Further exercises - 1st part Solutions PDF

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Politecnico di Milano – School of Industrial and Information Engineering

Logistics Management Prof. Marco Melacini and Alessandro Perego COLLECTION OF FURTHER EXERCISES – 1st PART (WAREHOUSES) TRADITIONAL WAREHOUSES - SOLUTIONS Exercise 1 1. Required storage capacity of the system The required storage capacity can be calculated from Table 1. Let j denotes the generic month and i the generic product family: 𝐹12

𝑆𝐶𝑟𝑒𝑞 = max ∑ 𝑆𝐶𝑖𝑗 = 8,050 𝑝𝑎𝑙𝑙𝑒𝑡 𝑙𝑜𝑎𝑑𝑠 𝑗

𝑖=𝐹1

2. Storage system configuration 𝐻 (𝑏𝑎𝑦 ℎ𝑒𝑖𝑔ℎ𝑡) = 𝐻𝑃𝐿 + 𝑏𝑒𝑎𝑚 ℎ𝑒𝑖𝑔ℎ𝑡 + 𝑐 = 1.4 + 0.1 + 0.15 = 1.65 𝑚

𝐿(𝑏𝑎𝑦 𝑤𝑖𝑑𝑡ℎ) = 𝑁𝑃𝐵 ∙ 𝐿𝑃𝐿 + (𝑁𝑃𝐵 + 1) ∙ 𝑑 + 𝑢𝑝𝑟𝑖𝑔ℎ𝑡 𝑤𝑖𝑑𝑡ℎ = 3 ∙ 0.8 + 4 ∙ 0.08 + 0.1 = 2.82 𝑚 𝐷 = 𝐷𝑃𝐿 +

0.16 𝑒 = 1.2 + = 1.28 𝑚 2 2

𝑀𝑜𝑑𝑢𝑙𝑒 𝑎𝑟𝑒𝑎 = 𝑀𝑜𝑑𝑢𝑙𝑒 𝑤𝑖𝑑𝑡ℎ ∙ 𝐿 = (2 ∙ 𝐷 + 𝐴𝑊) ∙ 𝐿 = (2 ∙ 1.28 + 2.5) ∙ 2.82 = 14.27 𝑚2 𝐻 𝑚𝑎𝑥𝑓𝑜𝑟𝑘

8 ⌉ = 5 𝑙𝑒𝑣𝑒𝑙𝑠 1.65 30 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 𝑃𝐿⁄𝑚𝑜𝑑𝑢𝑙𝑒 = 2.10 𝑃𝐿 ⁄𝑚2 = = 𝐴𝑈𝑅 = (2 ∙ 𝐷 + 𝐴𝑊) ∙ 𝐿 14.27 𝑀𝑜𝑑𝑢𝑙𝑒 𝑎𝑟𝑒𝑎 𝑁𝐿 = ⌈

𝐴𝑟𝑒𝑎 =

𝐻

⌉=⌈

𝑆𝐶𝑟𝑒𝑞 8,050 = = 3,833 𝑚2 𝐴𝑈𝑅 2.10

As the I/O is located at the corner of the warehouse, the optimal shape is a square: 𝑈 = 𝑉 = √𝐴𝑟𝑒𝑎 = 61.88 𝑚 The number of aisles can be calculated as: 𝑈 𝑁𝐴 = ⌈ ⌉ = 13 𝑎𝑖𝑠𝑙𝑒𝑠 𝑀𝑜𝑑𝑢𝑙𝑒 𝑤𝑖𝑑𝑡ℎ The number of bay columns NC: 𝑁𝐶 = ⌈

𝑆𝐶

2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐴 ∙ 𝑁𝐿 The real dimensions of the storage area are:

⌉ = 21 𝑐𝑜𝑙𝑢𝑚𝑛𝑠

𝑈𝑟𝑒𝑎𝑙 = 𝑁𝐴 ∙ 𝑀𝑜𝑑𝑢𝑙𝑒 𝑤𝑖𝑑𝑡ℎ = 13 ∙ (2 ∙ 1.28 + 2.5) = 65.8 𝑚 𝑉𝑟𝑒𝑎𝑙 = 𝑁𝐶 ∙ 𝐿 = 21 ∙ 2.82 = 59.2 𝑚

The real storage capacity is: 𝑆𝐶𝑟𝑒𝑎𝑙 = 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 ∙ 𝑁𝐴 = 2 ∙ 3 ∙ 5 ∙ 21 ∙ 13 = 8,190 𝑝𝑎𝑙𝑙𝑒𝑡 𝑙𝑜𝑎𝑑𝑠 3. Number of straddle reach trucks To assess the number of required trucks we need to assess the average time required to complete a single command cycle, which consists of two components: fixed and variable time. The fixed time is given (i.e. 55 s/cycle). As regards the variable time: 1

𝑉𝑇𝑆𝐶 =

𝑃

𝑆

+

𝑆𝐻 = The total time is𝑆: 𝑉

2∙(

𝑈𝑟𝑒𝑎𝑙 2 +

𝑉𝑟𝑒𝑎𝑙 + 𝑐𝑟𝑜𝑠𝑠 𝑎𝑖𝑠𝑙𝑒 𝑤𝑖𝑑𝑡ℎ) 2 𝑆𝐻

+

𝐻 ∙ (𝑁𝐿 − 1) 131 6.6 = 81.5 𝑠 + 0.3 = 2.2 𝑆𝑉

𝑇𝑆𝐶 = 𝑉𝑇𝑆𝐶 + 𝐹𝑇𝑆𝐶 = 81.5 + 55 = 136.5 𝑠

Based on the total time, the average throughput capacity of a single truck is: 𝑇𝐶𝑡𝑟𝑢𝑐𝑘 = 𝑈𝐹 ∙

3,600 3,600 = 1∙ = 26.36 𝑠𝑖𝑛𝑔𝑙𝑒 𝑐𝑜𝑚𝑚𝑎𝑛𝑑 𝑐𝑦𝑐𝑙𝑒𝑠/ℎ 𝑇𝑆𝐶 136.5

And the overall number of trucks required: 𝑁𝑡𝑟𝑢𝑐𝑘𝑠 = ⌈

𝑇𝐶𝑟𝑒𝑞 110 ⌉ = 5 𝑡𝑟𝑢𝑐𝑘𝑠 ⌉=⌈ 26.36 𝑇𝐶𝑡𝑟𝑢𝑐𝑘

𝑇𝐶𝑟𝑒𝑎𝑙 = 𝑇𝐶𝑡𝑟𝑢𝑐𝑘 ∙ 𝑁𝑡𝑟𝑢𝑐𝑘𝑠 = 26.36 ∙ 5 = 132 𝑠𝑖𝑛𝑔𝑙𝑒 𝑐𝑜𝑚𝑚𝑎𝑛𝑑 𝑐𝑦𝑐𝑙𝑒𝑠/ℎ 4. Labour cost for a single command cycle The labour cost for a single command cycle is: 𝐿𝑎𝑏𝑜𝑢𝑟 𝑐𝑜𝑠𝑡 =

136.5 [𝑠 ⁄𝑐𝑦𝑐𝑙𝑒 ] ∙ 12 [€⁄ ℎ] = 0.46 € ⁄𝑠𝑖𝑛𝑔𝑙𝑒 𝑐𝑜𝑚𝑚𝑎𝑛𝑑 𝑐𝑦𝑐𝑙𝑒 3,600 [𝑠 ⁄ℎ ]

Exercise 2 In this case, U is given (𝑈 = 𝑎𝑖𝑠𝑙𝑒 𝑙𝑒𝑛𝑔ℎ𝑡 ∙ 2 + 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑎𝑖𝑠𝑙𝑒 𝑤𝑖𝑑𝑡ℎ = 24 ∙ 2 + 6 = 54 𝑚), while V has to be calculated as follows: 24 # 𝑃𝑎𝑙𝑙𝑒𝑡 𝐿𝑜𝑎𝑑𝑠 = 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐶 ∙ 𝑁𝐿 = 2 ∙ 4 ∙ ∙ 5 = 240 𝑝𝑎𝑙𝑙𝑒𝑡 𝑙𝑜𝑎𝑑𝑠/𝑎𝑖𝑠𝑙𝑒 𝑎𝑖𝑠𝑙𝑒 4 𝑆𝐶𝑟𝑒𝑞

𝑁𝐴 =

#𝑃𝐿𝑠 𝑎𝑖𝑠𝑙𝑒

=

12,000 240

= 50 𝑎𝑖𝑠𝑙𝑒𝑠 (25 aisles per each side)

𝑆𝐶𝑟𝑒𝑎𝑙 = 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 ∙ 𝑁𝐴 = 2 ∙ 4 ∙ 5 ∙ 6 ∙ 50 = 12,000 𝑝𝑎𝑙𝑙𝑒𝑡 𝑙𝑜𝑎𝑑𝑠 𝑀𝑜𝑑𝑢𝑙𝑒 𝑤𝑖𝑑𝑡ℎ = 2 ∙ 𝐷 + 𝐴𝑊 = 2 ∙ 1.3 + 2.5 = 5.1 𝑚

𝑁𝐴 ∙ 𝑀𝑜𝑑𝑢𝑙𝑒 𝑤𝑖𝑑𝑡ℎ = 25 ∙ (2 ∙ 1.3 + 2.5) = 127.5 𝑚 2 To assess the required number of trucks we need to assess the average time required to complete a single command cycle, which consists of two components: fixed and variable time. The fixed time is given (i.e. 40 s/cycle). As regards the variable time, the formula to get P has to be adjusted according to the layout features: 𝑉=

𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑎𝑖𝑠𝑙𝑒 𝑤𝑖𝑑𝑡ℎ 𝑎𝑖𝑠𝑙𝑒 𝑙𝑒𝑛𝑔ℎ𝑡 𝑉 + + ) 2 2 2 6 24 127.5 2 ∙ (2 + 2 + 2 ) 2 ∙ (5 − 1) 157.5 𝑆 8 𝑃 + = + = = 98.26 𝑠 + = 𝑆𝐻 𝑆𝑉 0.3 2.2 0.3 2.2 𝑃 = 2∙(

𝑉𝑇𝑆𝐶 The total time is:

𝑇𝑆𝐶 = 𝑉𝑇𝑆𝐶 + 𝐹𝑇𝑆𝐶 = 98.26 + 40 = 138.26 𝑠

Based on the total time, the average throughput capacity of a single truck is: 𝑇𝐶𝑡𝑟𝑢𝑐𝑘 = 𝑈𝐹 ∙

3,600 3,600 = 25.3 𝑠𝑖𝑛𝑔𝑙𝑒 𝑐𝑜𝑚𝑚𝑎𝑛𝑑 𝑐𝑦𝑐𝑙𝑒𝑠/ℎ = 0.97 ∙ 138.26 𝑇𝑆𝐶

And the overall number of trucks required: 𝑁𝑡𝑟𝑢𝑐𝑘𝑠 = ⌈

𝑇𝐶𝑟𝑒𝑞 100 ⌉ = 4 𝑡𝑟𝑢𝑐𝑘𝑠 ⌉=⌈ 25.3 𝑇𝐶𝑡𝑟𝑢𝑐𝑘

𝑇𝐶𝑟𝑒𝑎𝑙 = 𝑇𝐶𝑡𝑟𝑢𝑐𝑘 ∙ 𝑁𝑡𝑟𝑢𝑐𝑘𝑠 = 25.3 ∙ 4 = 101.2 𝑠𝑖𝑛𝑔𝑙𝑒 𝑐𝑜𝑚𝑚𝑎𝑛𝑑 𝑐𝑦𝑐𝑙𝑒𝑠/ℎ 2

Exercise 3 1. Storage system configuration The system depth (V) is given (i.e. 30 m). As regards the system width: CASE a) Straddle reach trucks

𝐻 𝑚𝑎𝑥𝑓𝑜𝑟𝑘 9 ⌉ = ⌈ ⌉ = 6 𝑙𝑒𝑣𝑒𝑙𝑠 𝐻 1.6 𝑅𝑎𝑐𝑘 𝑙𝑒𝑛𝑔𝑡ℎ 30 𝑁𝐶 = ⌊ ⌋ = ⌊ ⌋ = 15 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 2 𝐿 3,600 𝑆𝐶𝑟𝑒𝑞 ⌉=⌈ ⌉ = 10 𝑎𝑖𝑠𝑙𝑒𝑠 𝑁𝐴 = ⌈ 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 2 ∙ 2 ∙ 6 ∙ 15 𝑈 = 𝑁𝐴 ∙ 𝑀𝑜𝑑𝑢𝑙𝑒 𝑤𝑖𝑑𝑡ℎ = 10 ∙ (2 ∙ 1.3 + 2.5) = 51 𝑚 𝑁𝐿 = ⌈

CASE b) Counterbalance forklift trucks

𝐻 𝑚𝑎𝑥𝑓𝑜𝑟𝑘 6 ⌉ = ⌈ ⌉ = 4 𝑙𝑒𝑣𝑒𝑙𝑠 𝐻 1.6 30 𝑅𝑎𝑐𝑘 𝑙𝑒𝑛𝑔𝑡ℎ ⌋ = ⌊ ⌋ = 15 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑁𝐶 = ⌊ 𝐿 2 𝑆𝐶𝑟𝑒𝑞 3,600 𝑁𝐴 = ⌈ ⌉=⌈ ⌉ = 15 𝑎𝑖𝑠𝑙𝑒𝑠 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 2 ∙ 2 ∙ 4 ∙ 15 𝑈 = 𝑁𝐴 ∙ 𝑀𝑜𝑑𝑢𝑙𝑒 𝑤𝑖𝑑𝑡ℎ = 15 ∙ (2 ∙ 1.3 + 3) = 84 𝑚 𝑁𝐿 = ⌈

2. Estimate the investment costs (i.e. truck, land and building costs) CASE a) Straddle reach trucks In order to assess the truck costs we need to calculate the number of trucks: 𝑈 𝑉 2 ∙ ( 3 + 2 + 𝑐𝑟𝑜𝑠𝑠 𝑎𝑖𝑠𝑙𝑒 𝑤𝑖𝑑𝑡ℎ) 𝐻 ∙ (𝑁𝐿 − 1) 74 𝑃 𝑆 8 = 53 𝑠 + = 𝑉𝑇𝑆𝐶 = + = + 0.5 𝑆 𝑆𝐻 𝑆𝑉 𝑆𝑉 2 𝐻 𝑇𝐶𝑡𝑟𝑢𝑐𝑘

𝑇𝑆𝐶 = 𝑉𝑇𝑆𝐶 + 𝐹𝑇𝑆𝐶 = 53 + 40 = 93 𝑠 3,600 3,600 = 38.7 𝑠𝑖𝑛𝑔𝑙𝑒 𝑐𝑜𝑚𝑚𝑎𝑛𝑑 𝑐𝑦𝑐𝑙𝑒𝑠/ℎ = = 93 𝑇𝑆𝐶 𝑇𝐶𝑟𝑒𝑞 150 ⌉ = 4 𝑡𝑟𝑢𝑐𝑘𝑠 𝑁𝑡𝑟𝑢𝑐𝑘𝑠 = ⌈ ⌉=⌈ 38.7 𝑇𝐶𝑡𝑟𝑢𝑐𝑘

𝐼0 = 𝐼𝐿𝑎𝑛𝑑 + 𝐼𝐵𝑢𝑖𝑙𝑑𝑖𝑛𝑔 + 𝐼𝑇𝑟𝑢𝑐𝑘𝑠 = 51 ∙ (30 + 5 + 5) [𝑚2 ] ∙ (500 + 370) [€ ⁄ 𝑚2 ] + 4 [𝑡𝑟𝑢𝑐𝑘𝑠] ∙ 30,000 [€ ⁄ 𝑡𝑟𝑢𝑐𝑘] = 1,894,800 € CASE b) Counterbalance forklift trucks 𝑈 𝑉 2 ∙ ( 3 + 2 + 𝑐𝑟𝑜𝑠𝑠 𝑎𝑖𝑠𝑙𝑒 𝑤𝑖𝑑𝑡ℎ) 𝐻 ∙ (𝑁𝐿 − 1) 96 4.8 𝑃 𝑆 = 60 𝑠 + = 𝑉𝑇𝑆𝐶 = + = + 0.4 𝑆𝐻 𝑆𝐻 𝑆𝑉 𝑆𝑉 2 𝑇𝐶𝑡𝑟𝑢𝑐𝑘

𝑇𝑆𝐶 = 𝑉𝑇𝑆𝐶 + 𝐹𝑇𝑆𝐶 = 60 + 40 = 100 𝑠 3,600 3,600 = 36 𝑠𝑖𝑛𝑔𝑙𝑒 𝑐𝑜𝑚𝑚𝑎𝑛𝑑 𝑐𝑦𝑐𝑙𝑒𝑠/ℎ = = 100 𝑇𝑆𝐶 𝑇𝐶𝑟𝑒𝑞 150 ⌉ = 5 𝑡𝑟𝑢𝑐𝑘𝑠 𝑁𝑡𝑟𝑢𝑐𝑘𝑠 = ⌈ ⌉=⌈ 36 𝑇𝐶𝑡𝑟𝑢𝑐𝑘

𝐼0 = 𝐼𝐿𝑎𝑛𝑑 + 𝐼𝐵𝑢𝑖𝑙𝑑𝑖𝑛𝑔 + 𝐼𝑇𝑟𝑢𝑐𝑘𝑠 = 84 ∙ (30 + 5 + 5) [𝑚2 ] ∙ (500 + 340) [€ ⁄ 𝑚2 ] + 5 [𝑡𝑟𝑢𝑐𝑘𝑠] ∙ 22,000 [€ ⁄ 𝑡𝑟𝑢𝑐𝑘] = 2,932,400 €

3

AUTOMATED WAREHOUSES - SOLUTIONS Exercise 1. 1. Storage system configuration We can start from sizing the generic bay: 𝐻 (𝑏𝑎𝑦 ℎ𝑒𝑖𝑔ℎ𝑡) = 𝐻𝑃𝐿 + 𝑏𝑒𝑎𝑚 ℎ𝑒𝑖𝑔ℎ𝑡 + 𝑐 = 1.5 + 0.1 + 0.15 = 1.75 𝑚

𝐿 (𝑏𝑎𝑦 𝑤𝑖𝑑𝑡ℎ) = 𝑁𝑃𝐵 ∙ 𝐿𝑃𝐿 + (𝑁𝑃𝐵 + 1) ∙ 𝑑 + 𝑢𝑝𝑟𝑖𝑔ℎ𝑡 𝑤𝑖𝑑𝑡ℎ = 3 ∙ 0.8 + 4 ∙ 0.08 + 0.09 = 2.81 𝑚 𝐷 = 𝐷𝑃𝐿 +

𝑒 0.2 = 1.3 𝑚 = 1.2 + 2 2

Warehouse dimensions: 22 𝑅𝑎𝑐𝑘 ℎ𝑒𝑖𝑔ℎ𝑡 ⌋=⌊ ⌋ = 12 𝑙𝑒𝑣𝑒𝑙𝑠 𝐻 1.75 𝑅𝑎𝑐𝑘 𝑙𝑒𝑛𝑔ℎ𝑡 90 𝑁𝐶𝑚𝑎𝑥 = ⌊ ⌋=⌊ ⌋ = 32 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝐿 2.81 16,000 𝑆𝐶𝑟𝑒𝑞 ⌉=⌈ ⌉ = 7 𝑎𝑖𝑠𝑙𝑒𝑠 𝑁𝐴𝑚𝑖𝑛 = ⌈ 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 2 ∙ 3 ∙ 12 ∙ 32 In order to minimize space and S/R machine travel distance, we can verify whether the minimum number of columns such that the required SC is satisfied is lower than 32: 𝑁𝐿𝑚𝑎𝑥 = 𝑁𝐿 = ⌊

𝑁𝐶 = ⌈

16,000 𝑆𝐶𝑟𝑒𝑞 ⌉=⌈ ⌉ = 32 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐴𝑚𝑖𝑛 2 ∙ 3 ∙ 12 ∙ 7

𝑆𝐶𝑟𝑒𝑎𝑙 = 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 ∙ 𝑁𝐴 = 2 ∙ 3 ∙ 12 ∙ 32 ∙ 7 = 16,128 𝑝𝑎𝑙𝑙𝑒𝑡 𝑙𝑜𝑎𝑑𝑠 Knowing the main parameters related to the warehouse design, we can proceed to the cycle time assessment (using the FEM 9851, as reported in the figure and table below), to make sure that the identified system configuration respects also the required Throughput Capacity:

Coordinates [m] y(I/O) P1 P2

X 0 17.98 59.95

Y 3.50 14.00 5.02

Average time to perform a single command cycle: 17.98 14 − 3.5 |;| |) = 26.25 𝑠 2 0.4 59.95 5.02 − 3.5 |) = 29.98 𝑠 𝑇(𝐼 → 𝑃2) = 𝑇(𝑃2 → 𝑂) = 𝑚𝑎𝑥 (| |;| 0.4 2 𝑇(𝐼 → 𝑃1) = 𝑇(𝑃1 → 𝑂) = 𝑚𝑎𝑥 (|

𝐴𝑉𝑇𝑆𝐶 = [𝑇(𝐼 → 𝑃1) + 𝑇(𝐼 → 𝑃2)] = 56.23 𝑠 𝐹𝑇𝑆𝐶 = 2 ∙ (5 + 8 + 10) = 46 𝑠 𝐴𝑇𝑆𝐶 = 102.23 𝑠/𝑆𝐶

4

Average time to perform a dual command cycle: 59.95 − 17.98 5.02 − 14.00 |) = 22.45 𝑠 |;| 0.4 2 𝐴𝑉𝑇𝐷𝐶 = [𝑇(𝐼 → 𝑃1) + 𝑇(𝑃1 → 𝑃2) + 𝑇(𝑃2 → 𝑂)] = 78.68 𝑠

𝑇(𝑃1 → 𝑃2) = 𝑚𝑎𝑥 (|

𝐹𝑇𝐷𝐶 = 3 ∙ (5 + 8) + 4 ∙ 10 = 79 𝑠 𝐴𝑇𝐷𝐶 = 157.68 𝑠/𝐷𝐶

Therefore, the Throughput Capacity of a single S/R machine is: 𝑇𝐶𝑆⁄

𝑅 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

%𝑆𝐶 0.6 + %𝐷𝐶) 2 2 + 0.4 = 3600 ∙ = 20.26 𝑃𝐿/ℎ = %𝑆𝐶 ∙ 𝐴𝑇𝑆𝐶 + %𝐷𝐶 ∙ 𝐴𝑇𝐷𝐶 0.6 ∙ 102.23 + 0.4 ∙ 157.68 3600 ∙ (

And the system TC is: 𝑇𝐶 = 𝑁𝐴 ∙ 𝑇𝐶𝑆⁄

𝑅 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

= 7 ∙ 20.26 = 141.8 𝑃𝐿𝑠/ℎ (higher than the target value = 130 PL/h)

2. Selection of the best storage location among those proposed for performing the dual command cycle To identify the best pallet location we apply the “No-cost zone” criteria. Apply the methodology consists in checking which (if any) of the four points A, B, C and D falls into the NCZ. Using the graphical method (i.e. drawing the NCZ made of 45° degree lines), results show that B is the only point falling into the NCZ, and therefore it is the most suitable candidate for optimizing the dual command cycle.

Time coordinates 45 40

A (3.7; 39.4)

35

D (32.8; 35)

30 [s]

P (27.2; 26.3) 25 20 B (13.1; 17.5) 15

C (23.4; 13.1)

10

I/O (0,8.8)

5

0 0

4

7

11

14

18

21

25

28

32

35

39

42

46

[s]

The same conclusion can be reached also analytically, by verifying the equation reported below for each of the four points (denoted as Q in the formula):

𝑉𝑇(𝐼 → 𝑃) = 𝑉𝑇 (𝐼 → 𝑄) + 𝑉𝑇(𝑄 → 𝑃)

𝑉𝑇(𝐼 → 𝑃) = 𝑚𝑎𝑥(|0 − 27.2|; |8.8 − 26.3|) = 27.2 𝑠

𝑉𝑇(𝐼 → 𝐴) + 𝑉𝑇(𝐴 → 𝑃) = 𝑚𝑎𝑥(|0 − 3.7|; |8.8 − 39.4|) + 𝑚𝑎𝑥(|3.7 − 27.2|; |39.4 − 26.3|) = 54.1 𝑠 𝑉𝑇(𝐼 → 𝐵) + 𝑉𝑇(𝐵 → 𝑃) = 27.2 𝑠 𝑉𝑇(𝐼 → 𝐶) + 𝑉𝑇(𝐶 → 𝑃) = 36.6 𝑠

𝑉𝑇(𝐼 → 𝐷) + 𝑉𝑇(𝐷 → 𝑃) = 41.5 𝑠

5

Exercise 2. 1. Dimension of the single bay 𝐻 (𝑏𝑎𝑦 ℎ𝑒𝑖𝑔ℎ𝑡) = 𝐻𝑃𝐿 + 𝑏𝑒𝑎𝑚 ℎ𝑒𝑖𝑔ℎ𝑡 + 𝑐 = 1.4 + 0.1 + 0.15 = 1.65 𝑚

𝐿 (𝑏𝑎𝑦 𝑤𝑖𝑑𝑡ℎ) = 𝑁𝑃𝐵 ∙ 𝐿𝑃𝐿 + (𝑁𝑃𝐵 + 1) ∙ 𝑑 + 𝑢𝑝𝑟𝑖𝑔ℎ𝑡 𝑤𝑖𝑑𝑡ℎ = 2 ∙ 0.8 + 3 ∙ 0.1 + 0.1 = 2 𝑚 𝐷 = 𝐷𝑃𝐿 +

0.2 𝑒 = 1.2 + = 1.3 𝑚 2 2

2. System configuration (i.e. number of the storage levels, columns of bays, and aisles) 𝑅𝑎𝑐𝑘 ℎ𝑒𝑖𝑔ℎ𝑡 30 ⌋=⌊ ⌋ = 18 𝑙𝑒𝑣𝑒𝑙𝑠 𝐻 1.65 100 𝑅𝑎𝑐𝑘 𝑙𝑒𝑛𝑔ℎ𝑡 ⌋=⌊ 𝑁𝐶𝑚𝑎𝑥 = ⌊ ⌋ = 50 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑊 2 𝑆𝐶𝑟𝑒𝑞 16,000 ⌉ = 5 𝑎𝑖𝑠𝑙𝑒𝑠 𝑁𝐴𝑚𝑖𝑛 = ⌈ ⌉=⌈ 2 ∙ 2 ∙ 18 ∙ 50 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 𝑁𝐿𝑚𝑎𝑥 = 𝑁𝐿 = ⌊

In order to minimize space and S/R machine travel distance, we can verify whether the minimum number of columns such that the required SC is satisfied is lower than 50: 𝑆𝐶𝑟𝑒𝑞 16,000 ⌉ = 45 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑁𝐶 = ⌈ ⌉=⌈ 2 ∙ 2 ∙ 18 ∙ 5 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐴𝑚𝑖𝑛

𝑆𝐶𝑟𝑒𝑎𝑙 = 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 ∙ 𝑁𝐴 = 2 ∙ 2 ∙ 18 ∙ 45 ∙ 5 = 16,200 𝑝𝑎𝑙𝑙𝑒𝑡 𝑙𝑜𝑎𝑑𝑠 3. Warehouse dimensions 𝑅𝐿 = 𝑁𝐶 ∙ 𝐿 = 45 ∙ 2 = 90 𝑚

𝑅𝐻 = 𝑁𝐿 ∙ 𝐻 = 18 ∙ 1.65 = 29.7 𝑚

𝑅𝐹𝑊 = 𝑁𝐴 ∙ 𝑀𝑜𝑑𝑢𝑙𝑒𝑤𝑖𝑑𝑡ℎ = 5 ∙ (2 ∙ 1.3 + 1.4) = 20 𝑚 4. Average single and dual command cycle time Knowing the main parameters related to the warehouse design, we can proceed to the cycle time assessment (using the FEM 9851, as reported in the figure and table below), to make sure that the identified system configuration respects also the required Throughput Capacity:

Coordinates [m] y(I/O) P1 P2

X 0 18 60

Y 6.6 20.9 7.81

Average time to perform a single command cycle: 18 20.9 − 6.6 |) = 23.83 𝑠 𝑇(𝐼 → 𝑃1) = 𝑇(𝑃1 → 𝑂) = 𝑚𝑎𝑥 (| | ; | 0.6 2 60 7.81 − 6.6 |) = 30 𝑠 𝑇(𝐼 → 𝑃2) = 𝑇(𝑃2 → 𝑂) = 𝑚𝑎𝑥 (| | ; | 0.6 2 𝐴𝑉𝑇𝑆𝐶 = [𝑇(𝐼 → 𝑃1) + 𝑇(𝐼 → 𝑃2)] = 53.83 𝑠

6

𝐹𝑇𝑆𝐶 = 2 ∙ (4 + 10 + 6) = 40 𝑠 𝐴𝑇𝑆𝐶 = 93.83 𝑠/𝑆𝐶

Average time to perform a dual command cycle: 60 − 18 20.9 − 7.81 |) = 21.82 𝑠 |;| 0.6 2 𝐴𝑉𝑇𝐷𝐶 = [𝑇(𝐼 → 𝑃1) + 𝑇(𝑃1 → 𝑃2) + 𝑇(𝑃2 → 𝑂)] = 75.65 𝑠 𝑇(𝑃1 → 𝑃2) = 𝑚𝑎𝑥 (|

𝐹𝑇𝐷𝐶 = 3 ∙ (4 + 6) + 4 ∙ 10 = 70 𝑠 𝐴𝑇𝐷𝐶 = 145.65 𝑠/𝐷𝐶

5. Throughput capacity of the overall system 𝑇𝐶𝑆⁄

𝑅 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

%𝑆𝐶 0.4 2 + %𝐷𝐶) 2 + 0.6 = 3600 ∙ = 23.05 𝑃𝐿/ℎ = %𝑆𝐶 ∙ 𝐴𝑇𝑆𝐶 + %𝐷𝐶 ∙ 𝐴𝑇𝐷𝐶 0.4 ∙ 93.83 + 0.6 ∙ 145.65 𝑇𝐶 = 𝑁𝐴 ∙ 𝑇𝐶𝑆 ⁄ 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 = 5 ∙ 23.05 = 115.3 𝑃𝐿/ℎ 3600 ∙ (

𝑅

Exercise 3. We can start from assessing the warehouse dimensions (the bay gross dimensions are given) 8 𝑅𝑎𝑐𝑘 ℎ𝑒𝑖𝑔ℎ𝑡 ⌋=⌊ ⌋ = 14 𝑙𝑒𝑣𝑒𝑙𝑠 𝐻 0.55 𝑅𝑎𝑐𝑘 𝑙𝑒𝑛𝑔ℎ𝑡 50 𝑁𝐶𝑚𝑎𝑥 = ⌊ ⌋ = ⌊ ⌋ = 21 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 2.3 𝑊 𝑆𝐶𝑟𝑒𝑞 8,500 𝑁𝐴𝑚𝑖𝑛 = ⌈ ⌉=⌈ ⌉ = 4 𝑎𝑖𝑠𝑙𝑒𝑠 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 2 ∙ 4 ∙ 14 ∙ 21 In order to minimize space and S/R machine travel distance, we can verify whether the minimum number of columns such that the required SC is satisfied is lower than 21: 𝑁𝐿𝑚𝑎𝑥 = 𝑁𝐿 = ⌊

𝑆𝐶𝑟𝑒𝑞 8,500 ⌉ = 19 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑁𝐶 = ⌈ ⌉=⌈ 2 ∙ 4 ∙ 14 ∙ 4 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐴𝑚𝑖𝑛

𝑆𝐶𝑟𝑒𝑎𝑙 = 2 ∙ 𝑁𝑃𝐵 ∙ 𝑁𝐿 ∙ 𝑁𝐶 ∙ 𝑁𝐴 = 2 ∙ 4 ∙ 14 ∙ 19 ∙ 4 = 8,512 𝑡𝑜𝑡𝑒𝑠 Knowing the main parameters related to the warehouse design, we can proceed to the cycle time assessment (using the FEM 9851, as reported in the figure and table below), to make sure that the identified system configuration respects also the required Throughput Capacity:

Coordinates [m] y(I/O) P1 P2

X 0 8.74 29.13

Y 1.10 5.13 1.8

7

Average time to perform a dual command cycle:

8.74 5.13 − 1.1 |) = 2.69 𝑠 |;| 1.5 4 29.13 1.8 − 1.1 |) = 7.28 𝑠 |;| 𝑇(𝐼 → 𝑃2) = 𝑇(𝑃2 → 𝑂) = 𝑚𝑎𝑥 (| 1.5 4 29.13 − 8.74 5.13 − 1.8 |) = 5.1 𝑠 𝑇(𝑃1 → 𝑃2) = 𝑚𝑎𝑥 (| |;| 1.5 4 𝐴𝑉𝑇𝐷𝐶 = [𝑇(𝐼 → 𝑃1) + 𝑇(𝑃1 → 𝑃2) + 𝑇(𝑃2 → 𝑂)] = 15.07 𝑠 𝐹𝑇𝐷𝐶 = 12 𝑠 𝐴𝑇𝐷𝐶 = 27.07 𝑠/𝐷𝐶 Therefore, the Throughput Capacity of a single S/R machine is: 3,600 3,600 𝑇𝐶𝑆 ⁄ 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 = = = 133 𝑡𝑜𝑡𝑒𝑠/ℎ 𝑅 𝐴𝑇𝐷𝐶 27.07 𝑇(𝐼 → 𝑃1) = 𝑇(𝑃1 → 𝑂) = 𝑚𝑎𝑥 (|

And the system TC is: 𝑇𝐶 = 𝑁𝐴 ∙ 𝑇𝐶𝑆⁄

𝑅 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

= 4 ∙ 133 = 532 𝑡𝑜𝑡𝑒𝑠/ℎ (higher than the target value = 450 totes/h)

8...