hw2 solutions

University of Toronto Department of Mechanical and Industrial Engineering

MIE468: Facility Planning (Winter 2014)

Homework #2 Solutions

Problem 1 All Ik values are rounded up.

𝐼𝑋𝐶 =𝑂𝑋

(1 − 𝑑𝑋𝐶)=

100,000

0.97= 103,093

𝐼𝑋𝐵 =𝑂𝑋

(1 − 𝑑𝑋𝐶)(1 − 𝑑𝑋𝐵)=

100,000

0.97 × 0.96= 107,389

𝐼𝑋𝐴 =𝑂𝑋

(1 − 𝑑𝑋𝐶)(1− 𝑑𝑋𝐵)(1 − 𝑑𝑋𝐴)=

100,000

0.97 × 0.96 × 0.95= 113,041

𝐼𝑌𝐶 =𝑂𝑌

(1 − 𝑑𝑌𝐶)=

200,000

0.97= 206,186

𝐼𝑋𝐴 =𝑂𝑌

(1 − 𝑑𝑌𝐶)(1 − 𝑑𝑌𝐴)=

200,000

0.97 × 0.95= 217,038

𝐼𝑋𝐵 =𝑂𝑌

(1 − 𝑑𝑌𝐶)(1 − 𝑑𝑌𝐴)(1 − 𝑑𝑌𝐵)=

200,000

0.97 × 0.95 × 0.96= 226,081

Setup times are identical for machines A, B and C for a particular product. The setup time for product X on each machine is 20 mins; the setup time for product Y on each machine is 40 mins. Assuming a single setup is needed to produce the annual requirement of a product on a machine, the number of machines required is determined as follows:

𝑭𝑨 =𝟎.𝟏𝟓 × 𝟏𝟏𝟑,𝟎𝟒𝟏

(𝟎.𝟖𝟓)(𝟏𝟔𝟎𝟎)(𝟎.𝟗𝟓)+

𝟎.𝟏𝟎 × 𝟐𝟏𝟕,𝟎𝟑𝟖(𝟎.𝟖𝟓)(𝟏𝟔𝟎𝟎)(𝟎.𝟗𝟓)

+𝟐𝟎 + 𝟒𝟎

(𝟔𝟎)(𝟏𝟔𝟎𝟎)× 𝑭𝑨

𝑭𝑨 = 𝟐𝟗.𝟗𝟒 → 𝟑𝟎 𝐦𝐚𝐜𝐡𝐢𝐧𝐞𝐬

𝑭𝑩 =𝟎.𝟐𝟓 × 𝟏𝟎𝟕,𝟑𝟖𝟗

(𝟎.𝟗𝟎)(𝟏𝟔𝟎𝟎)(𝟎.𝟗𝟎)+

𝟎.𝟏𝟎 × 𝟐𝟐𝟔,𝟎𝟖𝟏(𝟎.𝟗𝟎)(𝟏𝟔𝟎𝟎)(𝟎.𝟗𝟎)

+𝟐𝟎 + 𝟒𝟎

(𝟔𝟎)(𝟏𝟔𝟎𝟎)× 𝑭𝑩

𝑭𝑩 = 𝟑𝟖.𝟏𝟖 → 𝟑𝟗 𝐦𝐚𝐜𝐡𝐢𝐧𝐞𝐬

𝑭𝑪 =𝟎.𝟏𝟎 × 𝟏𝟎𝟑,𝟎𝟗𝟑

(𝟎.𝟗𝟓)(𝟏𝟔𝟎𝟎)(𝟎.𝟖𝟓)+

𝟎.𝟏𝟓 × 𝟐𝟎𝟔,𝟏𝟖𝟔(𝟎.𝟗𝟓)(𝟏𝟔𝟎𝟎)(𝟎.𝟖𝟓)

+𝟐𝟎 + 𝟒𝟎

(𝟔𝟎)(𝟏𝟔𝟎𝟎)× 𝑭𝑪

𝑭𝑪 = 𝟑𝟏.𝟗𝟒 → 𝟑𝟐 𝐦𝐚𝐜𝐡𝐢𝐧𝐞𝐬 If setups occur more frequently, then additional machines might be required due to the lost production cost by setups.

University of Toronto Department of Mechanical and Industrial Engineering

MIE468: Facility Planning (Winter 2014)

Problem 2

I =O

(1 − d) =7500.80

= 937.5 → 938

𝑭 =𝑺𝑸𝑬𝑯𝑹

=(𝟏𝟓𝟔𝟎) × 𝟗𝟑𝟖

�𝟏𝟓𝟐𝟎� × 𝟕 × 𝟏= 𝟒𝟒.𝟔𝟕 → 𝟒𝟓 𝐦𝐚𝐜𝐡𝐢𝐧𝐞𝐬

Alternatively, you could say that the loss of one hour per shift is reduction in the reliability:

𝑭 =𝑺𝑸𝑬𝑯𝑹

=(𝟏𝟓𝟔𝟎) × 𝟗𝟑𝟖

�𝟏𝟓𝟐𝟎� × 𝟖 × (𝟕𝟖)= 𝟒𝟒.𝟔𝟕 → 𝟒𝟓 𝐦𝐚𝐜𝐡𝐢𝐧𝐞𝐬

Problem 3 1

Machine 1 2 3 4 5 6

Part

1 1 1 1 2 1 1 3 1 1 1 4 1 1 5 1 1 6 1 1 7 1 1 1 8 1 1 9 1 1 1

2. DCA a. Tabulate Sums

Machine sum 1 2 3 4 5 6

Part

1 1 1 1 3 2 1 1 2 3 1 1 1 3 4 1 1 2 5 1 1 2 6 1 1 2 7 1 1 1 3 8 1 1 2 9 1 1 1 3

Sum 4 2 4 4 4 4

University of Toronto Department of Mechanical and Industrial Engineering

MIE468: Facility Planning (Winter 2014)

b. Sort Sums

Machine sum 2 6 5 4 3 1 Pa

rt

9 1 1 1 3 7 1 1 1 3 3 1 1 1 3 1 1 1 1 3 8 1 1 2 6 1 1 2 5 1 1 2 4 1 1 2 2 1 1 2

Sum 2 4 4 4 4 4 c. Sort Columns

Machine 6 1 4 5 3 2

Part

9 1 1 1 7 1 1 1 3 1 1 1 1 1 1 1 8 1 1 6 1 1 5 1 1 4 1 1 2 1 1

d. Sort Rows

Machine 6 1 4 5 3 2

Part

9 1 1 1 3 1 1 1 6 1 1 5 1 1 4 1 1 7 1 1 1 1 1 1 1 8 1 1 2 1 1

Group machines (6, 1, 4) and machines (5, 3, 2) 2. SCA Iteration 1 Machine pair SC Combine? {1,2} 0

{1,3} 0 {1,4} 3/5

University of Toronto Department of Mechanical and Industrial Engineering

MIE468: Facility Planning (Winter 2014)

{1,5} 0 {1,6} 3/5 {2,3} 1/2 {2,4} 0 {2,5} 1/2 {2,6} 0 {3,4} 0 {3,5} 1 Yes {3,6} 0 {4,5} 0 {4,6} 3/5 {5,6} 0 => Combine (3,5)

Iteration 2 Machine pair SC Combine?

{1,2} 0 {1,(3,5)} 0 {1,4} 3/5 Yes {1,6} 3/5 Yes {2,(3,5)} 1/2 {2,4} 0 {2,6} 0 {(3,5),4} 0 {(3,5),6} 0 {4,6} 3/5 Yes => Combine (1,4,6)

Iteration 3 Machine pair SC Combine?

{(1,4,6),2} 0 {(1,4,6),(3,5)} 0 {2,(3,5)} 1/2 Yes => Combine (2,3,5)

Group machines (1,4,6) and machines (2,3,5) 3

1.0

0.8

0.5

0.33

0

2 3 5 1 4 6

Part-Machine Relationship 1 2 3 4 5 6 7

1 1 0 0 1 0 1 0

2 0 1 1 0 1 0 0

3 0 0 0 1 0 1 0

4 0 1 1 0 0 0 0

5 0 0 1 0 0 0 1

Calculating the matrix

dij 1 2 3 4 5

1 0 6 1 5 5

2 6 0 5 1 3

3 1 5 0 4 4

4 5 1 4 0 2

5 5 3 4 2 0

Distance Matrix

dij 1 2 3 4 5

1 0 6 1 5 5

2 6 0 5 1 3

3 1 5 0 4 4

4 5 1 4 0 2

5 5 3 4 2 0

Variables:

Part, i 1 2 3 4 5 each part at most selected once

1 0 0 1 0 0 1

2 0 0 0 1 0 1

3 0 0 1 0 0 1

4 0 0 0 1 0 1

5 0 0 0 1 0 1

2

Objective 4.000001

family, j

number of families