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# Accounting Assignment On Finding The Economic Quantity (Math Problem Sample)

Instructions:

finding the economic quantity

source..Content:

Problem One

S=Cost of Shortage (Stock out or Lost sale) = $10 - $4 = $6

O=Cost of overage (Overstock) = $4 - $1.50 = $2.50

P<=S/(S + O) = 6/ (6 + 2.50) = 0.7059

Z value of 0.7059 is approximately 0.55

Therefore, the supermarket should purchase 250 + 0.55(34) = 268.7 or 269 boxes of lettuce.

Problem 20

* The economic quantity for Ben to order

The 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 = 2DSH

Substituting the values, we obtain,

Qopt = 2(5000)(10)0.203 = 408.25 or 408 bottles.

* Inventory level to place an order

dl = 100(3) =$300 cost per week.

L = √3(30)2 = 51.9615 or 52 bottles

Standard normal distribution, Appendix E for a 95 percent level, z = 1.65

R = dl + zL = 300 + 1.65(52) = 385.8 or 386

Ben orders 408 bottles when the on hand inventory level reaches 386 bottles.

Problem 27

Using your procedure, how many boxes of paper 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 = ᵭ (L+T) + Z*ð - I is a formula of finding optimum order quantity in fixed time period model.

Q is the Optimum order quantity.

ᵭ is the forecast average daily demand.

L is the lead time = 3days

T is the number of days between reviews = 14

Z is the standard deviation

ð is the standard deviation of usage in lead time and review period

I is the current inventory level = 60

Demand 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

ð= 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 = ᵭ (L+T) + Z*ð – I, we obtain,

Q = 13.69(3+14) + (2.05) (41.23) – 60

Q = 257.25 or 257 boxes.

Problem 1...

S=Cost of Shortage (Stock out or Lost sale) = $10 - $4 = $6

O=Cost of overage (Overstock) = $4 - $1.50 = $2.50

P<=S/(S + O) = 6/ (6 + 2.50) = 0.7059

Z value of 0.7059 is approximately 0.55

Therefore, the supermarket should purchase 250 + 0.55(34) = 268.7 or 269 boxes of lettuce.

Problem 20

* The economic quantity for Ben to order

The 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 = 2DSH

Substituting the values, we obtain,

Qopt = 2(5000)(10)0.203 = 408.25 or 408 bottles.

* Inventory level to place an order

dl = 100(3) =$300 cost per week.

L = √3(30)2 = 51.9615 or 52 bottles

Standard normal distribution, Appendix E for a 95 percent level, z = 1.65

R = dl + zL = 300 + 1.65(52) = 385.8 or 386

Ben orders 408 bottles when the on hand inventory level reaches 386 bottles.

Problem 27

Using your procedure, how many boxes of paper 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 = ᵭ (L+T) + Z*ð - I is a formula of finding optimum order quantity in fixed time period model.

Q is the Optimum order quantity.

ᵭ is the forecast average daily demand.

L is the lead time = 3days

T is the number of days between reviews = 14

Z is the standard deviation

ð is the standard deviation of usage in lead time and review period

I is the current inventory level = 60

Demand 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

ð= 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 = ᵭ (L+T) + Z*ð – I, we obtain,

Q = 13.69(3+14) + (2.05) (41.23) – 60

Q = 257.25 or 257 boxes.

Problem 1...

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