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Chapter 10 Dynamic Programming

Chapter 10 Dynamic Programming. Agenda for This Week. Dynamic Programming Definition Recursive Nature of Computations in DP Forward and Backward Recursion Basic Elements of DP Applications of DP. Dynamic Programming.

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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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