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TUTORIAL WEEK 12 (TOPIC : LAYOUT DECISION) 1. There are three work centers ( $A...
Jul 17, 2024
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Solution by Steps
step 1
Identify the distances between the spaces: d12=10d_{12} = 10 feet, d23=30d_{23} = 30 feet, d13=40d_{13} = 40 feet
step 2
Identify the trips per day between work centers from the table: A↔B=20A \leftrightarrow B = 20, A↔C=50A \leftrightarrow C = 50, B↔C=30B \leftrightarrow C = 30
step 3
Calculate the total distance for each possible assignment of work centers to spaces
step 4
For assignment A→1A \rightarrow 1, B→2B \rightarrow 2, C→3C \rightarrow 3: Total Distance=2×(20×10+50×40+30×30)=2×(200+2000+900)=2×3100=6200 feet \text{Total Distance} = 2 \times (20 \times 10 + 50 \times 40 + 30 \times 30) = 2 \times (200 + 2000 + 900) = 2 \times 3100 = 6200 \text{ feet}
step 5
For assignment A→1A \rightarrow 1, B→3B \rightarrow 3, C→2C \rightarrow 2: Total Distance=2×(20×40+50×30+30×10)=2×(800+1500+300)=2×2600=5200 feet \text{Total Distance} = 2 \times (20 \times 40 + 50 \times 30 + 30 \times 10) = 2 \times (800 + 1500 + 300) = 2 \times 2600 = 5200 \text{ feet}
step 6
For assignment A→2A \rightarrow 2, B→1B \rightarrow 1, C→3C \rightarrow 3: Total Distance=2×(20×10+50×40+30×30)=2×(200+2000+900)=2×3100=6200 feet \text{Total Distance} = 2 \times (20 \times 10 + 50 \times 40 + 30 \times 30) = 2 \times (200 + 2000 + 900) = 2 \times 3100 = 6200 \text{ feet}
step 7
For assignment A→2A \rightarrow 2, B→3B \rightarrow 3, C→1C \rightarrow 1: Total Distance=2×(20×30+50×10+30×40)=2×(600+500+1200)=2×2300=4600 feet \text{Total Distance} = 2 \times (20 \times 30 + 50 \times 10 + 30 \times 40) = 2 \times (600 + 500 + 1200) = 2 \times 2300 = 4600 \text{ feet}
step 8
For assignment A→3A \rightarrow 3, B→1B \rightarrow 1, C→2C \rightarrow 2: Total Distance=2×(20×40+50×30+30×10)=2×(800+1500+300)=2×2600=5200 feet \text{Total Distance} = 2 \times (20 \times 40 + 50 \times 30 + 30 \times 10) = 2 \times (800 + 1500 + 300) = 2 \times 2600 = 5200 \text{ feet}
step 9
For assignment A→3A \rightarrow 3, B→2B \rightarrow 2, C→1C \rightarrow 1: Total Distance=2×(20×30+50×10+30×40)=2×(600+500+1200)=2×2300=4600 feet \text{Total Distance} = 2 \times (20 \times 30 + 50 \times 10 + 30 \times 40) = 2 \times (600 + 500 + 1200) = 2 \times 2300 = 4600 \text{ feet}
Answer
The optimal assignment is A→2A \rightarrow 2, B→3B \rightarrow 3, C→1C \rightarrow 1 or A→3A \rightarrow 3, B→2B \rightarrow 2, C→1C \rightarrow 1 with a total distance of 4600 feet.
Key Concept
Optimal assignment
Explanation
The optimal assignment minimizes the total distance traveled by considering all possible assignments and selecting the one with the lowest total distance.
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