4 3 2 1 MTTF MTTF MTTF MTTF MTTF 1 1 1 1 1 MTTR MTTR - - PDF document

A Web Site receives 25 requests per seconds. A load balancer equally distributes incoming requests to 5 equal servers. The CPU service demand of a request is 20 msec, and the disk service demand is 50 msec. Assume that inter-arrival times of requests and service times are exponentially distributed. A server accepts at most 5 concurrent

• requests. The MTTF and the MTTR of a server are equal to 1000 hours and 10 hours,
• respectively. Calculate the average request response time, the throughput and the

percentage of requests rejected by the system.

• Data

sec 10 60 60 10 sec 1000 60 60 1000 5 server per requests concurrent Max sec 50 sec 20 /sec 25 5     = h = MTTR = h = MTTF = m = D m = D req = λ Server

DISK CPU

Number of faulty servers MTTF 5

MTTF 4 MTTF 1 MTTF 2

MTTF 3

MTTR 1 MTTR 1 MTTR 1 MTTR 1 MTTR 1

We can use the flow-in/ flow-out balance equations to calculate the probability

i

p that i server are working: MTTR p = MTTF p MTTR p = MTTF p MTTR p = MTTF p MTTR p = MTTF p MTTR p = MTTF p 1 1 1 2 1 3 1 4 1 5

5 4 4 3 3 2 2 1 1

1

5

= p

= i i



Performance of a single server Number of requests in a single server for scenarios with n<5 faulty servers: For each state, the service rate (throughput) (j) X0 depends on the number of requests in the server. Using MVA, we can calculate (j) X0 for each state:  N=1

             

DISK DISK CPU CPU DISK CPU DISK DISK CPU CPU

R X n R X n R R X m D R m D R ' 1 1 ' 1 1 1 ' 1 ' 1 1 sec 50 ) 1 ( ' sec 20 ) 1 ( '          

• n

 5  n  5  n  5  n  5  n  5 

) 1 ( X ) 2 ( X ) 3 ( X ) 4 ( X ) 5 ( X

 N=2

                 

DISK DISK CPU CPU DISK CPU DISK DISK DISK CPU CPU CPU

R X n R X n R R X n D R n D R ' 2 2 ' 2 2 ) 2 ( ' ) 2 ( ' 2 2 ) 1 ( 1 1 ) 2 ( ' ) 1 ( 1 1 ) 2 ( '            

•   

                  

DISK DISK CPU CPU DISK CPU DISK DISK DISK CPU CPU CPU

R X n R X n R R X n D R n D R ' 3 3 ' 3 3 3 ' 3 ' 3 3 ) 2 ( 1 2 ) 3 ( ' ) 2 ( 1 2 ) 3 ( '            

•  N=4

….

•  N=5

….

• We can use the flow-in/ flow-out balance equations to calculate

n

qi (probability that the server is in the state i assuming there are n faulty servers): ) ( X q = n λ q ) ( X q = n λ q ) ( X q = n λ q ) ( X q = n λ q ) ( X q = n λ q

n n n n n n n n n n

5 5 4 5 3 5 2 5 1 5

5 4 4 3 3 2 2 1

1

    

1

5

= q

= i n i



Hence, the throughput and the average response time of a single server when there are n faulty servers can be calculated as:

X(n) N(n) = R(n) q j = N(n) (j) X q = X(n)

= j j = j n j

 



5 1 5 1

Finally, the overall system throughput is:

) 5 (

4 1

n X(n) p = X

= n n





(for n=5 faulty servers no requests are served) and the average request response time:





4 1 5

) ( 1 1

= n n

n R p p = R

The number of rejected requests per second is 25 – X. The percentage of rejected requests is:

  100

25 X

• 25

 %