7.2 The Vidale-Wolfe Advertising Model
7.2.2 Solution Using Green’s Theorem When
To make use of Green’s theorem, it is convenient to consider timesτ and θ,where 0≤τ < θ≤T, and the problem:
max
J(τ , θ) = θ
τ
eưρt(πxưu)dt
(7.27)
238 7. Applications to Marketing subject to
˙
x=ru(1−x)−δx, x(τ) =A, x(θ) =B, (7.28)
0≤u≤Q. (7.29)
To change the objective function in (7.27) into a line integral along any feasible arc Γ1 from (τ , A) to (θ, B) in (t, x)-space as shown in Fig.7.4, we multiply (7.28) by dt and obtain the formal relation
udt= dx+δxdt r(1−x) ,
which we substitute into the objective function (7.27). Thus, JΓ1 =
Γ1
πx− δx r(1−x)
e−ρtdt− 1
r(1−x)e−ρtdx
.
Figure 7.4: Feasible arcs in (t, x)-space
Consider another feasible arc Γ2 from (τ , A) to (θ, B) lying above Γ1 as shown in Fig.7.4. Let Γ = Γ1−Γ2,where Γ is a simple closed curve traversed in the counter-clockwise direction. That is, Γ goes along Γ1 in the direction of its arrow and along Γ2 in the direction opposite to its arrow. We now have
JΓ=JΓ1−Γ2 =JΓ1 −JΓ2. (7.30)
7.2. The Vidale-Wolfe Advertising Model 239 Since Γ is a simple closed curve, we can use Green’s theorem to express JΓ as an area integral over the region R enclosed by Γ. Thus, treating x and tas independent variables, we can write
JΓ = ,
Γ
πx− δx r(1−x)
e−ρtdt− 1
r(1−x)e−ρtdx
=
R
∂
∂t
−e−ρt r(1−x)
− ∂
∂x
(πx− δx
r(1−x))e−ρt
dtdx
=
R
δ
(1−x)2 + ρ
(1−x) −πr e−ρt
r dtdx. (7.31)
Denote the term in brackets of the integrand of (7.31) by I(x) = δ
(1−x)2 + ρ
(1−x)−πr. (7.32)
Note that the sign of the integrand is the same as the sign of I(x).
Lemma 7.1 (Comparison Lemma) Let Γ1 and Γ2 be the lower and upper feasible arcs as shown in Fig.7.4. If I(x) ≥ 0 for all (x, t) ∈ R, then the lower arc Γ1 is at least as profitable as the upper arc Γ2. Analogously, ifI(x)≤0for all(x, t)∈R,thenΓ2 is at least as profitable as Γ1.
ProofIf I(x)≥0 for all (x, t)∈R,thenJΓ ≥0 from (7.31) and (7.32).
Hence from (7.30),JΓ1 ≥JΓ2.The proof of the other statement is similar.
2
To make use of this lemma to find the optimal control for the problem stated in (7.26), we need to find regions whereI(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 Exercise7.16) to get
x= 1− 2δ
−ρ±#
ρ2+ 4πrδ. (7.33)
To keep x in the interval [0,1],we must choose the positive sign before the radical. The optimal x must be nonnegative so we have
xs= max
1− 2δ
−ρ+#
ρ2+ 4πrδ,0
, (7.34)
240 7. Applications to Marketing where the superscriptsis used because this will turn out to be a singular trajectory. Sincexs is nonnegative, the control
us= δxs
r(1−xs) (7.35)
corresponding to (7.34) will always be nonnegative. Also sinceQ is as- sumed to be large,uswill always be feasible. Moreover, in Exercise7.17, you will be asked to show that xs = 0 and us = 0 if, and only if, πr ≤δ+ρ.
We now have enough machinery to obtain the optimal solution for (7.26) whenQis assumed to be sufficiently large, i.e.,Q≥us,whereusis given in (7.35). We state these in the form of two theorems: Theorem7.1 refers to the case in which T is large; Theorem 7.2 refers to the case in which T is small. To define these concepts, let t1 be the shortest time to go from x0 toxs and similarly let t2 be the shortest time to go from
Figure 7.5: Optimal trajectory for Case 1: x0≤xs and xT ≤xs xs toxT. Then, we say T is large if T ≥ t1+t2; otherwise T is small.
Figures 7.5, 7.6, 7.7, and 7.8 show cases for which T is large, while Figs.7.10and 7.11show cases for whichT is small. In Exercise7.21you are asked to determine whether T is large or small in specific cases. In the statements of the theorems we will assume that x0 and xT are such that xT is reachable fromx0.In Exercise7.15 you are asked to find the reachable set for any given initial condition x0.
In Figures 7.5, 7.6, 7.7, and 7.8, the quantities t1 and t2 are case dependent and not necessarily the same; see Exercise 7.20.
7.2. The Vidale-Wolfe Advertising Model 241 Theorem 7.1 Let T be large and let xT be reachable from x0. For the Cases1–4of inequalities relatingx0andxT toxs,the optimal trajectories are given in Figures 7.5, 7.6, 7.7, and 7.8, respectively.
Proof We give details for Case 1 only. The proofs for the other cases are similar. Figure7.9shows the optimal trajectory for Fig.7.5together with an arbitrarily chosen feasible trajectory, shown dotted. It should be clear that the dotted trajectory cannot cross the arc x0 to C, since u=Qon that arc. Similarly the dotted trajectory cannot cross the arc G to xT,becauseu= 0 on that arc.
We subdivide the interval [0, T] into subintervals over which the dot- ted arc is either above, below, or identical to the solid arc. In Fig.7.9
Figure 7.6: Optimal trajectory for Case 2: x0< xs and xT > xs
Figure 7.7: Optimal trajectory for Case 3: x0> xs and xT < xs
242 7. Applications to Marketing
Figure 7.8: Optimal trajectory for Case 4: x0> xs and xT > xs these subintervals are [0, d], [d, e], [e, f],and [f, T].BecauseI(x) is pos- itive for x > xs and I(x) is negative for x < xs, the regions enclosed by the two trajectories have been marked with a + or −sign depending on whether I(x) is positive or negative on the regions, respectively. By Lemma7.1, 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). 2
Figures7.5, 7.6, 7.7, and 7.8 are drawn for the situation whenT >
t1+t2. In Exercise 7.25, you are asked to consider the case when T = t1+t2.The following theorem deals with the case when T < t1+t2. Theorem 7.2 Let T be small, i.e.,T < t1+t2,and letxT be reach- able from x0. For the two possible Cases 1 and 2 of inequalities relating x0 and xT to xs, the optimal trajectories are given in Figs.7.10 and 7.11, respectively.
Proof The requirement of feasibility when T is small rules out cases where x0 and xT are on opposite sides of or equal to xs. The proofs of optimality of the trajectories shown in Figs.7.10 and 7.11are similar to the proofs of the parts of Theorem 7.1, and are left as Exercise7.25. In Figs.7.10 and 7.11, it is possible to have either t1 ≥ T or t2 ≥ T. Try sketching some of these special cases. 2
All of the previous discussion has assumed that Q was finite and sufficiently large, but we can easily extend this to the case whenQ=∞.
7.2. The Vidale-Wolfe Advertising Model 243
Figure 7.9: Optimal trajectory (solid lines)
Figure 7.10: Optimal trajectory when T is small in Case 1: x0 < xs and xT > xs
This possibility makes the arcs in Figs.7.5, 7.6, 7.7, 7.8, 7.9, and 7.10, corresponding tou∗ =Q,become vertical line segments corresponding to impulse controls. For example, Fig.7.6becomes Fig.7.12 when Q=∞ and we apply theimpulse control imp(x0, xs; 0) whenx0< xs.
Next we compute the cost of imp(x0, xs; 0) by assessing its effect on the objective function of (7.26). For this, we integrate the state equation in (7.26) from 0 toεwith the initial conditionx0 andutreated
244 7. Applications to Marketing
x0
x
xT
u* = 0 u* = Q
0 1
T – t2 t1 T
t xs
Figure 7.11: Optimal trajectory when T is small in Case 2: x0 > xs and xT > xs
Impulse Control
Impulse Control
Figure 7.12: Optimal trajectory for Case 2 of Theorem7.1for Q=∞
7.2. The Vidale-Wolfe Advertising Model 245 as constant. By using (A.7), we can write the solution as
x(ε) = x0e−(δ+ru)ε+ ε
0 e(δ+ru)(τ−ε)rudτ
=
x0− ru δ+ru
e−(δ+ru)ε+ ru δ+ru.
According to the procedure given in Sect.1.4, we must, for u, choose u(ε) so thatx(ε) isxs.It should be clear thatu(ε)→ ∞ asε→0.With F(x, u, τ) =πx(τ)−u(τ) andt= 0 in (1.23), we have the impulse
I = imp(x0, xs; 0) = lim
ε→0[−u(ε)ε].
It is possible to solve for I by lettingε→0,−u(ε)ε→I, u(ε)→ ∞,and x(ε) =xs in the expression forx(ε) obtained above. This gives
x(0+) =erI(x0−1) + 1.
Therefore,
imp(x0, xs; 0) =−1 rln
1−x0 1−xs
. (7.36)
We remark that this formula holds for any timet,as well ast= 0.Hence it can also be used at t= T to compute the impulse at the end of the period; see Fig.7.12and Exercise 7.28.