-- Feb 13 In-Class Exercise Thread

((resource:20190218_100911.jpg|Resource Description for 20190218_100911.jpg)) Incomplete steps: 1, 9, 10 Complete steps: 2, 3, 4, 5, 6, 7, 8 10 total steps needed.

-- Feb 13 In-Class Exercise Thread

1. T_1 = 17, T_inf = 8, P=2 T_p = 17/2 + 8 = 16.5 2. ((resource:InkedPFibDag_LI.jpg|Resource Description for InkedPFibDag_LI.jpg)) Incomplete: 1, 7, 9, 10, 11 Complete: 2, 3, 4, 5, 6, 8

-- Feb 13 In-Class Exercise Thread

((resource:IMG_1903.jpeg|Resource Description for IMG_1903.jpeg))

-- Feb 13 In-Class Exercise Thread

 Tp <= T1/p + Tinf
 p=2, T1 = 17, Tinf = 8
 T2 <= 17/2 + 8 = 16.5

Tp <= T1/p + Tinf p=2, T1 = 17, Tinf = 8 T2 <= 17/2 + 8 = 16.5 ((resource:exercise.png|Resource Description for exercise.png)) 10 steps total. Steps 1, 9, and 10 are incomplete.

-- Feb 13 In-Class Exercise Thread

Tp >= Ti / p + Tinf There are 17 nodes of work and a max depth of 8. Tp >= 17 / 2 + 8 Tp >= 16.5 Incomplete steps 1,9,10 Complete Steps 2,3,4,5,6,7,8 ((resource:inclass.png|Resource Description for inclass.png))

-- Feb 13 In-Class Exercise Thread

1. T_1 = 17, T_inf = 8, P=2 T_p = 17/2 + 8 = 33/2 2. ((resource:1881807400.jpg|Resource Description for 1881807400.jpg))

-- Feb 13 In-Class Exercise Thread

((resource:Sketch (2).png|Resource Description for Sketch (2).png))

-- Feb 13 In-Class Exercise Thread

''What would be the maximum run time according to Brent/Blumofe Leirsons's theorem?'' T1 = 17, Tinf = 8, Tmax <= T1/Tinf = 17/2 + 8 = 33/2 ((resource:PFibDag_marked.png|Resource Description for PFibDag_marked.png)) Complete Steps are 2, 3, 4, 5, 6, 7, 8; Incomplete Steps are 1, 9, 10. Total steps made is 10

-- Feb 13 In-Class Exercise Thread

1) Since T1 = 17 and T_infinity = 8, P = 2, per Brent Theorem, max run time = T1/p + T_infinity = 17/2 + 8 = 16.5; 2) PFA ((resource:inClassExecise_Fibonacci.pdf|Resource Description for inClassExecise_Fibonacci.pdf)) 3) Total # of steps = 10. Incomplete steps are 1, 9, and 10.