math.answers.com/math-and-arithmetic/Derive_EOQ_appllied_to_production_order
Preview meta tags from the math.answers.com website.
Linked Hostnames
8- 33 links tomath.answers.com
- 19 links towww.answers.com
- 1 link totwitter.com
- 1 link towww.facebook.com
- 1 link towww.instagram.com
- 1 link towww.pinterest.com
- 1 link towww.tiktok.com
- 1 link towww.youtube.com
Thumbnail

Search Engine Appearance
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]
Bing
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]
DuckDuckGo
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]
General Meta Tags
22- titleDerive EOQ appllied to production order? - Answers
- charsetutf-8
- Content-Typetext/html; charset=utf-8
- viewportminimum-scale=1, initial-scale=1, width=device-width, shrink-to-fit=no
- X-UA-CompatibleIE=edge,chrome=1
Open Graph Meta Tags
7- og:imagehttps://st.answers.com/html_test_assets/Answers_Blue.jpeg
- og:image:width900
- og:image:height900
- og:site_nameAnswers
- og:descriptionQ 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]
Twitter Meta Tags
1- twitter:cardsummary_large_image
Link Tags
16- alternatehttps://www.answers.com/feed.rss
- apple-touch-icon/icons/180x180.png
- canonicalhttps://math.answers.com/math-and-arithmetic/Derive_EOQ_appllied_to_production_order
- icon/favicon.svg
- icon/icons/16x16.png
Links
58- https://math.answers.com
- https://math.answers.com/math-and-arithmetic/80.00_increased_by_15_percent
- https://math.answers.com/math-and-arithmetic/Derive_EOQ_appllied_to_production_order
- https://math.answers.com/math-and-arithmetic/How_do_you_say_6005000000_in_words
- https://math.answers.com/math-and-arithmetic/How_many_20.5_by_20.5_inch_tiles_are_used_in_a_600_square_foot_area