Chapter 6 APPLICATIONS TO PRODUCTION AND INVENTORY Hwang, Fan, Erickson (1967) Hendricks et al (1971) Bensoussan et al (1974) Thompson-Sethi (1980) Stochastic Extension: Sethi-Thompson (1980) Kleindorfor et al (1975) Kleindorfor-Lieber (1975) Singhal (1981)

PRODUCTION-INVENTORY MODELS Deterministic model: The HMMS model I(t) = the inventory level at time t ( state variable ) P(t) = the production rate at time t (control variable) S(t) = the sales rate at time t (exogenous variable); assumed to be bounded and differentiable for t  0, T = the length of the planning period, = the inventory goal level, = the initial inventory level, = the production goal level, h = the inventory holding cost efficient; h >0, c = the production cost coefficient; c  0,  = the constant nonnegative discount rate;   0,

The objective function is: The inventory balance equation is Interpretation: We want to keep the inventory as close as possible to its goal level, and also keep the production rate P as close as possible to its goal level. The quadratic terms and impose “penalties” for having I or P not being close to its corresponding goal level, respectively.

Solution by the Maximum Principle The current-value Hamiltonian function: To apply the Pontryagin maximum principle, we get; From this we obtain the decision rule; The optimal control is; With the assumptions of a sufficiently large and a sufficiently small, we have and we can substitute (6.4) into (6.1) to obtain

The equation for the adjoint variable is We then use (6.7) to eliminate and (6.6) to eliminate from the resulting equation as follows: We rewrite this as where the constant  is given by

We can now solve (6.8) by using the standard method described in Appendix A. The auxiliary equation for (6.8) is which has two real roots note that m 1 0. We can therefore write the general solution to (6.8) as where Q(t) is a particular integral of (6.8). To get the other boundary condition we differentiate (6.11), substitute the result into (6.6), and solve for.

We obtain To do the latter we define two more constants: We now impose the boundary conditions in (6.11) and (6.12) and solve for a 1 and a 2 as follows:

If we recall that m 1 is negative and m 2 is positive, then when T is sufficiently large so that and are negligible, we can write Starting Correction is important only when t is small, Turnpike Expression is significant for all values of t, and the Ending Correction is important only when t is close to T.

Note that if b 1 =0, which means,then there is no starting correction. In other words, is a starting inventory that causes the solution to be on the turnpike initially. In the same way, if b 2 =0, then the ending correction vanishes in each of these formulas, and the solution stays on the turnpike until the end.

The Infinite Horizon Solution Note that as T , the ending correction disappears because a 2 defined in (6.16) becomes 0. We now have If S is bounded, then is bounded, and therefore, is bounded. Then for  > 0, By the sufficiency of the maximum principle conditions (Section 2.4), it can be verified that the limiting solution is optimal. If, the solution is always on the turnpike.

The triple represents a non-stationary turnpike. If I(0)  Q(0), then b 1  0 and the expressions (6.24) imply that the path of inventory and production only approach but never attain the turnpike. Note that the solution does not satisfy the sufficiency transversality condition when  = 0. In this case, all solutions give an infinite value (which is the worst value since we are minimizing) for the cost objective function. Our limiting solution also gives an infinite value for the objective function. It makes sense to define the limiting solution for  = 0 to also be the optimal infinite horizon solution for the undiscounted case.

A Complete Analysis of the Constant Positive S Case With Infinite Horizon Substituting in (6.8) and recognizing that S is a constant so that, we obtain Note that a 1 and a 2 are given in (6.15) and (6.16) in terms of b 1 and b 2, which from (6.13), (6.14), and (6.15) are

If I 0 =Q, then the optimal solution stays on the turnpike. If I 0  Q, then the optimal solution is given by Clearly if I 0  Q, this provides a nonnegative optimal production throughout. In case I 0 > Q, note first that P(t) increase with t and Furthermore, if,we have a negative value for P(0) which is infeasible. By (6.5), P*(0)=0. We can depict this situation in Figure 6.1. The time shown in figure is the time at which

Figure 6.1 Optimal Production and Inventory Levels

It should be obvious that For t <, we have We can substitute for I in equation (6.31) and solve for. Note that we can easily obtain as For t >, the solution is given by (6.19), (6.20) and (6.21) with t replaced by and.

Special Case of Time Varying Demands Let  = 0 and T < . Let a polynomial of degree 2p or 2p-1 so that S (2p+1)= 0, where S (k) denotes the kth time derivative of S with respect to t and where at least one of C 0 and C 1 is not zero. Case of seasonal demand

Example 6.1 Assume so that assume Solution. It is then easy to show from (6.34) that From (6.19) and (6.20),

Figure 6.2: Solution of Example 6.1 with I 0 =10

Figure 6.3: Solution of Example 6.1 with I 0 =50

Figure 6.4: Solution of Example 6.1 with I 0 =30

Example 6.2 As another example assume that Solution. Making use of formulas (6.34) and (6.36) we have the special particular integral

Continuous Wheat Trading Model T = the horizon time, x(t) = the cash balance in dollars at time t, y(t) = the wheat balance in bushels at time t, v(t) = the rate of purchase of wheat in bushels per unit time; a negative purchase means a sale, p(t) = the price of wheat in dollars per bushel at time t, r = the constant positive interest rate earned on the cash balance, h(y) = the cost of holding y bushels per unit time.

The state equations are: The control constraints are The objective function is

Solution by the Maximum Principle Introduce the adjoint variable 1 and 2, and define the Hamiltonian function The adjoint equations are: It is easy to solve (6.44) and (6.45) as From (6.43) the optimal control is

Complete Solution of a Special Case For this special case, we assume and Solution: For this case r =0, we have for all t from (6.46) so that the TPBVP is

Figure 6.5: The Price Trajectory (6.49)

The optimal policy (6.48) reduce to Since p(t) is increasing, short-selling is never optimal. Since the shortage cost is ½ unit per unit time, it never pays to store wheat more than 2 time units. Because y(0)=0, we have v*(t)=0 for 0  t  1. Clearly buying will continue at this rate until t =3, and no longer. Clearly we should start selling at t =3 + at the maximum rate of 1, and continue until a last sale time t**, given by Thus, v*(t)=0 in the interval [t**,6], which is also singular control. With this policy, y(t)>0 for all t  (t*,t**).

From (6.52), in the interval (t*,t**). In order to have a singular control in the interval [t**,6], we must have in that interval. Also, in order to have singular control in [0, t*], we must have in that interval. We can now conclude that and therefore t*= 2 and t**= 4. Thus from (6.52) and (6.53),

Figure 6.6: Adjoint Variable, Optimal Policy and Inventory in the Wheat Trading Model

The Wheat Trading Model with No Short- Selling ( y  0 ) Assume: and The statement of the problem is:

The Hamiltonian is The optimal control is Whenever y = 0, we must impose in order to insure that no short-selling occurs.Therefore, The Lagrangian The complementary slackness conditions:

Furthermore, the optimal trajectory must satisfy From this we see that 2 is always increasing except possibly at a jump time. Let be a time that there is no jump in the interval. Note from (6.66) that 1 (t)=1, t  (,3],then;

Using (6.66) with 1 = 1 in the interval (1,1.8] and v*=0 so that, we have Since h t =0, the jump condition in (4.29) for the Hamiltonian at reduces to We rewrite the condition as Since 1 (t)=1 for all t,

Substituting the values of from (6.57), from (6.70), and from above discussion, we obtain Furthermore, using (6.72) and (6.67), and the optimal control condition (6.60) holds, justifying our supposition that v*=-1 in this interval.

Figure 6.7: Adjoint Trajectory and Optimal Policy for the Wheat Trading Model

Decision Horizons and Forecast Horizons In some dynamic problems it is possible to show that the optimal decisions during an initial positive time interval are either partially or wholly independent of the data from some future time onwards. In such case, a forecast of the future data needs to be made only as far as that time to make optimal decisions in the initial Time interval. The initial time interval is called the decision horizon and the time up to which data is required to make the optimal decisions during the decision horizon is called the forecast horizon.

Horizons for the Wheat Trading Model t =1 is a decision horizon as well as a weak forecast horizon. Figure 6.8: Decision Horizon and Optimal Policy for the Wheat Trading Model

Change the terminal time to T=4 and define the price trajectory to be The optimal control is defined as: Defining a Lagrange multiplier  for the derivative of (6.74), i.e., for we form the Lagrangian Horizons for the Wheat Trading Model with Warehousing Constraint

where satisfy (6.63)-(6.65) and  satisfies Furthermore, the optimal trajectory must satisfy As before, 1 = 1 and 2 satisfies Let be the time of the last jump of the adjoint function 2 (t) before the terminal time T =4. Then, -2( -1)+7+1/2 = +1

Figure 6.9: Optimal Policy and Horizon for the Wheat Trading Model with Warehouse Constraint

To complete the maximum principle we must derive expressions for the Lagrange multipliers in the four intervals shown in Figure 6.9.

t =1 is a decision horizon and weak forecast horizon. = 17/6 is a strong forecast horizon. Example 6.3 Assume the price trajectory to be which is sketched in Figure 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 Figure 6.9.

Figure 6.10: Optimal Policy and Horizons for Example 6.3

Example 6.4 Assume the price trajectory to be which is sketched in Figure Note that if we had solved the problem with T = 1, then ; and if we had solved the problem with T = 17/6, then and. 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 time 17/6 can be found to be a forecast horizon which follows the decision horizon at time 1.

Figure 6.11: Optimal Policy and Horizons for Example 6.4

There are other ways to find strong forecast horizons; see Pekelman (1974, 1975, 1979), Lundin and Morton (1975), Morton (1978), Kleindorfer and Lieber (1979), Sethi and Chand (1979,1981), Chand and Sethi (1982, 1983,1990), Bensoussan, Crouhy, and Proth (1983), Teng, Thompson, and Sethi (1984), Bean and Smith (1984), Bes and Sethi (1988), Bhaskaran and Sethi (1987, 1988), Chand, Sethi, and Proth (1990), Bylka and Sethi (1992), Bylka, Sethi, and Sorger (1992), and Chand, Sethi, and Sorger (1992).