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Derive EOQ appllied to production order? - Answers

Q size to order : obej. to find best of Q to minimize the cost K:setup cost c:cost of one unit h: holding cost for one unit per one unit period of time landa = demand rate T cycle length = Q/landa The cost is = order cost + cost of all unit ordered + average holding cost C=K+cQ+(ThQ)/2 we divided all over time to be per unit time so the average cost is : G= (K+cQ)/T + (hQ/2) replace T=Q/landa G= ((K+cQ) / (Q/landa)) + (Qh/2) then by manipulating G= ((K* landa )/Q) + ( landa* c)+ (hQ/2) this formula represent period setup cost , period purchases cost and period holding cost respectively. because we want to find the minimize this formula gives we need to take the derivative -using calculus- to respect of Q: G' = (- k *landa / Q^2) + h/2 - give min and max - taking 2nd derivative G'' = (2*k*landa)/Q^3 for Q>0 => G''>0 => G' is the min having G' = 0 G' = (- k *landa / Q^2) + h/2 = 0 ( k *landa / Q^2) = h/2 => Q^2= (2*k*landa)/h by taking square root of both sides Q# (EOQ)= sqrt [(2*k*landa)/h]



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Derive EOQ appllied to production order? - Answers

https://math.answers.com/math-and-arithmetic/Derive_EOQ_appllied_to_production_order

Q size to order : obej. to find best of Q to minimize the cost K:setup cost c:cost of one unit h: holding cost for one unit per one unit period of time landa = demand rate T cycle length = Q/landa The cost is = order cost + cost of all unit ordered + average holding cost C=K+cQ+(ThQ)/2 we divided all over time to be per unit time so the average cost is : G= (K+cQ)/T + (hQ/2) replace T=Q/landa G= ((K+cQ) / (Q/landa)) + (Qh/2) then by manipulating G= ((K* landa )/Q) + ( landa* c)+ (hQ/2) this formula represent period setup cost , period purchases cost and period holding cost respectively. because we want to find the minimize this formula gives we need to take the derivative -using calculus- to respect of Q: G' = (- k *landa / Q^2) + h/2 - give min and max - taking 2nd derivative G'' = (2*k*landa)/Q^3 for Q>0 => G''>0 => G' is the min having G' = 0 G' = (- k *landa / Q^2) + h/2 = 0 ( k *landa / Q^2) = h/2 => Q^2= (2*k*landa)/h by taking square root of both sides Q# (EOQ)= sqrt [(2*k*landa)/h]



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https://math.answers.com/math-and-arithmetic/Derive_EOQ_appllied_to_production_order

Derive EOQ appllied to production order? - Answers

Q size to order : obej. to find best of Q to minimize the cost K:setup cost c:cost of one unit h: holding cost for one unit per one unit period of time landa = demand rate T cycle length = Q/landa The cost is = order cost + cost of all unit ordered + average holding cost C=K+cQ+(ThQ)/2 we divided all over time to be per unit time so the average cost is : G= (K+cQ)/T + (hQ/2) replace T=Q/landa G= ((K+cQ) / (Q/landa)) + (Qh/2) then by manipulating G= ((K* landa )/Q) + ( landa* c)+ (hQ/2) this formula represent period setup cost , period purchases cost and period holding cost respectively. because we want to find the minimize this formula gives we need to take the derivative -using calculus- to respect of Q: G' = (- k *landa / Q^2) + h/2 - give min and max - taking 2nd derivative G'' = (2*k*landa)/Q^3 for Q>0 => G''>0 => G' is the min having G' = 0 G' = (- k *landa / Q^2) + h/2 = 0 ( k *landa / Q^2) = h/2 => Q^2= (2*k*landa)/h by taking square root of both sides Q# (EOQ)= sqrt [(2*k*landa)/h]

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      Q size to order : obej. to find best of Q to minimize the cost K:setup cost c:cost of one unit h: holding cost for one unit per one unit period of time landa = demand rate T cycle length = Q/landa The cost is = order cost + cost of all unit ordered + average holding cost C=K+cQ+(ThQ)/2 we divided all over time to be per unit time so the average cost is : G= (K+cQ)/T + (hQ/2) replace T=Q/landa G= ((K+cQ) / (Q/landa)) + (Qh/2) then by manipulating G= ((K* landa )/Q) + ( landa* c)+ (hQ/2) this formula represent period setup cost , period purchases cost and period holding cost respectively. because we want to find the minimize this formula gives we need to take the derivative -using calculus- to respect of Q: G' = (- k *landa / Q^2) + h/2 - give min and max - taking 2nd derivative G'' = (2*k*landa)/Q^3 for Q>0 => G''>0 => G' is the min having G' = 0 G' = (- k *landa / Q^2) + h/2 = 0 ( k *landa / Q^2) = h/2 => Q^2= (2*k*landa)/h by taking square root of both sides Q# (EOQ)= sqrt [(2*k*landa)/h]
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