How good are SPT schedules for fair optimality criteria

HAL Id: hal-00341359

https://hal.archives-ouvertes.fr/hal-00341359

Submitted on 27 Mar 2014

HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Eric Angel, Evripidis Bampis, Fanny Pascual

To cite this version:

Eric Angel, Evripidis Bampis, Fanny Pascual. How good are SPT schedules for fair optimality criteria.

2nd Multidisciplinary International Conference on Scheduling : Theory and Applications (MISTA

2005), Jul 2005, New York, NY., United States. pp.244–257. �hal-00341359�

More formally, we denote by

the ordered collection

^(l1

;:::;l

n

)

and we as- sume that the indexing is chosen so that

^{l1 l2 ln}

. Our scheduling instance

^I

is completely defined by the system

^(;m)

, where

^m

is the num- ber of (identical) processors. Let

i

be a collection of processing times de- fined by

^{k = fln;ln 1;:::;ln m+1g}

,

^{k 1 = fln m;:::;ln 2m+1g}

,

^:::

,

1

=fl

n (k 1)m

;:::;l

1

g

, where

^{k =dn}

m

e

. The collection

i

is called the

^ith

rank of tasks with respect to

^(;m)

. Consider a schedule obtained by schedul- ing the tasks rank-by-rank in the order

1

;

2

;:::;

k

and such that no two tasks of the same rank are scheduled on the same processor. Notice that any permutation of the tasks of the same rank assures the optimality of the sched- ule with respect to the sum of completion times criterion. This is the family of schedules that we call SPT schedules in the sequel.

We are interested to know whether among the SPT schedules, which are the optimal schedules for the sum of completion times criterion, it is possible to obtain good solutions for the following optimality criteria:

- the maximum sum of completion times per machine:

^max

1im X

T

j 2P

i C

j

. This measure captures the wish of distributing as much as possible the total sum of completion times among the

^m

machines of the system.

- the global fairness [4]: For two vectors

^{X;Y 2V(I)}

, we write

^X4Y

if

^Xi

Y

i

for all

ⁱ

. The global approximation ratio of

^X

, denoted by

gbl

(X)

, is the smallest

such that

^{!X 4 !Y}

for all

^{Y 2 V(I)}

. Informally

gbl

(X)

is the smallest

for which

^!X

is an

-approximation, in the coordinate-wise sense, to every vector

^{Y 2V(I)}

. The best global approximation ratio achievable on the instance

^I

is then defined as

gbl

(I)= inf

X2V(I)

gbl (X):

- the individual fairness: The individual happiness factor compares the completion time of a task with the smallest possible completion time of the same task in any feasible schedule. More formally, we define

ind

(X)

to be the smallest

such that

^{X 4Y}

for all

^{Y 2V(I)}

. The best approximation ratio achievable on the instance

^I

is then defined as

ind

(I)= inf

X2V(I)

ind (X):

Our approach is in the same vein as the one of Bruno et al. [3] who con-

sidered the following question: among all optimal schedules for the sum of

P

1

P

₁

P

2

P

2

1 1

1

1 1

1 5 5

6 7 5

3

max. sum of completion times = 7

(total) sum of completion times = 11 (total) sum of completion times = 10

max. sum of completion times = 6

Figure 1. Instance for which the optimal solutions to the problems Minimize

P

n

j=1 C

j

and

MinMaxSCT

are different.

completion times, is it possible to compute one that minimizes the makespan?

(They proved that the problem is

^NP

-hard.) For a related problem, see also [6].

Organization of the paper and our contribution In Section 2, we consider the problem of minimizing the maximum completion time per machine (

^Min

Max SCT

). We first show that contrarily to the sum of completion times, the problem

^{Min Max SCT}

is

^NP

-hard. Furthermore, we show that an SPT schedule is a

^{(3 3=m+1=m2)}

-approximation algorithm for the

^{Min Max}

SCT

problem and that there are instances for which any SPT schedule cannot achieve an approximation guarantee better than

^{2 2}

m 2

+m

. In Section 3, we consider the global approximation ratio and we prove that SPT schedules have an approximation ratio of

^{2 1=m}

(and that no algorithm can have a better approximation ratio if

^m=2

). Philips et al. [5] presented a 3-approximation for the same problem when release dates are taken into account. For the in- dividual happiness factor however the performance ratio cannot be bounded by any constant but we prove that a SPT schedule obtains the best possible performance guarantee. Finally, we focus on a more restricted version of the individual happiness factor where the obtained solutions are compared with respect to the family of SPT schedules and we prove a 2-approximation bound.

2. The

^{Min Max SCT}

problem

We first remark that the problem of minimizing the sum of completion times and the

^{Min Max SCT}

problem are different. To see that consider the fol- lowing instance: three tasks of length

¹

and a task of length

⁵

on two identical processors (see Figure 1). The optimal solution for the problem of minimizing the sum of completion times consists in putting two tasks of length

¹

on a pro- cessor, and the other two tasks on the other processors (total sum of completion times

⁼¹⁰

, maximum sum of completion times per processor

⁼⁷

). The op- timal solution of

^{Min Max SCT}

consists in putting the three tasks of length

1

on a processor, and the task of length

⁵

on the other processor (total sum of

completion times

⁼¹¹

, maximum sum of completion time per processor

⁼⁶

).

2.1 Hardness

We prove in this section that the

^{Min Max SCT}

problem is

^NP

-hard.

Theorem 2.1 Min Max SCT

is

^NP

-hard.

Proof: Consider the decision version of the

^{Min Max SCT}

problem:

MinMaxSCT

problem: Given a system

^(;m)

, and a number

^k

. Does there exist a schedule such that its maximum sum of completion times is smaller than or equal to

^k

?

We will show that the

^partition

problem which is known to be

^NP

-hard [2] can be phrased in terms of problem

^{Min Max SCT}

.

Partition

problem: Given a collection

^C=fx1

;x

2

;:::;x

n

g

of integers.

Does there exist a partition

^(A;B)

of

^C

, i.e.

^A[B=C

and

^{A\B =;}

, such that

P

x2A x=

P

x2B x

?

Given an instance of

^partition

with a collection

^C=fx1

;x

2

;:::;x

n g

of non decreasing integers

^{x1 x2 xn}

, we define a system

^(;2)

and a limit

^k

such that the

^MinMaxSCT

instance admits a solution if and only the instance of

^partition

admits a solution. Let

^{k= 3}

2 (

P

n

i=1 x

i )

P

n

i=1 x

i

n i+1

. We now define the system

^(;2)

: we have

²ⁿ

tasks

^T1

;T

2

;:::;T

2n

to schedule on

²

processors

^P1

and

^P2

. Let

^li

denote the execution time of task

^Ti

. Let the execution times

^{= fl}1

;l

2

;:::;l

2n

g

be such that

^l2j 1

= P

j 1

i=1 x

i

n i+1

and

l

2j

= P

j

i=1 x

i

n i+1

, for

^{j2f1;2;:::;ng}

. For example, (if

ⁿ⁶

),

^{l1 =0; l2= x1}

n

;

l

3

= x

1

n

; l

4

= x

1

n +

x

2

n 1

;

l

5

= x

1

n +

x

2

n 1

; l

6

= x

1

n +

x

2

n 1 +

x

3

n 2

;

l

2n 1

= x

1

n +

x

2

n 1

++ x

n 1

2

;

l

2n

= x

1

n +

x

2

n 1

++ xn

1

2 +x

n

Let us now show that there is a solution of the

^{Min Max SCT}

problem if and only if there is a solution of

^partition

.

If a partition

^(A;B)

exists for

^C

then there is a solution of the

^{Min Max}

SCT

problem: we can obtain for our system

^(;2)

a schedule whose maxi- mum sum of completion times is

³

2 (

P

n

i=1 x

i )

P

n

i=1 x

i

n i+1

. This schedule can be obtained by assigning tasks

^T2j 1

and

^T2j

at rank

^j

: assign task

^T2j

to

P

1

and

^{T2j 1}

to

^P2

if

^{xj 2A}

, otherwise assign

^T2j

to

^P2

and

^{T2j 1}

to

^P1

. Indeed, the execution time of a task added at the

^jth

rank will be counted

n j+1

times in the sum of the completion times of its processor. For ex-

ample, the execution time of the last task of a processor will be counted only

once, whereas the execution time of the first task of a processor will be counted

n

times (because there are

ⁿ

tasks on each processor). Thus the contribution to the sum of completion times per processor of the task

^{T2j 1}

(resp.

^T2j

), assigned at the

^jth

rank, is

^{(n j+1)l}2j 1

(resp.

^{(n j+1)l}2j

=

(n j+1)(l

2j 1 +

x

j

n j+1

)

). The difference between these two contributions is then

^xj

: if

^xj

2A

(i.e.

^T2j

is assigned to

^P1

) then the contribution of the

^jth

task of

^P1

is equal to the contribution of the

^jth

task of

^P2

, plus

^xj

. Likewise, if

^xj

2B

(i.e.

^T2j

is assigned to

^P2

) then the contribution of the

^jth

task of

^P2

is equal to the contribution of the

^jth

task of

^P1

, plus

^xj

. Thus, the sum of the completion times of the tasks of

^P1

is

P

n

j=1 T

2j 1 +

P

x

j 2A

x

j

, and the sum of the completion times of the tasks of

^P2

is

P

n

j=1 T

2j 1

+ P

x

j 2B

x

j

. Moreover

P

n

j=1 T

2j 1

=(n 1) x1

n

+(n 2) x2

n 1 ++

x

n 1

2

=

x

1 x1

n +x

2 x2

n 1

++x

n 1 x

n 1

2

= P

n

i=1 x

i P

n

i=1 x

i

n i+1

. So the sum of the completion times of the tasks of

^P1

is

⁽

P

n

i=1 x

i P

n

i=1 xi

n i+1 )+

P

x

j 2A

x

j

, and the sum of the completion times of the tasks of

^P2

is

⁽

P

n

i=1 x

i

P

n

i=1 x

i

n i+1 )+

P

x

j 2B

x

j

. If there is partition of

^C

(i.e.

P

x

j 2A

= P

x

j 2B

= P

n

i=1 x

i

2

) then the maximum sum of completion times per processor is equal to

⁽

P

n

i=1 x

i P

n

i=1 xi

n i+1 )+

P

n

i=1 x

i

2

= 3

2 (

P

n

i=1 x

i )

P

n

i=1 x

i

n i+1

= k

. Thus, if there is a solution of

partition

, there is a solution of the

^{Min Max SCT}

problem.

Let us now show that if there is a solution of the

^{Min Max SCT}

problem, then there is also a solution of

^partition

. If the maximum sum of the comple- tion times per processor is smaller than or equal to

³

2 (

P

n

i=1 x

i )

P

n

i=1 x

i

n i+1

, then we have a SPT schedule. Indeed, the total sum of the completion times of a SPT schedule is

^S=2

3

2 (

P

n

i=1 x

i )

P

n

i=1 x

i

n i+1

, as we saw it above, and a schedule has a minimum total sum of completion times if and only if it is a SPT schedule [3]. Since the total sum of completion times cannot be greater than twice the maximum sum of completion times per processor, a schedule which is a solution of the

^{Min Max SCT}

problem is a SPT schedule. In any SPT schedule, the tasks of length

^l2i

and

^l2i 1

are at the

^ith

rank, because

l

1

< l

2 l

3

< :::l

2j 1

< l

2j l

2j+1

< l

2j+2

< l

n

. So, in any SPT schedule, the sum of completion times of

^P1

minus the one of

^P2

is equal to

P

T

2j 2P

1 x

j P

T

2j 2P

2 x

j

. If there is a solution of the

^{Min Max SCT}

problem with

^{k = 3}

2 (

P

n

i=1 x

i )

P

n

i=1 x

i

n i+1

then the sum of the completion times of

^P1

is equal to the sum of the completion times of

^P2

(otherwise the sum of completion times of all the tasks would be smaller than

^S

), and then

P

T2j2P1 x

j

= P

T2j2P2 x

j

: there is a partition of

^C

and we can construct this

partition

^(A;B)

by placing

^xj

in

^A

if

^T2j

is assigned to

^P1

and

^xj

in

^B

if

^T2j

is assigned to

^P2

.

²

2.2 Approximation

In this section, we show that an optimal algorithm for the sum of comple- tion times criterion gives a 3-approximate solution for the

^{Min Max SCT}

problem. Consider the following algorithm, denoted by SPT

greedy

:

Order tasks by non decreasing execution times. At each step

ⁱ

, for

^{1 i n}

, schedule the current task on the processor which has the smallest completion time.

This algorithm gives an optimal solution for the problem of minimizing the sum of completion times of the tasks. Let us show that this algorithm is a

(3 3

m +

1

m 2

)

-approximation algorithm for

^{Min Max SCT}

.

In order to let SPT

greedy

be deterministic when several processors have the smallest execution time, we will refer in the proofs to the following algorithm, which is a greedy SPT algorithm (Proposition 2.1 shows that we add at each step a task on a processor which has the smallest completion time):

Order tasks by non decreasing execution times. At each step

i

, for

^{1 i n}

, schedule the current task on the processor

P

imodm

.

Proposition 2.1

a) At the beginning of step

ⁱ

of SPT

greedy

, the processor

^Pimodm

is the processor which has the smallest completion time and the smallest sum of completion times.

b) At the end of step

ⁱ

of SPT

_greedy

, the processor

^Pimodm

is the processor which has the largest sum of completion times.

Proof:

a) We are at the beginning of step

ⁱ

. Let

be the processor

^Pimodm

, and let

^p

denote its completion time. Let us show that

has the smallest completion time:

– For each

^k>imodm

, processor

^Pk

has a completion time greater

than or equal to

^p

. Indeed

^Pk

has the same number of tasks as

and, for each

ⁱ

such that

¹ⁱ

number of tasks on

, the

ⁱ

-th task

of

^Pk

is greater than or equal to the

ⁱ

-th task of

, by construction.

– For each

^k<imodm

, processor

^Pk

has a completion time greater than or equal to

^p

. Indeed the number of tasks on

^Pk

is equal to the number of tasks on

plus one, and, for each

ⁱ

such that

^{1 i}

(number of tasks on

), the

⁽ⁱ⁺¹⁾

-th task of

^Pk

is greater than or equal to the

ⁱ

-th task of

, by construction.

We use the same reasoning to show that

has the smallest sum of com- pletion times.

b) We are at the end of step

ⁱ

. Let

be the processor

^Pimodm

, and let

^p

denote its sum of completion times at this step. Let us show that

has the largest sum of completion times:

– For each

^{k > imodm}

, processor

^Pk

has a sum of completion times smaller than or equal to

^p

. Indeed

^Pk

has one task less than

and, for each

ⁱ

such that

^{1 i<}

number of tasks on

, the

ⁱ

-th task of

^Pk

is smaller than or equal to the

⁽ⁱ⁺¹⁾

-th task of

, by construction.

– For each

^{k < imodm}

, processor

^Pk

has a sum of completion times smaller or equal to

^p

. Indeed

^Pk

has the same number of tasks as

and, for each

ⁱ

such that

^{1 i}

number of tasks on

, the

ⁱ

-th task of

^Pk

is smaller than or equal to the

ⁱ

-th task of

, by construction.

2

Proposition 2.2

Let

^OPT

be the maximum sum of completion times of a solution of

^{Min Max SCT}

. We have:

OPT

Min

⁽

P

n

j=1 C

j )

m

;

where Min

⁽

P

n

j=1 C

j

)

is the optimal value for the problem of minimizing the sum of completion times.

Proof: We will prove this proposition by contradiction. Let us suppose that we have

^{OPT < Min(}

P

n

j=1 Cj)

m

.

By definition each processor has a sum of completion times smaller than or equal to OPT:

^{8i2f1;:::;mg;}

P

T

j 2P

i C

j

OPT:

So,

P

n

j=1 C

j

= P

m

i=1 P

Tj2Pi C

j

mOPT

, and

^OPT

P

n

j=1 C

j

m

Min⁽

P

n

j=1 C

j )

m

;

a contradiction.

²

Theorem 2.2

The algorithm SPT

_greedy

achieves an approximation guaran- tee of

^{(3 3}

m +

1

m 2

)

for

^{Min Max SCT}

.

Proof: Before we add the last task

^Tn

, the processor

^Px

on which

^Tn

will be scheduled has the smallest sum of completion times, denoted by

^p

, and the smallest completion time, denoted by

^e

(Proposition 2.1 a). Let

^ln

denote the execution time of task

^Tn

. The processor on which the last task is scheduled has the largest sum of completion times (Proposition 2.1 b), so:

max

1im P

Tj2Pi C

j

= P

Tj2Px C

j

=p+e+l

n

P

n 1

j=1 C

j

m

+e+l

n

Since SPT

_greedy

gives an optimal solution of the minimum sum of comple- tion times problem, and since the completion time of

^Tn

in SPT

greedy

is

^e+ln

, we have:

max

1im P

T

j 2P

i C

j

Min^(Pn

j=1 C

j ) (e+l

n )

m

+e+l

n

Min^Pn

j=1 C

j

m

+(1 1

m )(e+l

n ):

Since

^e+ln

m

OPT

(because

^e

is the minimum completion time at step

^{n 1}

), and

^lnOPT

, we have:

e+l

n

= e+ l

n

m +

(m 1)ln

m

(1+ m 1

m

)OPT

(2

1

m )OPT

Moreover, since

^Min

P

n

j=1 Cj

m

OPT

(Proposition 2.2), we have:

max

1im P

T

j 2P

i C

j

OPT +(1 1

m )(2

1

m )OPT

(3

3

m +

1

m 2

)OPT:

2

Lower bound for SPT

greedy

. We now show that SPT

greedy

does not achieve an approximation guarantee better than

^{2 2}

m 2

+m

. Consider the following instance:

^m

processors,

^{m(m 1)}

tasks of length

¹

and a task of length

B= P

m

i=1 i=

m(m+1)

2

.

SPT

_greedy

will schedule

^{m 1}

tasks of length

¹

on each processor, and the

task of length

^B

will be the last task of the first processor (see Figure 2). The

max. sum of completion times = 11 P

P

2

P

1

P

2

P

3

P

₃

1 1

1 1

1 1 1

1

1 1

1 1 6

6

max. sum of completion times = 6

1

Figure 2. SPT

greedy

does not achieve an approximation guarantee better than (

^{2 2}

m 2

+m

):

example with m=3

maximum sum of completion times is then

P

m 1

i=1

i+(m 1)+B

, which is equal to

²⁽

P

m

i=1 i) 1

.

An optimal solution for

^{Min Max SCT}

would be the following one:

schedule

^m

tasks of length

¹

on each of the

^{(m 1)}

first processors, and the task of length

^B

on the last processor. The maximum sum of completion times in this solution is then

P

m

i=1 i

.

So the ratio between the maximum sum of completion times of these two schedules is

²⁽

P

m

i=1 i) 1

P

m

i=1 i

, which is equal to

^{2(1 1}

m 2

+m

)

, which tends towards

2

when

^m

gets large.

3. Fairness measures

In this section we will consider fairness measures in order to compare a schedule given by the greedy SPT algorithm, SPT

_greedy

, to any other schedule.

3.1 Global fairness measure

We will first use the fairness measure introduced in [4], namely the global approximation ratio. We prove that SPT

greedy

has a global approximation ratio of

^{2 1}

m

for the problem that we consider (and which is denoted by P

^k

all ), and that no algorithm can achieve a better global approximation ratio if

^m=2

.

Theorem 3.1

One has

gbl

(I) 2 1

m

for all instances

^I

of schedul- ing on

^m

identical parallel machines (P

^k

all), for any

^{m 1}

. Moreover the completion times vector

^X

of a schedule returned by SPT

greedy

verifies

gbl

(X)2 1

m

.

Proof: Let us consider an instance

^I

of tasks

^Ti

(

^{i 2 f1;:::;ng}

) ordered by increasing lengths. Let MR be the maximal ratio between a schedule

^XSPT

returned by SPT

_greedy

and any schedule

^Y

: MR

^{= max}Y

!

X

SPT

!

, where

^X

Y

means

^{maxi Xi}

Y

i

. So we have

^{8Y 2 V(I);}

!

XSPT

!

Y

MR, and then

!

X

SPT 4

MR

^!Y

. So MR

^=gbl(XSPT)

. We will show that MR cannot be greater than

2 1

m

. Let us consider that this ratio is the

^ith

coordinate of

!

X

SPT

, divided by the

^ith

coordinate of

^!Y

(

^i2f1;:::;ng

).

The worst ratio can be achieved if the completion time of the

^ith

task is as large as possible in the SPT

greedy

schedule. By construction, in a schedule returned by SPT

_greedy

, the

^ith

completion time is the completion time of the task

T

i

, and

^Ti

is started after the tasks from

^T1

to

^{Ti 1}

, and before the tasks from

T

i+1

to

^Tn

. Since SPT

_greedy

is a greedy algorithm, the worst completion time of the

^ith

task is achieved when the

^{(i 1)}

first tasks are completed when the

^ith

task start to be executed: in this case, the completion time of

^Ti

is

P

i 1

j=1 l

j

m +l

i

. So the worst completion time of

^Ti

in an SPT

_greedy

schedule is

P

i 1

j=1 lj

m +l

i

. Let us now find the minimal value which can be taken by the

^ith

coordinate of

^!Y

, an ordered completion time vector of a schedule of

^I

. Note that this value cannot be smaller than

^li

: indeed

^!Yi

is the

^ith

completion time and is then greater or equal to

^{(i 1)}

other completion times, and

^li

is the length of the

^ith

smallest task. Note also that

^!Yi

cannot be smaller than

P

i

j=1 l

j

m

: indeed this is the minimum completion time of the

ⁱ

smallest tasks (when no processor is idle). Therefore, we have:

MR

P

i 1

j=1 l

j

m +l

i

max

P

i

j=1 l

j

m

; l

i

:

Let

^A

denote

P

i

j=1 l

j

m

. We now consider the two possible cases:

Suppose that

^{li A}

, we have: MR

P

i 1

j=1 l

j

m +l

i

l

i

A l

i

m +l

i

l

i

A

l

i +1

1

m 2

1

m

Suppose that

^li

< A

, we have: MR

P

i 1

j=1 l

j

m +l

i

P

i

j=1 l

j

m

A l

i

m +l

i

A

1+

li

A (1

1

m )2

1

m

In both cases, we have MR

^{2 1}

m

. Since MR

⁼ gbl (X

SPT

)

, and since

X

SPT

is the completion time vector of the SPT

greedy

schedule of

^I

, we proved that a schedule

^X

returned by SPT

_greedy

verifies

gbl

(X)2 1

m

, and so that

gbl

(I)2 1

m

.

²

completion times vector: X2=(1, 1,3) P

P

2

P

1

P

2

1

1 1 1

2

2

completion times vector: X1=(1, 2, 2)

1

Figure 3. Example where we have

gbl (I)=

3

2

=2 1

m

.

Let us now show that it is not possible to have

gbl

(I) < 2 1

m

for all instances

^I

of (P

^k

all). Indeed, if

^{m = 2}

, it is not possible to have always

gl ob (I)<

3

2

:

Theorem 3.2

It is not possible to have

gbl (I) <

3

2

for all instances

^I

of scheduling on two identical parallel machines.

Proof: let us consider a system with two processors,

^P1

and

^P2

. Let us con- sider the following instance

^I

: a task

^T1

of length

¹

,

^T2

of length

¹

, and

^T3

of length

²

. Consider the two completion times vector

^{X1 =(1;2;2)}

(obtained by putting

^T1

and

^T2

on

^P1

and

^T3

on

^P2

: see Figure 3) and

^X2

= (1;1;3)

(obtained by putting

^T1

and

^T3

on

^P1

and

^T2

on

^P2

). We have:

^{gbl(X1) =2}

, and

gbl

(X

2 )=

3

2

. For each vector

^{Y 2V(I)}

, we have

^X!1 4

!

Y

or

^X!2 4

!

Y

. So for each vector

^{Y 2V(I)}

, we have

gbl

(Y) 3

2

, and so

gbl (I)=

3

2

.

²

3.2 Individual fairness measure

In this section, we will compare the completion time of each task in a so- lution given by SPT

greedy

to the best completion time this task could have in another schedule.

Theorem 3.3

ind

(I) 1 + n 1

m

for all instances

^I

of scheduling on

m

identical parallel machines (P

^{k ind all}

), for any

^{m 1}

. Moreover the completion times vector

^X

of a schedule returned by SPT

greedy

verifies

ind

(X)1+ n 1

m

.

Proof: Let us consider an instance

^I

of tasks

^Ti

(

^{i 2 f1;:::;ng}

) ordered by non decreasing lengths. It is easy to see that

ind

(X

SPT

)

is the maximal ratio between a schedule

^XSPT

returned by SPT

greedy

and any schedule

^Y

. We will show that

ind

(X

SPT

)

cannot be greater than

^{1+ n 1}

m

. Let us consider that this ratio is the

^ith

coordinate of

^XSPT

, divided by the

^ith

coordinate of

^Y

(

^i2f1;:::;ng

).

The worst ratio can be reached if the completion time of

^Ti

is as large as

possible in the SPT

greedy

schedule. Since SPT

greedy

is a greedy algorithm, this