MASSACHUSETTS INSTITUTE OF TECHNOLOGY Introduction to Optimization (Spring 2004) Midterm Solutions

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1 MASSAHUSTTS INSTITUT OF THNOLOGY Introduction to Otimization (Sring 2004) Midterm Solutions Please note that these solutions are much more detailed that what was required on the midterm. Aggregate midterm statistics are at the end. Problem 1 a) According to the rules of simlex, we ivot in a variable whose 1 st row coefficient is negative. For the tableau given the only variable with a negative coefficient is x 3. b) This question was to test the concet of the min ratio test. The key here is that when taking the ratios for each row RHS oefficient of row /ivot column coefficient of row, we only consider ratios where the ivot column coefficient of row i is non negative. In the examle given this is only true for the row where x 5 is the basic variable. Thus x 5 will leave the basis. There is not a tie when erforming the min ratio test. Problem 2 a) For the objective value of the roblem to be unbounded, we have to have a non-basic column with a negative cost coefficient, and non-ositive elements everywhere in this column. Since column x 4 contains ositive elements, it follows that a (in column x 3 ) must be negative. Hence: a < 0 and c 0 and d 0 and e 0. b) We must ensure that x 4 cannot be ivoted out, and x 3 can be ivoted in. Thus b 0, and a < 0. To be able to ivot x2 out, the corresonding element of the basic row must be ositive. Thus e > 0. Finally, e must win the min-ratio test against c, thus f 3 a < 0 and e > 0 and b 0 and < e c Note that e does not have to win the min-ratio test against d, since d < 0. c) Since b > 0, we cannot ivot in column x 4, thus x 3 is the only otential column, and we need a < 0. To have a degenerate ivot, we have to have a zero in the RHS the ivot row, and this is only ossible if f = 0. Therefore, we will need to ivot on e, thus e > 0. Once f = 0, element e will automatically win the min-ratio test, since all other elements on the RHS are ositive, so we do not need to imose conditions on c or d. The answer is a < 0 and e > 0 and f = 0. Page 1 of 9

2 d) An alternative otimal solution is obtained when a non-basic variable has a zero reduced cost, and since b > 0, the answer is a = 0. Note that we do not need to imose conditions on c, d, or e. ven if all these elements were negative, the roblem would not be unbounded, since a = 0, and not a < 0. We would be able to increase x 3 forever, and would obtain an infinite number of otimal solutions of same cost. Problem 3 a) The change in rofit is within the allowable range of change for the cost coefficient of cars. Therefore the objective value changes to 33420, and all the variables stay the same; that is x 1 =88, x 2 =27.6 and m 1 =98. b) We see that the number of machines available is a binding constraint. The shadow rice is 350 and the allowable decrease is 1.6; therefore, the objective would decrease by $350. c) We are not using all the steel that is available; we would therefore not ay anything for an extra ton of steel. d) We see that the shadow rice of car orders is -20. Furthermore the allowable decrease is 3. The rofit would therefore increase by $40. e) When asked if arco should roduce any jees, the answer is yes (we will always want to roduce at least a art of a jee). When considering all the constraints we notice that we can only roduce a art of a jee, but full credit was given for ricing out as if we would be able to roduce a whole jee, that is the rofit would increase by: $ *$400=$120. Problem 4 Let x 3 = number of jees roduced, then the modified roblem looks like: Max z = 300x x x 3-50m 1 s.t 0.8x 1 + x x 3 - m 1 0 m x x 2 + 2x x 1 + 3x 2 + 4x x 1 -.3x 2 +.7x 3 0 x 1 - x 3 3 -x 1 + x 3 3 x 1 0, x 2 0, x 3 0, m 1 0 Page 2 of 9

3 Problem 5 a) When we draw the feasible region, we find that there are three basic feasible solutions; that is, (1, 3), (2.5, 1.5) and (5, 0). Out of those (1, 3) is otimal with an objective value of 8. The feasible region lies to the right and above the constraints. b) If we draw the second constraint x 1 + x 2 4 as x 1 + x 2 4 +, we note that the otimal solution moves to (1, 3+ ), and the objective value becomes 8 +. The shadow rice is therefore 1. We see that as we make larger, the otimal solution remains at the intersection of the second and third constraint. On the other hand, if < -1, the otimal solution remains at (1, 2). Therefore the allowable range for the right hand side is from 3 to infinity. c) The cost coefficient for x 1 can increase to infinity, and (1,3) remains the otimal solution. (This makes intuitive sense since when the cost coefficient of x 1 becomes very large we will try to make x 1 as small as ossible). On the other hand if we decrease the cost coefficient to 1, the isocost lines become arallel with the second constraint and all oints on the line between (1, 3) and (2.5, 1.5) become otimal. If we decrease the cost coefficient to a value lower than 1,` then (1,3) is no longer otimal. The current otimal solution therefore stays otimal for values between 1 and infinity. Problem 6 a) The shadow rices are the reduced costs of the corresonding slack variables, which can be read from the otimal simlex tableau. The answer is: 0 for 8x 1 + 6x2 + x3 48 (slack variable s 1 ) 10 for 4x1 + 2x x3 20 (slack variable s2) 10 for x + 1.5x + 0.5x 8 (slack variable s 3 ) An alternative solution, which doesn t use the reduced costs of the slack variables directly is as follows. Imagine that we add the column to the RHS in the original tableau. The column becomes in the final tableau, and thus the shadow rice of the first constraint is 0. The shadow rices of the other two constraints are comuted similarly. b) Adding to the rofit er table is the same as increasing the cost coefficient of x 2 to 30+. Thus, in the original tableau, the cost row entry of x 2 will be -30-, and in the final tableau it will be 5-. To stay otimal, we need the non-basic variables to have nonnegative elements in the cost row, thus the only requirement is 5 0 5, and the answer is 5. Page 3 of 9

4 c) hanging the rofit er desk to 60+ means increasing the cost coefficient of x 1 to 60+. The cost row entry of x 1 will be -60- in the original tableau and - in the final tableau. However, because x 1 is a basic variable, its cost row needs to be zero for the tableau to be otimal. Therefore, we add times row 3 of the final tableau, which is (0, 1, 1.25, 0, 0, -0.5, 1.5, 2), to the cost row of the final tableau. The new cost row is (0, , 0, 0, , , ). For otimality we need the cost row coefficients to be non-negative in all columns excet the objective, and thus we obtain the system of equations: , and the answer is / 3 d) We list the decrease in resources caused by roducing one nightstand, and the imact according to the shadow rices ecrease Shadow Price/Unit Imact -1.5 cubic ft. wood hrs. building work hrs. finishing work Total imact -20 By roducing one nightstand, we loose -$20 due to resource consumtion. Since on the other hand, we earn $25, roducing nightstands as art of the otimal solution would be rofitable, and the answer is YS. Problem 7: a) A basic feasible solution is another name for a corner oint. Thus from the diagram the corner oints are and. b) If we are trying to maximize y we are trying to ick the oint that is highest on the y axis and still feasible. Thus the otimal oints are and. c) This was robably the hardest question of the test. Over 85% of the class icked the wrong answer. In the grahical method for otimizing, one moves the isorofit line arallel until it touches the feasible region, and moving it farther makes it leave the feasible region. But there is no isorofit line that can make B and simultaneously otimal. But, there is a way out of this aarent contradiction. The cost could be 0, in which case the entire feasible region is otimal. This is the only ossibility in this case. Thus since A is a feasible oint it is otimal as well. Many students tried to use the fact that if there are two otimal solutions every oint on the line connecting the two otimal solutions must be otimal. They then argued that since A was not on Page 4 of 9

5 this line it was not otimal. The flaw in the reasoning here is that every feasible oint is otimal. An alternative way of seeing that all oints are otimal is as follows: if the line segment joining B and is otimal, then an interior oint is otimal. We learned in recitation 3 that if an interior oint is otimal then all feasible oints are otimal and the objective function is a constant. Hence A must be otimal as well. Problem 8: Part a: Answer: True We have seen many examles in class, on homework, and even one in the exam itself, where slack variables remain in the basis when the simlex algorithm terminates. When we convert an LP to standard form, we add a slack variable to each inequality constraint. If at the otimal oint one of the original LP inequality constraints is not satisfied at equality, then the slack variable added for that constraint will remain in the basis when simlex terminates. Part b: This question was taken directly from recitation 3. The answer to this question is True. The idea we wanted you to get at was that if two corner oints are otimal then all oints along the segment that connects them are otimal. However, surrisingly enough in both classes students came u with the same counter examle, which forced us to start discussing convexity. The counter examle students came u with was: Max y ST the feasible region bellow (The blue region): y x Page 5 of 9

6 Thus students stated that this LP had two otimal solutions (The to oint of each of the two triangles). The flaw in this reasoning is subtle, which is why I think both classes came u with the same examle). The reason why this examle doesn t hold is because the feasible region above is not a ossible feasible region for an LP. One cannot have an intersection of inequality constraints that leads to this feasible region. More generally, the feasible region for an LP is always convex. The easiest way to think of a convex region is one where if you take any two oints in the region and connect them with a line, every oint on this line lies in the feasible region. Thus the region in the icture above is not convex and thus it would be not ossible for this region to be exressed in LP constraints. Part c: This question was taken directly from recitation 3. False: onsider the following LP. Max y Subject To: y 2 x 0 y 0 This LP has two corner oints (0,0) and (0,2). However any oint (x, 0) where x 0, is an otimal oint. Thus this LP has an infinite number of otimal oints but has only one BFS that is otimal (0, 0). Part d: Answer: False Two simle counterexamles are: 1. An infeasible LP has no corner oint. 2. The LP Max x + y s.t. x = 1 The feasible region for this LP is a line which has no corner oint. Page 6 of 9

7 Problem 9 a) If Bill were to invest $1 in, he would get $0.7 if N is elected; if Bill were to invest $1 in, he would get $0.9 if B is elected. Therefore, Bill would invest in, and get at least $1.1 for each $1 invested. b) The nonlinear rogram is max s.t. min { , , } +,, + 0. = 1 After converting to an LP, we obtain the answer: max s.t. z +,, z 1.5 z 1.1 z = c) Let be the roortion of money invested in stock. If B is elected, the ayoff is (1-). If K is elected, the ayoff is (1-). The worst case ayoff is the lower of these two. The best worst case ayoff is when these two ayoffs are equal, that is when (1 ) = (1 ) 0.2 = 0.6 =. 3 Hence, the answer is to invest 1/3 of $500,000 in stock, and 2/3 in stock. The following lot illustrates the reasoning above. return on investment Page 7 of 9

8 Statistics The mean, median, and standard deviation are rovided both in oints and in ercent, given that the test was out of 110. Points % Mean Median Standard deviation Highest score The following histogram illustrates the distribution of grades. Bin Frequency For examle, bin 90 contains all the grades between 80 and 90. Page 8 of 9

9 The following is a chart of the histogram. Histogram Frequency Frequency Bin Page 9 of 9