Fairness Guarantee provided by BSFQ

Definitions and Notations

Notations:

p_fⁱ = the i^th packet of flow f
A(p_fⁱ) = the arrival time of the i^th packet of flow f
L(p_fⁱ) = the departure (leave) time of the i^th packet of flow f
len(p_fⁱ) (also: l_fⁱ) = the length (#bytes) of the the i^th packet of flow f
r_f = the reserved rate (weight) of flow f
VT(t) = the system virtual clock (time) at real time t
Recall that:

Two important times:

VT(A(p_fⁱ)) = the system virtual clock (time) at the moment when packet p_fⁱ arrives
VT(L(p_fⁱ)) = the system virtual clock (time) at the moment when packet p_fⁱ leaves

V(p_fⁱ) = the virtual time stamp of the i^th packet of flow f
Recall that:

Recall: fairness

Fairness definition of Golestani: (See: click here )

A scheduling algorithm is fair if:

if f and g are continuously backlogged in [t₁,t₂], then: ∃ constant c: | w_f(t₁,t₂) - w_g(t₁,t₂) | < c i.e.: | W_f(t₁,t₂) W_g(t₁,t₂) | ∃ constant c: | --------- - --------- | < c | r_f r_g |

Showing that BSFQ guarantees fairness
- We will show the difference in normalized service received using the BSFQ scheduler by 2 different backlogged flows is bounded:
- But before we can do so, we need to prove some preliminary facts....

A relationship between V(p_fⁱ) and VT(L(p_fⁱ))

V(p_fⁱ) = the (virtual) time stamp of packet p_fⁱ
Packet p_fⁱ is inserted into some time bin [xΔ, (x+1)Δ):

Because packet p_fⁱ is in time bin [xΔ, (x+1)Δ), we conclude that:

xΔ ≤ V(p_fⁱ) < (x+1)Δ ........ (1)

When the packet p_fⁱ is about to leave the BSFQ server, the current bin must be time bin [xΔ, (x+1)Δ):
The virtual time at the moment that packet p_fⁱ departs is equal to:

Therefore:

xΔ ≤ V(p_fⁱ) < (x+1)Δ ...... (1) VT( L(p_fⁱ) ) = x Δ ...... (2)
==> VT( L(p_fⁱ) ) ≤ V(p_fⁱ) < VT( L(p_fⁱ) ) + Δ ...... (3)

Lemma 2: time stamp values when a flow is backlogged

When a flow f is backlogged, the time stamp V(p_fⁱ) of packet p_fⁱ is equal to:

len(p_fⁱ) V(p_fⁱ) = V(p_f^i-1) + -------- ........ (4) r_k

Proof:

If flow f is backlogged, then:
Therefore:
The virtual time function VT(t) of BSFQ is a monotonic increasing (step) function:
(I.e., the function values never decreases)
Property of a monotonic increasing function:

Therefore:

Because A(p_fⁱ) < L(p_f^i-1) , we have that:

Therefore:

VT( A(p_fⁱ) ) < VT( L(p_f^i-1) ) ......... (5)

VT( L(p_fⁱ) ) ≤ V(p_fⁱ) < VT( L(p_fⁱ) ) + Δ ...... (3)
Applied to packet p_f^i-1: VT( L(p_f^i-1) ) ≤ V(p_f^i-1) < VT( L(p_f^i-1) ) + Δ ...... (3a)

===> VT( A(p_fⁱ) ) < V(p_f^i-1) ......... (5A)

Therefore:

len(p_fⁱ) V(p_fⁱ) = max ( VT( A(p_fⁱ) ) , V(p_f^i-1) ) + --------- (use 5A) r_f len(p_fⁱ) = V(p_f^i-1) + --------- (Q.E.D.) r_f

Theorem 2: BSFQ is fair

Theorem 2:

If flows f and g are backlogged during time interval [t₁, t₂], then:

+- -+ | | | len_f^max len_f^max | | w_g(t₁, t₂) - w_g(t₁, t₂) | ≤ 2 × | ------ + ------ + Δ | | | | r_f r_g | +- -+

Where:

len_f^max = max packet length of packets of flow f
len_g^max = max packet length of packets of flow g

Proof:

The service W_f(t₁, t₂) to flow f in interval [t₁, t₂] is given by the shaded area:

A lower bound for the service W_f(t₁, t₂) is:

I.e.:

W_f(t₁, t₂) ≥ len(p_f^k1+1) + .... + len(p_f^k2)
len(p_f^k1+1) + .... + len(p_f^k2) <==> w_f(t₁, t₂) ≥ ------------------------------- .......... (6) r_f

An upper bound for the service W_f(t₁, t₂) is:

I.e.:

W_f(t₁, t₂) ≤ len(p_f^k1) + .... + len(p_f^k2+1)
len(p_f^k1) + .... + len(p_f^k2+1) <==> w_f(t₁, t₂) ≤ ------------------------------- .......... (7) r_f

We will first find an upper bound on w_f(t₁, t₂) :

From the timing in this figure:

and the fact that VT(t) is non-decreasing, we have that:

VT(t₁) ≤ V( L(p_f^k1) ) and: VT(t₂) ≥ V( L(p_f^k2) )

Therefore:

VT(t₂) - VT(t₁) ≥ V( L(p_f^k2) ) - V( L(p_f^k1) ) ....... (8)

Using Equation (3): (click here)

VT( L(p_fⁱ) ) ≤ V(p_fⁱ) < VT( L(p_fⁱ) ) + Δ ...... (3)
We have: V(p_f^k2) < VT( L(p_f^k2) ) + Δ <==> VT( L(p_f^k2) ) > V(p_f^k2) - Δ and: VT( L(p_f^k1) ) ≤ V(p_f^k1)

Apply the result to Equation (8):

VT(t₂) - VT(t₁) ≥ V( L(p_f^k2) ) - V( L(p_f^k1) ) ....... (8) ≥ [V(p_f^k2) - Δ] - V(p_f^k1) = V(p_f^k2) - V(p_f^k1) - Δ ....... (9)

Because flow f is backlogged during interval [t₁, t₂], we have:

len(p_f^k2) V(p_f^k2) = V(p_f^k2-1) + ------------ r_f len(p_f^k2-1) V(p_f^k2-1) = V(p_f^k2-2) + ------------ r_f .... len(p_f^k1+1) V(p_f^k1+1) = V(p_f^k1) + ------------ r_f
Therefore: [V(p_f^k2) - V(p_f^k2-1)] + [V(p_f^k2-1) - V(p_f^k2-2)] + ... + [V(p_f^k1+1) - V(p_f^k1)] len(p_f^k1+1) + ... + len(p_f^k2) = ----------------------------- r_f len(p_f^k1+1) + ... + len(p_f^k2) <==> V(p_f^k2) - V(p_f^k1) = ------------------------------- r_f

Apply the result to Equation (9):

VT(t₂) - VT(t₁) ≥ V(p_f^k2) - V(p_f^k1) - Δ ....... (9) len(p_f^k1+1) + ... + len(p_f^k2) = ------------------------------- - Δ r_f
len(p_f^k1+1) + ... + len(p_f^k2) <==> ------------------------------- ≤ VT(t₂) - VT(t₁) + Δ r_f

Apply the result to Equation (7) and we can finally formulate an upperbound for w_f(t₁, t₂) :

len(p_f^k1) + .... + len(p_f^k2+1) w_f(t₁, t₂) ≤ ------------------------------- .......... (7) r_f len(p_f^k1) + len(p_f^k1+1) + .... + len(p_f^k2) + len(p_f^k2+1) = ---------------------------------------------------- r_f len(p_f^k1) len(p_f^k2+1) = VT(t₂) - VT(t₁) + Δ + --------- + ----------- r_f r_f
Or: len(p_f^max) w_f(t₁, t₂) ≤ VT(t₂) - VT(t₁) + 2 --------- + Δ ..... (10) r_f

A lower bound for w_f(t₁, t₂) :

From the timing relationship:

and the fact that VT(t) is non-decreasing, we have:

V( Lp_f^k1-1) ) ≤ V(t₁) and: V(t₂) ≤ V( Lp_f^k2+1) )

Therefore:

V(t₂) - V(t₁) ≤ V( Lp_f^k2+1) ) - V( Lp_f^k1-1) ) ........ (10)

Using Equation (3): (click here)

VT( L(p_fⁱ) ) ≤ V(p_fⁱ) < VT( L(p_fⁱ) ) + Δ ...... (3)
We have: VT( L(p_f^k2+1) ) ≤ V(p_f^k2+1) and: V(p_f^k1-1) < VT( L(p_f^k1-1) ) + Δ <==> VT( L(p_f^k1-1) ) > V(p_f^k1-1) - Δ

Apply the result to Equation (10):

VT(t₂) - VT(t₁) < V( L(p_f^k2+1) ) - V( L(p_f^k1-1) ) ....... (10) < V(p_f^k2+1) - [V(p_f^k1-1) - Δ] = V(p_f^k2+1) - V(p_f^k1-1) + Δ ....... (11)

Because flow f is backlogged during interval [t₁, t₂]:

we have:

len(p_f^k2+1) V(p_f^k2+1) = V(p_f^k2) + ------------ r_f len(p_f^k2) V(p_f^k2) = V(p_f^k2-1) + ------------ r_f .... len(p_f^k1+1) V(p_f^k1+1) = V(p_f^k1) + ------------ r_f But: len(p_f^k1) V(p_f^k1) = max( VT(A(p_f^k1)), V(p_f^k1-1) ) + ------------ r_f
Therefore: [V(p_f^k2+1) - V(p_f^k2)] + [V(p_f^k2) - V(p_f^k2-1)] + ... + [V(p_f^k1+1) - V(p_f^k1)] len(p_f^k1) + ... + len(p_f^k2+1) = ----------------------------- r_f len(p_f^k1) + ... + len(p_f^k2+1) <==> V(p_f^k2+1) - V(p_f^k1) = ------------------------------- ....... (12) r_f

We must consider 2 cases to complete the proof:

V(p_f^k1-1) ≥ VT(A(p_f^k1))
V(p_f^k1-1) < VT(A(p_f^k1))

Case 1: V(p_f^k1-1) ≥ VT(A(p_f^k1))

Then we have:

len(p_f^k1) V(p_f^k1) = V(p_f^k1-1) + ------------ r_f

Apply the result to Equation (11):

VT(t₂) - VT(t₁) ≤ V(p_f^k2+1) - V(p_f^k1-1) + Δ ....... (11) = V(p_f^k2+1) - V(p_f^k1-1) + V(p_f^k1) - V(p_f^k1) + Δ = [V(p_f^k2+1) - V(p_f^k1)] + [V(p_f^k1) - V(p_f^k1-1)] + Δ

len(p_f^k1+1) + ... + len(p_f^k2+1) V(p_f^k2+1) - V(p_f^k1) = ------------------------------- ....... (12) r_f

len(p_f^k1+1) + ... + len(p_f^k2+1) ==> VT(t₂) - VT(t₁) ≤ ------------------------------- + [V(p_f^k1) - V(p_f^k1-1)] + Δ r_f

len(p_f^k1) V(p_f^k1) = V(p_f^k1-1) + ------------ r_f

len(p_f^k1+1) + ... + len(p_f^k2+1) len(p_f^k1) ==> VT(t₂) - VT(t₁) ≤ ------------------------------- + ---------- + Δ r_f r_f len(p_f^k1) + ... + len(p_f^k2+1) = ------------------------------ + Δ ...... (13) r_f

Apply the result to Equation (6) and we can finally give an lower bound for Case 1:

len(p_f^k1+1) + .... + len(p_f^k2) w_f(t₁, t₂) ≥ ------------------------------- .......... (6) r_f len(p_f^k1) + .... + len(p_f^k2+1) len(p_f^k1) len(p_f^k2+1) = ------------------------------- - ---------- - ---------- r_f r_f r_f

Use Equation (13)

len(p_f^k1) len(p_f^k2+1) w_f(t₁, t₂) ≥ VT(t₂) - VT(t₁) - Δ - ---------- - ---------- r_f r_f len(p_f^max) ≥ VT(t₂) - VT(t₁) - 2 ---------- - &Delta r_f

Case 2 is similar, so I will spare you the proof :-)

The proof is completed as follows:

We have found a lower bound and an upper bound for w_f(t₁, t₂) :

len(p_f^max) len(p_f^max) VT(t₂) - VT(t₁) - 2 ---------- - Δ ≤ w_f(t₁, t₂) ≤ VT(t₂) - VT(t₁) + 2 ---------- + Δ r_f r_f

The lower bound and an upper bound for w_g(t₁, t₂) of flow g is then:

len(p_g^max) len(p_g^max) VT(t₂) - VT(t₁) - 2 ---------- - Δ ≤ w_g(t₁, t₂) ≤ VT(t₂) - VT(t₁) + 2 ---------- + Δ r_g r_g

Therefore:

len(p_f^max) len(p_g^max) w_f(t₁, t₂) - w_g(t₁, t₂) ≤ 2 ---------- + Δ + 2 ---------- + Δ r_f r_g and: len(p_g^max) len(p_f^max) w_g(t₁, t₂) - w_f(t₁, t₂) ≤ 2 ---------- + Δ + 2 ---------- + Δ r_g r_f

Thus: