### Chapter 6 APPLICATIONS TO PRODUCTION AND INVENTORY

```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 m1< 0 and m2 > 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 a1 and a2 as follows:
If we recall that m1 is negative and m2 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 b1=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 b2=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 a2 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
turnpike.
, the solution is always on the
The triple
represents a
non-stationary turnpike. If I(0)  Q(0), then b1  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 a1 and a2 are given in (6.15) and (6.16) in
terms of b1 and b2, which from (6.13), (6.14), and
(6.15) are
If I0=Q, then the optimal solution stays on the turnpike.
If I0 Q, then the optimal solution is given by
Clearly if I0  Q, this provides a nonnegative optimal
production throughout. In case I0 > 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 C0 and C1 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 I0=10
Figure 6.3: Solution of Example 6.1 with I0=50
Figure 6.4: Solution of Example 6.1 with I0=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
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
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
so that the TPBVP is
for all t from (6.46)
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 ShortSelling ( 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 ht =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 (6.70), and
above discussion, we obtain
from (6.57),
from
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
Horizons for the Wheat Trading Model with
Warehousing Constraint
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
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 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 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 6.11.
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).
```