pleass show work solving provlem three, below are examples of another problem

Time Cost Example 3 Determine the project's cost and duration alternatives. Times (durations) are in weeks and costs in $Ks. Perform three cycles of the process: determine the crash design, the normal design, and the next two designs (a total of 4 project designs). Activity Code NTCT NCCC Prec. Design setup D 12 8 200 400 . Technology setup T 9 6 260 360 . Equipment setup E 10 9 100 110 . Validation V 4 3 900 1,000 D, T, E Main Rollout M 11 9 360 440 V Protocols P 10 6 120 200 V PM Problem (Excercise 4). Determine the project's cost and duration alternatives. Times (durations) in days, costs in $K. Perform three cycles of the process: determine the crash design, the normal design, and the next two designs (a total of 4 project designs). Code NT CC Prec. 100 B, C mo 000 B, C Provide your answer in the following format 1. Crash design duration and cost: (d, $m) 2. Normal design duration and cost: (d, $m) 3. Normal design critical path: X-Y-Z 4. What activity (ies) will be crashed in this iteration: Y 5. First crash duration and cost: (d, $m) 6. First crash critical path (s): X-Y-Z; R-C-V 7. What activity(ies) will be crashed in this iteration: Y, C 8. Second crash duration and cost: (d, $m) This is how your answer (what you will type into Canvas) should look like (this is Exercise 4 in the Problem set): 1.(36,$679) Time Cost Example 4 Determine the project's cost and duration alternatives. Perform three cycles of the process: determine the crash design, the normal design, and the next two designs (a total of 4 project designs). Times in days, costs in $K. Code Prec. Rate NT 19 c F G 3 Sum NC =345) SUM CC = (679) All crash B 110 22 D 22 26 D TO 10 I 16 36 Duration loto 49 30 1722 30 36 Cost 679 Normal | A 10 12 112 10 12 S=0 B 12 28 11329 S8=1 1221 B19 112 21 28 37 G 3740 29 30 273 38 41 SD = 1 So=1 JE/21 41 ( Duration 20/21 41) Soro CP: A-C-E IF 28 361 Cost = 345 Sp=5 Scoo 18 33 41 Crash C Alternatives Crash C Crash A 25 13 lost increases + 345 Now 352 new Duration 41-1 = 40 crashiNG no a B 112 28 5 2937_ JG37 40 40 112 20 0112 20 116/12 28 19/28 371 11210 12 Se=0 SD-O SAO F 20 40 2020 401 Sr: F129 36 1832 40) Sf = 4 Alternatives Crash A (in both) 25 Crash B&C 7.5+ 13 = 20.5 Crash DAC 24 +13=37 Crash Gec 40 +13=53 cert 345 Soo Duration 40 CP: AC-E A-B-D-G Crash Crash Crash C Crash A 25 40 Alternatives crash C 13 lost increases + 345 new Duration Now I 350l crashiNG 41-1 = 40 NO S AB 12 28 L JD 2037 G137.40|| AIO 12 116/12 28 928 371 112 10 12 Se=0 SD- Soo SAO 12 20 E 20 40 Duration 120/20 401 Spo 40 F2936 CP: A-CE 1832 40) A-B-D-G Sp=4 Alternatives Crash A (in both) 25 Crash Crash B&C 7.5+ 13 = 20.5 Crash DAC 24 +13=37 Crash Gec 40 +13=53 345 112 20 Crash 1832 (crashino ( 41-1=40 B/ 12 28 A0 2 16 12 28 G37 40 92837 7 9/28 37 11210 12 3 3740 So=0 SAO WC 12 20 E 20 40 18 12 20 120/20 40 Duration SEO 40 +28 36 CP. ACE A-B-D-G Sp=4 Alternatives Crash A (in both) 25 Crash Crash Bec7.5+13=20.5 Crash DEC 24 +13=37 Crash Gece 40+13=53 Cost increases to 345 82 ses to 13 C crashin c) SUMMARY 20.5 (crashing BSC 41 (normal) $ 345 New Cost = 378.5 40 (crash c) $ 359 New Duration=39 39 (cras BLC) 5379.5 36 (all crash) $679