運籌學第五版課后答案,運籌作業(yè)

1、精品文檔 頁 41 2 3 40用圖解法找不到滿足所有約束條件的公共范圍，所以該問題無可行解 頁 54-6 -5 -4 -3 -2 -11 2 3 4-1-2無界解.精品文檔1.2（b）( )約束方程的系數(shù)矩陣 A= 1 2 3 42 1 1 2P1 P2 P3 P4是否可行解P1 P4 -1/3 011/6P2 P3 0 1/2 2 0P2 P4 0 -1/2 0 2P3 P4 0 0 1 1是否是55最優(yōu)解A=(0 1/2 2 0)T和(0 0 1 1)T49頁13題設Xij為第i月租j個月的面積minz=2800 x11+2800 x21+2800 x31+2800 x41+4500 x

2、12+4500 x22+4500 x32+6000 x13+6000 x23+7300 x14s.t.x11+x12+x13+x1415x12+x13+x14+x21+x22+x2310 x13+x14+x22+x23+x31+x3220 x14+x23+x32+x4112Xij0用excel求解為：.3Z. 3.精品文檔50頁14題設a1，a2，a3, a4, a5分別為在A1, A2, B1, B2, B3加工的產(chǎn)品數(shù)量，b1，b2，b3分別為在A1, A2, B1加工的產(chǎn)品數(shù)量，c1為在A2，B2上加工的產(chǎn)品數(shù)量。則目標函數(shù)為maxz= (1.25-0.25)( a1+a2+a3)+(

3、2-0.35) b3+( 2.8-0.5)c1 -0.05 (a1+b1)-0.03 (a2+b2+c1)- 0.06 (a3+b3)-0.11(a4+c1)-0.05a5=0. 95a1+0. 97a2+0. 94a3+1.5b3+2.1c1-0.05b1-0.11a4-0.05a5s.t.5a1+10b160007a2+b2+12c1100006a3+8a340004a4+11c170007a54000a1+a2-a3-a4-a5=0b1+b2-b3=0a1，a2，a3, a4, a5, b1，b2，b3, c10用lindo求解得：LP OPTIMUM FOUND AT STEP6OBJ

4、ECTIVE FUNCTION VALUE1)16342.29VARIABLEVALUEREDUCED COST0.0000009.6400000.0000000.00000015.9000000.2300000.0000000.0000000.000000A1A2A3B3C1B1A4A5B21200.0000000.000000285.71429410000.0000000.0000000.000000342.857147571.42858910000.000000ROW SLACK OR SURPLUSDUAL PRICES0.1680001.5000000.0750000.0000000

5、.0085710.110000-1.5000002)3)4)5)6)7)8)0.0000000.0000000.0000005628.5712890.0000000.0000000.000000.對偶問題為：maxz=2y1+3y2+5y3s.t.y1+2y2+y323y3+y2+4y324y1+3y2+3y3=4minz=2x1+2x2+4x3s.t.81頁2.12.ROW SLACK OR SURPLUS.精品文檔最大生產(chǎn)計劃為A生產(chǎn)5個單位，C生產(chǎn)3個單位b)LP OPTIMUM FOUND AT STEP2OBJECTIVE FUNCTION VALUE1)27.00000VALUEV

6、ARIABLEREDUCED COST0.0000002.0000000.0000000.000000X1X25.0000000.0000003.0000000.000000X3X1,X2,X3ROW SLACK OR SURPLUSDUAL PRICES0.2000000.6000000.0000002)3)4)0.0000000.0000000.000000NO. ITERATIONS=2RANGES IN WHICH THE BASIS IS UNCHANGED:OBJ COEFFICIENT RANGESALLOWABLE ALLOWABLEVARIABLECURRENTCOEFINC

7、REASE1.8000002.0000001.0000000.000000DECREASE0.600000INFINITY1.500000INFINITYX1X23.0000001.0000004.0000000.000000X3X1,X2,X3RIGHTHAND SIDE RANGESALLOWABLEROWCURRENTRHSALLOWABLEDECREASE15.0000007.500000INFINITYINCREASE23445.00000030.0000000.00000015.00000015.0000000.000000.可知A產(chǎn)品的利潤變化范圍【6. 8,2.4】,上述計劃不

8、變。6x1+3x2+5x3+8x4453x1+4x2+5x3+2x430 x1,x2,x3 ,x40用lomdo求解為LP OPTIMUM FOUND AT STEP 0OBJECTIVE FUNCTION VALUE1) 27.50000.6x1+3x2+5x3453x1+4x2+5x3-y30 x1,x2,x3,y0用lomdo求解為LP OPTIMUM FOUND AT STEP 0OBJECTIVE FUNCTION VALUE.a）.設yi=x=0y1x=3y2x=5y2x=7y4y1+y2+y3+y4=1x1+x8=1.設xij為第i種泳姿用第名運動員minz=.EF.712587

9、334642671587346426.精品文檔最小部分樹 16582641728254582641728254最小部分樹 32.精品文檔 頁 8028紅色曲線為使用一年賣出藍色曲線為使用兩年賣出綠色曲線為使用三年賣出紫色曲線為使用四年賣出或或.精品文檔8028由圖可知，若摩托車最多使用三年，答案仍然不變6.14b（v2,1）（v5,1）（v4,1）.精品文檔根據(jù)題意,先給發(fā)點 vs 標號，由?。╲s,v2）得對 v2 v1 v3 v4 對 v4 v5 vt 反向追蹤找出增廣鏈最大流為 （v3,1）（v2,1）（v4,1）（vs,2）（v1,1）最大流 13，無最小割集 頁-7.2 表 .精品文

10、檔KD7834LAJHMBE9512FCG6194頁7.3b假定最初事件在時刻零出現(xiàn)，則有t t t(1,2)=t t(1,3)=(1,4)=0ESESES(1,2)=(1,2)+t(1,2)=2EFESt(1,3)=4EFt(1,4)=8EFt(2,5)=2ESt(2,5)=5EFt(5,8)=8ESt(5,8)=15EFt(8,9)=15ESt(8,9)=20EFt(5,7)=5ESt(5,7)=7EF.精品文檔tttttttttttt(4,7)=8(4,7)=11(4,6)=8(4,6)=10(3,6)=4(3,6)=12(6,7)=12(6,7)=13(7,9)=13(7,9)=15E

11、SEFESEFESEFESEFESEFES(6,9)=12(6,9)=19EF最短周期為20t t t(6,9)=(8,9)=(7,9)=20LFLFLFtttttttttLS(8,9)=15LS(7,9)=16LS(6,9)=13LF(5,8)=15LS(5,8)=8LF(5,7)=16LS(5,7)=14LS(4,7)=10LF(4,7)=13.精品文檔ttttttttttttttLS(6,7)=15LF(6,7)=16LS(4,6)=10LF(4,6)=13LS(3,6)=5LF(3,6)=13LS(1,3)=1LF(1,3)=5LS(1,4)=0LF(1,4)=8LF(2,5)=8LS

12、(2,5)=5LF(1,2)=5LS(1,2)=3t t t t t R F(i,j)(i,j)LS123456782483831055553103100000885808888002332227758789060.精品文檔174531500150關鍵路線為.精品文檔1.柴油機生產(chǎn)計劃設x1,x2,x3,x4分別為一二三四季度計劃minz= 11.4x1+11.55x2+11.3x3+11.45x4s.t.x125x235x330 x410 x1+x2+x3+x4=70LP OPTIMUM FOUND AT STEP3OBJECTIVE FUNCTION VALUE1)796.2500VALUEVARIABLEREDUCED COST0.0000000.0000000.0000000.000000X1X2X3X425.0000005.00000030.00000010.000000ROW SLACK OR SURPLUSDUAL PRICES0.1500000.0000002)3