d.0.5A - B - 050 = 0. Department 3 has 2500 hours. Transfers are allowed to departments 2 and 4, and from departments 1 and 2. If A measures the labor hours allocated to department i and Ty the hours transferred from department i to department j, then a. Ti + Tas - Tes - Tea - As = 2500. b.A; - T1 - Tas + Tas + Ta = 2500. C C. Ta+ Tax - Tax - Tus + A: = 2500. C d.A; + Tu+ Ta - Tax - Ta = 2500