1. (a) Multiply by negative one to make the problem into a min: 170A 170B 172A 172B 172C Antonovics Foster Groves 80 88

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1 Econ 172A, W2001: Final Examination, Possible Answers There were 480 possible points. The first question was worth 80 points (40,30,10); the second question was worth 60 points (10 points for the first array; 10 points for one completely correct pivot; 10 points for knowing when and why to stop; 10 points for knowing how to write down the answer; and the remaining 20 points for getting the remaining details); the third question was worth 100 (44;26;10;10;10); the fourth question was worth 80 (10,10,10,10,10,15,15); the fifth question was worth 60 (1 point for each blank and 4 points for each short answer); the sixth question was worth 100 (ten points per part). People did badly on the third question. Weights were adjusted to give lots of credit to people who did well on the first two parts. 1

2 1. (a) Multiply by negative one to make the problem into a min: 170A 170B 172A 172B 172C Antonovics Foster Groves Shachat Sobel Add an appropriate constant to each row to make the table nonnegative, with a zero in each row: 170A 170B 172A 172B 172C Antonovics Foster Groves Shachat Sobel Subtract a constant from the first two columns so that there will be a zero in every column: 170A 170B 172A 172B 172C Antonovics Foster Groves Shachat Sobel This table does not provide a zero cost assignment. If you delete the 172C and the Foster, Groves, and Shachat rows, then you can subtract 3 from all of the uncrossed entries (and add three to the double crossed entries) to get: 170A 170B 172A 172B 172C Antonovics Foster Groves Shachat Sobel From this array, we find many zero-cost matchings. For example, Antonovics - 172C; Foster - 172B; Groves - 172A; Shachat - 170B; Sobel - 170A. The average CAPE rating is 75%. 2

3 (b) In the second part of the problem, Antonovics and 170A leave the table. We can start with the table obtained after putting a zero in each row and column: 170B 172A 172B 172C Foster Groves Shachat Sobel This table has a zero-cost matching. Foster - 172B, Groves - 170B, Shachat - 172A, Sobel - 172C (Groves and Shachat can switch). The average CAPE, taking into account that Antonovics's replacement scores 50% is 66.6%. (c) Adding 50% to my approval scores does not change the optimal match, but it raises my approval rating 50% no matter what I teach. So the average approval rating will be 85% 3

4 2. Use the simplex algorithm to solve: max 15x 1 + 6x 2 + 9x 3 + 2x 4 subject to 2x 1 + x 2 + 5x 3 + 6x 4» 20 3x 1 + x 2 + 3x 3 + 3x 4» 24 7x 1 + x 4» 70 x 0 First, write the problem in the simplex array. The array introduces slack variables for the three constraints and uses these slack variables in the initial basis. Note that row zero reverses the signs of the coefficients of the objective function. Note also that I do not include the column for the basis element x 0. (This column does not change during the computation.) Row Basis x 1 x 2 x 3 x 4 x 5 x 6 x 7 Value (0) x (1) x (2) x 6 3 Λ (3) x I chose to pivot on the x 1 column (coefficient is negative in row zero). Row 2 wins" the minimum ratio test, so it is the pivot row. Therefore I pivot on the starred entry. Row Basis x 1 x 2 x 3 x 4 x 5 x 6 x 7 Value (0) x Λ (1) x (2) x (3) x Now I pivot on the x 2 column. The first row leads to the minimum ratio, and therefore I pivot x 5 out of the basis. The computation leads to: Row Basis x 1 x 2 x 3 x 4 x 5 x 6 x 7 Value (0) x (1) x (2) x (3) x Now Row 0 is non-negative, so we have a solution: x 0 =132;x 1 =4;x 2 = 2;x 3 = x 4 = 0 (and the slack variables take on the values: x 5 = x 6 =0 and x 7 = 42. 4

5 3. In the game, Blotto has three strategies: to attack with 0, 1, or 2 companies. The enemy has two strategies: to attack with 0 or 1 company. If Blotto attacks with 2 and his enemy attacks with 1, then both leave their camp undefended, so both camps fall, so the payoff is zero. If Blotto attacks with 2 and his enemy attacks with 0, then Blotto wins. Blotto also wins if both attack with one. Otherwise, neither side wins. The payoff matrix is: The security level for Blotto is zero. The security level for the enemy is 1 (that is, if Colonel Blotto knew the enemy's strategy, then he would certainly be able to defeat the enemy). It makes no sense for Blotto not to attack. That is, attacking with 0 companies is dominated. Eliminating that strategy, it should be clear that it is best for both players to randomize equally over their remaining strategies. If the enemy could spy, that means the enemy could make its choice of how to attack contingent on what Colonel Blotto does. One strategy would be: Attack with one company whenever Blotto leaves his camp undefended. Otherwise, don't attack. This strategy yields a payoff of zero. The payoff matrix for the game (with a spy) is quite large if you include all three of Blotto's original strategies (the enemy's strategy specifies whether to attack if Blotto attacks with 0, 1, or 2 companies. There are a total of 8 such strategies). The game simplifies if you take into account that Blotto will always either attack with 1 or 2 companies. In this case, the enemy has four strategies (00: never attack; 01: attack only if Blotto attacks with 2 companies; 10: attack only if Blotto attacks with 1 company; 11: always attack). The payoff matrix is:

6 4. (a) The bakery earns 310 by producing 5 loaves of whole wheat bread and 120 loaves of oatmeal and rye bread. (b) Nothing (the white flour constraint is not binding). (c) $2 (the dual price of constraint that describes the oatmeal constraint). (d) The allowable increase on the coefficient of oatrye in the objective function is infinity. Hence doubling its price would not change the solution (it would increase profits). (e) Constraint 7's right hand side goes down by5. This is in the allowable range. Hence profits decrease by the value of the dual variable for that constraint, 2, times 5. Profits decrease by 10. (f) The baker was not producing white bread before. It is as if he or she is given an opportunity to produce a new product using.75 pounds of white flour, 2 ounces of yeast, and.5 "spaces" of the oven. The value of these ingredients (using the dual prices) is 1, which is less than the price of white bread (1.5). Hence it pays to produce white bread under these circumstances. (g) The value of the ingredients is 2.50 ($2 for the oven space and.25($2) for the oatmeal). The other ingredients are in excess supply and therefore are available without cost. Hence if the baker can sell the new bread for more than $2.50 it will be profitable to do so. 6

7 5. I solved a linear programming problem written in the form: max c x subject to Ax» b; x 0: Attached find the Excel Answer and Sensitivity report. (I deleted some irrelevant information.) In these reports, I replaced several values with letters ((a) through (ii)). Using the information in the table, replace as many question marks as possible with the correct information. You need not justify these answers (simply write the answers in the appropriate spaces, next to the question mark). If you do not have enough information to figure out one or more of the values, write NOT ENOUGH INFORMATION" next to the question marks. In addition to completing the tables, please answer the following questions. For these questions, short justifications are required. (a) How many variables are in the original problem? Four. (b) How many variables are in the dual? Three (since there are three primal constraints). (c) What is the objective function of the original problem? x 1 +2x 2 +3x 3 +4x 4 : (d) Would the solution to the problem change if the coefficient of x 2 in the objective function were decreased by 3 (and the rest of the problem remained unchanged)? No change to the solution, since the allowable decrease in 3:275 > 3. (e) What would the value of the problem be if the right hand side constant on the first constraint were 224 (and the rest of the problem remained unchanged)? The right-hand side increases by 100, this is within the allowable range. The value goes up by 100 times the dual variable. The dual variable is equal to Hence the value becomes (f) What would the value of the problem be if the right hand side constant on the second constraint were 50 (and the rest of the problem remained unchanged)? The right-hand side increases by 7, this is within the allowable range. The value goes up by 7 times the dual variable. The dual variable is equal to Hence the value becomes (g) What would the solution to the problem change if the coefficientofx 1 in the objective function was 5 (and the rest of the problem remained unchanged)? The solution would change because this increases the coefficient of x 1 by 4, which is greater than the allowable increase of

8 6. For each of the statements below, circle TRUE if the statement is always true, circle FALSE otherwise. No justification is required. (a) TRUE The dual of FALSE can be written min 4x 1 + x 2 subject to 2x 1 + x 2 6 x 2 + x 3 = 4 x 1 4 x 2 x 3 0 min 2y 1 + 4y 2 subject to 2y 1» 4 y 1 y 2» 1 y 1 y 2 0 : True. You need to do some work to confirm this. You can replace the second constraint in the primal by x 2» 4 and drop x 3 from the problem. Substitute z = x and replace all x 1 by z and the third constraint by z 0. The transformed version of the primal leads to the form of the dual written in the question. (b) TRUE FALSE If a linear programming problem has more constraints than variables it is not feasible. False. This one is not even close. (c) TRUE FALSE If a linear programming problem is not feasible, then it will continue to be infeasible if the objective function changes. True. Changing the objective function does not change the feasible set. (d) TRUE FALSE In the two-player zero-sum game below, Row's strategy 1 is dominated. LEFT CENTER RIGHT

9 This is false. In fact, in equilibrium the row player plays the probability distribution: ( 1 9 ; 8 9 ; 0; 0) and column plays ( 5 9 ; 4 ; 0; 0). 9 The next six parts refer to the linear programming problem (P) written in the form: max c x subject to Ax» b; x 0 (e) TRUE FALSE If (P) has a unique solution, then its dual has a unique solution. This is false. Uniqueness of one problem is not related to uniqueness of its dual. For a dumb example, consider max 0 subject to x» 1;x 0: This problem has many solutions (any feasible x), but its dual has the unique solution y =0. (f) TRUE FALSE If x Λ is a solution to (P), then x Λ would still be a solution to max kc x subject to Ax» b; x 0 for any k 1: True. Multiplying the objective function by a positive constant does not change the solution (although it multiplies the value of the solution by the same constant). (g) TRUE FALSE If (P) has a solution, then max c x subject to Ax» b; x 0 has a solution (c may be different fromc). False. The second problem is feasible, but changing the objective function may make the problem unbounded. Think of unconstrained problems, like max0x and max x (so that c = 0 and c =1). The first has a solution. The second is unbounded. (h) TRUE FALSE If (P) has a solution, then the dual of the problem: is feasible. max c x subject to Ax» b; x 1 4;x 0 9

10 True. The dual of the new problem is the dual of the original problem with an additional variable (for the new primal constraint). The dual of the original problem is feasible (by the duality theorem). So, take a point from the dual of the original problem and set the new dual variable equal to zero to get a feasible point in the new dual. (i) TRUE FALSE If (P) has a solution, x Λ, then there exists a y such thatb y = c x Λ. True, but the Duality Theorem the solution to the Dual exists and satisfies the equation. (j) TRUE FALSE If (P) has a solution, then the simplex algorithm finds it in a finite number of pivots. True. This is the essential property of the algorithm. 10