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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.

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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


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
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
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 , 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

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
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


Case of seasonal demand

Example 6.1


so that


Solution. It is then easy to show from (6.34) that

From (6.19) and (6.20),

Example 6.2 Asanother example assume that

Solution. Making use of formulas (6.34) and (6.36) we

have the special particular integral

continuous wheat trading model
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 principle1
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
Complete Solution of a Special Case

For this special case, we assume



For this case r =0, we have for all t from (6.46)

so that the TPBVP is

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


the wheat trading model with no short selling y 0
The Wheat Trading Model with No Short-Selling ( y0 )



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 (6.57),

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.

decision horizons and forecast horizons
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
Horizons for the Wheat Trading Model

t =1 is a decision horizon as well as a weak forecast


Figure 6.8: Decision Horizon and Optimal Policy for the Wheat Trading Model

horizons for the wheat trading model with warehousing constraint
Horizons for the Wheat Trading Model withWarehousing 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

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.

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.

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).