An investment company is considering to invest in 6 projects. The decision to invest in each project i (i=1,2,3,4,5,6) is represented by the binary variable Xi, which takes a value of 1 if the company decides to invest in project i, and a value of 0 otherwise. Write the appropriate constraints to be involved in the mathematical model whose solution gives the best investment portfolio, which will explain the following situations: a) If the company wants to invest in project 1, it must also invest in projects 2,3 and 4. + X3 + X4 b) The company cannot invest in projects 4 and 5 at the same time. + X5 c) The company should invest in at least 2 projects. X1 + X2 + X3 +X4 +X5 +X6 d) If the company wants to invest in project 2 or 4, it must also invest in project 1 or 6. + X2 + X6) The 4 workers of a company, Adam, Brian, Cody and Daniel, are going to be assigned to 4 different jobs. The cost of each assignment is shown in the following Table. Job I Job 2 Job 3Job 4 Adam 50 100 160 80 Brian 60 50 150 70 Cody 60 60 70 100 Daniel70 80 70 100 How should the four workers be assigned to the 4 jobs to minimize the total cost? Use Hungarian method to find the best matches. You are required to fill in the blanks below. If there are multiple answers to some questions, just pick either one of them. First step: row reduction: Job I Job 2 Job 3 Job 4 Adam Brian Cody Daniel Second step: column reduction Job Job 2 Job 3 Job 4 Adam Brian Cody Daniel and the optimal assignment is: Adam is assigned to Brian is assigned to Cody is assigned to Daniel is assigned to The activities that are necessary to build an equipment, their expected completion times and the variance of completion times, and total costs are listed as in the following table. ACTIVITIES IMMEDIATE EXPECTED VARIANCE COST ($) CRASH TIME CRASH COST PREDECESSORS ACTIVITY TIME (WEEKS) ($) (WEEKS) f A 15 2000 3 4000 B 3 12 4500 2 6500 C AB 3 11 15000 12 7000 D A 16 6.5 3000 15 4500 C D 14 1.5 14000 3 15000 E 2 3000 3500 14 1000 3 2500 For some questions, there might be multiple alternatives; if that is the case, just select one of them. a) Given the picture below, which node corresponds to which activity? 7 2 3 finish start 5 Node 1 corresponds to Node 2 corresponds to Node 3 corresponds to Node 4 corresponds to Node 5 corresponds to Node 6 corresponds to Node 7 corresponds to b) Compute the ES, EF, LS and LF of the activities. What is the project completion time? Which activities form the critical path? Activity A: ES = EF= LS = LF = Activity B: ES = EF = LS = LF = Activity C: ES = EF = LS = LF = Activity D: ES = EF = LS = LF = Activity E: ES = EF = LS = LF = Activity F: ES = EF = LS = LF = Activity G: ES = EF= LS = LF = Midterm 2 Makeup (page 3 of 4) Project completion time: Critical path activities: c) Suppose we want to complete this project in 18 weeks. Compute the unit crash cost (i.e. cost to decrease the duration of activity by 1 week) for the activities below. Which activity should be selected to achieve this target? Activity A unit crash cost: Activity B unit crash cost: Activity C unit crash cost: Activity D unit crash cost: Activity Eunit crash cost: Activity Funit crash cost: Activity G unit crash cost: Select as the activity to be crashed. d) What is the probability that the project can be completed within 20 weeks? (Hint: Project variance = sum(variances of the activities on the critical path), Z = (X-u)/o).) Project variance = Z value = (Express in 2 decimal units) Probability of completing the project within 20 weeks: % (Express as an integer value) USE DOT("") FOR DECIMAL SEPARATOR. Probability TABLE A: STANDARD NORMAL PROBABILITIES (CONTINUED) ,00 .01 .02 .03 .04 .05 .06 .07 .08 09 .5871 0.0 .5000 .5398 0.2 .5793 0.3 .6179 0.4 .6554 0.5 .6915 0.6 .7257 0.7 .7580 0.8 .7881 0.9 ,8159 1.0 .8413 1.1 -8643 1.2 .8849 .9032 1.4 .9192 1.5 .9332 1.6 .9452 1.7.9554 1.8 .9641 1.9 9713 .9772 2.1 .9821 2.2 .9861 2.3 .9893 2.4 .9918 2.5 .9938 2.6 .9953 2.7 .9965 2.8 .9974 2.9 .9981 3.0 .9987 3.1 9990 3.2 9993 3.3 .9995 3.4 .9997 .S040 .5438 .5832 .6217 .6591 .6950 .7291 .7611 7910 .8186 .8438 .8665 .8869 .9049 .9207 .9345 .9463 9564 9649 9719 .9778 9826 .9864 9896 9920 .9940 .9955 9966 9975 9982 .9987 9991 .9993 .9995 9997 5080 5120 .5478 .5517 .5910 .6253 .6293 .6628 .6664 .6985 .7019 .7324 .7357 .7642 .7673 .7939 .7967 .8212 8238 .8461 .8485 .8686 .8708 .8888 .8907 9066 .9082 .9222 .9236 .9357 9370 .9474 9484 9573 .9582 .9656 .9664 .9726 .9732 .9783 9788 .9830 .9834 9868 .9871 .9898 9901 .9922 9925 .9941 9943 .9956 .9957 .9967 .9968 9976 9977 .9982 .9983 9987 9988 .9991 .9991 .9994 .9994 .9995 -9996 .9997 .9997 .5160 .5557 5948 .6331 6700 .7054 .7389 .7704 .7995 .8264 8508 8729 .8925 .9099 .9251 .9382 .9495 .9591 .9671 .9738 .9793 .9838 .9875 .9904 .9927 .9945 .9959 .9969 .9977 .9984 .9988 .9992 .9994 9996 .9997 .5199 .5239 .3596 .5636 .5987 .6026 .6368 .6406 .6736 .6772 7088 .7123 .7422 .7454 .7734 .7764 .8023 .805! .8289 .8315 .8531 .8554 .8749 .8770 .8944 .8962 9115 9131 .9265 .9279 .9394 9406 9505 .9515 .9599 .9608 .9678 .9686 9744 .9750 9798 .9803 .9842 9846 .9878 .9881 .9906 .9909 9929 9931 .9946 .9948 .9960 9961 .9970 .9971 .9978 .9979 .9984 9985 .9989 9989 9992 .9992 .9994 .9994 .9996 9996 9997 .9997 .3279 $319 3359 5675 .5714 5753 .6064 6103 .6341 .6443 .6480.6517 .6808 .6844 .6879 7157 ,7190 7224 .7486 .7517 .7549 .7794 .7823 .7852 8078 .8106 8133 .8340 .8365 .8389 .8577 .8599 .8621 .8790 .8810 .8830 .8980 .8997 9015 9147 9162 9177 .9292 .9306 9319 9418 .9429 944) 9925 .9535 9545 .9616 9625 .9633 .9693 .9699 9706 .9756 .9761 9767 9808 .9812 9817 9850 .9854 9857 .9884 .9887 9890 9911 .9913 .9916 .9932 9934 9936 9949 .9951 .9952 9962 9963 9964 .9972 9973 9974 .9979 .9980 9981 .9985 .9986 9986 9989 .9990 .9990 9992 9993 9993 .9995 9995 9995 9996 9996 .9997 .9997 .9997 .9998 2.0 Beverly has decided to lease a hybrid car for as long as 3 years to save on gas expenses and to do her part to keep the environment clean. There are 3 options that she can choose from, which have different costs based on how often she uses the car: Low usage Medium usage High usage Car A $5,000 $8,000 $10,000 Car B $6,500 $7,500 $9,000 Car C$7,000 $7,000 $7,000 a) Use Minimin (optimistic) and Minimax (pessimistic) approaches to find out the car Beverly should select. Minimin: comparing cost for A, cost for B and cost for C, Beverly should select Minimax: comparing cost for A, cost for B and cost for C, Beverly should select b) Use Minimax Regret criterion to find out which car should be selected. (Hint: the regrets are extra costs Beverly needs to pay in each situation.) Regret table: Low usage Medium usage High usage Car A Car B Carc max regret of Car A: max regret of Car B: max regret of Car C: Beverly should select (if there are two correct answers, just select either one of them.) From now on, assume that the probability that Beverly uses the car with low frequency is 0.4, with medium frequency is 0.3, and with high frequency is 0.3. c) What decision would Beverly make if she wanted to minimize her expected cost? Expected cost(Car A) = Expected cost (Car B) = Expected cost (Car C) = Beverly should select Compute the expected value of perfect information (EVPI = min EMV - EvwPl). min EMV = empt.php?attempt=229258&cmid=91640&page=3# Midterm 2 Makeup (page 4 of 4) EVWPI = 0.4* +0.3* +0.3 EVPI =