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- University
- Politecnico di Milano
- Degree programme
- Management Engineering
- Subject
- Logistics Management
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Study material for Logistics Management, shared by the Studwiz community and reviewed by moderators.
Study material for Logistics Management, shared by the Studwiz community and reviewed by moderators.
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1 Logistics Management - Formulas 1 STORAGE SYSTEMS Storage Capacity assessment [FF = forecasting factor; i = product family, item; j = month/period] πππ = βππΆπππ #locations occupied by an item i for a period j [SC = starage capacity] - Random storage ππΆ = πππ₯π{πππ} β πΉπΉ -> maximum value that I have considering the sums of all the values of each month - Dedicated storage ππΆ = β(πππ₯π{πππ,π} β πΉπΉπ)π -> sum of the maximum values that I have for each month - Class based storage ππΆπ = β(πππ₯π{πππ,π} β πΉπΉπ)π Storage area - Bay design β draw and compute W, H, D πππ΅ = β ππππ¦ πππΏ β [#pallet locations in a bay] - Module design β π΄ππππ’ππ = (2π· + π΄π) β π - Number of levels β ππΏ= πππ{β π π» π» β ; β π΅π» π» β} [RH = rack height; BH = building height] - Area utilization rate β π΄ππ ππππ’ππ = ππΏπ ππππ’ππ π΄ππππ’ππ = 2βNPBβNL Amodule - Required area π΄ππ ππππ’ππ = ππΆ π΄ β π΄ = ππΆ π΄ππ ππππ’ππ - if the rack length is fixed (π = π πΏ) o ππΆ = β π πΏ πβ o ππ΄= β ππΆ 2βπππ΅βππΏβππΆβ o π = ππ΄β πππππ’ππ - If we do not know the rack length, according to the position of the I/O point β expected path π = 2 β ( π π + π 2) β ππππ‘ = π 2 β ππππ‘ o I/O in the middle of the storage front π = 4 π = 2 β ( π 4 + π 2) Uopt = 2 Vopt o I/O on the vertex of the storage front π = 2 π = 2 β ( π 2 + π 2) Uopt = Vopt o I/O distributed along the storage front π = 3 π = 2 β ( π 3 + π 2) Uopt = 1,5 Vopt o Number of aisles ππ΄= β ππππ‘ π΄π+2π·β o Number of bay columns ππΆ = β ππΆ 2βππ΄βπππ΅βππΏβ o πππππ = ππ΄β (π΄π+ 2π·) [π] o πππππ = ππΆβ π [π] o π΄ = π β π - Real storage capacity: ππΆππππ = 2 β ππ΄β ππΆβ πππ΅β ππΏ 2 NB: if we have a block stacking storage areas - ππππππ = β ππΆ #ππ‘πππ βππΏ/π π‘πππβ#πππππ /ππ‘ππβ - π = π΄π+ 2 β [ππππππ β (π·ππΏ + πππ π‘ππππππΏ) + πππ π‘ππππππΏ] - π = (πππΏ + πππ π‘ππππππΏ) β #πππππ - π΄ = π β π - ππππΆπΆ = ππππ ππ π π»,πππ ππ + πππππ π π»,ππππ + π π π o ππππ ππβ¦
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