RESEARCH ARTICLE HUMAN CAPITAL DEVELOPMENT FOR PROGRAMMERS USING OPEN SOURCE SOFTWARE Ami Mehra Indian Shool of Business, Hyderabad, INDIA {Ami_Mehra@isb.edu} Vijay Mookerjee Shool of Managemen, Uniersiy of Texas a Dallas, Rihardson, TX 758 U.S.A. {ijaym@udallas.edu} Appendix Proofs of Lemmas and Proposiions Proof of Lemma 1 The maximum alue of w needed o saisfy he IR onsrain of he programmer (alled w IR ) is obained from w IR T M ax(t). Here x(t) represens he smalles possible skill leel wih whih he programmer ends he onra. This skill leel ours when is hosen hroughou he onra duraion. Nex, noe ha he adjoin equaion for he o-sae ariable μ() is μ μ() onsan œ. y Thus μ mus ake he same alue hroughou he onra duraion. If μ > 1, hen > implying ha w mus be he maximum possible and hene he opimal alue of w > w IR, sine he domain of he wage premium w is R +. Howeer, suh a hoie of w resuls in a smaller ne alue for he firm han if he firm hose w w IR sine paying wages is a os o he firm and has no impa on programmer produiiy. This is a onradiion beause a subopimal alue of w gies a beer soluion o he firm. Hene i mus be ha μ # 1. If μ < 1, hen < w. On he oher hand, if μ 1, hen w an ake any possible posiie alue (singular soluion). Proof of Lemma We represen by z. Thus, using Equaion, we an wrie λ z + Kx aμ Nex, aking he deriaie of boh sides of Equaion wih respe o, we hae (6) z Kx + λ + λ( + x) + aμ( + x) x x MIS Quarerly Vol. 36 No. 1 Appendix/Marh 1 A1
x Subsiuing in he alue of from Equaion 1, and ha of λ from he adjoin equaion, λ ( K( 1 ) + λ( x + δx βx) + aμ( x + δx βx) ), we an rewrie he aboe equaion as λ Equaing he alues of λ from equaions 6 and 7, we obain K( β δ ) z ( β δ) ( β δ ) x x x ( β δ) ( β δ ) Kx z z + K( ) β δ ( ( β δ) ( β δ )) x x x x x x (7) (8) Noe ha he oeffiien of z as well as he erm independen of z are purely funions of x and and are independen of he onrols or. Hene, we represen hese wo erms by g(x,) and h(x,), respeiely. Now suppose ha z a wo differen insans and 1, where < 1, wihou loss of generaliy. We onsider an insan suh ha < < 1 and z(). The general soluion of he differenial equaion 8 is g x d g x d z ()exp (, σ ) σ exp (, σ ) σ hxsds (, ) + C where C is he onsan of inegraion. Using his equaion we an wrie he soluion beween and 1 as Sine z( 1 ), his an be rewrien as 1 g( x, σ) dσ g( x, σ) dσ g( x, σ) dσ z( )exp z( )exp ( h( x, s))exp ds 1 1 S g( x, σ) dσ g x d z ( ) exp 1 ( hxs (, )) exp S ( σ) σ ds (9) Similar o aboe, we onsider he soluion of he differenial equaion 8 beween and. This gies us g( x, ) g x d z ( ) exp ( hxs (, )) exp S ( σ) σ ds (1) From equaions 9 and 1, we noe ha z() mus hae wo differen alues a. Howeer, equaion 8 implies ha z() mus be a oninuous funion. Hene, i anno hae wo alues a any poin. Hene our supposiion ha z a wo non-oniguous insans mus be inorre. Sine z is oninuous in, i an be eiher posiie of negaie prior o and afer when i beomes zero. If z >, 1 and if z <,. This gies us he saemen of he lemma. Proof of Proposiion 1 Saemen A of he proposiion follows from Lemma. d H For Saemen B, noe ha and are required for a singular soluion sine he parial deriaie of he Hamilonian wih respe d o he onrol mus be zero and mus oninue o remain o be zero for some ineral. Also, λ H is he sandard adjoin ondiion. These hree equaions simplify o he following: Kx + ( λ + aμ) (11) A MIS Quarerly Vol. 36 No. 1 Appendix/Marh 1
( ) K + K( β δ) ( λ + aμ) x( β δ) ( βx δx) ( K 1 a ) λ ( ) + λ ( + δ β ) + μ ( + δ β ) x x x x x x (1) (13) Simulaneously soling equaions 11 and 1 we obain he soluions for x and λ, whereupon we an find λ. Subsiuing in alues of λ, λ, and x in Equaion 13, we obain he alue of for he singular soluion. For Saemen C, noe ha for μ, he IR onsrain is no binding. Furher, λ(t) sine he firm does no hae a salage alue of programmer skills. From Equaion (), we hae T Kx. From Lemma, we know ha does no hange sign in o T. Hene < in his ineral. When # μ # 1 (i.e., when IR binds), is eiher posiie or negaie in his ineral. Correspondingly, 1 or. Proof of Lemma 3 (A) In he proof of Lemma, from Equaion 9, we see ha he sign of he Hamilonian, z(), is opposie o he sign of funion h(x,) for < K K ( ) Kx ( ) Kx. Wih he simplified sae equaion, we hae hx (, ) + + 1 3 3 3. Using his we an easily show + 3x ha x() < x z() > 1 and x() > x z() <. By Lemma, i mus be ha z() has he same sign in he ineral o. Aordingly, i is suffiien o hek if x x. If ha is rue hen 1 and if no hen. < (B) Applying he mehod oulined in proof of Proposiion 1, we find he alues of, x, and λ in he singular region are indiaed by,, and, respeiely. x λ x ( < ). These alues Noe ha and are independen of μ. Consequenly, he opimal alues of in he pre-singular and he singular region are unaffeed by wheher he IR onsrain is binding or no. (C) In he region T, we know ha when he IR onsrain is no binding (Proposiion 1). Suppose ha 1 is a soluion when he IR onsrain is binding. The orresponding Hamilonian is H 1 w + λ( 1 + + 3 x x) + μ(w + a( 1 + + 3 x x)). Using he adjoin equaion λ H. This implies ha λ ( 3)( λ + aμ). Noe ha he sign of λ ru- 1 ially depends upon he alue of λ a (i.e., λ ). Also noe ha λ mus be equal o a T sine he firm has no salage alue for he skills of he programmer a T. If λ > or if λ < aμ, hen his is impossible (sine hen λ eiher monoonially inreases from a posiie alue or monoonially dereases from a negaie alue). Howeer his may be possible when λ < and λ + aμ > (now λ an inrease monoonially from a negaie alue and reah zero). I is easy o erify ha he seond ondiion always holds under ( ( 3)( ) ) K requiremens imposed on he parameers 1,, 3, and. The firs ondiion requires μ >. Noe a3( 13 + ) a3( 13 + ) from he disussion preeding Lemma 1 ha μ # 1. This requires K. This is ruled ou due ( 3)( ) o our resriion on marginal produiiy of programmers. Hene, 1 anno be a soluion in his region when he IR onsrain is binding. Hene,, he only remaining possibiliy is he soluion. Proof of Proposiion 1 + 3 The general soluion o equaion 5 is x + exp ( ) C, where C is he onsan of inegraion. Using x x a we obain C. We know ha x x a. Using his relaion, we ge MIS Quarerly Vol. 36 No. 1 Appendix/Marh 1 A3
( Log + + + + + x + x ( 3 ) ( 3 ) ( 3)( )( 3 ))( 1 3 ) ( )( 1( 3 ) + ( + 3 )). 3 This expression is independen of μ. Hene i is appliable irrespeie of wheher he IR onsrain is binding or no. From Lemma 3, we know ha () [, T]. Now we wrie he differenial equaion for λ in his region. Using he adjoin equaion, λ H, his is λ a μ + λ K (1) aμ K The general soluion for his equaion is λ + exp C, where C is he onsan of inegraion and is ransformed o he sale T λ ( ( ) ( ( ) + ( )( + )))( K a ) 3 3 3 μ Log ( )( + ) K 3 1 T 1( 3) > [, T ]. Sine, here is no salage alue of skills for he firm, we hae λ a. Using his we obain he expression for λ as a funion of. This expression is used o find λ a whih is hen equaed wih ha we obained in he proof for Lemma 3. This enables us o ge. Noe ha > requires. This ondiion makes he denominaor of he Log erm negaie. Thus we an hae a real alue of only when he numeraor of he Log erm is also negaie. This will happen only when K > a μ. From disussion preeding Lemma 1 we know ha μ # 1. Noe ha K > a beause of our resriion on marginal produiiy of programmers and he requiremen on parameers due o >. Hene a is real alued. Thus, sine K > a μ, >, whih is wha we needed o esablish. μ K a μ Proof of Proposiion 3 Consider a siuaion where he IR onsrain binds bu he firm does no pay any wage premium o he programmer (w ). In his siuaion, i M mus be ha ax(t) M. Using equaion 5 and he boundary ondiions x x a and x a T, and using he fa ha a ( ( ) ( ( ) ( )( )))( ) Log + + a M 3 3 3 1 a( )( ) in beween and T, we find an expression for IRB T + + + 1. Equaing his expression wih ha from proof of Proposiion, we ge an expression for μ in erms of M. I is algebraially edious, bu easy o show ha μ is inreasing in M. Using his propery of μ, we an show ha # μ # 1 for M 1 # M # M. Noe ha when μ 1, < and so w. Howeer, if M > M hen μ 1, and hene he firm an pay a posiie wage w. Proof of Proposiion Par 1 Suppose x > x. From Lemma 3, we know ha. Now eiher implying (), or else here mus be some < < [, T] T suh ha < < T where. Suppose ha suh a exiss. Noe ha sine he singular soluion is ruled ou, an be a mos A MIS Quarerly Vol. 36 No. 1 Appendix/Marh 1
a some insan bu no in an ineral. Then as per Lemma, () œ [,T]. Furher, as per Lemma 3, () œ [, ]. Hene, () hroughou he onra duraion is he only soluion. This also shows ha he firm neer uses he opion of raining he programmers een if he IR onsrain binds. Par Suppose x < x and ha here exiss some suh ha < < T where. As in Par 1, we know ha singular soluion is no possible, hene an be a mos a some insan bu no in an ineral. Then using Lemma 3, we know ha () 1 for [,] and () for [,T]. To find, we firs onsider he region < < T. Using he adjoin equaion, λ H and he end ondiion λ a T, we work ( T + ) K aμ exp ( K aμ) ou λ(). We represen x a by x. Using his as he boundary ondiion, and using he sae exp ( x 1) + 1 equaion 5, we find x (). Using he expressions for x() and λ(), we an work ou he expression for. Now ( T) exp (( 1 + exp ) K + aμ) uilizing, we find. x T ( )exp K + exp ( K a μ) 3 3 Now we onsider he region < <. Using he sae equaion 5 and he boundary ondiion x() x, we an work ou he expression for x(). Now, an be obained from soling This is he equaion in he saemen of he proposiion. x() x (15) To show ha is unique, we ake he deriaie of x w.r.. This is easily seen o be negaie. Nex we ake he deriaie of x() w.r. and his expression shows ha is minimum alue ours a x x. This minimum alue of he slope is found o be posiie. Thus he expressions on he lef hand and righ hand side of he equaion aboe hae opposie signs w.r... Thus mus be unique. Finally, if <, hen for # # T. To see how behaes as he IR onsrain binds, we rewrie equaion 15 as x() x and represen he LHS his equaion by ƒ. Then f f d f f d aking he deriaie w.r.. μ we hae +. I is easy o show ha < and > μ. Thus, i mus be ha. Clearly dμ μ dμ > as μ inreases, also inreases leading o a greaer raining period a he beginning of he onra duraion. This leads o higher skills a he onlusion of raining, hus enabling he firm o pay lower wages. Howeer, we know ha μ# 1. Hene, one μ reahes 1, no furher exension in is possible and any shorfall in IR onsrain mus hen be paid hrough a wage premium. MIS Quarerly Vol. 36 No. 1 Appendix/Marh 1 A5