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THE BRANCH AND BOUND (COMPLETE ENUMERATION) METHOD
It is a method of systematic enumeration, very often followed in OR. In this method all the possible
solutions are enumerated and the best (minimum) one is selected.
𝑬𝒙𝒂𝒎𝒑𝒍𝒆
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝑆𝑒𝑙𝑒𝑐𝑡 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑐𝑜𝑙𝑢𝑚𝑛:
7 + 5 + 2 + 23 = 37
∴ 𝐿𝑜𝑤𝑒𝑟 𝐵𝑜𝑢𝑛𝑑 = 37
∴ 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑙𝑜𝑤𝑒𝑟 𝑡ℎ𝑎𝑛 37
𝐴 → 1: 90 + 14 + 2 + 23 = 129 (𝐹)
𝐵 → 1: 69 + 5 + 2 + 23 = 99 (𝐹)
𝐶 → 1: 57 + 5 + 48 + 23 = 133 (𝐼)
𝐷 → 1: 7 + 5 + 2 + 73 = 87 (𝐼)
𝐷 → 1, 𝐴 → 2: 7 + 5 + 2 + 79 = 93 (𝐼)
𝑫 → 𝟏, 𝑩 → 𝟐: 𝟕 + 𝟏𝟒 + 𝟐 + 𝟕𝟑 = 𝟗𝟔 (𝑭)
𝐷 → 1, 𝐶 → 2: 7 + 93 + 48 + 73 = 221 (𝐼)
𝐷 → 1, 𝐴 → 2, 𝐵 → 3: 7 + 5 + 83 + 79 = 174 (𝐹)
𝐷 → 1, 𝐴 → 2, 𝐶 → 3: 7 + 5 + 2 + 86 = 100 (𝐹)
𝑨𝒏𝒔. : 𝑨 → 𝟒 (𝟕𝟑), 𝑩 → 𝟐 (𝟏𝟒), 𝑪 → 𝟑 (𝟐), 𝑫 → 𝟏 (𝟕)
𝑻𝒐𝒕𝒂𝒍 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒕𝒊𝒎𝒆 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 = 𝟗𝟔 𝒖𝒏𝒊𝒕𝒔
Machine 
Job 
1 2 3 4
A 90 5 48 73
B 69 14 83 86
C 57 93 2 79
D 7 77 75 23
37
129
99
133
87
93
96
221
174
100
B→2
F
F
F
F
F
I
I
I
I
I
optimum
𝑬𝒙𝒂𝒎𝒑𝒍𝒆
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝑆𝑒𝑙𝑒𝑐𝑡 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑐𝑜𝑙𝑢𝑚𝑛:
0 + 2 + 4 + 0 = 6
∴ 𝐿𝑜𝑤𝑒𝑟 𝐵𝑜𝑢𝑛𝑑 = 6
∴ 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑙𝑜𝑤𝑒𝑟 𝑡ℎ𝑎𝑛 6
𝑁𝑜𝑤, 𝑎𝑠 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑖𝑡𝑠𝑒𝑙𝑓 𝑖𝑠 𝑓𝑒𝑎𝑠𝑖𝑏𝑙𝑒
∴ 𝑖𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
M1 M2 M3 M4
J1 2 6 4 2
J2 0 5 5 4
J3 3 2 7 3
J4 1 4 6 0

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The branch and bound method

  • 1. THE BRANCH AND BOUND (COMPLETE ENUMERATION) METHOD It is a method of systematic enumeration, very often followed in OR. In this method all the possible solutions are enumerated and the best (minimum) one is selected. 𝑬𝒙𝒂𝒎𝒑𝒍𝒆 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑆𝑒𝑙𝑒𝑐𝑡 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑐𝑜𝑙𝑢𝑚𝑛: 7 + 5 + 2 + 23 = 37 ∴ 𝐿𝑜𝑤𝑒𝑟 𝐵𝑜𝑢𝑛𝑑 = 37 ∴ 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑙𝑜𝑤𝑒𝑟 𝑡ℎ𝑎𝑛 37 𝐴 → 1: 90 + 14 + 2 + 23 = 129 (𝐹) 𝐵 → 1: 69 + 5 + 2 + 23 = 99 (𝐹) 𝐶 → 1: 57 + 5 + 48 + 23 = 133 (𝐼) 𝐷 → 1: 7 + 5 + 2 + 73 = 87 (𝐼) 𝐷 → 1, 𝐴 → 2: 7 + 5 + 2 + 79 = 93 (𝐼) 𝑫 → 𝟏, 𝑩 → 𝟐: 𝟕 + 𝟏𝟒 + 𝟐 + 𝟕𝟑 = 𝟗𝟔 (𝑭) 𝐷 → 1, 𝐶 → 2: 7 + 93 + 48 + 73 = 221 (𝐼) 𝐷 → 1, 𝐴 → 2, 𝐵 → 3: 7 + 5 + 83 + 79 = 174 (𝐹) 𝐷 → 1, 𝐴 → 2, 𝐶 → 3: 7 + 5 + 2 + 86 = 100 (𝐹) 𝑨𝒏𝒔. : 𝑨 → 𝟒 (𝟕𝟑), 𝑩 → 𝟐 (𝟏𝟒), 𝑪 → 𝟑 (𝟐), 𝑫 → 𝟏 (𝟕) 𝑻𝒐𝒕𝒂𝒍 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒕𝒊𝒎𝒆 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 = 𝟗𝟔 𝒖𝒏𝒊𝒕𝒔 Machine  Job  1 2 3 4 A 90 5 48 73 B 69 14 83 86 C 57 93 2 79 D 7 77 75 23 37 129 99 133 87 93 96 221 174 100 B→2 F F F F F I I I I I optimum
  • 2. 𝑬𝒙𝒂𝒎𝒑𝒍𝒆 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑆𝑒𝑙𝑒𝑐𝑡 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑐𝑜𝑙𝑢𝑚𝑛: 0 + 2 + 4 + 0 = 6 ∴ 𝐿𝑜𝑤𝑒𝑟 𝐵𝑜𝑢𝑛𝑑 = 6 ∴ 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑙𝑜𝑤𝑒𝑟 𝑡ℎ𝑎𝑛 6 𝑁𝑜𝑤, 𝑎𝑠 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑖𝑡𝑠𝑒𝑙𝑓 𝑖𝑠 𝑓𝑒𝑎𝑠𝑖𝑏𝑙𝑒 ∴ 𝑖𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 M1 M2 M3 M4 J1 2 6 4 2 J2 0 5 5 4 J3 3 2 7 3 J4 1 4 6 0