Solutions to Second Midterm

Transcription

1 Operations Management 33:623:386:04/05 Professor Eckstein, Spring 2002 Solutions to Second Midterm Q1 Q2 Q3 Total Max Mean Median Min Standard Deviation This exam conformed to the standard pattern for second midterms, in that there were far fewer extreme low grades than on the first midterm. The overall mean score was just slightly below the typical second midterm result of about 82 it was just below 80. Also, the highest score was 94 (attained by two people), which is a little lower than usual. In both cases, the cause was problem 2 being harder than I intended. I should have included a hint for this problem. I gave this midterm a little later than last term, and I think that showed in that people were a bit rustier on integer programming, but did better on the simulation question. Details for the individual questions are below. 1

2 Q1: Product Mix with Machine Setups (Algebra Formulation) x i = units of product i = 1,..., 5 produced y i = 1 if we produce any of product i = 1,..., 5, and otherwise 0 z j = 1 if we set up machine j = 1, 2, 3, 4, and otherwise 0 Here is the solution that I originally had in mind: max 7.5x 1 + 8x x 3 + 9x x 5 [4.3x x x 3 4.5x x z z z z 4 ] ST 2.5x x x 5 480z 1 2.2x x 4 480z 2 2.0x x 3 420z 3 1.8x x x 5 420z 4 50y 1 x 1 300y 1 50y 2 x 2 300y 2 50y 3 x 3 300y 3 50y 4 x 4 300y 4 50y 5 x 5 300y 5 y 1,..., y 5 {0, 1} x 1,..., x 5 integer z 1, z 2, z 3, z 4 {0, 1} However, many of the correct answers were variations on the following theme: instead of the constraints 2.5x x x 5 480z 1 2.2x x 4 480z 2 2.0x x 3 420z 3 1.8x x x 5 420z 4, people did 2.5x x x x x x x x x x y 1 z 1 y 2 z 1 y 5 z 1 y 2 z 2 y 4 z 2 y 1 z 3 y 3 z 3 y 3 z 4 y 4 z 4 y 5 z 4 I accepted this solution and various similar ones for full credit (there are actually some technical reasons why it might be considered better than the one I originally had in mind, but they have to do with the difficulty of solving the problem, not its modeling aspects). With some thought, you could also use tighter constants than 300 in the logical upper bounds, but I didn t expect that. The most common problem here was having binary variables only for producing products or only for setting up the machines. Another common problem was assuming that product 2

3 1, for example, could be produced on machine 1 or machine 3, leading to x ij variables. The language and example in the problem were quite explicit that product 1 requires time on machine 1 and machine 3, and only one setup is needed for a machine even if it makes several products. I deducted 2 points for this misunderstanding, but most people who went in this direction ended up losing more due to additional mistakes. Here is the complete grading key: Aspect Points Units produced variables 2 Binary production or machine variables (at least one) 2 Definitions clear 2 Max 1 Revenue in objective function 1 Unit costs in objective function 1 Setup costs in objective function 2 Machine minutes constraints 4 Logical upper bounds 4 Logical lower bounds 4 Sensible way of constraining setups 7 (including another set of binary variables) Binary constraints in algebra (not just definitions) 2 Production amounts integer 1 Linearity 2 Total 35 Standard Deductions Got machine setup variables, rest of machine setup doesn t work Variable mix-up in logical upper and/or lower bounds Correct setup constraints, but additional weird stuff (depending on how weird) If statements instead of logical upper/lower bounds And vs. or confusion for machines, but otherwise OK And vs. or confusion, but constraints/costs also wrong Logical upper/lower bounds missing their binaries Similar major error in logical upper/lower bounds Used product fixed charge variables as if they were machine setups Aggregated upper/lower bounds without binaries Points 5 4 2/ Q2: Setting up Cellular Telephone Towers (Spreadsheet) This problem consisted of a set covering problems problem for basic service coverage, and modified set covering problem for good coverage, and a budget constraint. The tricky part was not double-counting the calling minutes in areas (like area 2) that might have good 3

4 coverage from more than one tower. The trick is to make the Covered Good cells E32:E39 binary changing cells less than or equal to C32:C39, respectively. Almost everybody missed this aspect of the problem, which turned out harder than I intended. Of those who did see it, most didn t get it all; I think one or two people got the entire ideas for E32:E39, but then made errors elsewhere in the problem. I was considering including a hint that E32:E39 were changing cells, but decided not to because I mistakenly thought that would make the problem too easy I should have put in the hint. Some people didn t read part (e) carefully it was slightly different from the practice questions. The question asked do you have to assume nonnegative and is it possible to assume linear model. The answers were no, assuming nonnegative is optional since all the variables are binary, and yes, you can assume linear model. If you just wrote assume nonnegative, I counted that as wrong for a one-point deduction. I counted assume linear model as correct. Below is my grading key. The portions that most people missed are in italics. Part Aspect Points a B32 = SUMPRODUCT(B4:G4,B$26:G$26) 3 b C32 = SUMPRODUCT(B17:G17,B$26:G$26) 3 c G34 = SUMPRODUCT(B12:G12,B26:G26) 3 d G38 = SUMPRODUCT(H4:H11,E32:E39) 3 e1 Target is G38 1 e2 Maximize 1 e3 Changing cells B26:G26 2 e4 Changing cells E32:E39 2 e5 B32:B39 >= 1 3 e6 G34 <= G39 3 e7 E32:E39 <= C32:C39 3 e8 B26:G26 binary 2 e9 E32:E39 binary 2 e10 Assume nonnegative is optional 1 e11 You can assume linear model 1 all Correct $ absolute references 1 all Correct syntax 1 all Proper use of cell references for data 1 Total 35 Part Standard Deductions Points e Additional changing cells 2 e Missed E32:E39 idea (= e4 + e7 + e9) 7 a B32 = SUMPRODUCT(B4:G4,B17:G17) 2 e5 + formulas in constraints (treat as syntax error) 1 4

5 Q3: Chickens and Eggs (Simulation Spreadsheet) People scored very well on this problem. I don t think it was easier than the similar problem last term, but with two simulation homework assignments under your belts, I guess it wasn t that difficult. Good! The most common mistake was forgetting SIMOUTPUT. Sometimes people used formulas that were more complicated than absolutely necessary, but I typically didn t deduct for that. Some people applied the INT function to the output of GENBINOMIAL that s unnecessary because GENBINOMIAL always returns a whole number (just a random one whose average value might not be integer). Some people put the INT inside the arguments to GENBINOMIAL in a very confused way. Note that INT only takes one argument, and the second (p) argument to GENBINOMIAL is a probability that is only integer in the rare cases that it is exactly 0 or 1. Part Aspect Points a B9 = PARAMETER(D16:D26,1,A9) 3 b B11 = GENTABLE(D2:D12,E2:E12) 3 c B12 = GENBINOMIAL(B11,B2) 3 d B13 = B3*B12 1 e B14 = GENBINOMIAL(B9,B4) 3 f B15 = MIN(B13,B14) 3 g B16 = B13 B15 2 h B17 = B14 B15 2 i1 B19 = B5*B9 1 i2 B20 = B16*B6 1 i3 B21 = B17*B7 1 j B22 = SIMOUTPUT(B19 + B20 B2,A22) 3 k 15 chickens (scenario 6) 2 all Syntax 1 all Proper use of cell references for data 1 Total 30 Part Standard Deductions Points j Forgetting SIMOUTPUT 2 c/e GENBINOMIAL(INT(n,p)) 2 5