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- StudyBlue
- Michigan
- University of Michigan - Ann Arbor
- Industrial Engineering
- Industrial Engineering 202
- Epelman
- IPexamples.xls

Nicholas B.

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Postal employee scheduling Inputs Monday Tuesday Wednesday Thrusday Friday Saturday Sunday Demands 17 13 15 19 14 16 11 Changing cells Starting on # Mon 6 Tue 2 Wed 3 Thu 7 Fri 1 Sat 3 Sun 1 Objective function 23 Constraints Number of employess working on various days who started on varions days Working on Mon Tue Wed Thu Fri Sat Sun Mon 6 6 6 6 6 Tue 2 2 2 2 2 Wed 3 3 3 3 3 Started on Thu 7 7 7 7 7 Fri 1 1 1 1 1 Sat 3 3 3 3 3 Sun 1 1 1 1 1 Total 18 13 15 19 19 16 15 >= >= >= >= >= >= >= Needed 17 13 15 19 14 16 11 Courses Inputs Subject requirements Math214 IOE310 EECS280 IOE265 IOE373 IOE366 EECS283 Math 1 1 1 1 IOE 1 1 1 1 EECS 1 1 1 Prerequisites: Math214 for IOE310 IOE265 for IOE366 Also, credit not given for both EECS280 and EECS283 Changining cells Course Math214 IOE310 EECS280 IOE265 IOE373 IOE366 EECS283 Take/Not take 0 0 1 1 1 1 0 Objective Total courses 4 Constraints Have Need Math req. 2 2 IOE req. 3 3 EECS req. 2 2 EECS280/283 credit 1 Prerequisites specified within solver Formultion Variables Math214 IOE310 EECS280 IOE265 Binary variables, =1 if the corresponding course is taken, 0 otherwise IOE373 IOE366 EECS283 Mathematical model min Math214+IOE310+EECS280+IOE265+IOE373+IOE366+EECS283 Number of courses taken subject to Math214+IOE310+EECS280+IOE265+IOE373>=2 At least 2 Math courses IOE310+IOE265+IOE373+IOE366>=3 At least 3 IOE courses EECS280+IOE373+EECS283>=2 At least 2 EECS courses IOE310<=Math214 Can't take IOE310 without Math214 IOE366<=IOE265 Can't take IOE366 without IOE265 EECS280+EECS283<=1 Do not take both EECS280 and EECS283 All variables binary Reducing Pollution Inputs Cost to Cost to Pollutant 1 removed Pollutant 2 removed build treat 1 ton per ton treated per ton treated Site 1 $120,000.00 $20.00 0.3 0.3 Site 2 $60,000.00 $30.00 0.5 0.2 Site 3 $40,000.00 $40.00 0.1 0.5 Total to remove: 80000 50000 Preliminary computations Maximal amount to be processed at each plant To remove 1 To remove 2 Site 1 266666.6666666667 166666.6666666667 266666.6666666667 Site 2 160000 250000 250000 Site 3 800000 100000 800000 Changing cells Build/don't build Amount to process* Pullutant 1 removed Pollutant 2 removed Site 1 1 100000 30000 30000 Site 2 1 100000 50000 20000 Site 3 0 0 0 0 Total removed: 80000 50000 Cost: 5180000 *Has to be 0 if corresponding binary variable is 0 Formulation: Variables: y1 - binary variable; 1 if station 1 is built, 0 otherwise y2 - binary variable; 1 if station 2 is built, 0 otherwise y3 - binary variable; 1 if station 3 is built, 0 otherwise x1 - amount of water (in tons) to process at station 1 x2 - amount of water (in tons) to process at station 2 x3 - amount of water (in tons) to process at station 3 Mathematical model: min 120,000y1+60,000y2+40,000y3+20x1+30x2+40x3 Cost of building stations and processing water subject to 0.3x1+0.5x2+0.1x3>=80,000 Sufficient amount of pollutant 1 removed 0.3x1+0.2x2+0.5x3>=50000 Sufficient amount of pollutant 2 removed x1<=266667y1 No water can be processed at plant 1 if it is not built (the coefficient of y1 on the RHS represents the maximum amount of water we may want to process there) x2<=250000y2 No water can be processed at plant 2 if it is not built x3<=800000y3 No water can be processed at plant 3 if it is not built x1,x2,x3>=0 y1,y2,y3 binary Investment This is a modification of the investment problem we previously considered. The modification is as follows: for each investement you put money in, you have to pay a brokerage fee of $50 at the time of purchase. Inputs Cash flows Investments Now Year 1 Year 2 Year 3 A -1 1.4 B -1 1.15 C -1 1.28 D -1 1.15 E -1 1.32 Upper bound on each investment 500 Amount available for investmemt 1000 Retun rate on annual CDs 0.06 Brokerage fee 50 Changing cells Effective Investment Amount Fee? Bound CD Amount Investment A 500 1 500 Now 450 Investment B 0 0 0 Year 1 0 Investment C 0 0 0 Year 2 0 Investment D 0 0 0 Investment E 427 1 500 Objective function Cash on hand in 3 years 1263.64 Constraints Upper bounds on investment specified within solver Cash balance constraints: Now Year 1 Year 2 Invested 1000 477 0 Available 1000 477 0 Formulation Variables xA amount invested in A yA xB amount invested in B yB Binary variables, representing xC amount invested in C yC whether you put money into corresponding investment, xD amount invested inD yD and hence whether you have to pay the fee. xE amount invested in E yE x0 amount put into a CD now (time 0) x1 amount put into a CD 1 year from now x2 amount put into a CD two years from now Mathematical model max 1.4xA+1.15xD+1.32xE+1.06x2 Cash on hand in year 3 subject to xA+xB+xC+x0+500yA+500yB+500yC=1000 Cash balance now (time 0) xE+x1+500yE=1.06x0+1.15xB Cash balance 1 year from now xD+x2+500yD=1.06x1+1.28xC Cash balance 2 years from now xA<=500yA xB<=500yB Upper bound on investments, xC<=500yC as well as forcing constraints xD<=500yD xE<=500yE All x-variables nonnegative All y-variabled binary

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