Find solution using Simplex method (BigM method)
MIN Z = x1 + x2
subject to
2x1 + 4x2 >= 4
x1 + 7x2 >= 7
and x1,x2 >= 0;

Solution:
Problem is

Min `Z` `=` `` `` `x_1` ` + ` `` `x_2`

subject to

`` `2` `x_1` ` + ` `4` `x_2` ≥ `4` `` `` `x_1` ` + ` `7` `x_2` ≥ `7`

and `x_1,x_2 >= 0; `

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '`>=`' we should subtract surplus variable `S_1` and add artificial variable `A_1`

2. As the constraint-2 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_2`

After introducing surplus,artificial variables

Min `Z` `=` `` `` `x_1` ` + ` `` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2` ` + ` `M` `A_1` ` + ` `M` `A_2`

subject to

`` `2` `x_1` ` + ` `4` `x_2` ` - ` `` `S_1` ` + ` `` `A_1` = `4` `` `` `x_1` ` + ` `7` `x_2` ` - ` `` `S_2` ` + ` `` `A_2` = `7`

and `x_1,x_2,S_1,S_2,A_1,A_2 >= 0`

Tableau-1	`C_j`	`1`	`1`	`0`	`0`	`M`	`M`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`A_2`	`RHS`	`"Ratio"=(RHS)/(x_2)`
`R_1` `M`	`A_1`	`2`	`(4)`	`-1`	`0`	`1`	`0`	`4`	`(4)/(4)=1``->`
`R_2` `M`	`A_2`	`1`	`7`	`0`	`-1`	`0`	`1`	`7`	`(7)/(7)=1`
	`Z_j`	`3M`	`11M`	`-M`	`-M`	`M`	`M`	`Z=11M`
	`C_j-Z_j`	`-3M+1`	`-11M+1``uarr`	`M`	`M`	`0`	`0`

Most Negative `C_j-Z_j` is `-11M+1`. So, the entering variable is `x_2`.

Minimum ratio is `1`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `4`.

Entering `=x_2`, Departing `=A_1`, Key Element `=4`

`R_1`(new)`= R_1`(old) `-: 4`

	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`A_2`	`RHS`
`R_1`(old) =	`2`	`4`	`-1`	`0`	`1`	`0`	`4`
`R_1`(new)`= R_1`(old) `-: 4`	`0.5`	`1`	`-0.25`	`0`	`0.25`	`0`	`1`

`R_2`(new)`= R_2`(old) - `7 R_1`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`A_2`	`RHS`
`R_2`(old) =	`1`	`7`	`0`	`-1`	`0`	`1`	`7`
`R_1`(new) =	`0.5`	`1`	`-0.25`	`0`	`0.25`	`0`	`1`
`7 xx R_1`(new) =	`3.5`	`7`	`-1.75`	`0`	`1.75`	`0`	`7`
`R_2`(new)`= R_2`(old) - `7 R_1`(new)	`-2.5`	`0`	`1.75`	`-1`	`-1.75`	`1`	`0`

Tableau-2	`C_j`	`1`	`1`	`0`	`0`	`M`	`M`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`A_2`	`RHS`	`"Ratio"=(RHS)/(S_1)`
`R_1` `1`	`x_2`	`0.5`	`1`	`-0.25`	`0`	`0.25`	`0`	`1`	`(1)/(-0.25)` (ignore, denominator is -ve)
`R_2` `M`	`A_2`	`-2.5`	`0`	`(1.75)`	`-1`	`-1.75`	`1`	`0`	`(0)/(1.75)=0``->`
	`Z_j`	`-2.5M+0.5`	`1`	`1.75M-0.25`	`-M`	`-1.75M+0.25`	`M`	`Z=1`
	`C_j-Z_j`	`2.5M+0.5`	`0`	`-1.75M+0.25``uarr`	`M`	`2.75M-0.25`	`0`

Most Negative `C_j-Z_j` is `-1.75M+0.25`. So, the entering variable is `S_1`.

Minimum ratio is `0`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `1.75`.

Entering `=S_1`, Departing `=A_2`, Key Element `=1.75`

`R_2`(new)`= R_2`(old) `-: 1.75`

	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`A_2`	`RHS`
`R_2`(old) =	`-2.5`	`0`	`1.75`	`-1`	`-1.75`	`1`	`0`
`R_2`(new)`= R_2`(old) `-: 1.75`	`-1.4286`	`0`	`1`	`-0.5714`	`-1`	`0.5714`	`0`

`R_1`(new)`= R_1`(old) + `0.25 R_2`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`A_2`	`RHS`
`R_1`(old) =	`0.5`	`1`	`-0.25`	`0`	`0.25`	`0`	`1`
`R_2`(new) =	`-1.4286`	`0`	`1`	`-0.5714`	`-1`	`0.5714`	`0`
`0.25 xx R_2`(new) =	`-0.3571`	`0`	`0.25`	`-0.1429`	`-0.25`	`0.1429`	`0`
`R_1`(new)`= R_1`(old) + `0.25 R_2`(new)	`0.1429`	`1`	`0`	`-0.1429`	`0`	`0.1429`	`1`

Tableau-3	`C_j`	`1`	`1`	`0`	`0`	`M`	`M`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`A_2`	`RHS`	`"Ratio"`
`R_1` `1`	`x_2`	`0.1429`	`1`	`0`	`-0.1429`	`0`	`0.1429`	`1`
`R_2` `0`	`S_1`	`-1.4286`	`0`	`1`	`-0.5714`	`-1`	`0.5714`	`0`
	`Z_j`	`0.1429`	`1`	`0`	`-0.1429`	`0`	`0.1429`	`Z=1`
	`C_j-Z_j`	`0.8571`	`0`	`0`	`0.1429`	`M`	`M-0.1429`

Since all `C_j-Z_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=1`

Min `Z=1`

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