Proofs for all the results presented in the body of the article are presented below: ProofofLemma1:Consider a Nash schedule S where x O i > P i

ONLINE SUPPLEMENT FOR Non-Cooperative Games for Subcontracting Operations by George L. Vairaktarakis Weatherhead School of Management Department of Operations Case Western Reserve University 10900 Euclid Avenue, Cleveland, OH 44106-735 email: gxv5@case.edu Proofs for all the results presented in the body of the article are presented below: ProofofLemma1:Consider a Nash schedule S where x O i > P i for some i M. Then, the workload of player i on M i is P i x O i and his makespan C i is attained on F. Hence, C i x O i. Let y i be the workload of player i processed on F during the time interval [P i x O i,ci ]. By definition, 0 <y i x O i Also,. C i P i x O i + y i because the y i units start on F after time P i x O i. Then, remove the y i/ units of workload from F and schedule them on M i ;let S be the resulting schedule and C i be the new makespan of player i. Clearly, C i C i y i <Ci and hence the makespan of player i in S is better than that in S; this is a contradiction to S being a Nash schedule. Proof of Theorem 1: For proof by contradiction, let S be a Nash schedule such that P i P i+1 but x O i >x O i+1 for some 1 i<m. We assume that i is the smallest index for which this property holds; we refertothisastheminimality assumption for index i. Then, consider the difference δ =(x O i xo i+1 )/. In S, playeri + 1 is processed before i on F because x O i+1 <xo i.lett, t be the completion times for player i +1 onmachines M i+1,frespectively. If t <tplayer i + 1 could increase his workload on F by δ =min{ t t,δ}. This would reduce his makespan by δ / and wouldn t change his processing order on F because x O i+1 + δ x O i+1 + δ<xo i while obviously x O i 1 xo i+1 (due to the minimality assumption) before the reallocation and hence x O i 1 xo i+1 + δ afterwards. On the other hand, if t >tplayer i + 1 could decrease his workload on F by δ and improve his makespan by δ /onm i and possibly have his workload scheduled earlier on F because x O i+1 δ < x O i+1. In either case player i + 1 could improve his objective contradicting that S is a Nash schedule. Therefore, it must hold that t = t and the Gantt diagram for M i, M i+1, F has the configuration depicted in Figure 3. Note that P i P i+1, x O i+1 <xo i imply P i x O i <P i+1 x O i+1. But then, player i can improve his makespan by reducing his workload x O i on F ; contradiction to the fact that S is Nash. This completes the first part of the theorem. An argument similar to proving t = t and Lemma 1 yield that player i attains his makespan on M i, i M. Equivalently, i k=1 x O k P i x O i. Proof of Theorem : Given Nash strategies x 1,x,...,x M for the players, the makespan C i of player i at equilibrium is C i =max{ i k=1 x k,p i x i } = P i x i i M due to Theorem 1. Hence, 1

Figure 3: Schedule configuration for players [i], [i + 1]. x 1 +...+ x i 1 +x i P i and x 1... x i due to IRO. Then, (i +1)x 1 P i or x 1 P i From i+1 for i 1. x 1 + x +...+ x i P i x i (5) and i = 1 we see that the larger x 1, the smaller the makespan C 1 = P 1 x 1 for player 1. Therefore, player 1 will select strategy x O 1 = min i 1 P i i +1. (6) For x 1 = x O 1 inequality (5) yields x O 1 + x +...+ x i P i x i x +...+x i P i x O 1 or ix P i x O 1 for i (due to IRO), or x P i x O 1 i. From (5) and i = we observe that the larger x the smaller the makespan C = P x of player. Therefore, x O = min i P i x O 1 i. (7) Iterative arguments and expressions (6), (7) yield the result. Proof of Lemma : We first consider property (a). Let S be a Nash schedule. We first prove x P i P i by revising S (if necessary) so that the property holds for every player. Suppose that there exists player i with x P i > P i who attains his makespan (say Ci )onf while the completion time on M i is t i = P i x P i < P i <xp i C i. Then, we can assign on M i the workload processed on F during the interval [t i,c i ]. This reallocation does not increase the makespan of player i and the revised subcontracted workload, say x i,isnomorethant i = P i x P i < P i, i.e., x i P i in the revised schedule. Hence, there exists a Nash schedule such that x P i P i for all i M lets be such a schedule. If P i p i max P i (i.e., pi max P i )thenmin{p i p i max, P i } = P i and (a) holds trivially. Hence, it suffices to consider the case where p i max > P i for some player i. Let job j N i attain p ij = p i max.ifjis processed exclusively on M i,wehavex P i <P i p ij < P i and (a) holds. If p ij is not processed on M i exclusively then the makespan of player i is C i p ij > P i since no overlapping is allowed in S. Lety 1, y be the workload portions of p ij processed on M i, F respectively. Then, exchange the y periods of time that F is busy with p ij. This exchange preserves the property that jobs of player i are not processed simultaneously on M i and F. Also, the makespan

of i does not increase because the y time units of p ij are processed at the same time just on a different machine. Also, the total workload of i processed on F does not increase and (according to IRP) i is not scheduled later on F. However, after the exchange the job j is processed entirely on M i and hence x P i P i p ij = P i p i max =min{p i p i max, P i } because p i max > P i. This concludes property (a). We now consider property (b). For contradiction, suppose S is a Nash schedule that satisfies property (a) and there exists player [i] with x P 1 +...+ x P [i] > P [i] x P [i]. (8) Then, the makespan of player [i] isc [i] = x P 1 +...+xp [i]. Define Δ = min{ 1 (C[i] P [i] +x P [i] ),xp [i] } > 0. Replace x P [i] by x [i] = x P [i] Δ. Following the reallocation of his workload, the makespan of player [i] becomes C [i] C [i] Δ <C [i] because Δ > 0. If Δ = x P [i],then x [i] = 0 and no overlapping is possible. Otherwise, one can start processing jobs in N [i] on F, one at a time, starting at time C [i] x P [i] continuing until time C [i] possibly preempting a single job say j N [i], continuing with job j on M [i] at time 0, continuing with the rest of the jobs in N [i] until time C [i], preventing any overlapping due to the fact that x [i] satisfies property (a). Therefore, S can be revised to satisfy property (b) for player [i] or any other player violating property (b). This completes the proof of the lemma. ProofofTheorem3: Let x P 1,x P,...,x P M be the player strategies in a Nash schedule S chosen so that they satisfy Lemma. We prove the theorem by establishing that: a) If P i P k and P i p i max P k p k max, thenx P i x P k and player k s workload precedes i s on F, and b) If P i P k and P i p i max <P k p k max,thenplayerk s workload precedes i s on F. The properties imply the result. We prove claim a) by contradiction. Suppose S is a Nash schedule such that a) holds but x P i <x P k. Then, according to IRP player i precedes k on F as in Figure 4a (where the makespans of i, k are attained on M i,m k respectively due to Lemma (b)). Since P i P k and P i p i max P k p k max we have min{ P i,p i p i max} min{ P k,p k p k max} and hence x P i <x P k min{ P k,p k p k max} min{ P i,p i p i max}. Therefore player i could (if it were beneficial) subcontract at least x P k units of workload to F. Let Δ= 1 (xp k xp i ) > 0. Suppose that i reallocates to F x i = xp i + Δ units of workload instead of x P i. As a result player i will reduce his makespan by Δ even though P may now reschedule player i later still before player k, as in Figure 4b. In all cases i s makespan decreases by Δ contradicting that S is a Nash schedule. This proves claim a). To prove claim b) we distinguish 3 cases based on the disposable workload of players i and k as follows: Case i) P i p i max P i and P k p k max P k. In this case we can show as in claim a) that x P i x P k and according to IRP player k precedes i on F. Note that in this case min{ P i,p i p i max} = P i and min{ P k,p k p k max} = P k. Case ii) P i p i max P i and P k p k max P k. This case is not possible because P i P i p i max <P k p k max P k 3

Figure 4: Schedule configurations for players i, k. implies P i <P k ; this is a contradiction to the assumption P i P k. Case iii) P i p i max < P i. For proof by contradiction, suppose that player i precedes k on F as in Figure 4a. We must assume that x P i <x P k since otherwise we could schedule the xp i units of player i immediately after x P k without affecting the makespan of either i or k while ordering the two players in nondecreasing order of their total processing. If x P i = P i p i max then player i cannot benefit by subcontracting more workload to P (due to the assumption that x i satisfies Lemma ) and (according to IRP) can be rescheduled immediately after player k without affecting anyone s makespan. If x P i <P i p i max then consider increasing x P i to x i = xp i +Δwhere Δ=min{P i p i max xp i, xp k xp i Then, i will still precede player k on F because } > 0. x i = x P i +Δ x P i + xp k xp i = xp k + xp i and hence the makespan of i would improve by Δ i contradicting that S is a Nash schedule. In all subcases, claim b) holds. This completes the proof of the theorem. ProofofTheorem4:Let x P i, i M be the player strategies in a pure Nash equilibrium that satisfy Lemma and let S be the associated Nash schedule. We prove the result by contradiction. Suppose that there exist players i, k in S whose workload is processed consecutively on F and are such that min{ P i,p i p i max} min{ P k,p k p k max} but x P i >x P k. Let Δ = x P i x P k. Consider increasing xp k to x k = xp k + Δ or decreasing xp i to x i = xp i Δ. This is doable because x P k <x P i min{ P i,p i p i max} min{ P k,p k p k max}. 4 <x P k

Both reallocations can be made without incurring overlapping as in Lemma. Changing either x P i or x P k will not affect the ordering of i, k on F. If the makespan of either i or k decreases, then S cannot be a Nash schedule. The configuration in Figure 5 is the only configuration where x i and x k yield worse schedules for both players. Figure 5: The equilibrium schedule S. This configuration, however, does not correspond to a Nash schedule because player k can reduce his workload on F by δ = 1 P k + x P min{ck k,x P k } > 0 and improve his makespan by δ. And since x P k δ<x P k min{ P k,p k p k max}, jobs can be rescheduled on M i so as to avoid overlapping; contradicting the fact that S was a Nash schedule before the reallocation. This completes the proof of the theorem. ProofofLemma3:Theorems 1 and 3 indicate that the nondecreasing order of P i s is an equilibrium order when overlapping or preemption is allowed. This means that the quasi-spt order stipulated by IRP does not affect the ordering of a player on F when overlaps are not allowed. Consider equilibrium schedules S O, S with strategies {x O i i M}, {xp i i M} when overlapping or just preemption are allowed respectively, where 1,..., M is an SPT order of P i s. If x P k xo k for every k M the result holds trivially. Otherwise, let k be the first index such that x P k <xo k. If there exists i<k with x P i x P k then IRP implies that xp k = P k p k max and the result holds. Therefore we assume that x P i <x P k <xo k for all i<k. Still, if xp k = P k p k max the result holds. The last observations and the choice of index k imply x O i x P i <x P k <P k p k max for every i<k. (9) We now consider the following subcases. Case i) k x P i < k x O i. Then, consider revising the strategy of player k to k x P k =min{x P k + (x O i x P i ),P k p k max}. Evidently, x P k < xp k <xo k because k (x O i x P i ) > 0and k k 1 x P k x P k + (x O i x P i )=x O k + (x O i x P i ) <x O k 5

because x O i <x P i for all i<k. Moreover, expressions (9) and IRP imply which suggests that the strategies x P 1 x P... x P [k 1] <xp k < x P k <x O k x P i =min{x O i,p i p i max} for i>k are feasible for players k +1,..., M. Since (according to IRO) x O i x O k for i>k, IRP will preserve the k-th position for player k when using strategy x P k. Consequently, the makespan of player k is attained on M k because his revised makespan C k = P k x P k is such that k 1 k k k x P i + x P k x P i + (x O i x P i )= x O i P k x O k due to Theorem 1 < P k x P k. And since P k x P k <P k x P k,therevisedstrategy xp k is more beneficial to player k than strategy xp k, which contradicts the assumption that S is a Nash schedule. Case ii) k x P i k x O i. Then, consider strategies {xo i : i M} in S O and consider revising the strategies of players 1,...,k to x O i = min{x P i,xo i + xo k xp k } i<k, and k 1 x O k = x P k <xo k. Clearly, x O i >x O i for i<kbecause by choice of k we have x P i >x O i for i<k.strategies x O i for i<k improve the makespan of players i<kbecause i x O j j=1 i x P j P i x P i <P i x O i for i<k. j=1 Also, by definition of x O i for i k and the fact that, in S we have x P 1 xp...<xp k,wegetthat The latter observation together with the fact that x O 1 x O... x O [k 1] xo k = x P k <x O k. k 1 x O i k 1 + x P k (x O i + xo k xp k k 1 )+xp k = k x O k imply that players k +1,..., M can subcontract in S O amounts x O k+1... xo M respectively. Since x P k <xo k xo i for i>k,players1,...,k maintain their processing priority on F even when using strategies x O i for i k. Then the fact that these strategies improve the makespan of players 1,...,k 1 contradicts that S O is a Nash schedule. In both cases we reached a contradiction because x P k <xo k. Therefore, in S we must have xp k min{p k p k max,x O k }. Equivalently, if P k p k max <x O k then xp k P k p k max, i.e., x P k = P k p k max. Otherwise x P k xo k. This completes the proof of the lemma. 6

Proof of Theorem 5: Recall from Theorem that x O k = min i k P i x O 1... xo k 1 i + k for k =1,,..., M. (10) Consider the inside min operator in expressions (1) for x P k (r). Its numerator is no less than the numerator in (10), and its denominator less or equal to i + k. And since Γ(0) collects all values k : P k p k max <x O k, we have that, in (1), xp k (r) returns a value no less than min{xo k,p k p k max}. When all i k are already elements in Γ(r 1), then this min operator is null and x P k (r)=p k p k max; i.e., every player subcontracts his entire disposable workload. For k M Γ(0) workloads x P k (1) = min k i M i/ Γ(r 1) P i x P 1 (r)... x P k 1 (r) j<k (P j p j max) j Γ(0) i + k n k,i (r 1) replicate the optimal strategies in Theorem assuming that the workloads of players in Γ(0) are fixed. This is because rule IRO coincides with IRP for players i M Γ(0) who subcontract less than P i p i max. Note that, when computing the best strategy x P k (1) for given k M, the number of players up until player i (k i) that have not yet subcontracted the maximum possible amount P i p i max and their workload is not otherwise fixed (i.e., players 1,,...,k 1) are precisely i (k 1) n k,i (0). Therefore, expressions (1) correctly capture the dynamics of (10) for this situation. Value n k,i (0) is computed using set Γ(0). However, Lemma (a) suggests that x P k (1) cannot exceed amount P k p k max also enforced in (1). If none of the players i M Γ(0) subcontracts amount P i p i max then strategies xp k (1) : k M Γ(0) are optimal due to Theorem and values x P k (r) remain unchanged for every k M and r =,..., M. If on the other hand some player i M Γ(0) subcontracts P i p i max, set Γ(1) includes at least one more player and relations (1) revise the optimal strategies of players not in Γ(1). After at most M iterations, either all players subcontract amount P i p i max, or the optimal strategies have been found. In both cases, strategies x P k ( M ) are optimal. This completes the proof of the theorem. Proof of Lemma 4: If the last job of player i processed on F completes after time P i x N i + min j Ai p ij,playeri would be better off rescheduling the smallest job in A i on M i and revising his strategy to x i = xn i min j Ai p ij. Such a reallocation will not worsen his makespan and it may result in an improvement because x i <xn i and hence earlier players may be forced to subcontract less on F. This proves part (a). The proof of part (b) is similar to Lemma (a). ProofofTheorem6:For proof by contradiction, let S be a Nash schedule such that P i >P j but player i precedes player j on F, as in Figure 6. Since two such players exist, we may assume without loss of generality that they are processed consecutively on F, i.e., i immediately precedes j. Letx i,x j be the subcontracted workloads for players i, j, respectively. According to IRN, we have w i w j. Obviously, f i (w i )=x i, f j (w j )=x j, and by definition of f( ), x i w i and x j w j.ifx j x i,then x j w i which suggests x j f j (w i ) f j (w j )=x j, i.e., f j (w i )=f j (w j )=x j. Then, knapsack sizes w i, w i would be used by IRN to order the subcontracted workloads of players i and j respectively, and since P j <P i,playerj would be scheduled before i. This is not the case in S; hence we must assume x i <x j. Then, the start time t of processing the workload of player j on F must be t<c j = P j x j, as depicted in Figure 6. Otherwise, player j would be better off processing his entire workload on M j and IRN would have scheduled player j before player i on F. If f i (w j )=x i, then knapsack sizes w j, w j would be used for the subcontracted workloads of i, j, and since P i <P j,playerj would again precede i according to IRN. Hence, it must be that f i (w j ) x i and since f i (w i )=x i, w i w j we have f i (w j ) >x i and x i w i <w j. 7

Figure 6: The nonpreemptive equilibrium schedule S. Suppose that the knapsack sizes satisfy w [k] =max j<k x N [j]. Then, w i <w j and x i <x j imply that w j = x j. Then, we rewrite f i (w j ) >x i as f i (x j ) >x i and since by definition of f( ) wehave f i (x j ) x j,weget x i <f i (x j ) x j. The last expression means that player i can draw from a knapsack of size x j and improve his makespan because t +(x j x i ) < C i (11) where C i = P i x i is the makespan of player i. Indeed, we saw that t<c j = P j x j and since P j <P i,wehavet<p i x j. Rearranging terms and subtracting x i from both sides yields (11) which means that strategy f i (x j ) results to a smaller makespan for player i compared to strategy x i, contradicting our assumption that S is a Nash schedule. This completes the proof of the theorem. In what follows we develop an iterative algorithm to find an equilibrium schedule with respect to IRN. Non-Preemptive Nash (NPN) Input : Processing time profiles {p ij : j N i } for i M Output : Equilibrium workloads x N i : i M for the non-preemptive problem Begin 1. Order players in SPT order of P i s and determine sets X i : i M, as defined in (3) Let x N i := 0 for i M, x N 0 := 0. For k := 1 to M do For i := k to M do begin Compute w k 1 =max j<k x N j 3. Set x N k := the smallest value x X k that solves problem IP k. end End When solving IP k in line 3, there may be more than one most profitable strategies for player k. We select the smallest because player k is indifferent amongst them and selecting the smallest will allow subsequent players to subcontract more. Knowing sets X i : i M, the discussion preceding Theorem 7 establishes that NPN produces a Nash schedule. The effort expended in line 1 of NPN is O( N i Pi )fori M resulting in O( i N i Pi ) total because there are no more than O(P i) possible strategies for player i and the feasibility of each is tested by solving a knapsack problem in O( N i P i ) time (see Martello and Toth, 1990). Therefore, the total effort required in line 1 of NPN is bounded by O( M max i N i P M ). Every visit of line 3 of NPN requires effort O( X k ) when the values f i (x) are stored appropriately. Accounting for the i and k-loops, the total effort expended is bounded by 8

O( M P k ). In general, it is expected that M < max i N i and hence the overall complexity of NPN is dominated by O(max i N i M P M ) which is also the complexity of finding the sets X i, i M by solving O( M P M ) knapsack problems. 9