Mixed integer programing:Use math solver program To solve this model a reduced example has been considered because of the
large number of constraints
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πβ1 π
πβ1
πππ ( β β π₯π1 πππ + β πΌππ )
π=1 π=1
π=1
πππ 17π11 + 15π21 + 17π31 + 14π41 + 12π51 + 21π11 + 22π21 + 13π31 + 19π41 + 20π51
+ 14π11 + 15π21 + 13π31 + 19π41 + 18π51
= πππ 53π11 + 54π21 + 46π31 + 56π41 + 55π51
π
β π₯ππ = 1 , π = 1, β¦ ,5
π=1
π₯11 +π₯21 +π₯31 +π₯41 + π₯51 = 1
π₯12 +π₯22 +π₯32 +π₯42 + π₯52 = 1
π₯13 +π₯23 +π₯33 +π₯43 + π₯53 = 1
π₯14 +π₯24 +π₯34 +π₯44 + π₯54 = 1
π₯15 +π₯25 +π₯35 +π₯45 + π₯55 = 1
π
β π₯ππ = 1 , π = 1, β¦ , π
π=1
π₯11 +π₯12 +π₯13 +π₯14 + π₯15 = 1
π₯21 +π₯22 +π₯23 +π₯24 + π₯25 = 1
π₯31 +π₯32 +π₯33 +π₯34 + π₯35 = 1
π₯41 +π₯42 +π₯43 +π₯44 + π₯55 = 1
π₯51 +π₯52 +π₯53 +π₯54 + π₯55 = 1
π
π
πΌππ + β π₯π,π+1 πππ + ππ,π+1 β πππ β β π₯π,π ππ+1,π β πΌπ+1,π = 0 ππππ = 1, . . ,4; π = 1, β¦ ,2
π=1
π=1
k= 1, i=1
πΌ11 + 17 β π₯12 + 15 β π₯22 + 17 β π₯32 + 14 β π₯42 + 12 β π₯52 + π12 β π11
β 21 β π₯11 β22 β π₯21 β13 β π₯31 β19 β π₯41 β20 β π₯51 + βπΌ21 = 0
For k=1, i=2
πΌ21 + 21 β π₯12 + 22 β π₯22 + 13 β π₯32 + 19 β π₯42 + 20 β π₯52 + π22 β π21
β 14 β π₯11 β15 β π₯21 β13 β π₯31 β19 β π₯41 β18 β π₯51 β πΌ31 = 0
For k=2 , i=1
πΌ12 + 17 β π₯13 + 15 β π₯23 + 17 β π₯33 + 14 β π₯43 + 12 β π₯53 + π13 β π12 β 21 β π₯12 β 22 β π₯22 β 13
β π₯32 β 19 β π₯42 β 20 β π₯52 β πΌ22 = 0
For k=2 , i=2
πΌ22 + 21 β π₯13 + 22 β π₯23 + 13 β π₯33 + 19 β π₯43 + 20 β π₯53 + π23 β π22 β 14 β π₯12 β 15 β π₯22 β 13
β π₯32 β 19 β π₯42 β 18 β π₯52 β πΌ32 = 0
For k=3 , i=1
πΌ13 + 17 β π₯14 + 15 β π₯24 + 17 β π₯34 + 14 β π₯44 + 12 β π₯54 + π14 β π13 β 21 β π₯13 β 22 β π₯23 β 13
β π₯33 β 19 β π₯43 β 20 β π₯53 β πΌ23 = 0
For k=3 , i=2
πΌ23 + 21 β π₯14 + 22 β π₯24 + 13 β π₯34 + 19 β π₯44 + 20 β π₯54 + π24 β π23 β 14 β π₯13 β 15 β π₯23 β 13
β π₯33 β 19 β π₯43 β 18 β π₯53 β πΌ33 = 0
For k=4 , i=1
πΌ14 + 17 β π₯15 + 15 β π₯25 + 17 β π₯35 + 14 β π₯45 + 12 β π₯55 + π15 β π14 β 21 β π₯14 β 22 β π₯24 β 13
β π₯34 β 19 β π₯44 β 20 β π₯54 β πΌ24 = 0
For k=4 , i=2
πΌ24 + 1 β π₯15 + 22 β π₯25 + 13 β π₯33 + 19 β π₯43 + 20 β π₯56 + π24 β π24 β 14 β π₯14 β 15 β π₯24 β 13
β π₯34 β 19 β π₯44 β 18 β π₯54 β πΌ34 = 0
π11 = 0 , π = 1, β¦ , π β 1
π21 = 0
πΌ1π = 0 ; π = 1, β¦ , π β 1
πΌ11 = 0
πΌ12 = 0
πΌ13 = 0
πΌ14 = 0
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