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Chapter 10 Dynamic Programming

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  1. Chapter 10Dynamic Programming

  2. Agenda for This Week • Dynamic Programming • Definition • Recursive Nature of Computations in DP • Forward and Backward Recursion • Basic Elements of DP • Applications of DP

  3. Dynamic Programming • An approach to problem solving that decomposes a large problem into a number of smaller problems for finding optimal solution • Used for problems that are in multistage in nature • No standard mathematical formulation

  4. Stagecoach Problem • Used to illustrate the features of DP • Concerns a mythical fortune seeker to go to west to join the gold rush in California during mid-19th century • A prudent man who was quite concerned about his safety • Could not go through some territories • Starting and final destination points are fixed • Had considerable choice as to which states to travel

  5. Cost Estimates 10 2 5 5 6 9 9 8 6 4 4 10 6 3 8 1 3 6 4 4 8 9 5 7 6 5 4 7 7

  6. Solving Problem • Select routes that minimizes total cost • One possible approach is to use trail and error method • DP starts with a small portion of the original problem and finds the optimal solution • Optimizes the nest stage using the solution from preceding one • Continues until all stages are solved • First decompose it into stages as delineated by the vertical dashed lines 10 2 5 5 6 9 9 8 6 4 4 10 6 3 8 1 3 6 4 4 8 9 5 7 6 5 4 7 7

  7. Recursive Nature of Computations in DP • Start with the smaller problem where he has completed his journey and has only one to go • Computations are done recursively • Solution of one subproblem is used as an input for next subproblem 10 2 5 5 6 9 9 8 6 4 4 10 6 3 8 1 3 6 4 4 8 9 5 7 6 5 4 7 7

  8. DP Concept • Stages are formed by decomposing the original problem into a number of subproblems • Shortest path context • Input— any node or state • Decision Problem— which arc? • Decision criteria—shortest distance • Output– any node or state Decision Problem Decision Criteria Input Output

  9. DP Notation • Xn=input to stage n (output from stage n+1) • dn=decision variable at stage n (state variable) • Stage transformation function • Return function—return of a stage dN dn d1 XN Xn-1 X1 X0 XN-1 Xn Stage N Stage n Stage 1 rN(xN,dN) rn(xn,dn) r1(x1,d1)

  10. Stage one • First stage begins with nodes that are exactly one arc away from final destination (CA) • Nodes 8 and 9 are the input nodes or states for stage 1 • Nodes 10 is the output nodes for stage 1 10 2 5 5 6 9 9 8 6 4 4 10 6 3 8 1 3 6 4 4 8 9 5 7 6 5 4 7 7

  11. Decomposing the Problem-Stage two • Second stage begins with all nodes that are exactly two arcs away from destination and ends with all nodes that are exactly one arc away • Nodes 5, 6, and 7 are the input nodes for stage 2 • Nodes 8 and 9 are the output nodes for stage 2 10 2 5 5 6 9 9 8 6 4 4 10 6 3 8 1 3 6 4 8 4 9 5 7 6 5 4 7 7

  12. Decomposing the Problem-Stage three • Third stage begins with all nodes that are exactly three arcs away from destination and ends with all nodes that are exactly two arcs away • Nodes 2, 3, and 4 are the input nodes for stage 3 • Nodes 5, 6 and 7 are the output nodes for stage 3 10 2 5 5 6 9 9 8 6 4 4 10 6 3 8 1 3 6 4 8 4 9 5 7 6 5 4 7 7

  13. Decomposing the Problem-Stage Four • Forth stage begins with all nodes that are exactly four arcs away from destination and ends with all nodes that are exactly three arcs away • Nodes 1 the input node for stage 4 • Nodes 2, 3 and 4 are the output nodes for stage 3 10 2 5 5 4 6 9 9 8 6 4 10 6 3 8 1 3 6 4 8 4 5 9 7 6 5 4 7 7 Stage 1 Stage 4

  14. Solving the Problem-Stage one Input Arc Output Shortest to 10 6 8 8 9 8-10 9-10 10 10 6 4 6 10 4 9 4

  15. Solving by DP Notation-Stage 1 dn Xn Xn-1 • X1=input to stage 1 (output from stage 2) • d1=decision variable at stage 1 (state variable) • Stage transformation function • Return function—return of a stage Stage n rn(xn,dn) 6 d1 x1 8 f1(x1)= r1(x1,d1) d1* 6 10 10 8 9 6 4 10 10 4 9 4

  16. Solving the Problem- Stage two 11 5 5 Input Arc Output Shortest 6 8 5 6 7 5-8 6-8 7-9 8 8 9 11 10 9 9 6 10 4 10 6 8 4 Min [(5+6),(9+4)]=11 Min [(4+6),(8+4)]=10 Min [(7+6),(5+4)]= 9 9 9 7 5 4 7

  17. Solving by DP Notation-Stage 2 r2(x2,d2)+ f1(x1) d2 x2 8 9 f2(x2) d2* 11 11 10 9 8 8 9 5 6 7 11 10 13 13 12 9 5 5 6 8 9 6 10 4 10 6 8 4 9 9 7 5 4 7

  18. Solving the Problem- Stage Three Input Arc Output Shortest 2 3 4 2-6 3-7 4-6 6 7 6 16 15 14 11 2 5 10 5 6 16 6 8 Min [(10+11),(6+10)]=16 Min [(9+11),(8+10), (6+9)]=15 Min [(11+6),(10+4),(7+9)]= 14 9 6 10 15 9 4 10 6 3 8 8 4 6 9 9 6 7 14 5 4 4 7 4 7

  19. Solving by DP Notation-Stage 3 r3(x3,d3)+ f2(x2) d3 x3 f3(x3) d3* 5 6 7 2 3 4 21 20 17 16 18 14 - 15 16 16 15 14 6 7 6 11 2 5 10 5 6 16 6 8 9 6 10 15 9 4 10 6 3 8 8 4 6 9 9 6 7 14 5 4 4 7 4 7

  20. Solving the Problem- Stage Four Input Arc Output Shortest 1 1-3 18 3 Min [(4+16),(3+15), (5+14)]=18 16 11 2 5 10 5 6 6 8 9 4 6 15 10 18 9 4 10 6 1 3 8 3 8 4 6 9 5 9 14 6 7 5 4 4 7 4 7

  21. Solving by DP Notation-Stage 4 d4 x4 r4(x4,d4)+ f3(x3) f4(x4) d3* 2 3 4 1 20 18 19 18 3 16 11 2 5 10 5 6 6 8 9 4 6 15 10 18 9 4 10 6 1 3 8 3 8 4 6 9 5 9 14 6 7 5 4 4 7 4 7

  22. Best Decision at Each stage Stage 4 3 2 1 Arc 1-3 3-7 7-9 9-10 10 2 5 5 6 9 9 8 6 4 4 10 6 3 8 1 3 6 4 4 8 9 5 7 6 5 4 7 7

  23. Dynamic EOQ Models • Dynamic EOQ models are different, because • Inventory level is reviewed periodically over a number finite periods • Demand per period, though deterministic, is dynamic (MRP situation) • No shortage is allowed and setup cost is incurred each time a production lot is started

  24. A Production and Inventory Control Problem • Making decision on a production quantity for each of periods so that the demand can be met at a minimum cost • N=no. of periods (stages) • Dn=demand during stage n • Xn= amount of inventory on hand at beginning of stage n • dn= production quantity for stage n • Pn=production capacity in stage n • Wn=storage capacity at the end of stage • Cn=production cost per unit in stage • Hn=holding cost per unit of ending inventory for stage n

  25. Example Capacity Cost/unit Month Demand Production Storage Production Holding 1 2 3 2 3 3 3 2 3 2 3 2 175 150 200 30 30 40 Beginning inventory for month 1is 1 units

  26. DP Formulation Note: Ending Inventory= Beginning Inventory +Production -Demand P2=2 P1=3 P3=3 D2=3 D1=3 W2=3 D3=2 W1=2 W3=2 d3=? d1=? d2=? Month 1 X2 Month 2 Month 3 X3=1 X1 X0 r3(x3,d3) r2(x2,d2) r1(x1,d1)

  27. Knapsack Problem • N items can be put into a knapsack • Each time has a certain weight and a value • Determine how many units of each item to place in the knapsack to maximize the total value • Constraint is placed on the maximum weight • Similar to a manger who must make a biweekly selection of jobs from each of four categories to process during a 2-week period

  28. Job Data for the Manufacturing Operation • Select certain jobs during the next 2 weeks such that all the jobs selected can be processed within 10 working days and the total value of the jobs is maximized

  29. Formulation • At stage 1, we must decide how many jobs from category 1 to process • At stage 2, we must decide how many jobs from category 2 to process, etc. • Value rating is a subjective score (scale 1 to 20) • xn=No. of days of processing time remaining at the beginning of stage n (state variable) • dn=No. of jobs processed from category n (decision variable at stage n)

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