6 Supplmntary Matrials APPENDIX A PHYSICAL INTERPRETATION OF FUEL-RATE-SPEED FUNCTION A truck running on a road with grad/slop θ positiv if moving up and ngativ if moving down facs thr rsistancs: arodynamic air rsistanc rolling rsistanc and grad rsistanc 62]. Th air rsistanc is th friction of air which is modld as F a v = 2 ρa f c d v 2 8 whr ρ is th air dnsity and A f th frontal ara of th truck and c d is drag cofficint of th truck s Tab. III for c d and A f and v is th spd of th truck. Th rolling rsistanc is th friction btwn th tirs and th ground which is modld as F r = c r mg cos θ 9 whr c r is th cofficint of rolling rsistanc friction cofficint btwn th tirs and th ground m is th truck mass and g is gravitational acclration. Th grad rsistanc is th forc of th gravity on th opposit dirction of truck movmnt i.. F g = mg sin θ. 2 Thn th tractiv forc is F t v = F a v + F r + F g 2 which yilds to th powr consumption P f v = F t v v = 2 ρa f c d v 3 +c r cos θ+sin θmgv. 22 W can rgard P f v as th powr dmand to mov th truck on th road with constant spd v. To provision such powr dmand th intrnal combustion ngin ICE nds to convrt ful into mchanical nrgy. Thr ar a sstantial numbr of modls for ICE 49]. For th purpos of this physical intrprtation w us th following rlationship s 49 Equation ] P f = fv LH η 23 whr fv is th ful rat consumption unit: gallon pr hour LH is th lowr hating valu of th ful unit: KJ pr gallon and η is th ful fficincy 8. Eq. 23 givs th ful-rat-spd function fv as follows fv = P f LH η = 2 ρa f c d v 3 + c r cos θ + sin θmgv LH η 24 which shows that th ful-rat-spd function is polynomial with spd v and also strictly convx. Thrfor such physical intrprtation justifis our assumption for th ful-rat-spd function in Sc. II-A. 8 Th unit of powr dmand P f would b KW. W can appropriatly mak all units consistnt. APPENDIX B PROOF OF LEMMA W can prov this lmma by using th continuous Jnsn s inquality. For any spd profil v : ] R + ovr road/dg th incurrd ful consumption is f vtdt and th travlld distanc is vtdt. As w rquir that th truck must pass dg with xactly hours w must hav t vtdt = D. 25 Sinc f is convx according to th continuous Jnsn s inquality 63 Ch. 2.4] w hav t f t vtdt f vtdt D = f 26 which mans t D f vtdt f 27 with quality whn vt = D for all t ]. APPENDIX C PROOF OF LEMMA 2 Sinc th ful-rat-spd function f v is a polynomial function and thus twic diffrntiabl with rspct to v w can thus obtain th first and scond-ordr drivativ of c = f D with rspct to i.. and c = f D D f D 28 c = f D D t 2 D t 2 f D + D f D D t 2 ] = D2 t 3 f D. 29 Sinc f is strictly convx ovr th spd limit rgion w hav f D > and thus w conclud that c > 3 which provs that c is strictly convx with rspct to ovr t lb t ]. For th scond part of this lmma w first obsrv that c is a diffrntiabl and thus continuous and strictly incrasing function. Thus w will considr th following thr cass. Cas c t lb : In this cas w know that c is strictly incrasing ovr t lb t ] and w can st ˆ = t lb. Cas 2 c t lb c t : In this cas w can find a ˆ c t lb c t such that c ˆ = du to th continuity of c. Cas 3 c t : In this cas w know that c is strictly dcrasing ovr t lb t ] and w can st ˆ = t.
7 In all thr cass w obtain that c is first strictly dcrasing ovr t lb ˆ ] and thn strictly incrasing ovr ˆ t ]. Not that ˆ could b on th boundary of t lb t ] as shown in Cas and Cas 3. APPENDIX D PROOF OF LEMMA 3 First sinc p and t p is a fasibl solution to PASO w hav OPT cp t p. Scond sinc Algorithm 2 rturns in lin 3 th path cost will b no gratr than som c N thus w hav cp t p c = which clarly implis that min{ c + N + } N c = c + p. Thn w hav cp t p = c = c ] + c = cp t p which yilds to cp t p cp t p N = U + n + U + n + = U + n + = U + Lδ. APPENDIX E PROOF OF LEMMA 4 For PASO lt us dnot p t p as an optimal solution. Namly p is an optimal path and t p is th corrsponding optimal travl tim st. For ach dg p w must hav min{ c + N + } = c +. Suppos not. Thn which mans c + > N + c c > N = U + n + > U OPT. This is a contradiction to c c t = OPT. Thn w hav cp t p = c = min{ c + N + } = c ] + ] c + OPT + n U + n U + + n = N. 3 Hr is a critical stp which is diffrnt from Lmma 3 in 5] for RSP problm. For ach dg p may not b a rprsntativ point in vctor τ. Howvr w can considr th rprsntativ point = τ i whr i c which incurs th sam ful cost i.. c = c. Clarly w also hav cp t p N and whr t p { : p }. Thrfor path p and travl tim t p must b xamind by Algorithm 2 which complts th proof of th first part i.. Algorithm 2 must rturn a fasibl path p and travl tim t p. Morovr w hav cp t p cp t p = cp t p. 32 From 3 w first not that cp t p OPT + n. 33 Scond sinc Algorithm 2 rturns in lin 3 w must hav cp t p c = which clarly implis that min{ c + N + } N c = c + p. W thn not that cp t p = c = min{ c + N + } = c + c = cp t p. 34 Insrting inqualitis 33 and 34 into 32 w obtain which mans cp t p OPT + n cp t p OPT + n OPT + Lδ. Th proof is comptd. APPENDIX F PROOF OF THEOREM 2 Th first part of this thorm dirctly follow th analysis of St -3 in Sc. III-C. Namly Algorithm 3 rturns a +ɛ- approximat solution for PASO in tim Omn log ξ + mn 2 log log UB mn log ξ + + mn2. 35 LB ɛ ɛ2 Now w prov th scond part of this thorm. Namly if w us LB = and UB = nc whr min E c t and C max E c t lb Algorithm 3 has tim complxity polynomial in th input siz of th problm PASO and thrfor is an FPTAS. According to 35 w only nd to show log log UB LB = log log nc is polynomial in th input siz.
Suppos that C max E c t lb = c t lb. For dg w should input all its proprtis i.. {D R lb R f } whr f is a polynomial function. Suppos that f x = a x k + a 2 x k2 + + a q x kq. Thn to input ful-rat-spd function f w only nd to input a k a 2 k 2 a q k q. Thrfor for dg w should input th following ral numbrs {D R lb R a k a 2 k 2 a q k q }. Th input siz for dg is I log D + R lb + R + a + k + a 2 + k 2 + + a q + k q whr is th machin ilon i.. th maximum rlativ rror of for rounding a ral numbr to th narst floating point numbr that can b rprsntd by a digital machin. Now lt us show that log log C is polynomial in I. According to th dfinition of th ful-tim function c in 2 w gt C c t lb log log = log log = log log tlb f D t lb = log log ] D f R = log log + log = log log log R D R log ] f R I + log D R f R ] f R + log Sinc R > and thus R ] a R = log I + log k + + a q R kq ] qai R log I + log ki Dfin i arg max a jr j q] kj ai R = log I + log q + log + log ki ki ki ] R log I + I + I + k i log Sinc log < log ] I + I + I + k i R log Sinc < log I + I + I + k ] i I log ] I + I + I + 2 I I = log I + log3 + 2 I log I + log3 2 I + 2 I = log I + I + 2 = OI which is thus polynomial in I. Thn log log nc = log log nc log log nc 8 = log log n + log C } 2 max {log log n log log C = max{olog log n OI } 36 which is polynomial in th input siz of PASO bcaus both Olog log n and OI ar polynomial in th input siz of PASO. W thus prov th scond part of this thorm. APPENDIX G PROOF OF LEMMA 6 Dfin function h = c + λ. Thn w can gt th first drivativ as h = c + λ. 37 Sinc c is a strictly convx and strict dcrasing function w know that c and also h is a strictly incrasing function and c < at intrval t lb t ]. W thn considr th following thr cass. Cas : If λ < c t w gt that c t + λ < and thus h h t < t t lb t ]. 38 This shows that h is strictly dcrasing at t lb t ] and th minimal valu is attaind at t λ = t. Cas 2: If c t λ c t lb thn w can gt that c λ t lb t ]. Clarly th monotonic incrasing proprty of h implis that h < at t lb c λ and h > at c λ t lb ]. This mans that th minimal valu is attaind at t λ = c λ. ] Cas 3: If λ > c t lb w gt that c t lb + λ > and thus h h t lb > t t lb t ]. 39 This shows that h is strictly incrasing at t lb t ] and th minimal valu is attaind at t λ = t lb.
9 APPENDIX H PROOF OF THEOREM 3 Lt us considr any two λ λ 2 with λ < λ 2. W nd to prov δλ δλ 2. Suppos that th optimal path at λ is p λ = p and th optimal path at λ 2 is p λ 2 = p 2 9. For any path p and any λ w dnot its optimal gnralizd path cost as W p λ w λ = and dnot its corrsponding path ful cost as C p λ and dnot its corrsponding path dlay T p λ c t λ + λt λ] 4 c t λ. 4 t λ. 42 Clarly w hav W p λ = C p λ + λt p λ. Basd on such notations w hav δλ = T p λ and δλ 2 = T p2 λ 2 and w nd to prov T p λ T p2 λ 2. Whn λ = λ th optimal path is p which mans that W p λ = C p λ + λ T p λ W p2 λ = C p2 λ + λ T p2 λ 43 Similarly whn λ = λ 2 th optimal path is p 2 which mans that W p2 λ 2 = C p2 λ 2 + λ 2 T p2 λ 2 W p λ 2 = C p λ 2 + λ 2 T p λ 2 44 Now w will us th fact that t λ minimizs w λ as dfind in 3. Sinc both t λ and t λ 2 ar fasibl i.. in th intrval t lb t ] w gt that W p2 λ = C p2 λ + λ T p2 λ Similarly w hav = 2 c t λ + λ t λ = 2 min t lb t t c + λ 2 c t λ 2 + λ t λ 2 = C p2 λ 2 + λ T p2 λ 2. 45 W p λ 2 = C p λ 2 + λ 2 T p λ 2 C p λ + λ 2 T p λ. 46 Insrting 45 into 43 w gt that C p λ + λ T p λ C p2 λ 2 + λ T p2 λ 2 which implis that λ T p λ T p2 λ 2 ] C p2 λ 2 C p λ. 47 Similarly insrting 46 into 44 w gt that C p2 λ 2 + λ 2 T p2 λ 2 C p λ + λ 2 T p λ 9 Paths p and p 2 could b th sam. which implis that λ 2 T p λ T p2 λ 2 ] C p λ C p2 λ 2. 48 Summing 47 and 48 w gt that λ λ 2 T p λ T p2 λ 2 ]. 49 Sinc w assum that λ < λ 2 w must hav T p λ T p2 λ 2. 5 APPENDIX I PROOF OF THEOREM 4 At th point λ th dual function has valu Dλ = λ T + min x min c + λ x X t E lb t t = λ T + min x c t λ + λ t λ x X E = λ T + c t λ + λ t λ ] λ = λ T + λ c t λ + λ = λ T + λ δλ + = λ T + λ T + = λ λ λ λ c t λ c t λ t λ c t λ. 5 On on hand w know that any dual function valu will b a lowr bound of OPT according to th wak duality. Thus Dλ OPT. 52 On th othr hand w know that p λ is a fasibl path and {t λ p λ } satisfis t λ = T. 53 λ Hr p λ and {t λ p λ } is a fasibl solution to PASO with th objctiv valu λ c t λ = Dλ which is an uppr bound of OPT i.. Dλ OPT. 54 Eq. 52 and 54 conclud that Dλ = OPT and p λ and {t λ p λ } is an optimal solution to PASO. APPENDIX J PROOF OF THEOREM 5 First if w lt total travl dlay b T = λ L t λ L > T w gt a rlaxd vrsion of PASO. According to Thorm 4 w know that LB = λ L c t λ L is th optimal solution of th rlaxd vrsion and thus w hav LB OPT.
Scond sinc λ U t λ U < T w know that p λ U and {t λ U : p λ U } is a fasibl solution to PASO. Thus UB = λ L c t λ U OPT. 2