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MATH320500001–Linear and Integer Programming17117INSTRUCTIONS:1. Answer ALLof the following questions. The full mark for this examination is 100.2. Time:3:45pm - 6:30pm, 2 hours and 45 minutes in total.3. Calculators are allowed, but they must not be pre-programmed or have stored text.4.Show all your work neatlyand part marks may be awarded. Quote names of theoremsused as appropriate.4.You are permitted to use the text, your notes and any material handed out inclass.5. If you feel you need more space use the back of the preceding page.Student Name:Student No.:HONG KONG BAPTIST UNIVERSITYPage:ofSEMESTER 1 MID-TERM EXAMINATION, 2016-2017Course Code:Section No.:Time Allowed:Hour(s)Course Title:Total Number of Pages:
MATH320500001–Linear and Integer Programming172171.(15 marks)Solve the following problem using the dual simplex method, and present the simplex tableaufor each of the iterations.Minimizez= 2x1+ 3x2+ 4x3+ 5x4subject to⎧⎪⎪⎨⎪⎪⎩x1−x2+x3−x4≥10,x1−2x2+ 3x3−4x4≥6,3x1−4x2+ 5x3−6x4≥15,x1, x2, x3, x4≥0.Solution:If we would have inequalities≤instead of≥, then the usual simplex would work nicely, Thetwo-phase method is more tedious. But since all coeﬃcients inz= 2x1+ 3x2+ 4x3+ 5x4are non-negative, we are fine for the dual simplex.Multiply the equations by−1 and add to each of the equations its own variable. Then weget the following tableau.x1x2x3x4x5x6x7x0=−z02345000x5-10-11-11100x6-6-12-34010x7-15-34-56001Choose Row 1 to pivot on. The ratio forx1is better than forx3, to pivot ona1,1. Afterpivoting, we getx1x2x3x4x5x6x7x0=−z-200527200x1101-11-1-100x6401-23-110x71501-23-301Now everyai,0fori >0 is nonnegative. So, the tableau is optimal.HONG KONG BAPTIST UNIVERSITYPage:ofSEMESTER 1 MID-TERM EXAMINATION, 2016-2017Course Code:Section No.:Time Allowed:Hour(s)Course Title:Total Number of Pages:
MATH320500001–Linear and Integer Programming173172.(9 marks)Write the dual of the following LP.maximizez=−17x2+ 18x4−8x5subject to−x1−13x2+ 45x3+ 16x5−7x6≥1073x3−18x4+ 30x7≤814x1−5x3+x6=−13−10≤x1≤ −2−3≤x2≤17x3≥16x4≤0x5unrestrictedx6≥16x7≥16Write all complementary pairs in this primal, dual pair.Solution:(i)minimizew= 107π1+ 81π2−13π3−10π4−2π5−3π6+ 17π7+ 16π8subject to−π1+ 4π3+π4+π5= 0−13π1+π6+π7=−1745π1+ 3π2−5π3+π8= 0−18π2≤1816π1=−8−7π1+π3≥030π2≥0π1,π4,π6,π8≤0π2,π5,π7≥0π3unrestricted(ii) Complementary pairs:(−π1