A Model of Optimal Corporate Bailouts - Faculty of Business and ...

In sum, when restarts are unprofitable, the firm’s optimal first-period investment strategy isρ ∗ = IN if Π| ρ2≥ 0 ,=OUT1OUT otherwise .Subcase 2: Firm Restarts Project With Different Manager. Now suppose that π ∗( 0, 0 ) ≥20, so that the firm restarts, and assume as well that the firm could commit to fire the incumbentmanager after an initial failure. Thus, at time 1, the manager continues to view this as asingle-period game (at least for her), with payoff ofM ≡ e 1·w 1 − c·e2 12 .Consequently, her first-period effort remains e ∗ 1 ( w 1 ) = w 1 /c. In contrast, the firm anticipatesoperating in the second period under a different manager. Its continuation profit is thusΠ| F F=FIRE( e 1 ) = e 1·(R − w 1 ) + (1 − e 1 )·π 2 ( 0, 0 ) R 2 − 4·c·I 2 · c − w1=+ w 1· R − w 1− I4·c 2 1 ,cwhich implies a profit-maximizing wage of w ∗ = R + I 1 2/2 − R 2 /(8·c). Thus, the manager exertsfirst-period effort ofe ∗ 1 = R + I 22·c− R28·c 2 ,yielding maximized profits of Π| F F=FIRE( e ∗ ) (Appendix equation A.1). Note that when the firm1can profitably restart, it optimally sets lower-powered first-period incentives than it wouldin the absence of the restart option. 16 Intuitively, restarting gives the firm a “second bite atthe apple,” which reduces the importance of first-period success. The firm therefore reducesfirst-period wages (and with it, managerial effort). From the incumbent manager’s perspective,the restart option makes her worse off, because it depresses her first-period wage. The restartbenefits accrue to the firm and the replacement manager through the second-period payoffs.Subcase 3: Firm Restarts Project With Same Manager. Finally, suppose once againthat π 2 ( 0, 0 ) ≥ 0, so that the firm restarts, but now the incumbent manager is retained. In thiscase, the manager’s payoff isM ≡ e 1·w 1 − c·e2 12 + (1 − e 1)·m ∗ 2 = e 1·w 1 − c·e2 12 + (1 − e 1)· R28·c ,16 To see this, compare the effort level in the abandonment case to the above expression:e ∗ 1 | ρ 2 =OUT = R2·c> R + I 22·c− R28·c 2 = e∗ 1 | ρ 2⇔ R2=IN4·c − I 2 > 0 .} {{ }π ∗ 2 ( 0,0 )15