Accounting Assignment On Finding The Economic Quantity - 1100 Words

Accounting Assignment On Finding The Economic Quantity (Math Problem Sample) Content: Problem OneS=Cost of Shortage (Stock out or Lost sale) = $10 - $4 = $6O=Cost of overage (Overstock) = $4 - $1.50 = $2.50P=S/(S + O) = 6/ (6 + 2.50) = 0.7059Z value of 0.7059 is approximately 0.55Therefore, the supermarket should purchase 250 + 0.55(34) = 268.7 or 269 boxes of lettuce.Problem 20 * The economic quantity for Ben to orderThe annual demand is 5000 bottles and holding cost of 20 percent of the purchase price which is $3. The cost of placing an order is $10.Use EOQ model to find out economic order quantity for Ben.523875188595Qopt = 2DSHSubstituting the values, we obtain,Qopt = 2(5000)(10)0.203 = 408.25 or 408 bottles. * Inventory level to place an orderdl = 100(3) =$300 cost per week.L = 3(30)2 = 51.9615 or 52 bottlesStandard normal distribution, Appendix E for a 95 percent level, z = 1.65R = dl + zL = 300 + 1.65(52) = 385.8 or 386Ben orders 408 bottles when the on hand inventory level reaches 386 bottles.Problem 27Using your procedure, how many boxes of pa per would be ordered if, on the day the sales person calls, 60 boxes are on hand.It refers to the fixed economic order quantity that minimizes the total ordering cost as well as holding cost of inventory of a firm. In this situation, fixed order quantity is utilized.Q = á µ ­ (L+T) + Z* - I is a formula of finding optimum order quantity in fixed time period model.Q is the Optimum order quantity.á µ ­ is the forecast average daily demand.L is the lead time = 3daysT is the number of days between reviews = 14Z is the standard deviation is the standard deviation of usage in lead time and review periodI is the current inventory level = 60Demand is 5000365 = 13.69 boxes.Calculating the standard deviation with lead time of 3days and review period of 14 days is as shown below.Lead time = 3days + 2weeks = 17 days= 17*10= 41.23 boxes.Using excel and applying NORMSINV (0.98) we get the value of Z as 2.05.Substituting the values in the above equation that is Q = á µ ­ (L+T) + Z* I, we obtai n,Q = 13.69(3+14) + (2.05) (41.23) 60Q = 257.25 or 257 boxes.Problem 1...