[CST-2] CSM 1993p9q3 & 1994p7q3

[Phebe Mann]

To: Tim Harris
>From : Phebe

93/9/3

work station (CPU)	 40ms	20 #
fileserver (CPU) 	  6ms	4 #
fileserver (disc)	 10ms	2 * 4 = 8 #

Total no. of devices = 20 + 4 + 8 = 32

Di = Vi* Si

Vi = no. of visits
Si = service time per visit

For CPU
6 ms for 4 servers

6/4 = 1.2 ms per fileserver

For disk
10 ms for 8 disks

10/8 = 1.25 ms per disk

For workstation
40 ms for 20 workstations

40/20 = 2 ms per workstation

Bottleneck is at 2 ms per workstation

Given
For disks and filserver
U(highest) : U(lowest) <= 3/2

1.25 ms * 3/2 = 1.875 ms
1.2 ms * 3/2 = 1.8 ms

workstation usage is balanced

What else ? Something missing here !
Any suggestions ?

U = N / (N + K - 1 )

where
U = utilisation
N = number of customers

U = XS, and N = XR

XS = XR / (XR = K - 1)

XSR + SK - S = R
(XS - 1) R = S (1-K)

R = (S ( 1- K))/(XS - 1 )

R = (M/X) - Z
where
R is the response time
X = throughput
M = number of customers
Z = sleep time

Thanks

Phebe

On Mon, 4 Jun 2001, Hanna Wallach wrote:

> Hi,
>
> I'm currently stuck on CSM questions 1993p9q3 part A, and 1994p7q3 parts b
> and c. Has anyone done these questions, and if so could you explain to me
> how to do them?
>
> Thanks,
>
> --
> Hanna                                                     \   /
>                                                        <:><|||>==-
>                                                           /   \
>
>
>
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