1. Z-Transforms

2. Definition
The z-transform of a discrete function p(i), i = 0, 1, 2, … is defined as
Gp(z) = Σ{i = 0 to }p(i)zi
Examples:
X = Binomial(n,p)
X = Geometric(p)

3. Properties If X is a r.v. with z-transform Gp(z), then
G’p(z)|z=1 = E[X]
G’’p(z)|z=1 = E[X(X-1)]
G(n)p(z)|z=1 = E[X(X-1) (X-2)…(X-n+1)]
Let X and Y be two discrete, independent r.v.’s with z-transforms X(z) and Y(z), then the z-transform of Z=X+Y is
Z(z) = X(z)Y(z)
If X, A, and B are discrete r.v.’s where X is equal to A (resp. B) with probability p (resp. 1-p), then
X(z) = pA(z) + (1-p)B(z)

4. Response Time in Geom/Geom/1
Consider a Geom(p)/Geom(q)/1 queue with p<q
We know that i =  i(1-), where  = p(1-q)/q(1-p)
We want the distribution of the system response time, i.e., P(Tk)
Consider job that arrives to find k jobs (with probability k)
Tk = S1 + S2 +…+Sk + Sk+1
So that Tk(z) = (zq/(1-z(1-q))k+1 – service time is geometric with parameter q
This implies T(z) = Σ{k=0 to } k(1-)(zq /(1-z(1-q))k+1

5. Response Time in Geom/Geom/1
This is a Geom((1-)q) distribution

6. Number of Arrivals in a Service Time
Consider a Geom(p)/Geom(q)/1 queue with p<q
What is the distribution (z-transform) of the number of arrivals during a service time
Let aj = P{j arrivals in service time S}, so that ÂS(z) = Σ{j=0 to }ajzj is the corresponding z-transform
ÂS(z) = Σ{j=0 to }(Σ{k=1 to }P{AS=j | S=k}(1-q)k-1q)zj
= Σ{j=0 to }(Σ{k=1 to }choose(k,j)pj(1-p)k-j(1-q)k-1q)zj
= Σ{k=1 to }(1-q)k-1q(Σ{j=0 to k}choose(k,j)(zp)j(1-p)k-j)
= Σ{k=1 to } (1-q)k-1q (zp+(1-p))k
= q (zp+(1-p))Σ{k=0 to }[(1-q)k(zp+(1-p))k]
= q (zp+(1-p))/1-[(1-q)(zp+(1-p))]
= Ŝ(zp+(1-p)) – recall that S is geometrically distributed

7. General Chain Analysis
Let a chain be characterized by fi+2=bfi+1+afi
We use z-transforms to find fi
fi+2 zi+2 =bfi+1zi+2 +afi zi+2
Σ{i=0 to }fi+2 zi+2 =bΣ{i=0 to } fi+1zi+2 +aΣ{i=0 to } fi zi+2
 F(z)-f1z-f0 =bz(F(z)-f0)+az2F(z)
 F(z)=(f0+z(f1-bf0)/(1-bz-az2)
We rewrite F(z) via partial fractions
F(z)=N(z)/D(z)=A/h(z)+B/g(z), where h(z)=(1-z/r0), g(z)=(1-z/r1), A=f0-B, B=(r0f0+(f1-f0b)r0r1)/(r0-r1), and r0 and r1 are the roots of D(z)=1-bz-az2

8. General Chain Analysis
When D(z)=az2+bz+1, then
D(z) =(1-z/r0)(1-z/r1)
where r0 and r1 are the (real) roots of D(z)
We can then expand A/(1-z/r0)+B/(1-z/r1) into a series and identify individual fi terms
A/(1-z/r0)=AΣ{i=0 to }(z/r0)i
B/(1-z/r1)=BΣ{i=0 to }(z/r1)i
So that fi = A/r0i+B/r1i