Chapter 13 DETERMINISTIC DYNAMIC PROGRAMMING

1. Chapter 13 DETERMINISTIC DYNAMIC PROGRAMMING Math 305 2008 We will cover 9.1-9.4 plus some material not in the text

2. Dynamic Programming Technique for making a sequence of interrelated decisions Problem solving strategies, e.g. find a route from here to LA • forward: enumerate all possibilities • backward: figure which ways one can get to the desired end General type of problem: consecutive stages • at each stage you are in one of a number of possible states • each state has one or more possible policies from which to choose • the policy you choose determines your state at the next stage Method • start with a solution for a small part of the problem • expand  A useful approach when you don’t want to check all possiblities

3. Prototype problem (stage coach) A traveller is going from east to west coast on a series of stage coaches that travel from one state to another and wants the safest route S/he selects a life insurance policy for each stage, reasoning that cheapest = safest

4. Approach 4 stages States at each stage: 1: 1 2: 2,3,4 3: 5, 6, 7 4: 8,9 First stage: travel from 1 to 2, 3, or 4 Second stage: travel from state at that stage to next stage Decision variables: xn (n=1,..,4) = destination on nth stage

5. Approach fn(s,xn) = total cost of best policy for remaining stages given you are in state s at stage n and select policy xn (n = stage, s = state, xn = decision) fn(s) = cost of best policy given you are in state s at stage n = min fn(s, xn) over possible choices for xn fn(s, xn) = c s xn + fn+1(xn) (c s xn = cost of policy from s to xn) goal: find f1(1) f1(1) = min of f1 (1,2) = 2 + f2(2) f1 (1,2) = 5 + f2(3) f1 (1,4) = 1 + f2(4)

6. Start at Stage 4 f2*(2) = ? ... Better: go backward Find the best policy from stage 4 on, stage 3 on,... Stage 4: states 8 and 9 → we want f4*(8) and f4*(9) f4(8,10) = 1 = f4*(8) f4(9,10) = 4 = f4*(9)

7. Stage 3 States 5, 6, 7 → we want f3*(5), f3*(6). f3*(7)  f3(5,8) = 7 + f4*(8) = 8 f3(5,9) = 5 + f4*(9) = 9 f3(6,8) = 3 + f4*(8) = 4 f3(6,9) = 4 + f4*(9) = 8 f3(7,8) = 7 + f4*(9) = 8 f3(7,9) = 1 + f4*(9) = 5 So far this isn't any better than exhustive search

8. Stage 2 States 2, 3, 4 f2(2,5) = 10 + f3*(5) = 18 f2(2,6) = 12 + f3*(6) = 16 f2(3,5) = 5 + f3*(5) = 13 f2(3,6) = 10 + f3*(6) = 14 f3(3,7) = 7 + f3*(7) = 12 f2(4,6) = 15+f3*(6) = 19

9. Stage 1 f1(1,2) = 2 + f2*(2) = 18 f1(1,3) = 5 + f2*(3) = 17 f1(1,4) = 1 + f2*(4) = 19

10. Shortest route: 1 -> 3 3->7 7 -> 9 9->10

11. Computational Efficiency This method? 16 additions, 9 comparisons How else could we solve this? • list all paths (14) and total # additions (3 on each) => 42 • shortest route ? Compare n+1 stages with n choices at each stage except last Dynamic programming: n nodes with n additions each =n2 Exhaustive search: nn paths with n additions = nn+1 E.g. n=10: 100 versus 1011

12. 13.3 Inventory Theory What is inventory? • something that is produced • has a demand • needs to be stored until used • cookies at Dories, beer at the Fargo, flash drives at the bookstore What is an inventory policy? • when to order/produce more • how much at a time What costs are associated with inventory? • cost per unit (variable) • setup or ordering • holding • shortage

13. Inventory Theory What are we trying to optimize? Assumptions • time is broken into periods • production occurs at the beginning of the period • each period has an associated demand which is met from items held over from the last period and/or produced in the current period

14. 13.3 Example Demand for a product Costs c(x)=cost of producing x units=3 + 1x Other restrictions • at most 5 units can be produced each month • at most 4 units can be carried over to the next month • 0 units on hand at the beginning of month 1 What are the stages, states, and decisions? • stage: beginning of a month • state: entering inventory • decision: fn(s) = amt produced at beginning of n, given s units on hand

15. Stage 4 f4(i) = cost of entering period 4 with i units = cost of producing 4 - i units = c(4-s) f4(0) = set up + cost of 4 units = 3 + 4 f4(1) = set up + cost of 3 units = 3 + 3 Demand is 1 3 2 4

16. Stage 3 f3 (i) = min {c(x) + 0.5(x + i - 2) + f4(i+x -2) }       x=0..4ordering holding stage 4 on Demand is 1 3 2 4

17. Alternate Notation for Stage 3

18. Stage 2 f2 (i) = min {c(x) + 0.5(x + i - 3) + f2(i+x -3) }   for x = 0, 1, ...,5 x=0,1,2,3,4 Demand is 1 3 2 4

19. Alternate Notation for Stage 2

20. Stage 1 f1 (i) = min {c(x) + 0.5(x + i - 1) + f1(i+x - 1) }      for x = 0, 1, ...,5 x=0,1,2,3,4 Demand is 1 3 2 4 (text also finds f1 (i) for i = 1,2,3,4)

21. There must be an easier way(not in text) Only produce when entering inventory=0 • don't carry inventory to meet part of the demand if you will have to produce in the period Demand is 1 3 2 4

22. Network Representation 2,0 (i,j): i = period, j = beginning inventory 2,1 1, 0 5,0 2,2 2,3 2,4

23. 13.4 Resource Allocation I have 5 blocks of time to study and want to maximize the sum of my grades gi(xi) = grade in subject i given I study xi blocks. 3 decisions: xn = # blocks to spend on subject n stage = subject state = time available to allocate to remaining stages

24. Resource Allocation gi(xi) = grade in subject i given xi blocks Objective: max  gi(xi) subject to xi= 5  fn(s) = max effectiveness of s hours in stages n through 3 fn(s, xn) = gn(xn) + fn+1(s - xn) = grade from xn hours to subject n plus best effect of remaining hours in later stages fn(s) = max { gn(xn) + fn+1(s - xn)} xn=o..s

25. Resource Allocation f3(s) f2(s) f1(s) Solution: English 2, Econ 2, Physics 1

26. Continuous Functions Suppose we can allocate fractional units of time and grade is a continuous function of time spent English: g1(x) = 65 - x2 + 11x max at (5.5, 95) Econ: g2(x) = 80 - 2x2 + 13 x max at (3.25, 101.125) Physics: g3(x) = 55 + 7x  f3(s) = 55 + 7s x3 = s f2(s, x2) = g2(x2) + f3(s - x2) = 80 -2 x22 + 13x2 + 55 + 7(s - x2) = 135 - 2 x22 + 6x2 + 7s  df2/dx2 = -4x2 + 6 = 0 when x2 = 3/2  d2f2/dx22 = -4 -> max at x2 = 3/2 (if s ≥ 3/2) otherwise max at x2 = s

27. Continuous Functions Case 1, s < 3/2   x2 = s,   f2(s) = 80 - 2s2 + 13s + 55 + 7(s - s) = 135 + 13s - 2s2  Case 2, 3/2  s  f2(s) = 80 + 2(3/2)2 + 13(3/2) + 55 + 7(s - 3/2) = 150 + 7s Implications • if available time < 3/2, put it all into econ • if ≥ 3/2, put 3/2 into econ and surplus into physics (compare slopes of the two graphs before and after 3/2)

28. Continuous Functions f1(5, x1) = g1(x1) + f2(5- x1) case1, 5 - x1 < 3/2 (x1 > 3.5) f1(5, x1) = 65 - x1 2 + 11x + 135 + 13(5 - x1) - 2(5 - x1)2 df1/d x1 = -2 x1 + 11 - 13 - 4(5- x1) = 2 x1 - 22 < 0 for 3.5 < x1 5 -> max at x1 = 3.5 if 3.5 < x1 5. f1(5. 3.5) = 241.25 case 2, 5 - x1 3/2 (0 x1 3.5) f1(x1, 5) = 65 - x1 2 + 11 x1 + 150 + 7(5- x1) = 250 + 4 x1 - x1 2  df1/d x1 = 4 - 2 x1 = 0 at x = 2   d2f1/dx12 = -2 -> max at x = 2  f1(5) = f1(5,2) = 254

29. Continuous Functions Decision: x1 = 2 Thus 5 - x1 = 3 left for remaining stages 3 > 3/2 -> x2 = 3/2 -> x3 = 3 - 3/2 = 3/2 Solution: x1 = 2 x2 = 3/2 x3 = 3/2, sum of grades = 254 

30. Probabilistic Models The state at the next stage is not completely determined by decision at current stage, rather determines a probability distribution for the next stage.  Objective: maximize the expected value. Example: job interview • a job candidate has up to three interviews • at each, she will be offered a job which is terrific, good or fair • she must decide then whether to accept the job or interview again • a terrific job is worth 3 points, good: 2, fair: 1 Stages: interviews State: job status at stage n (T, G, or F) Decision: interview or accept

31. Probabilistic Models fn(s) = max expected value if in state s at stage n fn(s, xn) = max expected value if in state s at stage n and make decision xn (xn = i or a) T T T 0 G G G F F F

32. Stage 3 f3(T, i) = 3(.2) + 2(.5) + 1(.3) = 1.9 f3(T,a) = 3 -> f3(T) = 3, x3 = a f3(G, i) = 3(.2) + 2(.5) + 1(.3) = 1.9 f3(G,a) = 2 -> f3(G) = 2, x3 = a f3(F, i) = 3(.2) + 2(.5) + 1(.3) = 1.9 f3(F,a) = 1 -> f3(F) = 1.9, x3 = i

33. Stage 2 f2(T, i) = p(T)f3(T) + p(G)f3(G) + p(F)f3(F) = .2(3) + .5(2) + .3(1.9) = 2.17 f2(T,a) = 3 -> f2(T) = 3, x2 = a f1(i) = .2f2(T) + .5f2(G) + .3f2(F) = .2(3) + .5(2.17) + .3(2.17) = 2.336 Strategy at stage I: interview; II: interview in G or F.; III: interview in F Stage 1

34. Knapsack Problem (back to 13.4) A thief breaks into a house. Around the thief are various objects: a diamond ring, a silver candelabra, a Bose Wave Radio, a large portrait of Elvis Presley painted on a black velvet background (a "velvet-elvis"), and a large tiffany crystal vase. The thief has a knapsack that can only hold a certain capacity (8). Each of the items has a value and a size, and cannot hold all of the items in the knapsack. Which items should the thief take? There are three thieves: greedy, foolish and slow, and wise (ref for this example)

35. Knapsack Problem The greedy thief breaks into the window, and sees the items. He makes a mental list of the items, and grabs the most expensive item first. The ring goes in first, leaving a capacity of 7, and a value of 15. Next, he grabs the candelabra, leaving a knapsack of size 2 and a value of 25. No other items will fit in his knapsack, so he leaves. The foolish and slow thief climbs in the window, and sees the items. This thief was a programmer, downsized as part of the "dot-bomb" blowout. Possessing a solid background in boolean logic, he figures that he can simply compute all combinations of the objects and choose the best. So, he starts going through the binary combinations of objects - all 25 of them. While he is still drawing the truth table, the police show up, and arrest him. Although his solution would certainly have given him the best answer, it just took long to compute.

36. Knapsack Problem, Wise Thief The wise thief appears, and observes the items. He notes that an empty knapsack has a value of 0. He notes that a knapsack can either contain each item, or not. Further, his decision to include an item will be based on a quick calculation - either the knapsack with some combination of the previous items will be worth more, or else the knapsack of a size that will fit the current item was worth more. So, he does this quick computation, and figures out that the best knapsack he can take is made up of items 1,3, and 4, for a total value of 29

37. Generalized Resource Problem (p. 767) w units of a resource available T activities in which the resources can be allocated xt : the level at which activity t is implemented gt(xt ): # of units of the resource used by activity t rt(xt ): the resulting benefit States: each activity Stages: how much of the resource is available for remaining stages Decision: how much to use at this stage Formulation maximize Σrt(xt ) s.t. Σgt(xt ) ≤ w t = 1,...T t = 1,...T ft(d) = max benefit if d units are allocated to activities t through T ft(d) = max {rt(xt ) + ft+1(d - xt ) fT+1(d) = 0 xt

38. Wise Thief, Stages 4 and 3 f3(d) = max {r3(x3 ) + f4(d - x3 ) f3(3) = max {r3(0) + f4(3) = 0 + 0 r3(3) + f4(0) = 9 + 0} f3(4) = max {r3(0) + f4(4) = 4 r3(3) + f4(1) = 9} f3(7) = max {r3(0) + f4(7) = 4 r3(3) + f4(4) = 9 + 5 =14

39. Stage 2 f2(d) = max {r2(x2) + f43(d - x2) f2(3) = max {r2(0) + f3(3) = 4} f3(4) = max {r2(0) + f3(4) = 9} f3(5) = max {r2(0) + f3(5) = 3 r2(5) + f3(0) = 10} f3(7) = max {r2(0) + f3(7) = 14 r2(5) + f3(2) = 10} f3(8) = max {r2(0) + f3(8) = 14 r2(5) + f3(3) = 10 + 9 = 19}

40. Stage 1 f1(8) = max {r1(0) + f2(8) = 0 + 19} r1(1) + f2(7) = 15 + 14}

41. How is this an LP? xi = # item i max 15x1 + 10x2 + 9x3 + 5x4 s.t. x1 + 5x2 + 3x3 + 4x4 ≤ 8

42. Turnpike Theorem Let cj = benefit from item j, wj = weight of item j Order items by benefit per unit weight c1/w1 ≥ c2 /w2 ≥ c3 /w3 ... If there is a unique "best" item #1, e.g. c1/w1 > c2 /w2 when the max weight w ≥ w* = (c1w1)/(c1 - w1 ( c2 /w2 ) the optimal solution contains at least one of item 1 Thief problem: 15/1 > 9/3 > 10/5 > 5/4 w* = 15*1/(15 - 1*9/3) = 15/12 = 1.25 < 8 Why is this any use? • start with one ring and reduce computation Why turnpike • for a long trip you might go a little out of your way to maximize time on a turnpike.

43. Proof of Turnpike Theorem Without using any type 1 items we cannot do better than include w/w2 type 2 This would earn c2w/w2 Suppose we fill the knapsack with as many type 1 items as possible We can fit in at least (w/w1 ‑ 1) type 1 items These items would earn a benefit of c1(w/w1 ‑ 1) Thus if c1(w/w1 ‑ 1)c2w/w2 (1) there must be an optimal solution using a type 1 item (1) holds if w(c1/w1 ‑ c2/w2)c1 c1w1 or w ‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = w* c1 ‑ c2w1/w2 Thus if knapsack can hold at least w* pounds, there will be an optimal solution using at least one type 1 item