Math 5593 Linear Programming Problem Set 4

Transcription

1 Math 93 Linear Programming Problem Set 4 University of Colorado Denver, Fall 20 Solutions (October 28, 20) Solution 4. (Multiperiod Production Model) [steelt.mod] set PROD; param T > 0; # products # number of weeks param rate {PROD} > 0; param inv0 {PROD} >= 0; param avail {..T} >= 0; param market {PROD,..T} >= 0; # tons per hour produced # initial inventory # hours available in week # limit on tons sold in week param prodcost {PROD} >= 0; # cost per ton produced param invcost {PROD} >= 0; # carrying cost/ton of inventory param revenue {PROD,..T} >= 0; # revenue per ton sold var Make {PROD,..T} >= 0; # tons produced var Inv {PROD,0..T} >= 0; # tons inventoried var Sell {p in PROD, t in..t} >= 0, <= market[p,t]; # tons sold maximize Total_Profit: sum {p in PROD, t in..t} (revenue[p,t]*sell[p,t] - prodcost[p]*make[p,t] - invcost[p]*inv[p,t]); # Total revenue less costs in all weeks subject to Time {t in..t}: sum {p in PROD} (/rate[p]) * Make[p,t] <= avail[t]; # Total of hours used by all products # may not exceed hours available, in each week subject to Init_Inv {p in PROD}: Inv[p,0] = inv0[p]; # Initial inventory must equal given value subject to Balance {p in PROD, t in..t}: Make[p,t] + Inv[p,t-] = Sell[p,t] + Inv[p,t]; # Tons produced and taken from inventory # must equal tons sold and put into inventory Data (Multiperiod Production Model) [steelt.dat] data; param T := 4; set PROD := bands coils;

2 Math 93 Linear Programming Problem Set 4 (Solutions), UC Denver, Fall 20 2 param avail := ; param rate := bands 200 coils 40 ; param inv0 := bands 0 coils 0 ; param prodcost := bands 0 coils ; param invcost := bands 2. coils 3 ; param revenue: := bands coils ; param market: := bands coils ; AMPL Output [model steelt.mod; data steelt.dat; solve; display Make, Inv, Sell;] Presolve eliminates 2 constraints and 2 variables. Adjusted problem: 24 variables, all linear 2 constraints, all linear; 38 nonzeros linear objective; 24 nonzeros. iterations, objective 033 display Make, Inv, Sell; : Make Inv Sell := bands bands bands bands bands coils coils coils coils coils ; Solution 4.2 (Sensitivity and Shadow Prices in AMPL) (a) In AMPL, type model steel3.mod; data steel3.dat; solve; display Time, Make.rc; 2 iterations, objective Time = 4640 Make.rc [*] := bands.8 coils plate 0 ;

3 Math 93 Linear Programming Problem Set 4 (Solutions), UC Denver, Fall 20 3 As discussed in Section.6, AMPL interprets a constraint s name by itself as reference to the associated dual values or shadow prices, that indicate by how much the optimal objective value would improve (or worsen) if the constraint were relaxed (or tightened) by a small amount. Similarly, a variable s name appended by.lb,.ub, or.rc corresponds to the dual values or shadow prices of the constraint associated with that variable s lower bound (e.g., 0 for nonnegative variables, for free variables), its upper bound (e.g., 0 for nonpositive variables, for free variables), or its reduced cost (the objective coefficient in the optimal dictionary or tableau), respectively. Hence, the above display tells us that up to some (unknown) point, additional rolling time would produce another $4640 of extra profit per hour, higher market demand in bands would yield another $.8 per ton, lower (!) commitment of coils would yield an extra $ per ton (note that the shadow price is negative!), and (small) changes to either demand or commitment of coils would not affect the current profit; changes in the opposite directions would decrease the profit correspondingly. (b) Computing the profit rates (in dollars per hour) for both the reheat and rolling stage, we get profit rate ($/hour) bands coils plate reheat roll These rates indicate that while bands remain the most resource-efficient product to roll, coils and plates are more resource-efficient to reheat. In particular, the efficiency gains of coils and plates over bands for reheating clearly exceed their efficiency losses during rolling, which results in a higher production of plates that is compensated by a lower production of bands. (c) ampl: let avail["reheat"] := 36; solve; let avail["reheat"] := 37; solve; let avail["reheat"] := 38; solve; let avail["reheat"] := 38.; solve; 0 iterations, objective # $800 more than the initial profit iterations, objective # another additional $800 in profit iterations, objective # only $7.428 increase in profit 0 iterations, objective # no additional increase in profit (d) Also shown in the above plots, the following table lists the shadow prices of Time["reheat"] and the total profit for different integer values of avail["reheat"] (the exact break points of the piecewise linear shadow-price and total-profit curves can be computed using ranging).

4 Math 93 Linear Programming Problem Set 4 (Solutions), UC Denver, Fall 20 4 avail["reheat"] Time["reheat"] Total Profit When the available reheat time drops to (or below) hours, we receive the AMPL output let avail["reheat"] := ; solve; presolve: constraint Time[ reheat ] cannot hold: body <= cannot be >=.2; difference = -0.2 This tells us that our problem has become infeasible and even hints at the reason: production of our committed minimum quantities (000 tons bands, 00 tons coils, 70 tones plates) with reheat rates of 200 tons/hour each already require 220/200 =.2 hours of reheat time. Solution 4.3 (The Parametric Self-Dual Simplex Method): We begin to introduce two slack variables x 4 and x for the two inequality constraints and thus define the problem data as 0 2 A =, b =, c = [ ] T. 2 0 Next, we choose an initial decomposition of the columns of A into B = {4, } and N = {, 2, 3}, so B = B = 0, N = 0, c 2 B = [ 0 0 ] T, and cn = [ ] T. Then setting x = x 2 = x 3 = z 4 = z = 0, we compute the current basic primal and dual variables x x4 B = = B 2 b = and zn = T z z 2 z 3 = (B N) T c B c N = [ ] T x which are infeasible in x 4 and z. Hence, we let x B = ρ (, 0) T and z N = ρ 2 (, 0, 0) T and consider x 2 + ρ µ B + µ x B = and zn + µ z N = [ 2 + ρ 2 µ 6 0 ] T. Here ρ > 0 and ρ 2 > 0 are any two positive numbers that can be chosen (i) at random so to avoid cycling in the case of degeneracy, and (ii) to satisfy ρ < ρ 2 to force an initial dual pivot on x 4 : µ = min{µ : 2 + ρ µ 0, 2 + ρ 2 µ 0} = max{2/ρ, 2/ρ 2 } = 2/ρ if ρ 2 > ρ > 0. Randomly setting (ρ, ρ 2 ) = (0.2, 0.6) (but essentially any other choice will do as well), it follows that the above solution is primal-dual feasible and thus optimal for all µ µ = 2/ρ = 2/0.2 = 0. Since further decrease of µ is prevented by x 4, we then compute its corresponding dual pivot as z N = [ z z 2 z 3 ] T = (B N) T e 4 = { z j + z j µ } k = arg min : z j > 0 = arg min j N z j [ x4 x B = = B Ne x 3 = 2 T 2 0 j {,2,3} {4, 6, 0} = 3 ] [0 0 ] T = [ = [ ] T ].

5 Math 93 Linear Programming Problem Set 4 (Solutions), UC Denver, Fall 20 It follows that x 4 leaves and x 3 enters the basis, and we can update the primal and dual variables: x 3 t = x 4 x 4 = 2 = 2 x 3 t = x 4 x 4 = 0.2 = 0.2 z 4 s = z 3 z 3 = 0 = 0 z 4 s = z 3 z 3 = 0 = 0 x x B = 4 x x 2 0 B t x B = 2 = ] [ x x B = x x B t x B = ( 0.2) = zn = [ z z2 z3 ] T z N s z N = [ ] T z N = [ z z 2 z 3 ] T zn s z N = [ ] T. Hence, the new basic primal variables are (x 3, x ) = (2, ), the new dual variables are (z, z 2, z 4 ) = ( 2, 6, 0), and for the next iteration we let B = {3, } and N = {, 2, 4}. We continue as before: B = B 0 =, N =, c 2 0 B = [ 0 0 ] T, cn = [ ] T, x x3 B = = B 2 b =, and zn = T z z 2 z 4 = (B N) T c B c N = [ ] T x which is infeasible in x and z. Using x B = ( x 3, x ) = ( 0.2, 0.2) and z N = ( z, z 2, z 4 ) = (0.6, 0, 0), x 2 0.2µ B + µ x B = and zn + µ z + 0.2µ N = [ µ 6 0 ] T is optimal for 0 µ µ = max{/0.2, 2/0.6} =. Computing the dual pivot on x with µ = z N = [ ] T T z z 2 z 4 = (B N) T 0 e = = [ 2 ] T 2 { z j + µ z } { } j 6 k = arg min : z j > 0 = arg min = 2 j N z j j {2} 2 x3 x B = = B [0 ] T Ne x 2 = 0 = 2 2 we see that x leaves and x 2 enters the basis yielding the following new primal and dual variables: x 2 t = x x = 2 = 0. x 2 t = x x = = 0. z s = z 2 z 2 = 6 2 = 3 z s = z 2 z 2 = 0 2 = 0 x x B = 3 x x 2. B t x B = 0. = 2 0 ] x x B = 3 [ x x x B t x B = ( 0.) = x zn = [ z z2 z4 ] z N s z N = [ ] T T T 3 2 = 0 3 z N = [ z z 2 z 4 ] zn s z N = [ ] T. Since all primal and dual variables are now nonnegative, the current solution (x, x 2, x 3, x 4, x ) = (0, 0.,., 0, 0) and (z, z 2, z 3, z 4, z ) = (, 0, 0, 3, 3) is optimal with B = {3, } and N = {, 2, 4}.

6 Math 93 Linear Programming Problem Set 4 (Solutions), UC Denver, Fall 20 6 Solution 4.4 (The Simplex Method with Ranges): After solving for x and x 4 by hand (using substitution) or overkill (computers), the initial dictionary with basic variables x and x 4 is u ζ = 6 + 4x 2 2x 3 + 4x 2x 7 + 3x 8 = x = 4 x 2 x 3 + x + x 6 x 7 + x 8 = 4 0 x 4 = 3 + x 2 3x 3 + 4x + x 6 2x 7 + x 8 = 3 Adopting the largest-coefficient rule to find a first primal pivot, we find a tie between x 2 and x but may observe that x can be increased to its upper bound 2 without changing the current basis u ζ = 6 + 4x 2 2x 3 + 4x 2x 7 + 3x 8 = x = 4 x 2 x 3 + x + x 6 x 7 + x 8 = 6 0 x 4 = 3 + x 2 3x 3 + 4x + x 6 2x 7 + x 8 = Then choosing x 2 as next pivot, which is currently at its lower bound, we can increase its value to { } { x x 2 u 2 min i l i ui x } { } i 6 0 : ā i2 < 0 min : ā i2 > 0 = min 6,, = 4 ā i2 ā i2 ( ) in which case x 4 attains its upper bound and therefore leaves the basis in the updated dictionary u ζ = 8 + 0x 3 + 4x 4 2x 4x 6 + 6x 7 7x 8 = x = 7 4x 3 x 4 + x + 2x 6 3x 7 + 6x 8 = x 2 = 3 + 3x 3 + x 4 4x x 6 + 2x 7 x 8 = 4 Again using the largest-coefficient rule, we repeat the above computation for a primal pivot on x 3 { } { x x 3 u 3 min i l i ui x } { i 2 0 : ā i3 < 0 min : ā i3 > 0 = min 0, ā i3 ā i3 ( 4), 6 4 } = 0. 3 in which case x is reduced to zero and therefore replaced by x 3 in both basis and the new dictionary u ζ = 0. 2.x +.x x + x 6.x 7 2x 8 = x 2 = x + 0.2x 4 0.2x + 0.x 6 0.2x 7 0.x 8 =. 0 0 x 3 =.7 0.2x 0.2x 4 +.2x + 0.x 6 0.7x 7 +.x 8 = 0. The only remaining pivot candidate is now x 6 because x 4 and x are already at their upper bounds while all other variables that remain at their lower bounds have a negative objective coefficient, so { } { x x 6 u 6 min i l i ui x } { i : ā i6 < 0 min : ā i6 > 0 = min 0, 6. } 0 0., = ā i6 ā i and x 6 is set to and enters the basis while x 2 attains its upper bound 6 and thus leaves the basis u ζ = x + 2x 2 + x 4 + x x 7 x 8 = x 3 = x + x 2 0.x 4 +.x 0.x 7 + 2x 8 = 0 0 x 6 = 4. +.x + 2x 2 0.x x + 0.x 7 + x 8 =

7 Math 93 Linear Programming Problem Set 4 (Solutions), UC Denver, Fall 20 7 Because all variables that are currently at their lower bounds have a negative objective coefficient, and analogously, because all variables that have a positive objective coefficient are already at their upper bounds, we have found an optimal solution (x, x 2, x 3, x 4, x, x 6, x 7, x 8 ) = (0, 6,,, 2,, 0, 0) with an optimal objective value ζ = 24. There is a pretty good chance that this is the largest non-trivial LP (8 original variables, 2 original equality constraints, and 8 inequality constraints, or if written with slacks, 6 variables and 0 equality constraints) that you will ever solve by hand. Alternative Pivots Choices and Intermediate Dictionaries by Evan, Jenny, and Lauren The following series of pivot choices is remarkable in cleverly avoiding any easy updates that only switch between a nonbasic variable s lower and upper bound. Respect to the three Pivoteers for not giving up and working it out only on a bag of pretzels and two granola bars! Well done. Pivoting x and x 2 u ζ = 0 4x 6x 3 + 8x + 4x 6 6x 7 + 7x 8 = x 2 = 4 x x 3 + x + x 6 x 7 + x 8 = 4 0 x 4 = 7 x 4x 3 + x + 2x 6 3x 7 + 6x 8 = 7 2. Pivoting x 2 and x 6 u ζ = 6 2x 3 + 4x + 4x 2 2x 7 + 3x 8 = x 6 = 4 + x + x 3 x + x 2 + x 7 x 8 = 2 0 x 4 = + x 2x 3 + 3x + 2x 2 x 7 + 4x 8 = 3. Pivoting x 4 and x 8 u ζ = x 2 x x + 2 x 2 4 x x 4 = x 6 = x + 2 x 3 4 x x x 7 4 x 4 = 0 3 x 8 = 4 4 x + 2 x x 2 x x x 2 = 4. Pivoting x 8 and x u ζ = x x x x x x 4 = x 6 = x + 3 x x x x 7 3 x 4 = x = 3 3 x x x x x x 4 = 2 3. Pivoting x and x 3 u ζ = x + x x 8 + 2x 2 x 7 + x 4 = x 6 = x + 2 x + x 8 + 2x x 7 2 x 4 = 0 0 x 3 = x x + 2x 8 + x 2 2 x 7 2 x 4 = Of course, it is not difficult to see that this optimal dictionary describes the same solution as before. Please let me know if you have any questions, comments, corrections, or remarks!