Cls Notes 10-14 Separable

# Cls Notes 10-14 Separable - Problem maximize z =...

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Unformatted text preview: Problem maximize z = —3(x1—3)A2 — 2(x2m5)“2 — (x3—2)A4 subject to (xi—4V2 + 4x2 + 2(x3—1}"2 <: 30 2x1 + 3x2 + 5x3 <= 25 X1 >= 0, x2 >= 0, X3 >3 O Functions for Uariable 1H Functinns for Uariahle X2 Functions for Uariable H3 p MWOL/m Functions for Uariahle X3 MGR/LP (Revised 1/93) Date: 10/14/11 Time: 09:37:11 H:\MORLPS~1\separ.lpe MODEL DESCRIPTION *******afr*-k-k-k***-k-k~k‘k‘k-k‘k‘k‘k+*‘k******kﬁr*****~k***+*****~kk maximize z : —3(xlw3)“2 m 2(x2—5)“2 — (x3—2)A4 subject to (x1—4)“2 + 4x2 + 2(X3—1)A2 <: 30 2x1 + 3x2 + 5X3 <= 25 GRID(X1) 0, 1.5, 3, 4.5, 6} GRID(X2) = 0, 2, 4, 6} GRID(X3) : { 0, 0.5, 1, 2.5, 3.5,4.25,5} Grathrid ******-}r-k~k-k-k~k-k-k*‘k'k*‘k‘k‘k**-k**"kit*~k‘k~k******~k********~k~k~k*~k Separable Programming Initial Tableau BASIS Z X1.1 X1.2 X1.3 X1.4 X2.l X2.2 X2.3 X3.1 X3.2 X3.3 X3.4 X3.5 X3.6 Slackl SlackZ SOL. Z 1.00 —13.50 —4.50 4.50 13.50 ~16.00 —8.00 0.00 ~21.88 —8.13 —0.63 5.00 27.42 73.83 0 00 0 00 —93 00 Slackl 0.00 -6 50 —3.50 ~O 50 2 50 4.00 4 00 4.00 -3 00 —l.00 3 00 8 00 11.50 14 50 1.00 0.00 12 00 Slack2 0.00 2.00 2 00 2 00 2.00 3.00 3 00 3.00 5.00 5 00 5 00 5.00 5.00 5.00 0.00 1.00 25 00 bounds 1 50 1.50 1 50 1 50 2 00 2.00 2 00 0 50 0.50 1 50 1 00 0 75 0.75 AA“ Optimal Solution 0”“ Z : w2.7500 X1.1 = 1.5000 X1.2 = 1.5000 X1.3 = 0.0000 X1.4 = 0.0000 X1 = 3.0000 X2. X2. X2. Lul\>>—* X2 X3. X3. X3. X3. X3. X3. mmawmr—I X3 Slackl 5i [E H 33 [[ 11.8000 2. SEPARABLE PROGRAMMING Separable programming is concerned with optimization of a nonlinear 'objective function over a constraint set consisting of linear and/or nonlinear constraints. The method is extremely restrictive, however, since it applies only to functions which are separable; functions that can be written as the sum of functions of single variables. A function f(xl,x2,...,xn) is_said to be separable if I] f(5) = _{ 51(x1). Linear functions are obviously separable. Another example is _ 2 3 2 fﬂg) ~ x1 + 3x1 + x2 + &(x3—S) , where f (x ) = x2 + 3x 1 l l l’ 3 f2(x2) = x2, £3(x3) = 4(x3—5)2. _ For separable programming.to be applicable all of the functions, objective function and constraints, must he separable. Then the approach is to approximate each function term by a function consisting of piecewise linear segments. This is accomplished for each variable and associated functions by placing a grid on the feasible range of the variable. Then each function is replaced by linear segments which are- exact at the grid points. A different grid for each variable can be used. This procedure obviously has numerical inaccuracies built into it due to the function approximations. The grid for each variable can be refined to increase the accuracy at the expense of adding additional variables to the resulting linear programming problem. A segmentation, grid, is placed on each variable x . Then all i ‘the functions, objective and constraints, of this variable are approx- imated on the specified grid. Let a:, k=0,l,..., Ki, be the grid selection for variable xi. Then this variable is to be replaced by K a set of variables xi, xi, ..., xii representing the increments deter— mined by the grid. Thus, see Figure 1, Ki xi= i x‘i‘. <1) k=l _ where 0 < K? < ag — aF—l for k=l,...,K.. (2) ‘- 1— 1 J. 1 a0 al 2 3 4Hﬁ_¢ i i '31 a; a1 q———xi-db1r——-x{———->‘-—x§—b1-x-—5 w x1 Figure 1. Grid segmentation for each variable xi. The functions for variable x fi(xi), are then approximated via the ii 012 K grid {ai,ai,ai....,aii} by connecting the function values at this point Consider the separable programming problem, 11 mi a: 15(2) 1 1 1 1 subject to Z gji(xi) :_bj for j=1,...,m where each constraint gj(§) = X gji(xi). Then the resulting i linearized approximation is K 1 k k- n i f.(a ) ~ f.(a. } ' 0 1 i 1 1 k max E {f (a_) + i _____.___._1:dn_n_ x_} i=1 i 1 k=1 a}? - a1? 1 1 1 l K k k-l :1 i g (a )— g (a. ) O 31 i 31 1 k s.t. Efg (a) + ): Jib i= j i k=l a: - ai-l x 3 for j=l,...,m, - a for k=l,...,Ki and i=1,...,n, k l i at its upper bound. The linearized separable programming problem can be solved for a local solution, no global solutions guaranteed without convexity . kwl and each varlable x cannot become nonzero until each xi,...,x. ' by straight lines (linear segments). This is illustrated in Figure 2. The function fi(xi) is then written as K k _ k-l x a - k k—l - 5. i 1 1 Fl ai _ 31 1 (3) where 0 x: a: — a§hl for k:l,...,Ki. Note that for the linearized l 1 approximation to be as drawn in Figure 2, variable xi must be at its 1 0 upper bound value of (ai—ai) before variable xi can take on any value. 1 2 3 ! Also for variable xi to become nonzero, both xi and xi'must be at their upper limits, etc. Thus, for any variable x: to come into play in the approximation, all of its predecessor segments must be at their respec» tive upper bounds. The same concept is true for replacing a Variable in the linear programming basis. If all the variables with higher indicies within its segmentation are not zero, then the variable cannot be removed from the basis. Otherwise, the linear approximation as determined by the grid segmentation would be destroyed. ' ' ' ' f Figure 2. Piecewise linear approximation of i(xi) over 0 l 2 3 4 the grid {31,ai,ai,ai,ai . restrictions on the functional forms f and g, by augmenting the linear simplex algorithm to enter variables within a segmentation K 1 2 1 {x.,xi,....xi }, 1 into the basis in'the proper order and allowing other segment variables to enter only when all predicious are at their upper bounds. This is an easy extension of the linear programming logic. This is called a restricted basis linear programming method. Example 1. Consider the problem max x2 - 2(x2—2)2 + x l 3 2 s.t. x1 + ﬁxz + 6x3 j_16 2 3x1 + 6 x2 + 2x3 5‘30 £2.51- The x variable is already linear so it need not be approximated. For 3 an analysis of the constraints, the variables x1 and x2 have upper bounds of'le and 2, réspectiﬁely. For illustration purposes, x is l segmented by the grid {0,1,1.5,2[3.2} and x2 by the grid {0,0.5,l,2}. Thus, the x approximations are: l The x2 approximations are: The linear approximation problem becomes: 1 2 3 4 l 2 3 max x0 le + 2.5xl + 3.5xl.+ 5.2xl + 7x2 + 5x2 + 2x2 + x3 — 8 1 2 3 4 1 2 3 + + s.t. x1 x1 + x1 + x1 + 2x2 + 6x2 + 12x2 6x3 :_16 l 2 3 h l 2 3 3x1 + 7.5xl + 10.5):l + 15.6):l + 6x2 + 6x2 +_6x2 + 2x3 §_30 Oixiil, oixigoo, oixiioo, 033411;). The restricted basic linear programming solution of this example is given in Table l. The implicit upper bound linear programming metﬁod is used to reduce the number of constraints from nine to twa. This upper bounding procedure makes the separable programming method much more computationally tractable since there are a large number oﬁ bounded variable constraints. The optimal solution to the problem is x1 = 1, x1 = 1/2, x1 = 1/2, x1 = 0.568, 2 3 x: = 1/2, x2 = 1/2. x2 = 0, x3 = l.572, with x1=2.568, and x2 6.524, whereas the true function value at the solution point is 6.167. =1. The linearized objective function value is k The optimal solution for this problem to two decimal places is E_ = (2.643, 0.98, 1.58) with 133:6.49. It should be noted in the optimal tableau of this example problem (Table 1) that the segmentation variables in the basis at their upper bounds for variable x1 have nonoptimal cost indicators. Hence these variables want to leave the basis while retaining the fourth segment in the basis. This, of course, is not acceptable and the restricted basis methodology must be employed to keep the variables basic. Convex Separable Functions. If the functions f and g_are such that a convex problem exists (f is concave for maximization and_g is convex) then the restricted basis aspect of the linear simplex algorithm need not be utilized. The variables will enter into the basis in the proper sequence due to the order relationships placed on the slopes of the segmentations resulting from the convexity restrictions. The slopes of the objective function variables will be decreasing within a segu mentation while the convex constraint coefﬁicients will be increasing. Hence the best variable to enter is segment one, then segmen two, etc. “uoﬂusﬂom Hmﬁwumo man mwﬂwﬁh mwmmn mnu Ousﬁ ma mam WK mcﬂuOPHm mnznom ZOHHDAOW umvHMﬁh =vMVO .ausﬁm Buommmmuu wan map use Sumo msﬁuauﬁumpsm mgu mvnnop nmman Hausa um Mx.mx.wx.MK.MK mmanmwnm> mnﬁnmucm ZOHHDAom unawabom uwmmb uHoﬁHmEH mnﬁmn a anEme mo maﬁusaom mnwaﬁwumoum HmmnHH mammm mwuuwuummm H ma emu. 10 th.H owe. HHo.i «no.1 OmH.o Hma.o No~.I mam.| mnm.a mmm.: mmm.: mno.o Hmo.~ mom.¢ wmscﬂuuou H «HamH 099.: mmo.t mHo.I mmv.t mmo.l mmo.l th.i mMH.I mqo.| MOE/LP (Revised 1/93) Date: 10/ 9/06 Time: 08:05:58 G: \CURRY\ SURVEY~1\MORLPS~1 \ separz . lpe MODEL DESCRIPTEON **************************************************** maximize z : —3(x1~3)“2 m 2(x2—5)‘2 — (x3—2)“4 subject to (X1~4)“2 + 4X2 + 2(X3~1)“2 <= 30 2x1 + 3x2 + 5x3 <= 25 GRID(x1) = { 0, 3. 6} GRID(X2) = i 0. 2. 4. 6} GRID(X3) : { O, 0.5, l, 2.5, 5} “Grathrid“ ********************k*************************i***** Separable Programming Initial Tableau BASIS Z X1.l Kl.2 X2.1 X2.2 X2.3 X3.l X3.2 X3.3 X3.4 Slackl Slack2 SOL. Z 1 00 "9 00 9 00 -l6.00 v8 00 0 00 —21 88 —8 13 WD 63 32 37 0.00 0.00 —93 00 Slackl 0.00 —5 00 1 00 4.00 4 00 4.00 —3 00 ~1.00 3 00 ll 00 1.00 0 00 12.00 Slack2 0.00 2.00 2 00 3.00 3 00 3.00 S 00 5.00 5.00 S 00 0.00 1.00 25 00 bounds 3.00 3.00 2.00 2.00 2.00 0.50 BASIS Z ~Xl 1 X1 2 ~X2.l ~X2 2 X2 3 ~X3 l ~X3 2 X3 3 X3 4 Slackl Slackz SOL Z l 00 8 75 9 25 15.62 7 62 0 37 21 25 7 50 0 00 33 00 0 00 0 12 ~2 75 Slacki 0 00 6.20 —0.20 —2 20 -2 20 2 20 6 00 4 00 0.00 8.00 1 00 —0 60 11 80 X3.3 0 00 -0.40 0.40 ~0 60 —0.50 0 60 —l 00 ml 00 1.00 1.00 0 00 0.20 0 40 bounds 3.00 3.00 2 00 2 00 2 00 0 50 0 50 1.50 2 50 AA“ Optimal Solution “*A Z = —2.7500 Xl.l = 3.0000 Xl.2 = 0.0000 X1 = 3.0000 X2.l = 2.0000 X2.2 2 2.0000 X2.3 = 0.0000 X2 = 4 0000 X3.l : 0.5000 X3.2 : 0.5000 X3.3 = 0.4000 X3.4 = 0.0000 X3 = 1.4000 1] H 5.1 (13 O O O Slackl ...
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## This note was uploaded on 10/26/2011 for the course ISEN 620 taught by Professor Curry during the Fall '10 term at Texas A&M.

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Cls Notes 10-14 Separable - Problem maximize z =...

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