. In view of the above discussion we can determine the number of group preference schedules for which pairwise simple majority voting yields a group preference ranking by determining the total number of group preference schedules and subtracting the number with at least one 3-cycle. Since each of the 3 individuals has 4! = 24 choices for an individual preference ranking, there are 24 3 = 3, 824 group preference schedules. A B C X Y Z Y Z X Z X Y Table S2. X Z Y Figure S 2.4 With 4 alternatives there are C(4, 2) = 6 pairwise comparisons to consider. A 3-cycle consists of 3 alternatives and an order: for example, the alternatives X, Y, and Z and the order X Y Z X. By checking the possibilities, it can be verified that a group preference schedule with 4 alternatives can have at most two distinct 3-cycles. (Two 3-cycles are distinct unless they have the same alternatives and the same order.) We adopt a fairly direct approach to the counting problem. We begin by determining the number of group preference schedules which have the 3-cycle X Y Z X. Suppose the group preference schedule among the alternatives X, Y, and Z is as shown in the table above (reproduced from page 52 in the text). Then, the question becomes: in how many different ways can the alternative W be included in this table to yield a group preference schedule with all 4 alternatives? The answer to this question is 4 3 = 64, but to solve the exercise we need a deeper analysis of the situation. It is useful to introduce a graphical way to display group preference rankings. In Figure S 2.4, which is called a directed graph, we show the 3 alternatives and the simple majority

2.3 Social Choice 37 X X X X Z W Y Z W Y Z W Y Z W Y (a) (b) (c) (d) X X X X Z W Y Z W Y Z W Y Z W Y (e) (f) (g) (h) Figure S 2.5 preference ranking X Y Z X. In that figure a directed arc from X to Y has the meaning X Y. If we include the alternative W, then we have a figure similar to Figure S 2.4 for each of the simple majority preferences between X and W, Y and W, and Z and W. There are 2 possible results for each of these 3 comparisons, so there are 8 possible figures, shown in Figure S 2.5. By listing the possible options, we see that there are 5 different ways to add a W to each column in Table S2. so that the resulting group preference ranking is that shown in Figure S 2.5a, and 5 different ways to add a W to each column in Table S2. so that the resulting group preference ranking is that shown in Figure S 2.5b. In each case the resulting group preference schedule has a 3-cycle X Y W X in addition to the 3-cycle X Y Z X. Similarly, there are 5 different ways to add a W in each column in Table S2. so that the resulting group preference ranking is that shown in Figure S 2.5c, and 5 different ways to add a W to each column in Table S2. so that the resulting group preference ranking is that shown in Figure S 2.5d. In each case the resulting group preference schedule has a 3-cycle X W Z X in addition to the 3-cycle X Y Z X. Finally, there are 5 different ways to add a W in each column in Table S2. so that the resulting group preference ranking is that shown in Figure S 2.5e, and 5 different ways to add a W to each column in Table S2. so that the resulting group preference ranking is that shown in Figure S 2.5f. In each case the resulting group preference schedule has a 3-cycle Y Z W Y in addition to the 3-cycle X Y Z X. Note that in each of these 30 situations, there are two 3-cycles in the resulting group preferences. Since there are a total of 64 ways to add a W in each column in Table S2., and 30 of them result in two 3-cycles, there are 64 30 = 34 ways to add that W in such a way that there is only a single 3-cycle, the original X Y Z X. These are the situations depicted in Figures S 2.5g and S 2.5h; in 7 cases the alternative W is preferred to all the others (Figure S 2.5g) and in 7 cases each

38 CHAPTER 2 Model Building: Selected Case Studies of the others is preferred to W (Figure S 2.5h). There are 6 different group preference schedules for X, Y, and Z with the 3-cycle X Y Z X, and 6 with the 3-cycle X Z Y X; a total of 2 for the alternatives X, Y, and Z. Similarly, there will be 2 different group preference schedules for each of the other sets of 3 alternatives there are 4 such sets. For each preference schedule with 3 alternatives, there are 34 ways to add the fourth alternative so that there is a single cycle. For each preference schedule with 3 alternatives there are 30 ways to add the fourth alternative so that there are 2 cycles. However, each of the latter preference schedules will also be counted again if we begin with a preference schedule including alternative W, say X, Y, W, and consequently we must be careful to avoid double counting in the total. There are C(4, 3) = 4 ways to select 3 alternatives from 4 to begin this counting process, and consequently the total number of group preference schedules which contain (at least one) a 3-cycle is 4 2 34 + 4 2 30 = 632 + 720 = 2352. 2 Consequently, there are 3, 824 2352 =, 472 group preferences for which pairwise simple majority voting results in a group preference ranking. Note. A frequently used criteria for defining an alternative as the winner is that the alternative be the top choice of a majority of the individuals. With this definition of a winning alternative, there are 3456 different group preference rankings with a winner. Indeed, there are 4 possible winners, and for each choice of a winner there are 864 different group preference rankings with that winner. Of the 3456 group preference rankings with a winner, 480 also have a cycle. 4. Suppose there are three individuals, there are four alternatives, and the decision process is to determine pairwise group preferences using simple-majority vote. Find examples of preference schedules for which (a) There is an alternative that is a majority winner, and there is no intransitivity. (b) There is an alternative that is a majority vote winner, and there is intransitivity among the remaining three alternatives. (c) There is intransitivity among all four alternatives. (d) There is intransitivity among three (but not all four) alternatives and there is no alternative that is a majority vote winner.

2.3 Social Choice 39 Solution. (a) A B C (b) A B C X X Y X X W Y W W Y Z X Z Y Z Z W Y W Z X W Y Z (c) A B C (d) A B C X Y Z X Y Z Y Z W Y Z X Z W X Z X W W X Y W W Y 5. Suppose there are seven individuals, there are three alternatives, and the decision process is to determine pairwise group preferences using simple-majority vote. If individual preference rankings are selected at random, find the probability that this method yields a group preference ranking and a majority vote winner. Solution. If there is a majority winner, then a majority of the 7 voters must prefer one of the remaining alternatives over the other. Consequently, there is a group preference ranking for there to be a majority winner, at least 4 of the voters must select an individual preference ranking with the same most preferred alternative. 7 individuals can select individual preference rankings with the same most preferred alternative in 3 2 7 = 384 ways With 7 individuals, the number of different group preference schedules in which exactly 6 individuals have the same most preferred alternative is C(7, ) 3 2 6 2 2 = 5376 With 7 individuals, the number of different group preference schedules in which exactly 5 individuals have the same most preferred alternative is C(7, 2) 3 2 5 (2 2) 2 = 32, 256 With 7 individuals, the number of different group preference schedules in which exactly 4 individuals have the same most preferred alternative is C(7, 3) 3 2 4 (2 2) 3 = 07, 520 Therefore, there are 45,536 different group preference schedules that result in a majority winner and a group preference ranking. There are a total of 6 7 = 279, 936 different group preference schedules. Consequently, if individual preferences rankings are selected at random, then the result will be a majority winner and a group preference ranking with probability.520.