Solution idea D=Demand= 24000 S=Ordering/Setup cost= 2400 H=Holding/Carrying Cost= 3 U= Usage rate=24000/240= 100 P= Production rate= 600 Optimal Run Size: Q0= √2DS/H√p/(p-u) Q0= √(2240002400)/3√600/(600-100) Q0= √115200000/3√600/500 Q0= √38400000* √1.20 Q0= 6196.77*1.0954 Q0= 6787.94 ≅6788 Minimum total annual cost for carrying and setup cost. = Carrying Cost + Set up Cost =( I max/2)H+ ( D/Q0)S Where, I max= Q0/p ((p-u)) I max= 6788/600 (600-100) I max= 11.31 (500) I max= 5655 TC min= Carrying Cost+ Setup Cost TC min= (I max/2) H+ (D/Q0) S TC min= (5655/2) 3 + (24000/6788) 2400 TC min= (2827.5) (3) + (3.53) (2400) TC min= 8482.5 + 8472 TC min= 16954.5 Cycle time for the Optimal Run Size. Q0/U=6788/100 = 67.88 ≅ 68 days Run time Q0/p=6788/600 = 11.31 days