6.3 Decision Horizons and Forecast Horizons
6.3.2 Horizons for the Wheat Trading Model with No Short-Selling and a Warehousing Constraint
In order to give an example in which a strong forecast horizon occurs, we modify the example of Sect.6.2.4by adding the warehousing constraint y≤1 or
1−y≥0, (6.82)
changing the terminal time to T = 4, and defining the price trajectory to be
p(t) =
⎧⎪
⎨
⎪⎩
−2t+ 7 for t∈[0,2), t+ 1 for t∈[2,4].
(6.83)
6.3. Decision Horizons and Forecast Horizons 215
Sell Do Nothing Buy
Decision Horizon
Price Shield
Sell Weak Forecast
Horizon
Figure 6.8: Decision horizon and optimal policy for the wheat trading model
The Hamiltonian of the new problem is unchanged and is given in (6.66). Furthermore, λ1 = 1. The optimal control is defined in three parts as:
v∗(t) = bang[−1,1;λ2(t)−p(t)] when 0< y <1, (6.84) v∗(t) = bang [0,1;λ2(t)−p(t)] wheny= 0, (6.85) v∗(t) = bang[−1,0;λ2(t)−p(t)] wheny= 1. (6.86) Defining a Lagrange multiplier η1 for the derivative of (6.82), i.e., for
−y˙=−v≥0,we form the Lagrangian
L=H+μ1(v+ 1) +μ2(1−v) +ηv+η1(−v), (6.87) where μ1, μ2,and η satisfy (6.70)–(6.72) and η1 satisfies
η1 ≥0, η1(1−y) = 0, η˙1≤0. (6.88) Furthermore, the optimal trajectory must satisfy
∂L
∂v =λ2−p+μ1−μ2+η−η1 = 0. (6.89)
216 6. Applications to Production and Inventory As before, λ1= 1 and λ2 satisfies
λ˙2 = 1/2, λ2(4−) =p(4) +γ1−γ2= 5 +γ1−γ2, (6.90) where
γ1 ≥0, γ1y(4) = 0, γ2 ≥0, γ2(1−y(4)) = 0. (6.91) Let us first tryγ1 =γ2 = 0.Let ˆtbe the time of the last jump of the adjoint function λ2(t) before the terminal time T = 4.Then,
λ2(t) =t/2 + 3 for ˆt≤t <4. (6.92) The graph of (6.92) intersects the price trajectory att= 8/5 as shown in Fig.6.9. It also stays above the price trajectory in the interval [8/5,4]
so that, if there were no warehousing constraint (6.82), the optimal de- cision in this interval would be to buy at the maximum rate. However, with the constraint (6.82), this is not possible. Thus ˆt > 8/5, since λ2 will have a jump in the interval [8/5,4].
Figure 6.9: Optimal policy and horizons for the wheat trading model with no short-selling and a warehouse constraint
6.3. Decision Horizons and Forecast Horizons 217 To find the actual value of ˆtwe must insert a line of slope 1/2 above the minimum price att= 2 in such a way that its two intersection points with the price trajectory are exactly one time unit (the time required to fill up the warehouse) apart. Thus using (6.83), ˆtmust satisfy
−2(ˆt−1) + 7 + (1/2)(1) = ˆt+ 1, which yields ˆt= 17/6.
The rest of the analysis for determiningλ2 including the jump con- ditions is similar to that given in Sect.6.2.4. Thus,
λ2(t) =
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
t/2 + 9/2 fort∈[0,1), t/2 + 29/12 fort∈[1,17/6), t/2 + 3 fort∈[17/6,4].
(6.93)
This makes γ1 =γ2= 0 the correct guess.
Given (6.93), the optimal policy is given by (6.84)–(6.86) and is shown in Fig.6.9. To complete the maximum principle we must derive expressions for the Lagrange multipliers in the four intervals shown in Fig.6.9.
Interval[0,1) : μ2=η=η1 = 0, μ1=p−λ2 >0;
v∗ =−1, 0< y∗ <1.
Interval[1,11/6) : μ1 =μ2 =η1= 0, η=p−λ2 >0, η˙ ≤0;
v∗ = 0, y∗ = 0.
Interval[11/16,17/6) : μ1 =η=η1 = 0, μ2 =λ2−p >0;
v∗ = 1, 0< y∗ <1.
Interval[17/6,4] : μ1 =μ2 =η= 0, η1 =λ2−p >0, η˙1≤0, γ1 =γ2= 0;
v∗ = 0, y∗ = 1.
In Exercise 6.17 you are asked to solve another variant of this problem.
For the example in Fig.6.9we have labeledt= 1 as a decision horizon and ˆt = 17/6 as a strong forecast horizon. By this we mean that the
218 6. Applications to Production and Inventory optimal decision in [0,1] continues to be to sell at the maximum rate regardless of the price trajectory p(t) fort > 17/6. Because ˆt= 17/6 is a strong forecast horizon, we can terminate the price shield at that time as shown in the figure.
In order to illustrate the statements in the previous paragraph, we consider two examples of price changes after ˆt= 17/6.
Example 6.3 Assume the price trajectory to be
p(t) =
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
−2t+ 7 fort∈[0,2), t+ 1 fort∈[2,17/6), 25t/7−44/7 fort∈[17/6,4],
which is sketched in Fig.6.10. Note that the price trajectory up to time 17/6 is the same as before, and the price after time 17/6 goes above the extension of the price shield in Fig.6.9.
Figure 6.10: Optimal policy and horizons for Example 6.3
6.3. Decision Horizons and Forecast Horizons 219 SolutionThe newλ2 trajectory is shown in Fig.6.10, which is the same as before fort <17/6,and after that it isλ2(t) =t/2+6 fort∈[17/6, 4].
The optimal policy is as shown in Fig.6.10, and as previously asserted, the optimal policy in [0,1) remains unchanged. In Exercise 6.17 you are asked to verify the maximum principle for the solution of Fig.6.10.
Example 6.4 Assume the price trajectory to be
p(t) =
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
−2t+ 7 fort∈[0,2), t+ 1 fort∈[2,17/6),
−t/2 + 21/4 fort∈[17/6,4], which is sketched in Fig.6.11.
Figure 6.11: Optimal policy and horizons for Example 6.4 Solution Again the price trajectory is the same up to time 17/6, but the price after time 17/6 is declining. This changes the optimal policy
220 6. Applications to Production and Inventory in the time interval [1,17/6),but the optimal policy will still be to sell in [0,1).
As in the beginning of the section, we solve (6.90) to obtain λ2(t) = t/2+5/4 for ˆt1 ≤t≤4,where ˆt1 ≥1 is the time of the last jump which is to be determined. It is intuitively clear that some profit can be made by buying and selling to take advantage of the price rise between t= 2 and t = 17/6.For this, the λ2(t) trajectory must cross the price trajectory between times 2 and 17/6 as shown in Fig.6.11, and the inventory y must go to 0 between times 17/6 and 4 so that λ2 can jump downward to satisfy the ending condition λ2(4−) = p(4) = 13/4. Since we must buy and sell equal amounts, the point of intersection of theλ2trajectory with the rising price segment, i.e., ˆt1−α,must be exactly in the middle of the two other intersection points, ˆt1 and ˆt1−2α, of λ2 with the two declining price trajectories. Thus, ˆt1 and α must satisfy:
−2(ˆt1−2α) + 7 +α/2 = (ˆt1−α) + 1, (ˆt1−α) + 1 +α/2 = −tˆ1/2 + 21/4.
These can be solved to yield ˆt1 = 163/54 and α = 5/9. The times tˆ1,ˆt1−α,and ˆt1−2α are shown in Fig.6.11. The λ2 trajectory is given by
λ2(t) =
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
t/2 + 9/2 fort∈[0,1), t/2 + 241/108 fort∈[1,163/54), t/2 + 5/4 fort∈[163/54,4].
Evaluation of the Lagrange multipliers and verification of the maximum principle is similar to that for the case in Fig.6.9.
In Sect.6.3we have given several examples of decision horizons and weak and strong forecast horizons. In Sect.6.3.1 we found a decision horizon which was also a weak forecast horizon, and it occurred exactly when y(t) = 0. We also introduced the idea of a price shield in that section. In Sect.6.3.2we imposed a warehousing constraint and obtained the same decision horizon and a strong forecast horizon, which occurred when y(t) = 1.
Note that if we had solved the problem with T = 1,then y∗(1) = 0;
and if we had solved the problem with T = 17/6, then y∗(1) = 0 and y∗(17/6) = 1. The latter problem has the smallest T such that both y∗ = 0 and y∗ = 1 occur for t > 0, given the price trajectory. This is one of the ways that timet= 17/6 can be found to be a forecast horizon
Exercises for Chapter 6 221 along with the decision horizon at time t= 1.There are other ways to find strong forecast horizons. For a survey of the literature, see Chand et al. (2002).
Exercises for Chapter 6
E 6.1 Verify the expressions for a1 and a2 given in (6.16) and (6.17).
E 6.2 Verify (6.27). Note thatρ= 0 is assumed in Sect.6.1.4.
E 6.3 Verify (6.29). Again assumeρ= 0.
E 6.4 Given the demand function
S =t(t−4)(t−8)(t−12)(t−16) + 30, ρ= 0,Iˆ= 15, T = 16, and α= 1,obtain Q(t) from (6.27).
E 6.5 Complete the solution of Example6.2 in Sect.6.1.4.
E 6.6 For the model of Sect.6.1.6, derive the turnpike triple by using the conditions in (6.39).
E 6.7 Solve the production-inventory model of Sect.6.1.6 for the pa- rameter values listed on Fig.6.4, and draw the figure using MATLAB or another suitable software.
E 6.8 Give an intuitive interpretation of (6.55).
E 6.9 Assume that there is a transaction costcv2whenvunits of wheat are bought or sold in the model of Sect.6.2.1. Derive the form of the optimal policy.
E 6.10 In Exercise6.9, assume T = 10, x(0) = 10, y(0) = 0, c= 1/18, h(y) = (1/2)y2, V1 = V2 = ∞, r = 0, and p(t) = 10 +t. Solve the resulting TPBVP to obtain the optimal control in closed form.
E 6.11 Set up the two-point boundary value problem for Exercise 6.9 with c= 0.05, h(y) = (1/2)y2, and the remaining values of parameters as in the model of Sect.6.2.3.
E 6.12 Use Excel, as illustrated in Sect.2.5, to solve the TPBVP of Exercise 6.11.
222 6. Applications to Production and Inventory E 6.13 Show that the solution obtained for the problem in Sect.6.2.3 satisfies the necessary conditions of the maximum principle. Conclude the optimality of the solution by showing that the maximum principle conditions are also sufficient.
E 6.14 Re-solve the problem of Sect.6.2.3with V1 = 2 andV2 = 1.
E 6.15 Compute the optimal trajectories forμ1, μ2,andηfor the model in Sect.6.2.4.
E 6.16 Solve the model in Sect.6.2.4 with each of the following condi- tions:
(a) y(0) = 2.
(b) T = 10 and p(t) = 2t−2 for 3≤t≤10.
E 6.17 Verify that the solutions shown in Figs.6.10and6.11satisfy the maximum principle.
E 6.18 Re-solve the model of Sect.6.3.2 withy(0) = 1/2 and with the warehousing constraint y≤1/2 in place of (6.82).
E 6.19 Solve and interpret the following production planning problem with linear inventory holding costs:
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩ max
J =
T
0 −[hI+c 2P2]dt
subject to
I˙=P, I(0) = 0, I(T) =B; 0< B < hT2/2c, P ≥0 and I ≥0.
(6.94)
E 6.20 Re-solve Exercise6.19with the state equation ˙I(t) =P(t)−S(t), where I(0) =I0≥0 and I(T) is not fixed. Assume the demand S(t) to be continuous intand non-negative. Keep the state constraintI ≥0,but drop the production constraint P ≥0 for simplicity. For specificity, you may assumeS =−sinπt+C with the constantC≥1 and T = 4.(Note that negative production can and will occur when initial inventory I0 is too large. Specifically, how large is too large depends on the parameters of the problem.)
Exercises for Chapter 6 223 E 6.21 Re-solve Exercise6.19 with the state equation ˙I(t) =P(t)−S, where S >0 and h > 0 are constants, I(0) = I0 > cS2/2h, and I(T) is not fixed. Assume that T is sufficiently large. Also, graph the optimal P∗(t) and I∗(t), t∈[0, T].
Chapter 7
Applications to Marketing
Over the years, a number of applications of optimal control theory have been made to the field of marketing. Many of these applications deal with the problem of finding or characterizing the optimal advertising rate over time. Others deal with the problem of determining the optimal price and quality over time, in addition to or without advertising. The reader is referred to Sethi (1977a) and Feichtinger et al. (1994a) for comprehensive reviews on dynamic optimal control problems in advertising and related problems. In this chapter we discuss optimal advertising policies for two of the well-known models called the Nerlove-Arrow model and the Vidale-Wolfe model.
To describe the specific problems under consideration, let us assume that a firm has some way of knowing or estimating the dynamics of sales and advertising. Such knowledge is expressed in terms of a differential equation with either goodwill or the rate of sales as the state variable and the rate of advertising expenditures as the control variable. We assume that the firm wishes to maximize an objective function (the criterion function) which reflects its profit motives expressed in terms of sales and advertising rates. The optimal control problem is to find an advertising policy which maximizes the firm’s objective function.
The plan of this chapter is as follows. Section 7.1 will cover the Nerlove-Arrow model as well as a nonlinear extension of it. Section 7.2 deals with the Vidale-Wolfe advertising model and its detailed analy- sis using Green’s theorem in conjunction with the maximum principle.
The switching-point analysis for this problem is a good example of the reverse-time construction technique used earlier in Chaps.4 and 5. Ex-
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225
226 7. Applications to Marketing tensions of these models to multi-state problems are treated in Turner and Neuman (1976) and Srinivasan (1976).