AM 121: Intro to Optimization Models and Methods Fall 2017

Transcription

1 AM 121: Intro to Optimization Models and Methods Fall 2017 Lecture 8: Sensitivity Analysis Yiling Chen SEAS Lesson Plan: Sensitivity Explore effect of changes in obj coefficients, and constraints on the optimal solution see a connection to duality Approaches: Geometric intuition Use AMPL to get sensitivity information Use basis to tableau eqns Jensen & Bard: 4.1 1

2 Sensitivity Analysis What happens if the data change slightly? E.g., a change in right-hand side value E.g., a change in objective coefficient E.g., a new decision variable E.g., modifying the entries in the column corresponding to a variable Important to understand the robustness of a solution. First Approach: Graphical Intuition 2

3 Example 1: Wooden toys Make toy soldiers and toy trains $3 profit per soldier, $2 profit per train Skilled labor of two types (carpentry and finishing) Soldier: 2 hours finishing, 1 hour carpentry Train: 1 hour finishing, 1 hour carpentry 100 finishing hours, 80 carpentry hours each week Unlimited demand for trains, at most 40 soldiers purchased each week Goal: maximize profit (Use LP, assume solution will be integral) x 1 = number of soldiers produced x 2 = number of trains produced max z =3x 1 +2x 2 s.t. 2x 1 + x 2 apple 100 finishing x 1 + x 2 apple 80 carpentry x 1 apple 40 soldier demand x 1,x 2 0 3

4 x 2 2x 1 +x (finishing) isoprofit 60 A B x 1 40 (soldiers) C x 1 +x 2 80 (carpentry) x 1 How would optimal solution change if objective coe cient or RHS values change? 4

5 Sensitivity to Objective Coefficient x 2 2x 1 +x (finishing) isoprofit 60 A B x 1 40 (soldiers) Q: Let c 1 be contribution to profit of a soldier. For what values does current basis remain optimal? C x 1 +x 2 80 (carpentry) x 1 5

6 Isoprofit line is: c 1 x 1 +2x 2 = constant x 2 = c 1 2 x 1 + constant/2 c 1 =) slope is 2 Slope of carpentry constraint is -1 isoprofit lines flatter than this if c 1 /2 > 1, or c 1 < 2 New optimal solution would be at A Slope of finishing constraint is -2 isoprofit lines steeper than this if c 1 /2 < 2, or c 1 > 4. New optimal solution would be at C =) Basis remains optimal for 2 apple c 1 apple 4 Would still manufacture 20 soldiers and 60 trains Of course profit changes! If c 1 = 4, profit will be 4(20)+2(60) = $200 Sensitivity to Right-hand side 6

7 x 2 2x 1 +x (finishing) isoprofit 60 A B x 1 40 (soldiers) D Q: Let b 1 be number of available finishing hours. For what values does current basis remain optimal? C x 1 +x 2 80 (carpentry) x 1 Current basis remains optimal as long as intersection of carpentry and finishing constraints remains feasible. Consider these two binding constraints: See that when b 1 < 80 then x 1 < 0 and basis infeasible. When b 1 > 120 then x 1 > 40 and basis infeasible. =) Basis remains optimal for 80 apple b 1 apple 120. both 2x 1 + x 2 = b 1 x 1 + x 2 = 80 =) x 1 = b 1 80 Within the range, decision and objective value changes: if b 1 = then x 1 = 20 + and x 2 = 60 z = (Do you see why?) 7

8 Shadow prices Definition. The shadow price on the i th constraint is the amount by which objective value is improved if RHS b i is increased by 1 (while current basis remains optimal.) For example, we know that if b 1 =100+, then x 1 =20+, x 2 =60 and z =3x 1 +2x 2 =180+ Shadow price of the finishing constraint is $1 it is $0 for 3 rd (non-binding) constraint on demand of soldiers. We Shadow will see price that: == Dual value == reduced cost on associated slack variable shadow price = optimal dual value corresponding to constraint = reduced cost on slack var in optimal primal tableau Second Approach: Using AMPL Sensitivity Report 8

9 Example 2: Furniture Make desks, tables and chairs Profit of $60, $30 and $20 respectively Have 48 lumber, 20 finishing hours, 8 carpentry hours. Goal: maximize profit Amount of each resource needed to make each type of furniture Desk Table Chair Lumber Finishing 4 hrs 2 hrs 1.5 hrs Carpentry 2 hrs 1.5 hrs 0.5 hrs (Use LP, assume solution will be integral) Example 2: Furniture x 1 desks x 2 tables x 3 chairs max z = 60x x x 3 s.t. 8x 1 +6x 2 + x 3 + x 4 = 48 lumber 4x 1 +2x x 3 + x 5 = 20 finishing 2x x x 3 + x 6 = 8 carpentry x 1,...,x 6 0 Optimal (primal) tableau: B = {4, 3, 1} x =(2, 0, 8, 24, 0, 0) z = 280 z +5x x x 6 = 280-2x 2 +x 4 + 2x 5-8x 6 = 24-2x 2 +x 3 + 2x 5-4x 6 = 8 x x 2-0.5x x 6 = 2 9

10 AMPL: Step 1 (furniture.mod) # AMPL script for the Furniture model. set PROD := 1..6 # Decision variables (production program) var X {j in PROD} >= 0 # Objective function maximize Obj: 60*X[1] + 30*X[2] + 20*X[3] # Constraints subject to Lumber: 8*X[1]+6*X[2]+1*X[3]+1*X[4]=48 subject to Finishing: 4*X[1]+2*X[2]+1.5*X[3]+1*X[5]=20 subject to Carpentry: 2*X[1]+1.5*X[2]+0.5*X[3]+1*X[6]=8 end Data hard coded in the model file here. Not a good practice for large LP instances. AMPL: Step 2 (furniture.run) reset reset data model furniture.mod option solver './cplex' option cplex_options 'sensitivity primalopt' option presolve 0 solve display X > furniture.sens Prevents solver using algebraic manipulation to simplify and remove variables, constraints before solving display _varname, _var.rc, _var.down, _var.current, _var.up > furniture.sens display _conname, _con.dual, _con.down, _con.current, _con.up > furniture.sens 10

11 AMPL: Step 3 ampl: include furniture.run --> Now look at furniture.sens file X [*] := reduced cost (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. changes in obj fcn coeff s : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' USEFUL VALUES 2 'X[2]' -5-1e 'X[3]' 'X[4]' slack vars, less 5 'X[5]' -10-1e interesting 6 'X[6]' -10-1e dual (shadow price) changes in RHS values : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing Carpentry USEFUL VALUES 11

12 Questions (answer using AMPL s report) If desks (x 1 ) were selling for $10 more per desk, how much more profit would we make? -> What if desks sold for $30 more per desk? -> If we have 2 fewer finishing hours, how would the profits change? -> If we have 3 more feet of lumber, how would the profits change? -> Interpreting shadow prices reduced cost changes in obj fcn coeff s : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e dual (shadow price) changes in RHS values : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing Carpentry Q: How much would you be willing to pay for one additional carpentry hour? A: Since profits $60, $30, $20 incorporate costs of material, you d pay up to $10 more than regular cost (since profit = $10) 12

13 Careful: Multiple changes at once The valid range of changes is only applicable when a single change is made. What if we want to consider sensitivity to multiple changes at once? Case 1: If changes are only in obj coefficients on nonbasic variables and the RHS for non-binding constraints, simultaneous change within ranges ok. Case 2: Otherwise, need to use the 100% rule: Defines the allowable simultaneous changes to the obj coeffs of basic variables. Defines the allowable simultaneous changes to RHS values of binding constraints. The 100% rule Changes in objective coefficients 13

14 X [*] := (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e <- to 70 <- to 18 OK : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing Carpentry X [*] := (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e <- to 70 <- to 22 Not OK : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing Carpentry

15 100% rule for objective coe cient changes if change is made on c j to one or more basic variables then need P j r j apple 1 where r j is the ratio change for variable x j with respect to its valid range. Let c j denote change in objective value coe cient for variable x j, and D j denote the allowable decrease (if c j < 0) or increase to c j (if c j > 0). r j = c j /D j E.g., in furniture example, if desks profit of now desks bring $70 and chairs $18 the current solution remains optimal because r P 1 = /20 = 0.5,r 3 = /5 =0.4, r 2 =0, rj =0.9 < 1. But, if tables bring $33 and desks $58, r 1 = /4 =0.5,r 2 = /5 =0.6,r 3 =0 and P r j =1.1 > 1. so Invalid solution changes might (basis change. could change). The 100% rule Changes in RHS values 15

16 X [*] := (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing <- to 22 3 Carpentry <- to 9 OK X [*] := (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing <- to 17 3 Carpentry <- to 9 Not OK 16

17 100% rule for RHS changes if change is made on b i for one or more binding constraints then need P i r i apple 1 where r i is the ratio change for RHS b i. Let b i denote change in RHS for constraint i 2 {1,...,m}. Let D i denote the allowable decrease (if b i < 0) or increase to b i (if b i > 0). r i = b i /D i E.g., in furniture example, if have 22 finishing hours and 9 carpentry hours then r 1 = 0 r 2 = /4 =0.5, r 3 = 9 8 /2 =0.5. OK, because P r i = 1. Third Approach: Use Basis to Tableau Eqns and Optimal Primal Tableau 17

18 ORIGINAL PROBLEM FINAL TABLEAU Recall: Furniture example x 1 desks x 2 tables x 3 chairs max z = 60x x x 3 s.t. 8x 1 +6x 2 + x 3 + x 4 = 48 4x 1 +2x x 3 + x 5 = 20 2x x x 3 + x 6 =8 x 1,...,x 6 0 B = {4, 3, 1} x =(2, 0, 8, 24, 0, 0) z = 280 Optimal (primal) tableau: z +5x x x 6 = 280-2x 2 +x 4 + 2x 5-8x 6 = 24-2x 2 +x 3 + 2x 5-4x 6 = 8 x x 2-0.5x x 6 = 2 18

19 Variables x 1,..., x 6. An optimal tableau looks like: z + 5x x x 6 = 280 = 24 = 8 = 2... with anything values in in the constraint matrix that leave the basic variables isolated. In considering whether change in LP data will cause optimal basis to change, we determine how changes a ect RHS and row 0 of optimal tableau Need b 0 (for feasibility) and c 0 (for optimality) Review: Tableau from a Basis Given an optimal (primal) basis, B, then: RHS: b=a 1 B b Dual solution: y T = c T B A 1 (last lecture) B Objective value: z = c T B A 1 B b = yt b Nonbasic obj coe : c B T 0 =(c T B A 1 B A B 0 ct B ) 0 For nonbasic j, c j = c T B A 1 B A j c j = y T A j c j Immediate observations: (a) (b) c j = y j for slack variable x j since c j = 0 and A j = e j (i.e., the j th unit vector) dual variable y j is shadow price on RHS of constraint j (since z = y T b) ß can read optimal dual directly from optimal primal tableau ß the optimal dual value gives shadow price 19

20 Aside: Reading A B -1 from the final tableau Furniture example x 1 desks x 2 tables x 3 chairs max z = 60x x x 3 s.t. 8x 1 +6x 2 + x 3 + x 4 = 48 4x 1 +2x x 3 + x 5 = 20 2x x x 3 + x 6 =8 x 1,...,x 6 0 B = {4, 3, 1} x =(2, 0, 8, 24, 0, 0) z = 280 Optimal (primal) tableau: dual values (and shadow prices) from reduced costs on slack vars z +5x x x 6 = 280-2x 2 +x 4 + 2x 5-8x 6 = 24-2x 2 +x 3 + 2x 5-4x 6 = 8 x x 2-0.5x x 6 = 2 20

21 Example Analyses A. Changing objective function coe cient of a non-basic variable B. Changing objective function coe cient of a basic variable C. Changing the RHS D. Changing the entries in column of a non-basic variable E. Introducing a new activity (decision variable) A. Changing objective coeff. of non-basic variable Consider x 2 (tables) and change in c 2 b = A 1 B b, so RHS does not change c T B 0 = ct B A 1 B A B 0 while basis B is unchanged ct B 0,so c j for j 6= 2 is unchanged Must check reduced cost on c 2 remains non-negative Can use c 2 = y T A 2 c 2 0 (notice y T constant) Require -> gives c 2 35 Can or, get also sensitivity read from optimal information primal directly tableau from the reduced cost c 2 in optimal tableau. Since c 2 =5(whenc 2 = 30), then basis is optimal while c 2 apple 35 21

22 Furniture example x 1 desks x 2 tables x 3 chairs max z = 60x x x 3 s.t. 8x 1 +6x 2 + x 3 + x 4 = 48 4x 1 +2x x 3 + x 5 = 20 2x x x 3 + x 6 =8 x 1,...,x 6 0 B = {4, 3, 1} x =(2, 0, 8, 24, 0, 0) z = 280 Optimal (primal) tableau: z +5x 2 reduced cost + 10x x 6 = 280-2x 2 +x 4 + 2x 5-8x 6 = 24-2x 2 +x 3 + 2x 5-4x 6 = 8 x x 2-0.5x x 6 = 2 X [*] := Check against AMPL (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing Carpentry

23 B. Changing objective coeff. of basic variable x 1 and x 3 are basic variables RHS b = A 1 B b unchanged c T B 0 = ct B A 1 B A B 0 ct B 0 may change for multiple variables (c B changes) Must check reduced cost on every non-basic variable remains non-negative For example, suppose profit on x 1 (desks) increases by > 0 c B =(0, 20, 60 + ). See how y T changes 0 y T = c T BA 1 B = (y T not constant here) while basis B is A = Use c j = y T A j c j to analyze new reduced cost coe cients 0 1 c 2 = y T A 2 c 2 = A 30 = c 5 = y T A 5 c 5 = A 0 = c 6 = y T A 6 c 6 = A 0 = For non-negative reduced costs in row 0, need: =) =) apple =) 20/3 Overall: need 4 apple apple 20. While 60 4 apple c 1 apple , current basis remains optimal. x unchanged, but if c 1 := 70 then 2(10) additional revenue profit and z =

24 X [*] := Check against AMPL (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing Carpentry C. Changing the RHS c j = c T B A 1 B A j But b = A 1 B b changes. Check b 0 c j unchanged while basis B is unchanged to keep feasibility Consider b 2 = 20 + in furniture example 0 b A Need: =) =) =) apple 4 Overall, need 4 apple apple 4, and 16 apple b 2 apple 24 1 A (*) E ect on decision variables 0 given by 1 (*). E ect on objective value is 48 z = y T b = A = (y T from opt. tableau, constant) 8 24

25 X [*] := Check against AMPL (ampl: include furniture.run) Note: reduced cost (.rc) in AMPL is the negation of our reduced cost. : _varname _var.rc _var.down _var.current _var.up := 1 'X[1]' 'X[2]' -5-1e 'X[3]' 'X[4]' 'X[5]' -10-1e 'X[6]' -10-1e : _conname _con.dual _con.down _con.current _con.up := 1 Lumber e+20 2 Finishing Carpentry D. Changing column entries for non-basic variable Consider tables (x 2 ). Change to: profit sell for of $43, use lumber 5, finishing 2, carpentry 2. b = A 1 B b unchanged c j = c T B A 1 B A j c j see that only change is to the reduced cost of non-basic variable x 2 (only place A j and c j appear) Pricing out the activity. Check reduced cost remains 0 (note y T constant). new A 2, new c (y T from opt. tableau, constant here) c 2 = while basis B is A c 2 = = 3 See that the current basis is no longer optimal because c 2 < 0 25

26 E. Introducing a new activity (decision variable) Footstools (x 7 ). Profit Sell for of $15, use lumber 1, finishing 1, carpentry 1. b = A 1 B b unchanged c j = c T B A 1 B A j (y T from opt. tableau, constant here) while basis B is unchanged c j unchanged for all existing variables. Just need to price out the new activity. Check reduced cost remains 0 (again, y T constant). c 7 = A c 7 = = 5 Current basis remains optimal because c 7 > >= 00 Review: Sensitivity analysis with basis to tableau equations Change E ect on optimal solution Non-basic objective Reduced cost c j is function coe cient c j changed All reduced costs Basic objective function coe cient c j may change, obj. value changed RHS in ofopt. constraints tableau, RHS b i of a constraint and optimal also soln, obj. and value obj changed value Changing column entries for a non-basic on cost non-basic xvariable, j and Changes Changesreduced reduced cost variable x j or adding or also introduces the constraint new a new variable x ij reduced columncost. Āj Current basis still optimal if: Need reduced cost c j 0 Need reduced cost c i 0 for all i 2 B 0 Need RHS b i 0 on each constraint Reduced cost c j 0 26

27 Summary: LP Sensitivity Analysis Sensitivity analysis provides an understanding of the robustness of an LP solution Important that optimal basis does not change: Reduced costs remain non-negative, RHS values in optimal tableau remain non-negative Different approaches include: Geometric arguments AMPL s sensitivity report Basis to tableau equations 27