Question

10. Mills Manufacturing Company has a production and inventory control problem (such as that described in Section 21.4) for an armature the company manufactures as a component for a generator. The available data for the next three-month planning period are provided in the following table SELFtest Capacity Cost per Unit Month Demand Production Warehouse Production Holding 20 30 40 $2.00 $0.30 2 30 20 30 1.50 0.30 3 30 30 20 2.00 0.20 Use dynamic programming to find the optimal production quantities and inventory levels in each period for the Mills Manufacturing Company. Assume an inventory of 10 units on hand at the beginning of month 1 and production runs are completed in multiples of 10 units (i.e., 10, 20, or 30 units)

Answer 1

鼋3 CN rt) Stage 1 (March) x1 x3-10Stage 3 laniary) Stage 2 (February) r3(x3,d3) r2(x2,d2) r1(x1,d1)

Transformation functions take the form of ending inventory 5 beginning inventory 1 production 2 demand. Thus, we have

x3 = 10

x2 = x3 + d3- D3 = x3 + d3 - 20

x1 = x2 + d2 -D2 = x2 + d2 - 30

x0 = x1 + d1 - D1 = x1 + d1 - 30

The return functions for each stage represent the sum of production and holding costs for
the month. For example, in stage 1 (March), r1(x1, d1) = 2d1 + 0.2(x1 + d1 - 30) represents
the total production and holding costs for the period. The production costs are $2 per unit,
and the holding costs are $0.2 per unit of ending inventory. The other return functions are

r2(x2,d2) = 2d2 + 0.3(x2+d2-30) stage2,february

r3(x3,d3) = 1.5d3 + 0.3(x3+d3-20) stage 3,january

This problem is particularly interesting because three constraints must be satisfied
at each stage as we perform the optimization procedure. The first constraint is that the
ending
inventory must be less than or equal to the warehouse capacity. Mathematically,
we have

xn + dn + Dn $\leq$  Wn or xn + dn $\leq$ Wn + Dn

The second constraint is that the production level in each period may not exceed the production
capacity. Mathematically, we have

dn $\leq$ Pn

In order to satisfy demand, the third constraint is that the beginning inventory plus
production must be greater than or equal to demand. Mathematically, this constraint can
be written as

xn + dn $\geq$ Dn

Let us now begin the stagewise solution procedure. At each stage, we want to minimize
rn(xn, dn) + fn-1(xn-1) subject to the constraints given by above equations

Min r1(x1,d1) = 2d1 + 0.3(x1+d1-30)

so that

x1 + d1  $\leq$ 70 ware house constraint

d1 $\leq$ 30 production constraint

x1 + d1 $\leq$ 20 satisfy demand constraint

Combining terms in the objective function, we can rewrite the problem:

Min r1(x1,d1) = 2.3d1 + 0.3x1 - 9

so that

x1 + d1  $\leq$ 70

d1 $\leq$ 30

x1 + d1 $\leq$ 20

Following the tabular approach we adopted above, we will consider all possible
inputs to stage 1 (x1) and make the corresponding minimum-cost decision. Because
we are attempting to minimize cost, we will want the decision variable d1 to be as small
as possible and still satisfy the demand constraint.

The optimal production schedule is as follows:

Month	Beginning inventory	Production	Ending Inventory
1	10	20	10
2	10	20	0
3	0	30	0