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Program Evaluation and Review Technique - (Part - I) (in Hindi)
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This lesson throws light on program evaluation and review technique.

Harshit Aggarwal
Cleared UPSC ESE twice with Rank 63 and 90 in mechanical engg. Got 99 percentile in GATE. Cracked ONGC, BHEL,ISRO, SAIL, GAIL successfully

Unacademy user
Jagat Chaudhary
a year ago
thankyou girdhar ji. stay connected and keep learning. and Please rate this course :)
Jagat Chaudhary
a year ago
thankyou giridhar for donating blue hat. and I am creating lessons on Probability, Binomial theorem and Permutation combination from this week. On 1st september , you will get the notification for starting for these courses.. and we will complete them in next 2 weeks. keep watching and keep learning
network diagram is confusing. how activities B and F time is taken?
how u calculate probability

  2. ABOUT ME Graduated from NIT Nagpur in 2008 Cleared Engineering Services Examination (ESE-UPSC) Exam & Got the offer letter from most of the Maharatna and Navratna Companies * Cleared GATE Exam Rate, Review, Recommend, Share Follow me on Unacademy at: aggarwal

  3. PROGRAM EVALUATION AND REVIEW TECHNIQUE optinustk. time then essinislic tiue t. 6 e expecteel value achuih duretia 6 Oe.

  4. PROGRAM EVALUATION AND REVIEW TECHNIQUE Standand Nomal Vaiat 2-Ts-Te Te Expected Time of f,oject 0 Stondere Davia tion yu tee fios sha to end

  5. QUESTION Activity Immediate Optimistic Most Likely Pessimistic Predecessor Time (To) Time (Tm) Time (Tp) A project consist of 8 activities with the following relevant information. i) Determine the critical path and compute the expected project completion time. ii) Determine the probability of completing the project in 50 days. ii If a company agrees to complete the project in 50 days failing which it would pay Rs 200/day. What is the probability that a penalty but not exceeding Rs 1000 will be 20 25 26 al D,F G,H

  6. SOLUTION 14 32. E, 18 H2 2. 31 Kie 47 A-BE-47

  7. SOLUTION Activity Time Taken Te = (To + 4 Tm + Tp )/ 6 = (Tp-T0)/6 0 A (I-2) 4 6 B (2-3) C (2-4) D (3-6) E (3-5) 10 F (4-6) G (5-7) H (6-7) 1 (7-8) 4/6 10/6 8/6 10/6 16/6 8/6 8/6 2/6 2/6 20 20 18 10 25 26 7 7

  8. SOLUTION Expected ?mj.ct condekon tuue=47. 3.49 3.41 .19 Net Psb (8ses) o.otss (.959-.30t) Z- SS-r229 3.4a Prob (2.29): 0.989