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