2copies of a code module are to be executed in parallel on identical processors organized into two communicating clusters a and b. cluster a contains 8 processors and cluster b contains 4 processors. how many sets of processors can be used if cluster a has 2 failed processors and cluster b has 4 failed processors?

Number of sets of processors can be used is given by

A = {A1,A2,A3,A4,A5,A6} and B = Ф

Step-by-step explanation:

Since we have given that

Number of processors in cluster A = 8

A = {A1,A2,A3,A4,A5,A6,A7,A8}

Number of processors in cluster B = 4

B= {B1,B2,B3,B4}

Number of processors failed in cluster A = 2

Set A becomes

A = {A1,A2,A3,A4,A5,A6}

Number of processors failed in cluster B = 4

B = Ф

Hence, number of sets of processors can be used is given by

A = {A1,A2,A3,A4,A5,A6} and B = Ф

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

3

step-by-step explanation: