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practice_midterm2_v2_3133(2)

practice_midterm2_v2_3133(2) - ISYE3133C Engineering...

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Unformatted text preview: ISYE3133C - Engineering optimization Midterm 2 - April 2, 2008 NAME: GTID: SIGNATURE: Time: 50 minutes Instructions: WORK ALONE. Do not collaborate with anyone. Show ALL your work. You are NOT allowed to use any written notes or books. You are NOT allowed to use any electronic devices such as (but not limited to) calculators, PDAs, cell phones, laptops, etc. You will only receive credit for the work written on the exam itself or the extra pages I have provided. You will find the dual of a standard form problem at the end of the exam. You can use this in any of the question. The Exam has 3 mandatory questions plus an optional extra credit question. By signing above, you agree to abide by the letter and spirit of the GeorgiaTech Honor Code while taking this examination. Good luck! 1 1. (4 pts each for a total of 20 pts) For each of these statements, mark True or False (no justification needed): a) T - F - Whenever we convert a problem to standard form, we do not change the total number of decision variables involved. b) T - F - There is a one to one correspondence between the extreme points of a standard form linear program and its basic feasible solutions. c) T - F - Every transportation problem is balanced. d) T - F - In a basic feasible solution only the nonbasic variables can have value zero. e) T - F - In a standard form problem, if a basic feasible solution has all nonbasic variables with reduced costs less than or equal to zero, then the basic feasible solution is optimal. 2. (50 pts Total) a) Convert the following problem into standard form. Minimize Subject to z = 5x1 - x2 + x3 -4x1 + x2 + x3 2 (1) 3x1 + x2 + x3 4 (2) x1 - 4 x2 =3 (3) x1, x2, x3 0 b) Consider the following problem and the graph of its feasible region Maximize Subject to z = 2x1 + x2 x1 + x2 1 (1) 5x1 + 4x2 20 (2) x1, x2 0 x2 5 4 3 2 1 0 0 1 2 3 4 x1 The problem can be written in standard form as: Maximize Subject to z = 2x1 + x2 x1 + x2 - x3 = 1 (1) 5x1 + 4x2 + x4 = 20 (2) x1, x2, x3, x4 0 Find a basic feasible solution for the standard form problem. Indicate what is the basis and what are the associated primal and dual solutions. Indicate if the basis is primal/dual feasible/infeasible and why. Write the reduced costs of the non-basic variables. Identify the associated extreme point in the graph of the original problem. Is this solution optimal? Why? c) Consider the following standard form problem. Maximize Subject to z = 5x1 + 3 x2 + 2 x3 2x1 + x2 + x3 + x4 = 40 (1) x1 + x2 + 2x3 + x5 = 30 (2) x1, x2, x3, x4, x5 0 The optimal basis is B={1,2} with x1 =10, x2 =20, x3 =0, x5 =0, x6 =0. i. Why is this solution optimal? ii. Is this basis still optimal if we include a new variable x6 with objective coefficient 10 and whose coefficients for constraints (1) and (2) are 5 and 1 respectively? Justify your answer. iii. Is this basis still optimal if we include a new variable x6 with objective coefficient 4 and whose coefficients for constraints (1) and (2) are both 1? Justify your answer. 3. (30 pts) A hospital needs to purchase 3 gallons of a perishable medicine for use during the current month and 4 gallons for use during the next month. Because the medicine is perishable, it can only be used during the month of purchase. Two companies (Daisy and Laroach) sell the medicine. The medicine is in short supply. Thus, each month the hospital is limited to buying at most 5 gallons from each supplier. The companies charge the prices shown bellow. a. Formulate a transportation problem to minimize the cost of purchasing the needed medicine. Clearly indicate the supply and demand constraints. b. Draw the Graph or diagram of your transportation problem. Clearly label the nodes indicating if they are supply or demand nodes and their respective capacities or demands. Write the profit/cost per unit of flow besides each one of the arcs. c. Is the transportation problem from a)-b) balanced? Company Current Month's Price per Gallon (\$) 120 140 Next Month's Price per Gallon (\$) 150 130 Daisy Laroach Remember to describe (in words) the variables, constraints and objective function. 4. (Extra Credit Question: Up to 10 extra credit points) Remember the following problem from question 3 in Homework 3. Max s.t. 5 x1+ x2 5 (1) 2 x1- x2 4 (2) x1, x2 0 a) Find a direction of unboundedness which shows that the LP is unbounded. b) Write the problem in standard form and find a direction of unboundedness d for this problem. Show that Ad=0, d0 and cd>0. z=+x1+x2 redit Question You can use the following in any of the problems. normal max problem of Standard Form problem Credit Question The Dual normal max problem max n z= j=1 n cj xj cj xj j=1 n max s.t. z = s.t. n j=1 j=1 aij xj = bi aijx j 0 i x =b j i = 1, . . . , m i = 1, . .. .. ., , n m j = 1, j = 1, . . . , n (P) (P) is m xj 0 min w = min w = s.t. s.t. m i=1 bi y i bi y i i=1 m m i=1 aij yi cj aij yi cj j = 1, . . . , n j = 1, . . . , n (D) (D) (1) (1) i=1 hat that asible solution to (D) Also, the dual of (D) is (P). easible solution to (D) of (P) which can take the following values: optimal objective value he optimal objective value of = + if (P) is unbounded. (P) which can take the following values: = = + if (P) is unbounded. - if (P) is infeasible. s = - if (P) (P) has an optimal solution (i.e. (P) is feasible and is not unbounded). a number if is infeasible. is a number if (P) has an optimal solution (i.e. (P) is feasible and is not unbounded). + for any number d. d + for any number d. m i=1 d consider the following three possibilities for (P): uld consider the following three possibilities for (P): nbounded. By lemma 3 if the primal is unbounded then the dual is infeasible. Then this unbounded. By lemma if the primal y unbounded solution to (D). not happen because we3assumed that is is a feasible then the dual is infeasible. Then this nnot happen because we assumed that y is a feasible solution to (D). feasible. Then z = -. (2) holds because - is less than or equal to any number and nfeasible. Then z = -. (2) holds because - is less than or equal to any number and i is a number. i y i is a number. an optimal solution. Lemma 1 states that s an optimal solution. Lemma 1 states that z m bi y i . z i=1 bi y i . (2) (2) m m ...
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