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# Please give source code in C prograrmming........?

## Multi load balancing We will have 3 processors whose completion should be equal we have to show 20 -30 cases with the following limitations for different values We have to consider the following details Number of jobs - 4 to 7 Volume of jobs - 5 to 15 No of process – 3 Job units – 1 to 6 Number of jobs backlog 15 12 10 5 (volume of job) completion time B1=6 2 2 5 3(job units)T1 B2= 10 5 4 4 2T2 B3=15 4 3 51 T3 We have to write a program so that we will be getting the completion time for t1,t2,t3 First we have to solve for the 2 processors Solution First we consider 2 processor and strictly backlog should be in ascending order Number of jobs backlog 15 12 10 5 (volume of job) completion time B1=6 2 2 5 3(job units)T1 B2= 10 5 4 4 2T2 2/5=0.4 2/4=0.5 5/4=1.25 3/2=1.5 Since 1.5 is the largest ratio here we will split 5 as x and 5-x and now the constraint for x is x<=5 bcos if x>5 we get –ve value Number of jobs backlog 15 12 10 5 (volume of job) completion time B1=6 15* 2 12* 2 5 x 3(job units)T1 B2= 10 5 4 10*4 (5-x)*2T2 (* we assigned 15 to 2 since 2<5, 12*2 since 2<4) T1=6+15*2+12*2+(x*3) T2=10+10*4+(5-x)*2 ( T1=t2 so we get) 6+15*2+12*2+(x*3)= 10+4+10*4+(5-x)*2 6+30+24+3x=10+40+10-2x 60+3x=60-2x 5x=0 X=0 Substitute x in t1 and t2 equations We get T1=6+(15*2)+(12)*2+0*3 =6+30+24 =60 T2=10+10*4+(5-0)*2 =10+40+10 =60 Now we will consider for 3 processors as we have t1=t2 B3<t1=t2 Number of jobs backlog 15 12 10 5 (volume of job) completion time B1=6 2 2 5 3(job units)T1=60 B2= 10 5 4 4 2T2=60 B3=15 4 3 5 1 t3=b3=15 *[if 3rd processor is not activated t3=15 ie its backlog] For the first 2 columns take the ratios of 1st processor (B1)and 3rd processor(B3) ,for the last two columns take the ratio of 2nd processor(B2) and 3rd processor(B3) and also consider +ve load for 2nd processor(+ve load means the variables that have been assigned value) Number of jobs backlog 15 12 10 5 (volume of job) completion time B1=6 15*2 (12-x)*2 5 3(job units) T1=60 B2= 10 5 4 (10-y)*4 5* 2T2=60 B3=15 4 x* 3 y*5 1 t3=b3=15 60-2x=60-4y (since t1-x *p(i)=t2-y*q(k)) Thus y=0.5x T3+x*r(i)+y*r(k) <=t1-x*p(i) Or t2-y*q(k) Therefore 15+3x+5y<=60-2x 7.5x<=45 X<=6 Therefore y=3 The value of x & y are satisfied because x<12 & y<10 The complertion time for the 3rd processor Is T3=15+18+15=48 Substituting the new value of x and y in t1 and t2 we get T1=6+30+(12-6)*2=6+30+12=48 T2=10+(10-3)*4+5*2=10+28+10=48 We have t1=t2=t3=48

Looks a little involved for free online amatuer help. How about hiring a consultant or at least going to guru.com for the code?
Hmm ahaan! Comin' right up.

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