# CHAPTER 7 APPLICATIONS TO MARKETING. Chapter 7 p. 1/54

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1 CHAPTER 7 APPLICATIONS TO MARKETING Chapter 7 p. 1/54

2 APPLICATIONS TO MARKETING State Equation: Rate of sales expressed in terms of advertising, which is a control variable Objective: Profit maximization Constraints: Advertising rate to be nonnegative Chapter 7 p. 2/54

3 THE NERLOVE-ARROW ADVERTISING MODEL Let G(t) 0 denote the stock of goodwill at time t, with Ġ = u δg, G(0) = G 0, (1) where u = u(t) 0 is the advertising effort at time t measured in dollars per unit time. Sales rate S(t) is given by S = S(p,G,Z). (2) Assuming the rate of total production costs is c(s), we can write the total revenue net of production costs as R(p,G,Z) = ps(p,g,z) c(s). (3) The revenue net of advertising expenditure is therefore R(p,G,Z) u. Chapter 7 p. 3/54

4 THE NERLOVE-ARROW ADVERTISING MODEL The firm wants to maximize the present value of net revenue streams discounted at a fixed rate ρ, i.e., max u 0,p 0 { J = 0 } e ρt [R(p,G,Z) u]dt (4) subject to (1). Since the only place that p occurs is in the integrand, we can maximize J by first maximizing R with respect to price p holding G fixed, and then maximize the result with respect to u. Thus, R(p,G,Z) p = S + p S p c S S p = 0, (5) which implicitly gives the optimal price p (t) = p(g(t),z(t)). Chapter 7 p. 4/54

5 THE NERLOVE-ARROW ADVERTISING MODEL Defining η = (p/s)( S/ p) as the elasticity of demand with respect to price, we can rewrite condition (5) as p = ηc S(S) η 1, (6) which is the usual price formula for a monopolist, known sometimes as the Amoroso-Robinson relation. In words, the formula means that the marginal revenue (η 1)p/η must equal the marginal cost c S (S). See, e.g., Cohen and Cyert (1965, p.189). Chapter 7 p. 5/54

6 THE NERLOVE-ARROW ADVERTISING MODEL Defining π(g,z) = R(p,G,Z), the objective function in (4) can be rewritten as max u 0 { J = 0 e ρt [π(g,z) u]dt }. For convenience, we assume Z to be a given constant and restate the optimal control problem which we have just formulated: max u 0 { J = 0 e ρt [π(g) u]dt } subject to Ġ = u δg, G(0) = G 0. (7) Chapter 7 p. 6/54

7 SOLUTION BY THE MAXIMUM PRINCIPLE H = π(g) u + λ[u δg], (8) λ = ρλ H π = (ρ + δ)λ G G, (9) lim t + e ρt λ(t) = 0. (10) The adjoint variable λ(t) is the shadow price associated with the goodwill at time t. Thus, the Hamiltonian in (8) can be interpreted as the dynamic profit rate which consists of two terms: (i) the current net profit rate (π(g) u) and (ii) the value λġ = λ[u δg] of the new goodwill created by advertising at rate u. Chapter 7 p. 7/54

8 SOLUTION BY THE MAXIMUM PRINCIPLE CONT. Equation (9) corresponds to the usual equilibrium relation for investment in capital goods; see Arrow and Kurz (1970) and Jacquemin (1973). It states that the marginal opportunity cost λ(ρ + δ)dt of investment in goodwill should equal the sum of the marginal profit ( π/ G)dt from increased goodwill and the capital gain dλ := λdt. Chapter 7 p. 8/54

9 SOLUTION BY THE MAXIMUM PRINCIPLE CONT. Defining β = (G/S)( S/ G) as the elasticity of demand with respect to goodwill and using (3), (5), and (9), we can derive (see Exercise 7.3) G = βps η[(ρ + δ)λ λ]. (11) We use (3.74) to obtain the optimal long-run stationary equilibrium or turnpike {Ḡ,ū, λ}. That is, we obtain λ = λ = 1 from (8) by using H/ u = 0. We then set λ = λ = 1 and λ = 0 in (9). Finally, from (11) and (9), Ḡ or also the singular level G s can be obtained as Ḡ = G s = βps η(ρ + δ). (12) Chapter 7 p. 9/54

10 SOLUTION BY THE MAXIMUM PRINCIPLE CONT. The property of Ḡ is that the optimal policy is to go to Ḡ as fast as possible. If G 0 < Ḡ, it is optimal to jump instantaneously to Ḡ by applying an appropriate impulse at t = 0 and then set u (t) = ū = δḡ for t > 0. If G 0 > Ḡ, the optimal control u (t) = 0 until the stock of goodwill depreciates to the level Ḡ, at which time the control switches to u (t) = δḡ and stays at this level to maintain the level Ḡ of goodwill. This optimal policy is graphed in Figure 7.1 for these two different initial conditions. Chapter 7 p. 10/54

11 FIGURE 7.1: OPTIMAL POLICIES IN THE NERLOVE-ARROW MODEL Case : ; Case 2: Chapter 7 p. 11/54

12 SOLUTION BY THE MAXIMUM PRINCIPLE For a time-dependent Z, however, Ḡ(t) = G(Z(t)) will be a function of time. To maintain this level of Ḡ(t), the required control is ū(t) = δḡ(t) + Ḡ(t). If Ḡ(t) is decreasing sufficiently fast, then ū(t) may become negative and thus infeasible. If ū(t) 0 for all t, then the optimal policy is as before. However, suppose ū(t) is infeasible in the interval [t 1,t 2 ] shown in Figure 7.2. In such a case, it is feasible to set u(t) = ū(t) for t t 1 ; at t = t 1 (which is point A in Figure 7.2) we can no longer stay on the turnpike and must set u(t) = 0 until we hit the turnpike again (at point B in Figure 7.2). However, such a policy is not necessarily optimal. Chapter 7 p. 12/54

13 TURNPIKE AND THE NATURE OF OPTIMAL CONTROL Chapter 7 p. 13/54

14 SOLUTION BY THE MAXIMUM PRINCIPLE For instance, suppose we leave the turnpike at point C anticipating the infeasibility at point A. The new path CDEB may be better than the old path CAB. Roughly the reason this may happen is that path CDEB is closer to the turnpike than CAB. The picture in Figure 7.2 illustrates such a case. The optimal policy is the one that is closest to the turnpike. This discussion will become clearer in Section 7.2.2, when a similar situation arises in connection with the Vidale-Wolfe model. For further details, see Sethi (1977b) and Breakwell (1968). Chapter 7 p. 14/54

15 A NONLINEAR EXTENSION The new optimal control problem is: max u 0 { J = 0 subject to Ġ = u δg, G(0) = G 0. } e ρt [π 1 (u) + π 2 (G) u]dt (13) H = π 1 (u) + π 2 (G) u + λ(u δg), (14) λ = ρλ π 2(G) + δλ, (15) H u = π 1(u) 1 + λ = 0. (16) Since π 1 < 0, we can invert π 1 function of λ. Thus, to solve (16) for u as a u = f 1 (λ). (17) Chapter 7 p. 15/54

16 A NONLINEAR EXTENSION CONT. The function f 1 in (17) permits us to rewrite the state equation in (13) and the adjoint equation (15) as the following initial value problem: { Ġ + δg = f1 (λ), G(0) = G 0, λ + (ρ + δ)λ = π 2 (G), λ(0) = λ 0. (18) We note that in Regions I and II, f 1 (λ) δg < 0, in Regions III and IV, f 1 (λ) δg > 0, in Regions I and III, (ρ + δ)λ π 2 (G) > 0, in Regions II and IV, (ρ + δ)λ π 2 (G) < 0. Chapter 7 p. 16/54

17 A NONLINEAR EXTENSION CONT. The above implies that in Region I, G decreasing and λ increasing, in Region II, G decreasing and λ decreasing, in Region III, G increasing and λ increasing, in Region IV, G increasing and λ decreasing. Chapter 7 p. 17/54

18 FIGURE 7.3: PHASE DIAGRAM OF SYSTEM (18) FOR PROBLEM (13) Chapter 7 p. 18/54

19 A NONLINEAR EXTENSION CONT. Because of these conditions it is clear that for a given G 0, a choice of λ 0 such that (λ 0,G 0 ) is in Regions II and III, will not lead to a path converging to the turnpike point ( λ,ḡ). On the other hand, the choice of (λ 0,G 0 ) in Region I when G 0 > Ḡ or (λ 0,G 0 ) in Region IV when G 0 < Ḡ, can give a path that converges to ( λ,ḡ). From a result in Coddington and Levinson (1955), it can be shown that at least in the neighborhood of ( λ,ḡ), there exists a locus of optimum starting points (λ 0,G 0 ). A detailed analysis leads to the saddle point path as shown darkened in Figure 7.3. Clearly the initial control u (0) = f 1 (λ 0 ). Chapter 7 p. 19/54

20 A NONLINEAR EXTENSION CONT. Given G 0 > Ḡ, we choose λ 0 on the saddle point path in Region I of Figure 7.3. Clearly the initial control u (0) = f 1 (λ 0 ). Furthermore, λ(t) is increasing and by (17), u(t) is increasing, so that in this case the optimal policy is to advertise at a low rate initially and gradually increase advertising to the turnpike level ū = δḡ. If G 0 < Ḡ, it can be shown similarly that the optimal policy is to advertise most heavily in the beginning and gradually decrease it to the turnpike level ū as G approaches Ḡ. Chapter 7 p. 20/54

21 A NONLINEAR EXTENSION CONT. Note that the approach to the equilibrium Ḡ is no longer via the bang-bang control as in the Nerlove-Arrow model. This, of course, is what one would expect when a model is made nonlinear with respect to the control variable u. Chapter 7 p. 21/54

22 THE VIDALE-WOLFE ADVERTISING MODEL Ṡ = au(1 S M ) bs. (19) Now we can rewrite (19) as x = S M. (20) r = a M, δ = b + M M. (21) ẋ = ru(1 x) δx, x(0) = x. (22) From now on we assume M, and hence δ and r, to be positive constants. Chapter 7 p. 22/54

23 THE VIDALE-WOLFE ADVERTISING MODEL CONT. The optimal control problem can be stated as follows: { max J = } T 0 e ρt (πx u)dt subject to ẋ = ru(1 x) δx, x(0) = x 0, the terminal state constraint x(t) = x T, and the control constraint 0 u Q. (23) Here Q can be finite or infinite and the target market share x T is in [0, 1]. Chapter 7 p. 23/54

24 SOLUTION USING GREEN S THEOREM WHEN Q IS LARGE To make use of Green s theorem, it is convenient to consider times τ and θ, where 0 τ < θ T, and the problem: subject to { max J(τ,θ) = θ τ } e ρt (πx u)dt (24) ẋ = ru(1 x) δx, x(τ) = A, x(θ) = B, (25) 0 u Q. (26) Chapter 7 p. 24/54

25 SOLUTION USING GREEN S THEOREM WHEN Q IS LARGE CONT. To change the objective function in (24) into a line integral along any feasible arc Γ 1 from (τ,a) to (θ,b) in (t,x)-space as shown in Figure 7.4, we multiply (25) by dt and obtain the formal relation udt = dx + δxdt r(1 x), which we substitute into the objective function (24). Thus, J Γ1 = Γ 1 {[ πx δx r(1 x) ] e ρt dt } 1 r(1 x) e ρt dx. Chapter 7 p. 25/54

26 SOLUTION USING GREEN S THEOREM WHEN Q IS LARGE CONT. Consider another feasible arc Γ 2 from (τ,a) to (θ,b) lying above Γ 1 as shown in Figure 7.4. Let Γ = Γ 1 Γ 2, where Γ is a simple closed curve traversed in the counter-clockwise direction. That is, Γ goes along Γ 1 in the direction of its arrow and along Γ 2 in the direction opposite to its arrow. We now have J Γ = J Γ1 Γ 2 = J Γ1 J Γ2. (27) Chapter 7 p. 26/54

27 FIGURE 7.4: FEASIBLE ARCS IN (t, x)-space Chapter 7 p. 27/54

28 SOLUTION BY THE MAXIMUM PRINCIPLE Since Γ is a simple closed curve, we can use Green s theorem to express J Γ as an area integral over the region R enclosed by Γ. Thus, treating x and t as independent variables, we can write J Γ = = = Γ {[ πx R R { t [ ] δx e ρt dt r(1 x) [ ] e ρt r(1 x) x δ (1 x) 2 + ρ (1 x) πr 1 r(1 x) e ρt dx [ (πx } ]} δx r(1 x) )e ρt dtdx ] e ρt dtdx. (28) r Denote the term in brackets of the integrand of (28) by I(x) = δ (1 x) 2 + ρ (1 x) πr. (29) Note that the sign of the integrand is the same as the sign of I(x). Chapter 7 p. 28/54

29 LEMMA 7.1 Lemma 7.1 (Comparison Lemma). Let Γ 1 and Γ 2 be the lower and upper feasible arcs as shown in Figure 7.4. If I(x) 0 for all (x,t) R, then the lower arc Γ 1 is at least as profitable as the upper arc Γ 2. Analogously, if I(x) 0 for all (x,t) R, then Γ 2 is at least as profitable as Γ 1. Proof. If I(x) 0 for all (x,t) R, then J Γ 0 from (28) and (29). Hence from (27), J Γ1 J Γ2. The proof of the other statement is similar. Chapter 7 p. 29/54

30 To make use of this lemma to find the optimal control for the problem stated in (23), we need to find regions where I(x) is positive and where it is negative. For this, note first that I(x) is an increasing function of x in [0, 1]. Solving I(x) = 0 will give that value of x, above which I(x) is positive and below which I(x) is negative. Since I(x) is quadratic in 1/(1 x), we can use the quadratic formula (see Exercise 7.15) to get x = 1 2δ ρ ± ρ 2 + 4πrδ. To keep x in the interval [0, 1], we must choose the positive sign before the radical. Chapter 7 p. 30/54

31 The optimal x must be nonnegative so we have x s = max { 1 } 2δ ρ + ρ 2 + 4πrδ, 0, (30) where the superscript s is used because this will turn out to be a singular trajectory. Since x s is nonnegative, the control u s = δx s r(1 x s ). (31) Note that x s = 0 and u s = 0 if, and only if, πr δ + ρ. Chapter 7 p. 31/54

32 FIGURE 7.5: OPTIMAL TRAJECTORY FOR CASE 1: x 0 x s AND x s x T Chapter 7 p. 32/54

33 FIGURE 7.6: OPTIMAL TRAJECTORY FOR CASE 2: x 0 x s AND x s x T Chapter 7 p. 33/54

34 FIGURE 7.7: OPTIMAL TRAJECTORY FOR CASE 3: x 0 x s AND x s x T Chapter 7 p. 34/54

35 FIGURE 7.8: OPTIMAL TRAJECTORY FOR CASE 4: x 0 x s AND x s x T Chapter 7 p. 35/54

36 THEOREM 7.1 Theorem 7.1 Let T be large and let x T be reachable from x 0. For the Cases 1-4 of inequalities relating x 0 and x T to x s, the optimal trajectories are given in Figures , respectively. Proof. We give details for Case 1 only. The proofs for the other cases are similar. Figure 7.9 shows the optimal trajectory for Figure 7.5 together with an arbitrarily chosen feasible trajectory, shown dotted. It should be clear that the dotted trajectory cannot cross the arc x 0 to C, since u = Q on that arc. Similarly the dotted trajectory cannot cross the arc G to x T, because u = 0 on that arc. Chapter 7 p. 36/54

37 FIGURE 7.9: OPTIMAL TRAJECTORY (SOLID LINE) Chapter 7 p. 37/54

38 PROOF OF THEOREM 7.1 We subdivide the interval [0,T] into subintervals over which the dotted arc is either above, below, or identical to the solid arc. In Figure 7.9 these subintervals are [0, d], [d, e], [e, f], and [f,t]. Because I(x) is positive for x > x s and I(x) is negative for x < x s, the regions enclosed by the two trajectories have been marked with + or sign depending on whether I(x) is positive or negative on the regions, respectively. By Lemma 7.1, the solid arc is better than the dotted arc in the subintervals [0, d], [d, e], and [f, T]; in interval [e,f], they have identical values. Hence the dotted trajectory is inferior to the solid trajectory. This proof can be extended to any (countable) number of crossings of the trajectories; see Sethi (1977b). Chapter 7 p. 38/54

39 THEOREM 7.2 Theorem 7.2 Let T be small, i.e., T < t 1 + t 2, and let x T be reachable from x 0. For the two possible Cases 1 and 2 of inequalities relating x 0 to x T and x s, the optimal trajectories are given in Figures 7.10 and 7.11, respectively. Proof. The requirement of feasibility when T is small rules out cases where x 0 and x T are on opposite sides of or equal to x s. The proofs of optimality of the trajectories shown in Figures 7.10 and 7.11 are similar to the proofs of the parts of Theorem 7.1, and are left as Exercise In Figures 7.10 and 7.11, it is possible to have either t 1 T or t 2 T. Try sketching some of these special cases. Chapter 7 p. 39/54

40 FIGURE 7.10: OPTIMAL TRAJECTORY WHEN T IS SMALL IN CASE 1: x 0 < x s AND x T < x s Chapter 7 p. 40/54

41 FIGURE 7.11: OPTIMAL TRAJECTORY WHEN T IS SMALL IN CASE 2: x 0 > x s AND x T < x s Chapter 7 p. 41/54

42 FIGURE 7.12: OPTIMAL TRAJECTORY FOR CASE 2 OF THEOREM 7.1 FOR Q = Impulse Control Impulse Control Chapter 7 p. 42/54

43 IMPULSE CONTROL The effect on the objective function of the impulse control imp(x 0,x s ) is x(ε) = ( x 0 ru ) e (δ+ru)ε + δ + ru ru δ + ru. We must choose u(ε) so that x(ε) is x s. It should be clear that u(ε) as ε 0. With F(x,u,τ) = πx(τ) u(τ) and t = 0 in (1.23), we have the impulse I = imp(x 0,x s, 0) = lim ε 0 [ u(ε)ε]. Letting ε 0, u(ε)ε I, and x(ε) = x s in the expression for x(ε) obtained above. This gives x(0+) = e ri (x 0 1) + 1, and imp(x 0,x s, 0) = 1 r ln [ 1 x0 1 x s ]. (32) Chapter 7 p. 43/54

44 SOLUTION WHEN Q IS SMALL H = πx u + λ[ru(1 x) δx] = πx δλx + u[ 1 + rλ(1 x)], (33) λ = ρλ L x L = H + µ(q u), (34) = ρλ + λ(ru + δ) π, (35) where λ(t) is a constant, as in Row 2 of Table 3.1, that must be determined. Furthermore, the Lagrange multiplier µ in (34) must satisfy µ 0, µ(q u) = 0. (36) Chapter 7 p. 44/54

45 SOLUTION WHEN Q IS SMALL CONT. From (33) we notice that the Hamiltonian is linear in the control. So the optimal control is where u (t) = bang[0,q;w(t)], (37) W(t) = W(x(t),λ(t)) = rλ(t)(1 x(t)) 1. (38) Chapter 7 p. 45/54

46 SOLUTION WHEN T IS INFINITE We now formulate the infinite horizon version of (23): max { J = 0 e ρt (πx u)dt } subject to ẋ = ru(1 x) δx, x(0) = x 0, 0 u Q. (39) When Q is small, i.e., Q < u s, it is not possible to follow the turnpike x = x s, because that would require u = u s, which is not a feasible control. Intuitively, it seems clear that the "closest" stationary path which we can follow is the path obtained by setting ẋ = 0 and u = Q, the largest possible control, in the state equation of (39). Chapter 7 p. 46/54

47 This gives SOLUTION WHEN T IS INFINITE CONT. x = and correspondingly we obtain λ = rq rq + δ, (40) π ρ + δ + rq (41) by setting u = Q and λ = 0 in (35) and solving for λ. More specifically, we state the following theorems which give the turnpike and optimal control when Q is small. Let us define ˆx and µ such that W(ˆx, λ) = r λ(1 ˆx) 1 = 0 and L u ( x,ū, λ, µ) = W( x, λ) µ = 0. Thus, ˆx = 1 1/r λ, (42) µ = r λ(1 x) 1. (43) Chapter 7 p. 47/54

48 THEOREM 7.3 Theorem 7.3 When Q is small, the following quantities form a turnpike. { x,q, λ, µ} (44) Proof. We show that the conditions in (3.73) hold for (44). The first two are obvious. By Exercise 7.28 we know x ˆx, which, from definitions (42) and (43), implies µ 0. Furthermore ū = Q, so (36) holds and the third condition of (3.73) also holds. Finally because W = µ from (38) and (43), it follows that W 0, so the Hamiltonian maximizing condition of (3.73) holds with ū = Q, and the proof is complete. Chapter 7 p. 48/54

49 FIGURE 7.13: OPTIMAL TRAJECTORIES FOR x(0) < ˆx Chapter 7 p. 49/54

50 FIGURE 7.14: OPTIMAL TRAJECTORY FOR x(0) > ˆx τ τ Case(b) Case (a) Chapter 7 p. 50/54

51 THEOREM 7.4 Theorem 7.4 When Q is small, the optimal control at time τ is given by: (a) If x(τ) ˆx, then u (τ) = Q. (b) If x(τ) > ˆx, then u (τ) = 0. Proof. (a) We set λ(t) = λ for all t τ and note that λ satisfies the adjoint equation (35) and the transversality condition (3.70). By Exercise 7.28 and the assumption that x(τ) ˆx, we know that x(t) ˆx for all t. The proof that (36) and (37) hold for all t τ relies on the fact that x(t) ˆx and on an argument similar to the proof of the previous theorem. Chapter 7 p. 51/54

52 PROOF OF THEOREM 7.4 Figure 7.13 shows the optimal trajectories for x(0) < ˆx and two different starting values x(0), one above and the other below ˆx. Note that in this figure we are always in Case (a) since x(τ) ˆx for all τ 0. (b) Assume x(0) > ˆx. For this case we will show that the optimal trajectory is as shown in Figure 7.14, which is obtained by applying u = 0 until x = ˆx and u = Q thereafter. Using this policy we can find the time t 1 at which x(t 1 ) = ˆx, by solving the state equation in (39) with u = 0. This gives t 1 = 1 δ ln x(0) ˆx. (45) Chapter 7 p. 52/54

53 PROOF OF THEOREM 7.4 CONT. Clearly for t t 1, the policy u = Q is optimal because Case (a) applies. We now consider the interval [0,t 1 ), where we set u = 0. Let τ be any time in this interval as shown in Figure 7.14, and let x(τ) be the corresponding value of the state variable. Then x(τ) = x 0 e δτ in the interval [0,t 1 ]. With u = 0 in (35), the adjoint equation on [0,t 1 ) becomes λ = (ρ + δ)λ π. So we solve the adjoint equation with λ(t 1 ) = λ and obtain λ(τ) = π ρ + δ + ( λ π ) e (ρ+δ)(τ t1), τ [0,t 1 ]. (46) ρ + δ In Exercise 7.31, you are asked to show that the switching function W(t) defined in (38) is negative in the interval [0,t 1 ) and W(t 1 ) = 0. Chapter 7 p. 53/54

54 PROOF OF THEOREM 7.4 CONT. Therefore by (37), the policy u = 0 used in deriving (45) and (46) satisfies the maximum principle. This policy "joins" the optimal policy after t 1 because λ(t 1 ) = λ. In this book the sufficiency of the transversality condition (3.70) was stated under the hypothesis that the derived Hamiltonian was concave; see Theorem 2.1. In the present example, this hypothesis does not hold. However, as mentioned in Section 7.2.3, for this simple bilinear problem it can be shown that (3.70) is sufficient for optimality. Because of the technical nature of this issue we omit the details. Chapter 7 p. 54/54

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