The choice of an optimal LCSCR contract involves the choice of an x L. such that the supplier chooses the LCS option when x xl

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1 EHNIA APPENDIX AMPANY SIMPE S SHARIN NRAS Proof of emma. he choice of an opimal SR conrac involves he choice of an such ha he supplier chooses he S opion hen and he R opion hen >. When he selecs he S opion and his innae cos realizaion is, he supplier s choice of cos reducion is deermined b: Maimize α... implies: α... implies ha hen he chooses he S opion, he supplier s profi under he S opion a he cu-off value of innae coss,, is: α α α..3 α Because he supplier is indifferen beeen he o opions hen his innae cos realizaion is and since his profi is alas zero under he R opion,.3 implies ha: α α... implies ha he buer s procuremen cos hen is: α α α α α..5 rom. and.5, he buer s procuremen cos hen α α α α : α α α..6 he reducion in epeced procuremen coss from he SR conrac ih cos reimbursemen fracion α relaive o he R conrac is: rom.7: α α d..7 α

2 α α f α d α α f..8.8 implies α as f > α..9 Because / f is monoonicall increasing,.9 implies he unique value of ha maimizes α is: α minimum{ α, }.. Proof of emma. he buer s problem of minimizing he epeced cos of securing one uni of he good is: Minimize, d. subjec o: u for all, ; and. u u for all,,,.3 here u,., and.5. Differeniaing.5 ih respec o provides: d d..6 Defining u u, i follos from.3,., and.6 ha a he soluion o he buer s problem:

3 3 d d u..7 Since.7 implies he supplier s uili declines ih, he buer opimall ses u. herefore: d u ~ ~..8.7 and.8 impl ha he supplier s epeced uili under he opimal conrac is: d f du u d u..9 he firs equali in.9 follos from inegraion b pars. he second equali in.9 holds because u u u. Because he agen s uili is he difference beeen he pamen he receives and he coss he incurs: u... implies he epeced pamen o he supplier is: d u d...9 and. impl: d f d.. he derivaive of. ih respec o is: f..3 Because and,.3 can be rien as:..

4 Because. implies hen, i follos ha f epeced procuremen coss are minimized hen for all.. implies ha hen, epeced procuremen coss iniiall decline and hen increase ih. herefore, he opimal cos reducion is inerior, and is deermined b..5.5 implies: hen..6 Proof of emma 3. e denoe he reducion in he buer s epeced procuremen coss under he opimal conrac relaive o he cos reimbursemen conrac. o analze, i is necessar o consider o cases. In he firs case case I, is relaivel large, and so. onsequenl, he supplier is opimall induced o suppl sricl posiive effor levels for small s and o suppl no effor for large s. In he second case ase II, is relaivel small, and so he supplier is induced o suppl a sricl posiive level of effor for all realizaions of. ase I: or. f d 3. d 3. d d d 3.5

5 follos from 3. above b leing. Noice ha varies from o as varies from o. 3. follos from.6. ase II: or. d f 3.8 d d 3. d 3. d Noice again ha 3. follos from 3.9 b leing. Also noice ha varies from o as varies from o. 3. follos from.6. I follos from.7 and. ha he epeced gain from he opimal PR conrac relaive o he R conrac is:

6 * * d. 3.5 here minimum {, }. 3.6 * o analze he epeced gain in 3.5, i is convenien o consider o cases. In he firs case ase I, is relaivel large, and so. onsequenl, is inerior, and so he agen chooses he fied price conrac for lo s and he cos reimbursemen conrac for high s. In he second case, is relaivel small, and so he agen chooses he fied price conrac for all realizaions of. ase I: or. I follos from 3.5 and 3.6 ha: d 3.7 d 3.8 d follos from 3.7 b leing. Noice ha varies from o as varies from o. ase II: or. I follos from 3.5 and 3.6 ha: d 3. 6

7 7 d 3. d 3.3 d , 3.6 here. Again, 3.3 follos from 3. b leing. Noice ha varies from o as varies from o. o complee he proof, e analze separael he hree cases implied b he analsis immediael above and he analsis in he proof of emma. ase. or. ase here corresponds o ase I in boh he analsis of he opimal conrac and he analsis of he PR conrac. herefore, from 3.7 and 3.:. 3.7 ase., or,. ase here corresponds o ase II in he analsis of he opimal conrac and o ase I in he analsis of he opimal PR conrac. herefore, from 3. and 3.: 3.8

8 reveals ha / declines ih in his case if D is an increasing funcion of. D is an increasing funcion of because: D. 3.3 ase 3., or,. ase 3 here corresponds o ase II in he analsis of boh he opimal conrac and he PR conrac. herefore, from 3. and 3.6:. 3.3 Noice ha / increases ih because he numeraor of his erm increases ih hile he denominaor decreases ih. Also noice ha since, in ase 3, > >. herefore, i follos from 3.3 ha / is a decreasing funcion of in his case. Proof of emma. hen. herefore, from 3.7: limi limi.. he las equali in. follos from Hopial s Rule. Proof of Proposiion..7 and. impl ha, relaive o he cos reimbursemen conrac, he epeced gain from he opimal SR conrac given α, α, is:

9 α α d. P. o analze his epeced gain, e need o consider o cases. In he firs case ase I, is relaivel large, and so he supplier chooses he S opion for lo s and he R opion for high s. In he second case ase II, is relaivel small, and so he agen chooses he S opion for all realizaions of. ase I: α or. α α α α α d. α P. α α d α α d P.3 α α P. α α P.5 α α. P.6 P.3 follos from P. b leing α from o as varies from o α.. Noice ha varies ase II: α or. α α d. α P.7 α α α α d α α α α α d P.8 P.9 9

10 d α P. α P. α, P. here / α. Again, P.8 follos from P.7 b leing α varies from o α as varies from o.. Noice ha or given, and, he opimal SR conrac is derived b choosing α opimall. ase I: or. α α α Since α α α. P.3, he sign of he parial derivaive of ih respec o α α has he same sign as he derivaive of α α, hich is α α. Seing his derivaive equal o zero reveals ha he value of α ha uniquel maimizes α is: α. P. Subsiuing P. ino P.3 provides: *. P.5 α ase II:, or,. ase IIA: α. α α α. P.6

11 Again, he parial derivaive ih respec o α of he epression in has he same sign as he α epression α α. When α : α he inequali in P.7 holds because e are in ase II.. P.7 When α, α >. herefore, he value of α ha uniquel maimizes in his range is: α α. P.8 Subsiuing P.8 ino P.6 provides ase IIB: >. P.9 α. α α. P. α he sign of he parial derivaive ih respec o α of he epression for in P. is he α same as he sign of he derivaive of derivaive is negaive because: α α, hich is α. his α P. α α P.

12 . P.3 P.3 holds because e are in ase II. P. holds because e are in ase IIB. he negaive derivaive implies he opimal α ill never lie in his range. ase III: or,. ase IIIA: α. α α α. P. he parial derivaive ih respec o α of he epression for in P. has he same sign α as he derivaive of α α, hich is α α. his derivaive is posiive because: α α > > α P.5. P.6 he las equivalence in P.6 holds, given ha e are in ase III. he las equivalence in P.5 holds because e are in ase IIIA. he posiive derivaive implies he opimal α ill no lie in he specified range. ase IIIB: α. α α α. P.7 he parial derivaive of he epression for α ih respec o α in P.7 has he same sign as he derivaive of α α Noice ha hen α,, hich is α.

13 3 α. P.8 Also noice ha α hen α. onsequenl, he opimal value of α in his range is he soluion o: α. herefore, he unique maimizer of α in his case is: α. P.9 Subsiuing P.9 ino P.7 provides:. P.3 I is no useful o prove ha for given, is a non-increasing funcion of. he proof proceeds b eamining hree cases: ase I: or. In his case:. P.3 ase II:, or,.. P.3

14 P.3 implies i suffices o sho ha is an increasing funcion of. his fac follos because he derivaive of his funcion is:. P.33 ase III: or,.. P.3 Since is a decreasing funcion of hile is an increasing funcion of, P.3 implies / is a decreasing funcion of. he foregoing analsis implies ha / aains is smalles value in ase I immediael above. onsequenl, he proof follos if: e >. P.35 o sho ha P.35 holds, firs recall ha e lim. No i ill be useful o sho ha ln ln is an increasing funcion of on,. o do so, e ill sho ha is derivaive is declining ih, and is non-negaive a. } { ln d d ln ln. P.36 As, ln ln. P.37 I remains o sho ha he derivaive of ln is negaive on,. his derivaive is:

15 5. P.38 P.38 implies is an increasing funcion of, hose limi as is e. onsequenl, e for,. hus: / > e. Proof of Proposiion. We firs prove ha / is a non-increasing funcion of. he proof proceeds b analzing four disinc cases. ase I: or. rom 3., P.3, and 3.6:,, and. onsequenl:. P.

16 6 ase II:, or,. rom 3., P.3, and 3.:,, and /. onsequenl:. P. is a decreasing funcion of if and onl if is an increasing funcion of. his is he case if and onl if R is a decreasing funcion of, here R. P.3 Noice ha > R for all relevan. herise, i ould follo from P. ha E E >, hich canno be he case. he derivaive of R ih respec o on, is: S. P.

17 7 We ill no sho ha S for, b shoing ha i is sricl conve and he values of is ending poins are non-posiive. Noice, firs, ha S hen. P.5 Ne observe ha S is conve because: S, and so > S. P.6 No noe ha hen : S, hich is nonposiive because: ln ln ln Z. P.7 Noice ha ln ln ln Z. herefore, o sho ha Z ln for all, i suffices o sho ha Z for all. rom P.7: ln ln ln ln Z. P.8 P.8 implies is a decreasing funcion of on,. ase III:, or,. rom 3., P.9, and 3.:,, and. onsequenl:

18 8. P.9 o sho is a decreasing funcion of, i suffices o sho W is an increasing funcion of, here: W. P. W. P. ase IV: or. rom 3.7, P.5, and 3.:,, and. onsequenl:. P. Noice ha he epression in P. is independen of. In summar, e have shon is a non-increasing funcion of in all four cases.

19 9 Because declines as increases, i suffices o sho ha > in ase IV above. herefore, from P., o complee he proof, i suffices o sho: > for. P.3 P.3 holds if and onl if: > P. > P.5 M. P.6 rom he proof of Proposiion, is increasing in and is bounded above b e. Also noice ha: ln.69 ln.69. P.7 P.7 implies is a decreasing funcion of. We ill no demonsrae ha M over all relevan inervals in hich. Iniiall, suppose.. Because e.78, and because is decreasing in, e no ha in his range: M. P.8

20 . 3. No suppose.3,. 3.. Because is increasing in, in his range. Also, since is decreasing in, in his range. herefore, M / 3.99/. 3 in his range No suppose.,.3.3. Because is increasing in, in his range. Also, since is decreasing in, in his range. herefore, M.97.73/ / in his range. inall, suppose,... Because is increasing in, in his range. Also, since is decreasing in, ¾..75 in his range. herefore, M.65.75/ 3.96/ in his range. he simulaions in Appendi B reflec he folloing analsis. rom A3 in Appendi A: d f. B Maimizing ih respec o provides: / f. B Subsiuing ino provides: * d, B3 f

21 here, from B: rom A6 in Appendi A: * * / f. B * * d. B5 Maimizing ih respec o rom A in Appendi A: * provides: * * / f. B α α α d. B5 Maimizing α ih respec o provides: α / f. B6 Solving B6 for α provides: α. B7 f Maimizing α ih respec o α provides: α d. B8 B8 follos in par from he fac ha d d. Subsiuing B7 ino B8 reveals: / f d /. B9 he simulaions idenif he opimal values of from B9 and α from B7.