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Excersise 7-2: Harley Davidson



Harley has decided to implement Just In Time (JIT) at the motorcycle assembly plant. As

part of this initiative it has reduced the number of engines loaded on each truck to 100. If

each truck trip still costs $1,000, how does this decision impact annual costs at Harley?

What should the cost of each truck be if a load of 100 engines is to be optimal for Harley?







Part I



Inputs

Demand, R = 109,500 (assuming 365 operating days)

Fixed cost per order, S = $ 1,000

Cost per engine, C = $ 500

Holding cost, h = 20%

Order quantity 100



Outputs

# of orders per year 1,095

Annual Order Cost 1,095,000

Cycle inventory (units) 50

Annual Holding Cost $ 5,000

Total Inventory Cost $ 1,100,000





Part II



Inputs

Demand, R = 109,500 (assuming 365 operating days)

Fixed cost per order, S = $ 1,000

Cost per engine, C = $ 500

Holding cost, h = 20%

EOQ 100



Outputs

Fixed cost per order, S = $ 4.57



# of orders per year 1,095

Annual Order Cost 5,000

Cycle inventory (units) 50

Annual Holding Cost $ 5,000

Total Inventory Cost $ 10,000









CR - 12/4/2011 3:01 AM 50900ff8-dabf-441e-8a81-1805eb03c412.xls - VALUES

Page 1

Excersise 7-2: Harley Davidson



Harley has decided to implement Just In Time (JIT) at the motorcycle assembly plant. As part of this initiative it has reduced the number of engines loaded on each truck to 100. If each

truck trip still costs $1,000, how does this decision impact annual costs at Harley? What should the cost of each truck be if a load of 100 engines is to be optimal for Harley?









Part I



Inputs

Demand, R = =300*365 (assuming 365 operating days)

Fixed cost per order, S = 1000

Cost per engine, C = 500

Holding cost, h = 0.2

Order quantity 100



Outputs

# of orders per year =C13/C17

Annual Order Cost =C20*C14

Cycle inventory (units) =C17/2

Annual Holding Cost =C15*C22*C16

Total Inventory Cost =C21+C23





Part II



Inputs

Demand, R = =300*365 (assuming 365 operating days)

Fixed cost per order, S = 1000

Cost per engine, C = 500

Holding cost, h = 0.2

EOQ 100



Outputs

Fixed cost per order, S = =((C34^2)*C33*C32)/(2*C30)



# of orders per year =C30/C34

Annual Order Cost =C39*C37

Cycle inventory (units) =C34/2

Annual Holding Cost =C32*C41*C33

Total Inventory Cost =C40+C42









CR - 12/4/2011 3:01 AM Page 2 50900ff8-dabf-441e-8a81-1805eb03c412.xls - FORMULAS



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