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- Politecnico di Milano
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- Management Engineering
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- Logistics Management
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- Exam Β· Full exam
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xx 05 20 simulation sol
Full exam for Logistics Management in the Management Engineering degree programme at Politecnico di Milano. The document covers: 01. Exam simulation 07.05.2020 1 EXERCISE 1 PART 1 We have to design the bay. The dimensions after the computations will be: - πΏ = 0,1 β 4 + 0,8 β 2 = 2 π - π» = 0,1 + 0,15 + 1,5 = 1,75 π - π· = 1,2 + 0,2 2 = 1,3 π The next step is to determine the number of the levels ππΏ
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Full exam for Logistics Management in the Management Engineering degree programme at Politecnico di Milano. The document covers: 01. Exam simulation 07.05.2020 1 EXERCISE 1 PART 1 We have to design the bay. The dimensions after the computations will be: - πΏ = 0,1 β 4 + 0,8 β 2 = 2 π - π» = 0,1 + 0,15 + 1,5 = 1,75 π - π· = 1,2 + 0,2 2 = 1,3 π The next step is to determine the number of the levels ππΏ
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01. Exam simulation 07.05.2020 1 EXERCISE 1 PART 1 We have to design the bay. The dimensions after the computations will be: - πΏ = 0,1 β 4 + 0,8 β 2 = 2 π - π» = 0,1 + 0,15 + 1,5 = 1,75 π - π· = 1,2 + 0,2 2 = 1,3 π The next step is to determine the number of the levels ππΏ = πππ{β 8,5 1,75β ; β 7,5 1,75β} = πππ{4; 5} = 4 πππ£πππ We know from the text that the length of the aisle is fixed. So, I do not have to determine the optimal shape because I have these constraints. Since the lengths of aisle is fixed (50m), the number of columns is fixed. So, the number of aisles is just a consequence: - ππΆ = 50 2 = 25 ππππ’πππ [ππ‘βπππ€ππ π, ππΆ = β ππΆ 2βππ΄βπππ΅βππΏβ ] - ππ΄ = β ππΆ 2βππΆβπππ΅βππΏβ = β 4.000 2β4β2β25β = 10 πππ πππ The problem is easier than the traditional one solved in class: we do not have to find the optimal shape. Now we have to find U: - πΏππππ’ππ = 2 β π· + π΄π = 2 β 1,3 + 2,8 = 5,4 π - π = ππ΄ β πΏππππ’ππ = 10 β 5,4 = 54 π Now we go through the computation of the cycle time: πππΆ = πΉπ + ππ = πΉπ + [ π ππ» + π ππ ] = = πΉπ + 2 β ( π 2 + π 2 + ππππππ π πππ ππ) β 1 ππ» + 2 β π» β ( ππΏβ1 2 ) β 1 ππ = = 50 + 2 β ( 54+50 2 + 3) β 1 2 + 1,75 β 4β1 0,4 = 50 + 55 + 13,125 = 118,125π ππ ππΆπ‘ππ’ππ = UF β 3600 TSC = 1 β 3600 118,125 = 30,48 π ππ/β # π‘ππ’ππ = β π‘βπππ’πβππ’π‘ πππππππ‘π¦ ππππ’ππππ ππΆπ‘ππ’ππ β = β 160 30,48β = β5,25β = 6 π‘ππ’πππ // this was the first part; it is worth 24 points out of 30. If the mistake in the formula is -2 or -3 points, for a computation error is -1, for worst error of concepts more points will be removed (like if we would have used the AUR here -10 points would have been removed). PART 2 Since the I/O point is in the corner, the iso-time curve will be something like the one in the drawing on the right. If the I/O point would have been in the middle, in this case we would have doneβ¦
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