1. Continuous Dynamic Programming

2. Continuous DP
Suppose we have a hiring problem where the employment level needed for each season is given by
cost of excess employment = $2,000 per season
cost to hire/fire = $200 x (# hired/fired)2

3. Continuous DP
Let sn = employment level at start of season n xn = employment level decision Ex: Suppose at start of 3 (winter) we had 250 employees & we decide to keep 220
250
220 s3 s4

4. Continuous DP
Let sn = employment level at start of season n xn = employment level decision sn xn n
n+1

5. Continuous DP
Let rn = min level required for season n (xn - sn) = # hired or fired (xn - rn) = excess over min required

6. Continuous DP
Cost at stage (season) n cn = 200(xn - sn)2 + 2,000(xn - rn) fn(sn, xn) = cn + fn+1*(xn)

7. Continuous DP
Feasible range for season n = # required in current season n to # required in previous season n-1

8. Continuous DP
Feasible range for season n = # required in current season n to # required in previous season n-1

9. Continuous DP
Season 4 At n=4 (spring), we must have 255 since this min requirement is the max min for all seasons. Then xn = s4 cn = 200(255 - s4)2 + 2,000( )

10. Continuous DP
Season 4 At n=4 (spring), we must have 255 since this min requirement is the max min for all seasons. Then xn = s4 cn = 200(255 - s4)2 + 2,000( )

11. Continuous DP
Stage 3 (winter) start at s3 = employment decision at stage < x3 - s3 < 255 f s x n 3 2 4
200
000
255 ( , ) * = - +

12. Continuous DP Stage 3 (winter) f s x 200 000 255 ( , ) = - +4
200
000
255 ( , ) * = - +
But, s4 = x3 f s x n 3 2
200
000
255 ( , ) = - +

13. Continuous DP Now, min f3(sn,xn) f s x 200 000 255 ( , ) = - + ¶ = - +
400 2
000
255 ( , )

14. Continuous DP x s 250 2 = + Now, min f3(sn,xn) f s x 200 000 255 ( , )- +
Now, min f3(sn,xn) = - + f s x n 3
400 2
000
255 ( , ) x s 3
250 2 = +

15. Continuous DP x s 250 2 = + f s x 50 250 1 000 150 260 ( , ) = - + 3 *

16. Continuous DP Stage 2 (autumn) s2 = x1 220 < s2 , 255 f s x 200 0003
200
000
240 ( , ) * = - + = - +
200 2
000
240 50
250
260 3 ( ) , x s + - 1
000
150 3 , ( ) s

17. Continuous DP Stage 2 (autumn) s2 = x1 220 < s2 , 255 f s x 200 0003
200
000
240 ( , ) * = - + = - +
200 2
000
240 50
250
260 3 ( ) , x s + - 1
000
150 3 , ( ) s
but s2 = x1

18. Continuous DP f s x 2
200
000
240 50
250
260 1
150 ( , ) = - +

19. Continuous DP ¶ = - + f x s 400 000 100 250 260 1 ( ) , f s x 200 000
240 50
250
260 1
150 ( , ) = - + = - + f x s 2
400
000
100
250
260 1 ( ) ,

20. Continuous DP x s 240 3 = + ¶ = - + f x s 400 000 100 250 260 1 ( ) ,
400
000
100
250
260 1 ( ) , x s 2
240 3 = +