Question

2) Assuming a server takes 6 msec to get a request for work, dispatch it, and do the rest of the necessary processing if the data needed are in a cache. If a disk operation is needed, as is the case 5% of the time, an additional 100 msec is required, during which time the thread sleeps. [10 points) (a) How many requests/sec can the server handle if it is single threaded? (b) How many requests/sec can the server handle if it is multithreaded?

Answer 1

Given:-

Time taken by server to get a request (cache hit)                              = 6 msec
Time taken by server to get request when cache miss occurs        = 100 msec

Percentage of time cache miss occurs                                    = 5% = 5/100
Percentage of time cache hit occurs                                       = 100-5 = 95% = 95/100

a) Single – Threaded Case

In single-threaded case, cache hit and cache miss both occur. So we need to take average of both the cases.

Weighted Average =

(Percentage of time cache hit occurs) * (Time taken by server to get a request (cache hit) )

+

(Percentage of time cache miss occurs) * (Time taken by server to get request when cache miss occurs)

Therefore, from given values,

Weighted Average = ( 5/100 )*( 6 ) + ( 95/100 )*( 100 )
= (30/100) + (95)
    = 95.3

Mean Request Time = 95.3 msec

Number of tasks server can handle per sec
= 1 second / Mean Request Time

= 1 second / 95.3 msec

= 1 second / 0.0953 second                        [Since, 1 sec = 1000 msec]

= 10.5

Therefore Single-Threaded server can handle 10.5 Requests/Second.

b) Multi-threaded Case

In multi-threaded server, there is no waiting for disk so all the waiting is skipped, i.e., cache miss doesn’t occur.

So every request can be processed in 6 msec.

Number of tasks server can handle per sec:-
= 1 second / Request Time

= 1 second / 6 msec                                       [Since, 1 sec = 1000 msec]

= 1 second / 0.006 second

= 166.66

Therefore Multi-Threaded server can handle 166.6 Requests/Second.

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