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NOTE! The answer to problem 3 on the white test is wrong. There are three basic feasible solutions and they are: (1/3, 5/3, 0, 0, 5/3, 0) and (1/8, 0, 0, 5/8, 10/8, 0) and (1/8, 0, 5/64, 0, 85/64, 0).

In problem 4b) for both tests there are ten integer pairs in the region of feasible solutions. Seven of them easy to find (one or both components are zero) and three require either being good with fractions, a calculator, or plotting carefully on graph paper or with a ruler.

The average on the white test was 55.81 and the standard devation was 16.63. To compute your post-belled mark, take your score, subtract 55.81, multiply by 10, divide by 16.63, and add 69. Take the maximum of this number and your original score. This number is 20% of your course mark.

The average on the yellow test was 56.93 and the standard devation was 13.00. To compute your post-belled mark, take your score, subtract 56.93, multiply by 10, divide by 13.00, and add 69. Take the maximum of this number and your original score. This number is 20% of your course mark.

After belling, the average score is 69 with a standard deviation of 10.

And here are two nice websites about the Prisoner's Dilemma: site 1, and site 2.

The white and green tests were statistically indistinguishable. The average score was 67.16 and the standard deviation was 19.24. To compute your post-belled mark, take your score, subtract 67.16, multiply by 10, divide by 19.24, and add 69. Take the maximum of this number and your original test score. This number is 20% of your course mark. (Note to students who had scores below 50 before the belling --- you need help with this material! Please come see me or your TA for help. Or find a study buddy and work

After belling, the average score is 69 with a standard deviation of 10.

I will have office hours on Monday April 28 1pm-4pm.

Here is the final exam and here are the answers to the final exam.

NOTE: the answer to 3b is incomplete. The only The only way to know that a basic feasible solution is optimal is to compute the objective row. Take x_22 entering and x_24 departing, resulting in a basic feasible solution with basic variables x_11, x_12, x_22, x_23, x_32, and x_34. Solving the dual problem, you find: v1=0, v2=1, v3=3, w1=3, w2=4, w3=4, and w4=1. Using these values to compute the objective row, you find: obj_13 = -1, obj_14 = -3, obj_21 = -2, obj_24 = -4, obj_31 = 0, and obj_33 = -1. Since the objective row is nonpositive, the basic feasible solution *is* optimal, as claimed.