[CST-2] CSM 1993p9q3 & 1994p7q3

[Phebe Mann]

To : Tim Harris
>From : Phebe

94/7/3

(b)
all service rates are equal - balanced system

p1 = p2 + p3 =1

mu = mu(xa) = mu(xb) = mu(ya) = mu(yb)

mu(p1) = mu(p2)=mu(p3)
p1 = p2 = p3 = 1/3
There is no distinction between the 2 customers

pA = p(A is busy) = p2 + p3 = 2/3
pB = p(B is busy) = p1 + p2 = 2/3

states : 0,2 -> 1,1 -> 2,0
	 0,2 <- 1,1 <- 2,0
	 p1	p2	p3

all are mu ->
N =2 K=3

U = N/(N+K-1) = 2/3


(c)
states : 0,3 -> 1,2 -> 2,1 -> 3,0
	 0,3 <- 1,2 <- 2,1 <- 3,0
	 p1	p2	p3	p4

p1 = p2 = p3 = p4 = 1/4
pA = p2 + p3 + p4 = 3/4 = 1 - p1 = 1 - 1/4

N customers, p1 = 1/(N+1)
pA = 1 - 1/(N+1) = N/(N+1)

U = N/(N+K-1) = N/(N+1)


HTH
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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