Engineering Optimization: Theory and Practice, Fourth Edition

568 Dynamic Programming

wherex 1 andx 2 indicate the number of drilling machines manufactured in thefirst

month and the second month, respectively. To solve this problem as a final value

problem, we start from the second month and go backward. IfI 2 is the inventory at

the beginning of the second month, the optimum number of drilling machines to be

manufactured in the second month is given by

x∗ 2 = 201 −I 2 (E 1 )

and the cost incurred in the second month by

R 2 (x∗ 2 , I 2 )= 8 I 2 + 05 x 2 ∗+ 0. 2 x∗ 22

By using Eq. (E 1 ), R 2 can be expressed as

R 2 (I 2 )= 8 I 2 0 + 5 ( 120 −I 2 ) + 0. 2 ( 120 −I 2 )^2 = 0. 2 I 22 − 09 I 2 + 8808 (E 2 )

Since the inventory at the beginning of the first month is zero, the cost involved in the

first month is given by

R 1 (x 1 ) = 50 x 1 + 0. 2 x 12

Thus the total cost involved is given by

f 2 (I 2 , x 1 ) =( 50 x 1 + 0. 2 x^21 )+( 0. 2 I 22 − 09 I 2 + 8808 ) (E 3 )

But the inventory at the beginning of the second month is related tox 1 as

I 2 =x 1 − 08 (E 4 )

Equations(E 3 ) nd (Ea 4 ) ead tol

f=f 2 (I 2 ) =( 50 x 1 + 0. 2 x 12 ) + 0. 2 (x 1 0 − 8 )^2 − 09 (x 1 − 08 )+ 8880

= 0. 4 x^21 − 27 x 1 + 71 , 360 (E 5 )

Sincefis a function ofx 1 only, the optimum value ofx 1 can be obtained as

df

dx 1

= 0. 8 x 1 − 27 =0 or x∗ 1 = 09

Asd^2 f (x 1 ∗)/dx 12 = 0. 8 > 0 , the value ofx∗ 1 corresponds to the minimum off.Thus

the optimum solution is given by

fmin= f(x 1 ∗) =$ 14 , 120

x∗ 1 = 0 and 9 x∗ 2 = 101