Mid-term Exam of Operations Research

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1 Mid-term Exam of Operations Research Economics and Management School, Wuhan University November 10, 2016 ******************************************************************************** Rules: 1. No electronic device (for either computation or communication) allowed. 2. No document or paper allowed. 3. No exit of the exam within the rst 30 mins. 4. No allowance for participation of exam after 30-min late. 5. All conclusions/deductions need appropriate justications. New variable needs introduction. ********************************************************************************* Ex. 1 [LP Modeling] The Transportation Department of Wuhan University considers reoptimizing its campus-bus schedule. An OR team is employed to carry out an analysis and to give reasonable recommendations. The study seeks a minimum number of buses that can handle the transportation needs within campus. After gathering necessary information, the OR team noticed that minimum number of buses needed uctuated with the time of the day and that the required number of buses could be approximated by constant values over successive 4-hour intervals. To carry out the required daily maintenance, each bus can operate 8 successive hours a day only and is operated only once a day. The following table summarizes the team's ndings. Question [12.5'+12.5=25']. (a)[12.5'] A non-stop duration of 8 hours is called a shift. Suppose that the Transportation Department will purchase a number of buses for the plan, and each bus costs A RMB. Write a linear programming to determine the optimal bus plan that will meet the demand. 1

2 (b)[12.5'] A non-stop duration of 4 hours is called an interval. Suppose that the Transportation Department will rent buses for the plan. The rent rates are as follows: each bus/interval costs B 1 RMB, each bus/2 succesive intervals costs B 2 RMB, where B 1 > B 2 /2. Write a linear programming to determine the optimal bus plan that willl meet the demand. Solution. (a) Consider the following 6 succesive shifts: 1) 12:00 AM-8:00 AM; 2) 4:00 AM-12:00 Noon; 3) 8:00 AM-4:00 PM; 4) 12:00 Noon-8:00 PM; 5) 4:00 PM-12:00 Midnight; 6) 8:00 PM-4:00 AM. [2'] Let x 1,..., x 6 be the number of buses planned for shift 1,..., shift 6 respectively, and let z be the purchasing cost of all these buses. [2'. Most students fail to specify the shifts and the decision variables in a fully correct way] The linear programming problem is [8.5'. "-1'" for answer without x i 0; "-2'" for answer with x 1 4 the rst ineauqlity; "-2" or more for wrongly specied decision variables x i ] Min z = A(x 1 + x 2 + x 3 + x 4 + x 5 + x 6 ) x 1 + x 6 4 s.t. x 1 + x 2 8 x 2 + x 3 10 x 3 + x 4 7 x 4 + x 5 12 x 5 + x 6 4 x 1, x 2, x 3, x 4, x 5, x 6 0 (b) Consider the following 6 succesive intervals: 1) 12:00 AM-4:00 AM; 2) 4:00 AM-8:00 AM; 3) 8:00 AM-12:00 Noon; 4) 12:00 Noon-4:00 PM; 5) 4:00 PM-8:00 PM; 6) 8:00 PM-12:00 Midnight. [2'] Let x ij be the number of buses rented at the beginning of interal i and run for j succesive intervals, for i = 1, 2,..., 6 and j = 1, 2. Let z be the rental cost for the plan. [2] The linear programming problem is [8.5'] s.t. 6 Min z = B 1 x i1 + B 2 6 i=1 i=1 x i2 x 11 + x 12 + x 62 4 x 21 + x 22 + x 12 8 x 31 + x 32 + x x 41 + x 42 + x 32 7 x 51 + x 52 + x x 61 + x 62 + x 52 4 x ij 0, i = 1,..., 6, j = 1, 2 Ex. 2 [LP: simplex method, sensitivity analysis and duality] Consider the following linear programming: Max z = x 1 + 2x 2 9x 3 + 8x 4 36x 5 2

3 s.t. 2x 2 x 3 + x 4 3x 5 40 x 1 x 2 + 2x 4 2x 5 10 x 1, x 2, x 3, x 4, x 5 0 Question [15'+15'+15'+10'=55']. (a)[15'] Solve the problem using the simplex method. Write out the simplex tableau and specify the entering/exiting variable (or the pivot number) in each step. (b)[15'] Compute the new optimal solution (z, x 1,..., x 5, s 1, s 2 ) if the constant "10" on the RHS of the second constraint is reduced to 4. (c)[15'] Write down the dual of the above problem. Solve the dual graphically and check your solution with the optimal simplex tableau in primal program. (d)[10'] Solve the primal using the complementary slackness conditions. Solution. (a) Adding s 1, s 2 as slack variables into the constaints to obtain the LP in canonical form: [1'. Sucient to introduce s 1, s 2 0] Max z = x 1 + 2x 2 9x 3 + 8x 4 36x 5 2x 2 x 3 + x 4 3x 5 + s 1 = 40 s.t. x 1 x 2 + 2x 4 2x 5 + s 2 = 10 x 1, x 2, x 3, x 4, x 5, s 1, s 2 0 The initial simplex tableau is [3'. "-1'" for each error] It is not optimal since "-1,-2,-8<0". [0.5. Most students did not write it explicitly] We observe that the entering variable is x 4 and the leaving variable is s 2. [1'. It is okey if you specify the pivot number correctly] The next simplex tableau is [3'] Eq. BV z x 1 x 2 x 3 x 4 x 5 s 1 s 2 RHS (0) z (1) s (2) x It is not optimal since "-6<0". [0.5'] We observe that the entering variable is x 2 and the leaving variable is s 1. [1'] The next simplex tableau is [3'] 3

4 Eq. BV z x 1 x 2 x 3 x 4 x 5 s 1 s 2 RHS (0) z (1) x (2) x / It is now optimal since all the coecients "1.8, 0, 6.6, 0, 23.2, 2.4, 2.4" in the objective equation are nonnegative [0.5']. The optimal solution is (x 1, x 2, x 3, x 4, x 5, s 1, s 2 ) = (0, 14, 0, 12, 0, 0, 0) and the optimal value is z = 124 [0.5']. (b) We rst compute the optimum range for the RHS constant "10" (corresponding the s 2 ). By adding to 10 on the RHS, the same operations of simplex method give us the optimal simplex tableau as above except for the last column, which is now "( , , )". To have the optimal basis unchanged in the perturbed problem, we should have and Equivalently, [ 30, 70] is the optimum range [0.5']. Now "10" is reduced to 4, we have = 4 10 = 6 which is still within the optimum range so the optimal basis and the shadow prices remain unchanged. [3'+2'=5'. Most students do not have this argument. In this case, you can obtain this 5 point only if you point out that "15.2, 9.6>0" so the perturbed nal simplex tableau remains optimal. Some students solve the new problem directly using the simplex method. Points are given if the computation is correct. ] The new optimal solution is (z, x 2, x 4) = ( ( 6), ( 6), ( 6) ) = (107.2, 15.2, 9.6) [8'. "2'" for z's value, "3'" for x 2 or x 4 's value. ] and [2'. Most fail to mention. ] (x 1, x 3, x 5) = (0, 0, 0). (c) Let y 1 and y 2 be the dual variables of the two constraints. The dual LP is written as [5'] The graphical solution to the dual program is shown in Figure 1. [4'. Important to point out each line's equation.] 4

5 We see that the optimal solution is the intersection of the lines "2y 1 y 2 = 2" and "y 1 +2y 2 = 8", which is (y1, y 2 ) = (2.4, 2.8) [3'. "-1.5" if you do not mention or we can not read from the graph 'which two lines are they']. Check with the optimal simplex tableau of the primal program: (2.4, 2.8) are (s 1, s 2 )'s coecients in the optimum tableau's objective equation [3'. It is ne if you say "at optimum v = 40y y 2 = 124" as in the primal problem]. (d) For (y1, y 2 ) = (2.4, 2.8), we see that only the two constraints "2y 1 y 2 2" and "y 1 +2y 2 8" are satised with equality. This imples that the dual variables x 1, x 3, x 5 are all equal to zero [3'. It is important to show that all other three inequalities do not bind]. Moreover, both y1 = 2.4 and y 2 = 2.8 are stictly positive, this imples that 2x 2 x 3 + x 4 3x 5 40 and "x 1 x 2 + 2x 4 2x 5 10" are satised with equality [3']. We put them together to obtain { 2x 2 + x 4 = 40 x 2 + 2x 4 = 10 and (x 2, x 4 ) = (14, 12) [2'+2'=4'. Assigning "2'" only if the solution is obtained without correct argument],which are the same as obtained in (a) by the simplex method. Ex. 3 [Optimal Transportation] Use the VAM and MODI methods to nd an optimal solution to the following transportation problem [10'+10'=20']. 5

6 Solution. Phase I: looking for an initial BF solution using the VAM method [6'. Partial point is given if you obtain a dierent initial BF solution from here, by using some other method than the VAM method.] The above transportation is not optimal since K 22 = 2 < 0.[3'] [Put 2' for specifying the loop being used for a new BF solution.] Phase II: looking for an optimal solution(transportation) using the MODI method. The modied transportation is [5'. Partial point is given for minor error in the nal solution.] We verify that it is optimal [3']. Conclusion: The optimal transportation is x 12 = 5, x 14 = 15, x 22 = 10, x 23 = 15, x 31 = 10, x 34 = 5, and all other x ij = 0; The optimal transportation cost is = 460 [1']. 6

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