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510
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

U
thanku soo much sir..bs appe trust krke chal r only apko hi follow krna hai ab or ibps nikalna h ...is br
network diagram is confusing. how activities B and F time is taken?
As
how u calculate probability
pls elaborate more..its like ur trying to teach urself...
RP
sir how to find probablility for any value of Z
sir plz explain how you have calculated probability!
1. COURSE : INDUSTRIAL ENGINEERING LESSON: PROGRAM EVALUATION AND REVIEW TECHNIQUE-PART-1)

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: https://unacademy.in/user/harsh.t 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

9. THANK YOU