NAME Givon An electronic store stock irni sl g an order fort

NAME: Givon: An electronic store stock« irni s»l: g an order fortbe, no carrying cotst for those computcif4% 12 Spa_/com puttst. Required Caiculate the optimal ordor uantity, arus thc, tota cof te rriociel I shortages are ollowed, tre orcder eost is iedticeci by 2s s. and ttna she itia shornaae cost is 12 Spa/unit. Calculate th ptimal order quartity. the opUmai Short c) What ls the cycle time for either option? d) Ir orders are delivered in 72 hours, arnd the store ia open 7 iays tweek.whatlE age quantity, and tho nw tota cost of the model. the lead time demand? Which option is cheape? Why? How are savings achieveci e)

Solution

It is given that,

Monthly demand = 100 units

Annual demand, D = 100*12 = 1200 units

Carrying cost (H) = 12

Ordering cost (K) = 50

(a) Optimal order quantity (Q) = (2DK/H)

= (2*1200*50/12)

= 100 units

Total cost of the model = Ordering cost + Carrying cost = (D/Q)*K + (Q/2)*H

= (1200/100)*50 + (100/2)*12

= $ 1200 per year  

(b) The new parameters of the inventory model are :

Ordering cost, (K) = 25

Shortage cost (B) = 12

Optimal order quantity (Q) = [(2DK/H)*(H+B)/B]

= [(2*1200*25/12)*(12+12)/12]

= 100 units

Optimal shortage quantity (S) = Q*H/(H+B)

= 100*12/(12+12)

= 50 units

Total cost of the model = (Q - S)²*H/(2Q) + D*K/Q + S²*B/(2Q)

= (100 - 50)²*12/(2*100) + 1200*25/100 + 50²*12/(2*100)

= $ 600 per year

(c) Cycle time = Q/D = 100/1200 = 1/12 year = 1 month

(d) Lead time = 72 hours = 3 days

Lead time demand = Daily demand * Lead time in days = (1200/365)*3 = 9.9 ~ 10 units

(e) The option with shortages is cheaper. The savings are achieved by reducing the order cost with planned shortages.