t i s m a d e u p o f m a n y components which in turn is made up of many parts. Each part has a cost associated to is, and each component and each product has a manufacturing cost. How much does it take to produce a given product?
|component| // add component costs component.parts.each do |part| // add part costs end end // print product price end Product.each do |product| // add product costs product.components.each do |component| // add component costs component.parts.each do |part| // add part costs end end // print product price end Product.each do |product| // add product costs product.components.each do |component| // add component costs component.parts.each do |part| // add part costs end end // print product price end
to describe the performance or complexity of an algorithm. Big O specifically describes the worst-case scenario, and can be used to describe the execution time required or the space used (e.g. in memory or on disk) by an algorithm.
costs product.components.each do |component| // add component costs component.parts.each do |part| // add part costs end end // print product price end
costs product.components.each do |component| // add component costs component.parts.each do |part| // add part costs end end // print product price end
product_id, products.name AS product_name, products.manufacture_cost + sum(component_costs.cost)) AS product_manufacture_cost FROM products INNER JOIN product_components ON product_components.product_id = products.id INNER JOIN ( SELECT components.id AS component_id, (components.manufacture_cost + sum(cost)) AS cost FROM components INNER JOIN component_parts ON components.id = component_parts.component_id INNER JOIN parts ON parts.id = component_parts.part_id GROUP BY components.id ) AS component_costs ON component_costs.component_id = product_components.component_id GROUP BY products.id, products.name ORDER BY products.id
FROM components INNER JOIN component_parts ON components.id = component_parts.component_id INNER JOIN parts ON parts.id = component_parts.part_id GROUP BY components.id
sum(component_costs.cost)) AS product_manufacture_cost FROM products INNER JOIN product_components ON product_components.product_id = products.id INNER JOIN ( … ) AS component_costs ON component_costs.component_id = product_components.component_id GROUP BY products.id, products.name ORDER BY products.id