EE 424 Introduction to Optimization Techniques

Transcription

1 EE 44 Introduction to Optimization Techniques Homework No.7, Due date : November, A company is planning its advertising strategy for next year for its three major products. Since the three products are quite different, each advertising effort will focus on a single product. In units of millions of dollars, a total of 6 is available for advertising next year, where the advertising expenditure for each product must be an integer greater than or equal to 1. The vice-president for marketing has established the objective: Determine how much to spend on each product in order to maximize total sales. The following table gives the estimated increase in sales (in appropriate units) for the different advertising expenditures: Solve this problem by dynamic programming. Solution) X n : Advertising expenditures for product n, n = 1,, S n : Amount of budget still available for remaining product (n,,) P n (XX nn ) : Increase sale when X n millions dollar spend on product f n (SS nn ) = i) n = max {PP nn (XX nn ) + ff nn+1 (SS nn XX nn )} 1 XX nn SS nn s f (SS ) X ii) n =

2 f (SS, XX ) f (SS ) or or iii) n = 1 f 1 (SS 1, XX 1 ) f 1 (SS 1 ) or X x 1 Optimal Solution is X 1 = 1, XX =, XX = or X 1 =, XX =, XX = 1. Re-solve the Local Job Shop employment scheduling problem (Example in the lecture note) when the total cost of changing the level of employment from one season to the next is changed to $100 times the square of the difference in employment levels. Solution) Season Spring Summer Autumn Winter Spring Requirements x n : employment level for each stage x 1 summer x :autumn x :winter x 4 :spring, xx 4 = 55 r n : minimum employment requirement for stage n r 1 = 0, rr = 40, rr = 00, rr 4 = 55 Cost for stage n = 100(x n xx nn 1 ) + 000(xx nn rr nn ) S n = x nn 1

3 4 Minimize [100(x i x i 1 ) + 000(x i rr ii )] ii=1 subject to r i x i 55 (ii = 1,,,4) f n (SS nn, x n ) = 100(x n x n 1 ) + 000(x n rr nn ) + min [100(x i x i 1 ) + 000(x ii rr ii ) rr ii x i 55 f n (SS nn ) = min rr nn x nn 55 ff nn(ss nn, x n ) 4 ii=nn+1 = min { 100(x n S n ) + 000(x ii rr ii ) + ff nn+1 (x n )} nn x nn 55 i) n = 4 S 4 f 4 (SS 4 ) x 4 00 S (55 S 4 ) 55 ii) n = f (SS ) = min 00 x 55 { 100(x S ) + 000(x 00) + 100(55 x ) } f (SS, x ) x = 00(x SS ) (55 x ) = 0 f (SS, x ) x = = 400 > 0 (positive) x = SS + 45 iiii mmmmmmmmmmmmmmmmmm for all S [40,55] f (SS ) = ff (SS, x ) = 100 SS + 45 SS SS + 44 = 5(45 S ) + 5(65 SS ) (SS 155) S + 45 S f (SS ) x 40 S 55 5(45 S ) + 5(65 SS ) (SS 155) S + 45 iii) n = f (SS, x ) = 100(x SS ) + 000(x 40) + 5(45 x ) + 5(65 x ) (x 155) (0 S 55) f (SS, x ) x = 00(x SS ) (45 x ) 50(65 x ) = 0

4 x = SS + 5 f (SS, x ) x = = 00 > 0, (pppppppppppppppp) 40 x SS 70 So, in 47.5 SS 55, x = SS +5 is minimizer, In 0 S 47.5, x = 40 SS x and f (SS,x ) x = 00x 00SS S f (SS ) x 0 S < (40 S ) S [(5 SS ) + (85 SS ) + 180(SS 05)] S + 5 iv) n = 1 f 1 (SS 1, x 1 ) = 100(x 1 SS 1 ) + 000(x 1 0) [(5 x 1) + (85 x 1 ) + 180(x 1 05)] (47.5 x 1 55) = 100(x 1 SS 1 ) + 000(x 1 0) + 100(40 x 1 ) (0 x 1 < 47.5) f 1 (SS 1, x 1 ) = 00(x 1 SS 1 ) (5 x 1) 00 9 (85 x 1) (47.5 x 1 55) x 00(x 1 SS 1 ) (40 x 1 ) (0 x 1 < 47.5) (0 + SS 1 ) = 4.5 (0 x x 1 = 1 < 47.5) SS = 7.5 (47.5 x ) Second case is impossible, so x 1 = 4.5 x 1 = 4.5, x = 40, x = 4.5, x 4 = 55 S = 4.5, SS = 40, SS 4 = 4.5 f 1 (55) = 100(4.5 55) + 000(4.5 0) + 100(40 4.5) = 16500

5 . Consider following nonlinear programming problem : maximize z = x 1 + xx + 4xx xx subject to x 1 + xx + xx 4 x 1 0, xx 0, xx 0 Use dynamic programming to solve this problem S n : amount of remaining resource x n : Some resources f (SS, xx ) = 4xx xx (0 xx SS ) f (SS, xx ) = xx + ff (SS xx ) (0 xx SS ) f 1 (SS 1, xx 1 ) = xx 1 + ff (SS 1, xx 1 ), (0 xx 1 SS 1 = 4) i) n = max 4xx xx 0 xx SS f (SS, xx ) xx = 4 xx = 0 f (SS, xx ) xx = < 0 S f (SS ) x 0 S 4S SS S S 4 4 ii) n = max xx + ff (SS xx ) 0 xx SS Case 1) 0 S xx max xx + 4(SS xx ) (SS xx ) 0 xx SS

6 f (SS, xx ) xx = 4 + (SS xx ) = 0 xx = SS 1 f (SS, xx ) xx = < 0 In, ss 1 0, x = SS 1, ff (SS ) = SS + 1 (1 SS 4) ff (SS, 0) = ff (SS ) = 4SS SS ff(ss, SS ) = SS In 0 S 1, 4S SS > SS. So, x = 0, ff (SS ) = 4SS SS (SS SS ) Case ) S xx 4 max 0 xx SS xx + 4 This function is monotonically increasing. So the larger x maximizer x xx SS SS xx 4 max(0, SS 4) xx min(ss, SS ), 0 xx SS x = SS, ff (SS ) = SS But f (SS ) = SS < SS + 1 So, this case is wrong S f (SS ) x 1 S 4 S + 1 S 1 0 S 1 4S SS 0 iii) n = 1 max 0 xx 1 SS 1 xx 1 + ff (SS 1 xx 1 ) = max 0 xx 1 xx 1 + ff (4 xx 1 ) Case 1) 0 4 x 1 1 max 0 xx 1 xx 1 xx 1 + 4(4 xx 1 ) (4 xx 1 ) = max xx 1 + 8xx 1 xx 1 f 1 (SS 1, xx 1 ) xx 1 = 4xx = 0 xx 1 = f (SS, xx ) xx = 4 < 0 f 1 (SS 1 ) = 8 Case ) 1 4 x 1 4

7 max 0 xx 1 xx 1 xx 1 + (4 xx 1 ) + 1 = max xx 0 xx 1 1 4xx f 1 (SS 1, xx 1 ) xx 1 = 4xx 1 4 = 0 xx 1 = 1 iiii mmmmmmmmmmmmmmmmmm f (SS, xx ) xx = 4 < 0 f 1 (SS 1, 0) = 9, ff 1 SS 1, 15 = (eeeeee pppppppppp) At x 1 = 0, ff 1 haaaa mmmmmmmmmmmmmm ppaavvmmmm 9 x 1 = 0, xx =, xx = 1, SS = SS 1 xx 1 = 4, SS = SS xx 1 = 1 x 1 = 0, xx =, xx = 1, zz = 9 aaaaaa ppppppiimmaavv iippvvmmppiippmm 4. Consider the following linear programming problem : maximize 15x xx subject to x 1 + xx 6 x 1 + xx 8 x 1 0, xx 0 a) Use dynamic programming to solve this problem. S n : amount of remaining resources x n : level of activity n S n = (RR 1, RR ) S 1 = (6, 8), SS = (6 xx 1, 8 xx 1 ) = (RR 1, RR ) f (RR 1, RR, xx ) = 10xx f 1 (6, 8, xx 1 ) = 15xx 1 + max 10xx 1 xx RR 1 xx RR xx 0 f n (RR 1, RR ) = max xx nn ff nn (RR 1, RR, xx nn ) f (RR 1, RR ) = max xx RR 1 xx RR xx 0 10xx (1)

8 ff 1 (6,8, x 1 ) = 15xx 1 + ff (6 xx 1, 8 xx 1 ) f 1 (6,8) = max {15xx 1 + ff (6 xx 1, 8 xx 1 )} xx 1 6 xx 1 8 xx 1 0 () i) n = Solve (1) x RR 1, xx RR, xx 0 0 x min RR 1, RR (R 1, RR ) f (RR 1, RR ) x R 1 0, RR 0 10 min RR 1, RR min RR 1, RR ii) n = 1 Solve (), x 1 6, xx 1 8, xx xx 1 8 max 0 xx xx min 6 xx 1, 8 xx 1 10 min 6 xx xx 1, 8 xx 1 =, 0 xx 1 10(8 xx 1 ), xx 1 8 max max 15xx xx 1, max 15xx 0 xx 1 xx xx 1 max 10xx = 50, (xx 1 = ) 0 xx 1 max 15xx xx = 50, (xx 1 = ) At x 1 =, the optimal value 50 R 1 = 6 xx 1 = 4, RR = 8 xx 1 = x = min 4, = x 1 =, xx =, mmmmxxiimmmmmm iiii 50 b) Use the Simplex method to solve the problem. Convert the original problem to the standard form minimize 15x 1 10xx subject to x 1 + xx + xx = 6 x 1 + xx + xx 4 = 8

9 x 1, x, xx, xx 4 0 a 1 a a a 4 b c T At (,1), pivot a 1 a a a 4 b c T At (1,), pivot a 1 a a a 4 b c T At x 1 =, xx =, the maximum value is 50

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