Upper Bound

In this section, we determine an upper bound onZ.

Definition 3.5. Lettnbe the end of the latest non-busy non-displacing interval fordbeforetd, if

The following two lemmas have been proved previously for both GEDF [41] and GFIFO [78] for ordinary sporadic task systems without self-suspensions. Note that the value ofLAG(d, td, S) + B(D, td, S)depends only on allocations in the PS scheduleP S and allocations to jobs ind∪D

in the actual scheduleS by timetd. The PS schedule is not impacted by self-suspensions. Also,

Property (P) alone is sufficient for determining how much work any job ind∪Dother thanτl,j

completes beforetd. For these reasons, Lemmas 3.1 and 3.2 continue to hold for task systems with

self-suspensions.

Lemma 3.1. LAG(d, td, S)≤LAG(d, tn, S) +Pτk∈DHδk(1−uk), wheret∈[0, td].

Proof. By (2.5), we have

LAG(d, td, S) ≤ LAG(d, tn, S) +A(d, tn, td, P S)

−A(d, tn, td, S). (3.2)

We split[tn, td)intoznon-overlapping intervals[tpi, tqi),1≤i≤z, such thattn=tp1, tqi−1 =tpi,

andtqz =td. Each interval[tpi, tqi)is either busy or non-busy displacing ford, by the selection of tn. We assume that the intervals are defined so that for each non-busy displacing interval[tpi, tqi), if

a task inDHexecutes in[tpi, tqi)then it executes continuously throughout[tpi, tqi); we letαidenote

the set of such tasks.

We now bound the difference between the work performed in the PS schedule and the GSA scheduleSacross each of these intervals[tpi, tqi). The sum of these bounds will give us a bound on

the total allocation difference throughout[tn, td). Depending on the nature of the interval[tpi, tqi),

two cases are possible.

Case 1. [tpi, tqi) is busy. Since in S all processors are occupied by jobs in d, we have A(d, tpi, tqi, P S)−A(d, tpi, tqi, S)≤Usum(tqi, tpi)−m(tqi−tpi)≤0.

Case 2. [tpi, tqi) is non-busy displacing. The cumulative utilization of all tasksτk ∈ αi is

P

τk∈αiuk. The carry-in jobs of these tasks do not belong tod, by the definition ofd. Therefore, the

All processors are occupied at every time instant in the interval[tpi, tqi), because it is displacing.

Thus,A(d, tpi, tqi, S) = (tqi −tpi)(m− |αi|). Therefore, the allocation difference for jobs ind

throughout the interval is

A(d, tpi, tqi, P S)−A(d, tpi, tqi, S) ≤ (tqi −tpi) ((m− X τk∈αi uk)−(m− |αi|)) = (tqi −tpi) X τk∈αi (1−uk). (3.3)

For each taskτkinDH, the sum of the lengths of the intervals[tpi, tqi)in which the carry-in

job ofτkexecutes continuously is at mostδk. Thus, summing the allocation differences for all the

intervals[tpi, tqi)given by (3.3), we have

A(d, tn, td, P S)−A(d, tn, td, S) ≤ z X i=1 X τk∈DH (tqi−tpi)(1−uk) ≤ X τk∈DH δk(1−uk). (3.4)

Setting this value into (3.2), we get LAG(d, td, S) ≤ LAG(d, tn, S) +A(d, tn, td, P S)− A(d, tn, td, S)≤LAG(d, tn, S) +Pτk∈DHδk(1−uk).

Lemma 3.2. lag(τi, t, S)≤ui·x+ei+ui·sifor any taskτiandt∈[0, td].

Proof. Letdi,k be the deadline of the earliest pending job ofτi,τi,k, in the scheduleSat timet. If

such a job does not exist, thenlag(τi, t, S) = 0, and the lemma holds trivially. Letγibe the amount

of workτi,k performs beforet.

By the selection ofτi,k, we have

lag(τi, t, S) = X h≥k

lag(τi,h, t, S)

= X

h≥k

Given that no job executes before its release time,A(τi,h,0, t, S) =A(τi,h, ri,h, t, S). Thus,

lag(τi, t, S) = A(τi,k, ri,h, t, P S)−A(τi,k, ri,k, t, S)

+X

h>i

(A(τi,h, ri,h, t, P S)

−A(τi,h, ri,h, t, S)). (3.5)

By the definition of P S, A(τi,k, ri,h, t, P S) ≤ ei, and Ph>kA(τi,h, ri,h, t, P S) ≤ ui · max(0, t₋di,k). By the selection ofτi,k,A(τi,k, ri,k, t, S) =γi, andPh>kA(τi,h, ri,h, t, S) = 0.

By setting these values into (3.5), we have

lag(τi, t, S)≤ei−γi+ui·max(0, t−di,k). (3.6)

There are two cases to consider.

Case 1.di,k ≥t. In this case, (3.6) implieslag(τi, t, S)≤ei−γi, which implieslag(τi, t, S)≤ ui·x+ei+ui·si.

Case 2. di,k < t. In this case, becauset≤tdanddl,j =td,τi,k is not the jobτl,j. Thus, by

Property (P),τi,k has tardiness at mostx+ei+si, so t+ei −γi ≤ di,k +x+ei +si. Thus, t₋di,k ≤x+γi+si. Setting this value into (7.4), we havelag(τi, t, S)≤ui·x+ei+ui·si.

Lemma 3.3 below upper boundsLAG(d, tn, S).

Definition 3.6. LetE_sums be the total execution cost of all self-suspending tasks inτ. LetEsum be

the total execution cost of all tasks inτ. LetS_sums be the total suspension length of all tasks inτ. Let us_maxbe the maximum utilization of any self-suspending task inτ.

Definition 3.7. Let U_Lc be the sum of themin(m₋1, c)largest computational task utilizations,

wherec is the number of computational tasks. LetE_Lc be the sum of themin(m−1, c) largest

computational task execution costs.

Proof. By summing individual task lags attn, we can bound LAG(d, tn, S). If tn = 0, then LAG(d, tn, S) = 0, so assumetn>0. Consider the set of tasksβ={τi :∃τi,vindsuch thatτi,v

is enabled att−_n}. Given that the instantt−_n is non-busy non-displacing, at mostm₋1computational

tasks in β have jobs executing at t−_n. Due to suspensions, however, β may contain more than m−1tasks. In the worst case, all suspending tasks inτ have a suspended enabled job att−_n and min(m−1, c)computational tasks have an enabled job executing att−_n. If taskτidoes not have an

enabled job att−

n, thenlag(τi, tn, S)≤0. Therefore, by (2.5), we have LAG(d, tn, S) = X τi:τw i,v∈d lag(τi, tn, S) ≤ X τi∈β lag(τi, tn, S) {by Lemma 3.2_} ≤ X τi∈β (ui·x+ei+ui·si) ≤(U_sums +U_Lc)_·x+E_sums +Ec_L +us_max·S_sums .

The demand placed by jobs inDaftertdisB(D, td, S) = Pτk∈DH(ek−δk). Thus, by (3.1)

and Lemmas 3.1 and 3.3, we have the following upper bound:

Z ≤ (U_sums +U_Lc)·x+E_sums +E_Lc +us_max·S_sums

+ X

τk∈DH

(δi(1−uk) + (ek−δk))

≤ (U_sums +U_Lc)·x+E_sums +E_Lc +us_max·S_sums

+Esum. (3.7)