2016-Midterm2-Questions\u2014\u2014answer - HONG KONG BAPTIST UNIVERSITY Page 1 SEMESTER 1 MID-TERM EXAMINATION 2016-2017 Course Code MATH3205 Section No

2016-Midterm2-Questionsu2014u2014answer - HONG KONG...

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MATH3205 00001 Linear and Integer Programming 17 1 17 INSTRUCTIONS: 1. Answer ALL of 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 neatly and part marks may be awarded. Quote names of theorems used as appropriate. 4. You are permitted to use the text, your notes and any material handed out in class . 5. If you feel you need more space use the back of the preceding page. Student Name: Student No.: HONG KONG BAPTIST UNIVERSITY Page: of SEMESTER 1 MID-TERM EXAMINATION, 2016-2017 Course Code: Section No.: Time Allowed: Hour(s) Course Title: Total Number of Pages:
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MATH3205 00001 Linear and Integer Programming 17 2 17 1. (15 marks) Solve the following problem using the dual simplex method, and present the simplex tableau for each of the iterations. Minimize z = 2 x 1 + 3 x 2 + 4 x 3 + 5 x 4 subject to x 1 x 2 + x 3 x 4 10 , x 1 2 x 2 + 3 x 3 4 x 4 6 , 3 x 1 4 x 2 + 5 x 3 6 x 4 15 , x 1 , x 2 , x 3 , x 4 0 . Solution: If we would have inequalities instead of , then the usual simplex would work nicely, The two-phase method is more tedious. But since all coe cients in z = 2 x 1 + 3 x 2 + 4 x 3 + 5 x 4 are 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 we get the following tableau. x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 0 = z 0 2 3 4 5 0 0 0 x 5 -10 -1 1 -1 1 1 0 0 x 6 -6 -1 2 -3 4 0 1 0 x 7 -15 -3 4 -5 6 0 0 1 Choose Row 1 to pivot on. The ratio for x 1 is better than for x 3 , to pivot on a 1 , 1 . After pivoting, we get x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 0 = z -20 0 5 2 7 2 0 0 x 1 10 1 -1 1 -1 -1 0 0 x 6 4 0 1 -2 3 -1 1 0 x 7 15 0 1 -2 3 -3 0 1 Now every a i, 0 for i > 0 is nonnegative. So, the tableau is optimal. HONG KONG BAPTIST UNIVERSITY Page: of SEMESTER 1 MID-TERM EXAMINATION, 2016-2017 Course Code: Section No.: Time Allowed: Hour(s) Course Title: Total Number of Pages:
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MATH3205 00001 Linear and Integer Programming 17 3 17 2. (9 marks) Write the dual of the following LP. maximize z = 17 x 2 + 18 x 4 8 x 5 subject to x 1 13 x 2 + 45 x 3 + 16 x 5 7 x 6 107 3 x 3 18 x 4 + 30 x 7 81 4 x 1 5 x 3 + x 6 = 13 10 x 1 ≤ − 2 3 x 2 17 x 3 16 x 4 0 x 5 unrestricted x 6 16 x 7 16 Write all complementary pairs in this primal, dual pair. Solution: (i) minimize w = 107 π 1 + 81 π 2 13 π 3 10 π 4 2 π 5 3 π 6 + 17 π 7 + 16 π 8 subject to π 1 + 4 π 3 + π 4 + π 5 = 0 13 π 1 + π 6 + π 7 = 17 45 π 1 + 3 π 2 5 π 3 + π 8 = 0 18 π 2 18 16 π 1 = 8 7 π 1 + π 3 0 30 π 2 0 π 1 , π 4 , π 6 , π 8 0 π 2 , π 5 , π 7 0 π 3 unrestricted (ii) Complementary pairs: ( π 1
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