Tight lower bounds for semi-online scheduling on two uniform machines with known optimum

Abstract

This problem is about scheduling a number of jobs on two uniform machines with given speeds 1 and $s\ge 1$, so that the overall finishing time, i.e., the makespan, is earliest possible. We consider a semi-online variant (introduced for equal speeds) by Azar and Regev, where the jobs are arriving one after the other, while the scheduling algorithm knows the optimum value of the corresponding offline problem. One can ask how close any possible algorithm could get to the optimum value, that is, to give a lower bound on the competitive ratio: the supremum over ratios between the value of the solution given by the algorithm and the optimal offline solution. We contribute to this question by constructing tight lower bounds for all values of s in the intervals $[\frac{1+\sqrt{21}}{4},\frac{3+\sqrt{73}}{8}]\approx [1.3956,1.443]$ and $[\frac{5}{3},\frac{4+\sqrt{133}}{9}]\approx [\frac{5}{3},1.7258]$, except a very narrow interval.

Appendix: Proof of Lemma 1

The technical proofs of the Lemmas can be easily carried out using a computer algebra system. We prepared a Maple worksheet that is available upon request from the authors.

First, we present some lower and upper bounds on r(s).

Lemma 2

(i) If s is small, regular or medium,

$$\begin{aligned} r(s)\le & {} r_{2}(s). \end{aligned}$$

(33)

$$\begin{aligned} r(s)\le & {} \frac{11s+9}{9s+5}\le \frac{9s+7}{9s+1}. \end{aligned}$$

(34)

$$\begin{aligned} r(s)< & {} \min \left\{ \frac{3s+1}{2s+1},\frac{4s+3}{3s+2}\right\} < s. \end{aligned}$$

(35)

$$\begin{aligned} r(s)< & {} \frac{s+2}{s+1}\le \frac{2s+1}{s+1}. \end{aligned}$$

(36)

$$\begin{aligned} r(s)\ge & {} \frac{63s+53}{45s+41}\ge \frac{s+1}{2}. \end{aligned}$$

(37)

$$\begin{aligned} r(s)> & {} \frac{4}{3}>\frac{3s+2}{2s+2}. \end{aligned}$$

(38)

$$\begin{aligned} r(s)\ge & {} \frac{2s+1}{2s}. \end{aligned}$$

(39)

(ii) If s is small,

$$\begin{aligned} r(s)= & {} r_1(s)\le \frac{5s+2}{4s+1}\le \frac{2}{s}. \end{aligned}$$

(40)

(iii) If s is regular or medium,

$$\begin{aligned} r(s)\ge & {} \frac{11s+8}{8s+6}\ge \frac{5s+6}{4s+4}. \end{aligned}$$

(41)

$$\begin{aligned} r(s)\ge & {} \frac{6s+5}{6s+1}\ge \frac{2}{s}. \end{aligned}$$

(42)

$$\begin{aligned} r(s)\le & {} \frac{3s+5}{2s+4}\le \frac{5s+4}{3s+4}. \end{aligned}$$

(43)

(iv) If s is smaller regular or smaller medium,

$$\begin{aligned} r(s) \le \frac{18s+16}{16s+7}. \end{aligned}$$

(44)

(v) If $s \le s_0$ and s is smaller regular,

$$\begin{aligned} r(s)\le \frac{17s+14}{15s+6}. \end{aligned}$$

(45)

(vi) If $s \ge s_0$ and s is smaller regular, or s is smaller medium,

$$\begin{aligned} r(s)\ge \frac{7s+7}{7s+2}. \end{aligned}$$

(46)

(vii) If s is bigger medium,

$$\begin{aligned} r(s)\ge \frac{4s+4}{13-3s}. \end{aligned}$$

(47)

(vii) If s is bigger regular,

$$\begin{aligned} r(s)\ge \frac{4s+3}{6s-3}. \end{aligned}$$

(48)

Proof

Note that $s > 0$ in all cases. Hence in all inequalities below, we never multiply with negative numbers, thus the sense of the inequalities does not change.

• (i)

  • Equation (33) The inequality $r_{1}(s) \le r_{2}(s)$ is equivalent to $(12s+10)(4s+5)-(6s+6)(9s+7) = -6s^{2}+4s+8 \ge 0$, which holds for $0 < s \le \frac{1+\sqrt{13}}{3} \approx 1.5352$. Hence it holds for all $s \le q_{2}$. The inequality $r_{3}(s)\le r_{2}(s)$ is equivalent to $(12s+10)(16s+7)-(18s+16)(9s+7)=30s^{2}-26s-42\ge 0$, which holds for $s\ge \frac{13+\sqrt{1429}}{30}=q_{3}$. The inequality $r_{4}(s)\le r_{2}(s)$ is equivalent to $(12s+10)(3s+10)-(8s+7)(9s+7)=-36s^{2}+31s+51\ge 0$, which holds for $0 < s\le \frac{31+\sqrt{8305}}{72}=q_{5}$. Hence, $r(s) \le r_{2}(s)$ by the definition of r(s).

  • Equation (34) The inequality $r_{2}(s)\le \frac{11s+9}{9s+5}$ is equivalent to $(12s+10)(9s+5)-(11s+9)(9s+7)=9s^{2}-8s-13\le 0$. The inequality $\frac{11s+9}{9s+5}\le \frac{9s+7}{9s+1}$ is equivalent to $(9s+1)(11s+9)-(9s+5)(9s+7)=18s^{2}-16s-26\le 0$. Both inequalities hold for $0 < s\le \frac{4+\sqrt{133}}{9}=q_{7}$.

  • Equation (35) The inequality $r_{2}(s)<\frac{3s+1}{2s+1}$ is equivalent to $(12s+10)(2s+1)-(3s+1)(9s+7)=-3s^{2}+2s+3<0$, which holds for $s>\frac{1+\sqrt{10}}{3}\approx 1.3874$. Hence it holds for all $s \ge q_{1}$. The inequality $r_{2}(s)<\frac{4s+3}{3s+2}$ is equivalent to $(12s+10)(3s+2)-(4s+3)(9s+7)=-1-s<0$, which holds for all $s \ge 1$. The inequality $\frac{3s+1}{2s+1}<s$ is equivalent to $(3s+1)-s(2s+1)=-2s^{2}+2s+1<0$, which holds for $s>\frac{1+\sqrt{3}}{2}\approx 1.3660$. Hence it holds for all $s \ge q_{1}$. In combination with (33) we thus obtain (35).

  • Equation (36) The inequality $r_{2}(s)<\frac{s+2}{s+1}$ is equivalent to $(12s+10)(s+1)-(s+2)(9s+7)=3s^{2}-3s-4< 0$, which holds for $0< s<\frac{3+\sqrt{57}}{6}\approx 1.7583$. Hence it holds for all $s \le q_{7}$. The inequality $\frac{s+2}{s+1}\le \frac{2s+1}{s+1}$ holds for any $s\ge 1$. In combination with (33) we thus obtain (36).

  • Equation (37) The inequality $r_{1}(s)\ge \frac{63s+53}{45s+41}$ is equivalent to $(45s+41)(6s+6)-(63s+53)(4s+5)=18s^2-11 s-19\ge 0$, which holds for $s\ge \frac{11+\sqrt{1489}}{36}\approx 1.37743$. Hence it holds for all $s \ge q_{1}$. The inequality $r_{2}(s)\ge \frac{63s+53}{45s+41}$ is equivalent to $(12s+10)(45s+41)-(63s+53)(9s+7)=-27s^{2}+24s+30\ge 0$, which holds for $0 < s\le \frac{4+\sqrt{133}}{9}=q_{7}$. The inequality $r_{3}(s)\ge \frac{63s+53}{45s+41}$ is equivalent to $(18s+16)(45s+41)-(63s+53)(16s+7)=-198 s^2+169s+285\ge 0$, which holds for $0 < s\le \frac{169+\sqrt{254281}}{396}\approx 1.70016$. Hence it holds for all $q_{3}\le s \le q_{4}$. The inequality $r_{4}(s)\ge \frac{63s+53}{45s+41}$ is equivalent to $(8s+7)(45s+41)-(63s+53)(3s+10)=171 s^2-146 s-243\ge 0$, which holds for $s\ge \frac{73-\sqrt{46882}}{171}\approx 1.69311$. Hence it holds for all $q_{4}\le s \le q_{5}$. By the definition of r(s), we have $r(s)\ge \frac{63s+53}{45s+41}$. The inequality $\frac{63s+53}{45s+41}\ge \frac{s+1}{2}$ is equivalent to $2(63s+53)-(s+1)(45s+41)=-45 s^2+40 s+65\ge 0$, which holds for $0 < s\le \frac{4+\sqrt{133}}{9}=q_{7}$.

  • Equation (38) Since $\frac{63s+53}{45s+41}>\frac{4}{3}$ for any $s>1$, and $\frac{4}{3}>\frac{3s+2}{2s+2}$ for any $0<s<2$, (38) is obtained in combination with (37).

  • Equation (39) The inequality $r_{1}(s)\ge \frac{2s+1}{2s}$ is equivalent to $2s(6s+6)-(2s+1)(4s+5)=4s^{2}-2s-5\ge 0$, which holds for $s\ge \frac{1+\sqrt{21}}{4}=q_{1}$. The inequality $\frac{63s+53}{45s+41} \ge \frac{2s+1}{2s}$ is equivalent to $2s(63s+53)-(2s+1)(45s+41)=36 s^2-21 s-41\ge 0$, which holds for $s> \frac{7+\sqrt{705}}{24}\approx 1.39799$. Hence it holds for all $s \ge \frac{5}{3}$. By the definition of r(s) and (37), (39) can be proved.

• (ii)

  • Equation (40) The inequality $r_{1}(s)\le \frac{5s+2}{4s+1}$ is equivalent to $(4s+1)(6s+6)-(5s+2)(4s+5)=4s^{2}-3s-4\le 0$, which holds for $0 < s\le \frac{3+\sqrt{73}}{8}=q_2$. The inequality $\frac{5s+2}{4s+1} \le \frac{2}{s}$ is equivalent to $s(5s+2)-2(4s+1)=5s^{2}-6s-2\le 0$, which holds for $0 < s\le \frac{3+\sqrt{19}}{5}\approx 1.47178$. Hence it holds for all $s \le q_{2}$. Equation (40) is thus proved.

• (iii)

  • Equation (41) The inequality $\frac{63s+53}{45s+41} \ge \frac{11s+8}{8s+6}$ is equivalent to $(8s+6)(63s+53)-(11s+8)(45s+41)=9s^{2}-9s-10\ge 0$, which holds for $s\ge \frac{5}{3}$. The inequality $\frac{11s+8}{8s+6}\ge \frac{5s+6}{4s+4}$ is equivalent to $(11s+8)(4s+4)-(8s+6)(5s+6)=-4- 2s + 4s^2\ge 0$, which holds for $s\ge \frac{1+\sqrt{17}}{4}\approx 1.28078$. Hence it holds for all $s \ge \frac{5}{3}$. In combination with (37) we thus obtain (41).

  • Equation (42) The inequality $r_{2}(s) \ge \frac{6s+5}{6s+1}$ is equivalent to $(6s+1)(12s+10)-(6s+5)(9s+7)=18s^{2}-15s-25\ge 0$, which holds for $s\ge \frac{5}{3}$. The inequality $\frac{63s+53}{45s+41} \ge \frac{6s+5}{6s+1}$ is equivalent to $(6s+1)(63s+53)-(6s+5)(45s+41)=108s^{2}-90s-152\ge 0$, which holds for $s\ge \frac{15+\sqrt{2049}}{36}\approx 1.67405$. Hence it holds for all $q_{3}\le s \le q_{5}$. By the definition of r(s) and (37), $r(s) \ge \frac{6s+5}{6s+1}$ for any s is regular or medium. The inequality $\frac{6s+5}{6s+1} \ge \frac{2}{s}$ is equivalent to $s(6s+5)-2(6s+1)=6s^{2}-7s-2\ge 0$, which holds for $s\ge \frac{7+\sqrt{97}}{12}\approx 1.40407$. Hence it holds for all $s \ge \frac{5}{3}$, and thus (42) is obtained.

  • Equation (43) The inequality $r_{2}(s)\le \frac{3s+5}{2s+4}$ is equivalent to $(12s+10)(2s+4)-(3s+5)(9s+7)=-3s^{2}+2s+5\le 0$, which holds for $s\ge \frac{5}{3}$. The inequality $\frac{3s+5}{2s+4} \le \frac{5s+4}{3s+4}$ is equivalent to $(3s+4)(3s+5)-(2s+4)(5s+4)=4 - s - s^2\le 0$, which holds for $s\ge \frac{-1+\sqrt{17}}{2}\approx 1.56155$. Hence it holds for all $s \ge \frac{5}{3}$. In combination with (33) we thus obtain (43).

• (iv)

  • Equation (44) The inequality $r_2(s)\le \frac{18s+16}{16s+7}$ is equivalent to $(12s+10)(16s+7)-(18s+16)(9s+7)=30s^{2}-26s-42\le 0$, which holds for $0 < s\le \frac{13+\sqrt{1429}}{30}=q_{3}$. $r_3(s)\le \frac{18s+16}{16s+7}$ trivially holds when $q_{3}\le s\le q_{4}$.

• (v)

  • Equation (45) The inequality $r_{2}(s)\le \frac{17s+14}{15s+6}$ is equivalent to $(12s+10)(15s+6)-(17s+14)(9s+7)=27s^{2}-23s-38\le 0$, which holds for $0 < s\le \frac{23+\sqrt{4633}}{54}=s_0$.

• (vi)

  • Equation (46) The inequality $\frac{63s+53}{45s+41}\ge \frac{7s+7}{7s+2}$ is equivalent to $(63s+53)(7s+2)-(45s+41)(7s+7)=126s^2-105 s-181\ge 0$, which holds for $s\ge \frac{35+\sqrt{11361}}{84}\approx 1.68557$. Hence it holds for all $s \ge s_0$. Combining it with (37), we have $r(s)\ge \frac{7s+7}{7s+2}$.

• (vii)

  • Equation (47) The inequality $r_4(s)\ge \frac{4s+4}{13-3s}$ is equivalent to $(13-3s)(8s+7)-(4s+4)(3s+10)=-36s^2+31s+51\ge 0$, which holds for $0 < s\le \frac{31+\sqrt{8305}}{72}=q_5$. Hence, $r(s)\ge \frac{4s+4}{13-3s}$.

• (viii)

  • Equation (48) The inequality $r_2(s)\ge \frac{4s+3}{6s-3}$ is equivalent to $(6s-3)(12s+10)-(4s+3)(9s+7)=36s^2-31s-51\ge 0$, which holds for $s\ge \frac{31+\sqrt{8305}}{72}=q_5$. Hence, $r_2(s)\ge \frac{4s+3}{6s-3}$. $\square$

Now we are in a position to prove Lemma 1.

Proof of Lemma 1

• (i)

  • Equation (14) $B_4=4s - 2r - 3rs + 3>0$ is equivalent to $r(s)< \frac{4s+3}{3s+2}$, which is valid by (35). $B_4-T_4=(4s - 2r - 3rs + 3)-(r-1)=4s-3r-3rs+4<0$, since $r(s)>\frac{4}{3}$ by (38). $T_4-B_2=(r-1)-(s + 1 - sr)=sr+r-s-2<0$, since $r(s)<\frac{s+2}{s+1}$ by (36). $B_2-T_2=(s + 1 - sr)-r=s+1-sr-r<0$ as $r(s)>1$. $T_2=r\le s$ is valid by (35). Equation (14) is thus proved.

  • Equation (15) Since $c=1-D_2=1-(T_2-B_2)=B_2+(1-T_2)=B_2-T_4$ by (8) and (5), we have $c>0$ by (14). $c-B_3=(-rs-r+ s+2)-(2s - 2r - rs + 2)=r-s<0$, since $r(s)<s$ by (35). $B_3-T_3=(2s - 2r - rs + 2)-(sr-s)=-2rs-2r+3s+2<0$, since $r(s)>\frac{3s+2}{2s+2}$ by (38). $T_3-B_1=(sr-s)-(s + 1 - r)=sr+r-2s-1<0$, since $r(s)<\frac{2s+1}{s+1}$ by (36). $B_1-T_1=(s + 1 - r)-sr=s+1-r-sr<0$ as $r(s)>1$. (15) is thus proved.

  • Equation (16) By (39), $r(s)\ge \frac{2s+1}{2s}$. Hence, $b=2rs-2s-1\ge 0$. Equation (16) is thus proved.

• (ii)

  • Equation (17) The first inequality of (17) is due to (16). $(b+c)-B_4=(2rs-2s-1-rs-r+s+2)-(4s-2r-3rs+3)=4rs+r-5s-2\le 0$, since $r(s)< \frac{5s+2}{4s+1}$ by (40). By (9) and (16), $B_4 \le B_3$. $B_3-2c=(2s-2r-rs+2)-(-2rs-2r+2s+4)=rs-2\le 0$, since $r(s)\le \frac{2}{s}$ by (40). Equation (17) is thus proved.

  • Equation (18) Since $r(s)=\frac{6s+6}{4s+5}$ when s is small, $B_{3}+B_{4}=-4rs-4r+6s+5=r-1= T_{4}$, which is (18).

  • Equation (19) $(B_2+B_4)-(1+c)=(s+1-rs+4s-2r-3rs+3)-(1-rs-r+s+2)=-3rs-r+4s+1\le 0$, since $r(s)>\frac{4}{3}>\frac{4s+1}{3s+1}$ by (38). Equation (19) is thus proved.

• (iii)

  • Equation (20) $B_2-(B_3+B_4)=(s+1-rs)-(2s-2r-rs+2+4s-2r-3rs+3)=3rs+4r-5s-4\le 0$, since $r(s)\le \frac{5s+4}{3s+4}$ by (43). $(B_3+B_4)-(s-1)=(2s-2r-rs+2+4s-2r-3rs+3)-(s-1)=-4rs-4r+5s+6\le 0$, since $r(s)\ge \frac{5s+6}{4s+4}$ by (41). Equation (20) is thus proved.

  • Equation (21) $T_{4}-3B_4=(r-1)-(-9rs-6r+12s+9)=9rs+7r-12s-10\le 0$, since $r(s)\le r_2(s)=\frac{12s+10}{9s+7}$ by (33). $3B_{4}-B_{2}=(-9rs-6r+12s+9)-(s+1-rs)=-8rs-6r+11s+8\le 0$, since $r(s)\ge \frac{11s+8}{8s+6}$ by (41). Equation (21) is thus proved.

  • Equation (22) $e=\frac{1}{2}r-3s+3rs-\frac{5}{2}\ge 0$, since $r(s)\ge \frac{6s+5}{6s+1}$ by (42). $a+c=(rs+3r-2s-3)+(-rs-r+s+2)=2r-s-1\ge 0$, since $r(s)\ge \frac{s+1}{2}$ by (37). By the definition of f and d, we have $f\ge a+c \ge 0$, $d\ge e$ and $d\ge f$. Equation (22) is thus proved.

  • Equation (23) $d-B_4=(6rs+ 3r-7s-6)-(4s - 2r - 3rs + 3)=9rs+5r-11s-9\le 0$, since $r(s)\le \frac{11s+9}{9s+5}$ by (34). Equation (23) is thus proved.

  • Equation (24) $a-c=(rs+ 3r-2s-3)-(-rs-r+ s+2)=2rs+4r-3s-5\le 0$, since $r(s)\le \frac{3s+5}{2s+4}$ by (43). Equation (24) is thus proved.

  • Equation (25) $T_4-(b+2c)=(r-1)-(2rs-2s-1-2rs-2r+2s+4)=3r-4\ge 0$, since $r(s)> \frac{4}{3}$ by (38). Together with (15) and (16), (25) is thus proved.

• (iv)

  • Equation (26) From (44) we have $r(s)\le \frac{18s+16}{16s+7}$. Therefore, $c+B_2-2d-b=(-rs-r+ s+2)+(s+1-rs)-(12rs+ 6r-14s-12)-(2rs-2s-1)=-16rs-7r+18s+16\ge 0$. Equation (26) is thus proved.

• (v)

  • Equation (27) From (45) we have that $r(s)=r_2(s)\le \frac{17s+14}{15s+6}$. Therefore, $c+T_4-2d-b=(-rs-r+ s+2)+(r-1)-(12rs+ 6r-14s-12)-(2rs-2s-1)=-15rs-6r+17s+14\ge 0$. Equation (27) is thus proved.

• (vi)

  • Equation (28) From (46) we have that $r(s)\ge \frac{7s+7}{7s+2}$. Therefore, $c+B_2-d-B_3=(-rs-r+ s+2)+(s+1-rs)-(6rs+ 3r-7s-6)-(2s-2r-rs+2)=-7rs-2r+7s+7\le 0$. Equation (28) is thus proved.

• (vii)

  • Equation (29) Since $r(s)=\frac{8s+7}{3s+10}$ when s is bigger medium, $B_{4}=4s - 2r - 3rs + 3=8r-4s-4=4a+4c$. Equation (29) is thus proved.

  • Equation (30) From (47) we know $r(s)\ge \frac{4s+4}{13-3s}$. Therefore, $b-(8c+7a)=(2rs-2s-1)-(-8rs-8r+8s+16+7rs+21r-14s-21)=3rs-13r+4s+4\le 0$. Since $r(s)=r_4(s)=\frac{8s+7}{3s+10}\le \frac{8s+7}{15}$ for any $s\ge \frac{5}{3}$, $8c+7a-B_3=(-8rs-8r+8s+16+7rs+21r-14s-21)-(2s - 2r - rs + 2)=15r-8s-7\le 0$. Equation (30) is thus proved.

• (viii)

  • Equation (31) From (48) we have that $r_2(s)\ge \frac{4s+3}{6s-3}$. Therefore, $e-(a+c)=(\frac{1}{2}r-3s+3rs-\frac{5}{2})-(rs+ 3r-2s-3-rs-r+ s+2)=3rs-\frac{3}{2}r-2s-\frac{3}{2}\ge 0$. (31) is thus proved.

  • Equation (32) $c+T_4-2e-b=(-rs-r+ s+2)+(r-1)-(r-6s+6rs-5)-(2rs-2s-1)=-9rs-r+9s+7\ge 0$, since $r(s)\le \frac{9s+7}{9s+1}$ by (34). Equation (32) is thus proved.

$\square$