**3.4 Optimized LCC**

Obviously, the *LCC* may span many different networks in *DNE*, so *LCC* can be optimized through the network skip distance, network bandwidth, and the resource demand vector of *T*. Suppose that *NSET*= {*Ni* (*bwi*)| 1≤*i*≤*m*} is the network set in *LCC. bwi* indicates bandwidth of *Ni*, and *m* indicates the numbers of network. *RDV* (*w1, w2, w3, w4*) indicates the resource demand vector of task *T*. The process is described as follows:

1. Construct network matrix *NM*= (*sij*) (1≤*i*≤m, 1≤*j*≤*m*) that is composed of the factors of the network skip distance and bandwidth. The construction method for *NM* is as follows:

```
sii=1;
```
140 Grid Computing – Technology and Applications, Widespread Coverage and New Horizons

1. *MR*= (*rij*) (1≤*i*≤*n*, 1≤*j*≤*4*) is the resource matrix in *DNE*, where *n* is the numbers of all

5. Divide the cache nodes of *PCS* into several equivalence class *LCC1*∪*LCC2*∪…*LCCe* by

Suppose that *LCCS* = *LCC1*∪*LCC2*∪…∪*LCCe* is the set of all logical computer clusters which

Calculate total resource vector (*arCpu*, *armem*, *ardisk*, *arnet*) of all cache nodes of *S*, it is as follows:

*CN RSV r* ; *armem*<sup>=</sup> . . *mem*

*CN RSV r* ; *arnet*<sup>=</sup> . . *net*

*y1=arCpu/(arCpu +armem +ardisk+arnet); y2=armem/(arCpu +armem +ardisk+arnet);* 

*y3=ardisk/(arCpu +armem +ardisk+arnet); y4=arnet/(arCpu +armem +ardisk+arnet);* 

*CN in S*

*CN in S*

*CN RSV r* ;

*CN RSV r* ;

<1, and *x1*+

(*x1*| *ri1* – *rj1*|+*x2*| *ri2* – *rj2*|+*x3*| *ri*3 – *rj3*|+*x4*| *ri4* – *rj4*|),when *i*≠*j*, and 0<

*FT*=*FM* ⊙ *FM*; //⊙ is the operation theorem to take the maximum and minimum

6. Choose a *L*CC for T according to its resource demand vector by algorithm3.2.

are built through algorithm3.1, and *T* is a parallel cache mode task. **Algorithm 3.2. Choose LCC method for parallel cache mode task.**  1. Get the resource demand vector RDV (w1, w2, w3, w4) of T;

> {*arCpu*= . . *cpu CN in S*

*ardisk*= . . *disk CN in S*

**Algorithm 3.1 Partition method for LCC.** 

3. build the fuzzy equivalence matrix;

4. Calculate the *c*-cut matrix *FMc*;

do

Construct the vector Y(y1,y2,y3,y4); e=|y1-w1|+|y2-w2|+|y3-w3|+|y4-w4|;

3. Choose LCC as the logical computer cluster for T;

if mine<e then { LCC=S; mine=e };

LCCS=LCCS-{S}; };// end while

4. End.

**3.3 Choose LCC for parallel cache mode task** 

{ Get *S*∈*LCCS* ; /\* *S* is a logical computer cluster \*/

*fij*=1- 

Repeat do

FM=FT; End do;

*FMc* ;

mine= ;

};

2. While *LCCS*≠

*x2*+ *x3*+ *x4*=1, *xk*>0 (1≤*k*≤4);

If *FT*=*FM* then goto (4);

cache nodes; *T* is a parallel cache mode task.

2. Construct the fuzzy matrix *FM*= (*fij*) (1≤*i*≤*n*, 1≤*j*≤*n*), where *fii*=1;

*sij*=1- (*distance*( *Ni* – *Nj*)+ | *bwi* – *bwj* |*)*,when *i*≠*j*, and 0<<1;

Where,

When the network skip distance between *Ni* and *Nj* is 1, distance (*Ni*, *Nj*) =1;

When the network skip distance between *Ni* and *Nj* is 2, distance (*Ni*, *Nj*) =3;

When the network skip distance between *Ni* and *Nj* > 2, distance (*Ni*, *Nj*) =6;

When the bandwidth of *Ni* is 10MB, *bwi*=1;

When the bandwidth of *Ni* is 100MB, *bwi*=3;

When the bandwidth of *Ni* is 1000MB, *bwi*=6;

If *Ni.bwi*=1 and (*Nj.bwj*=3 or *Nj.bwj*=6) then distance(*Ni*, *Nj)*=6

If *Ni.bwi*=3 and (*Nj.bwj*=1 or *Nj.bwj*=6) then distance(*Ni*, *Nj)*=6

If *Ni.bwi*=6 and (*Nj.bwj*=1 or *Nj.bw*j=3) then distance(*Ni*, *Nj)*=6

2. Build the fuzzy equivalence matrix:

Repeat do

*FT*=*NM* ⊙ *NM*; //⊙ is the operation theorem to take the maximum and minimum If *FT*=*NM* then goto (4);

NM=FT;

End do;

