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Data envelopment analysis (DEA method) example ( Enter your problem)
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  2. Example-2

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1. Example-1





Data envelopment analysis (DEA method)
InuptOutupt
51,38169,119
60,45243,167
43,33173,158
53,43216,138
43,38155,161
44,35169,157


Solution:


DMU-1
Max z `= (169u1+119u2)/(51v1+38v2)`

`(169u1+119u2)/(51v1+38v2)<=1`

`(243u1+167u2)/(60v1+45v2)<=1`

`(173u1+158u2)/(43v1+33v2)<=1`

`(216u1+138u2)/(53v1+43v2)<=1`

`(155u1+161u2)/(43v1+38v2)<=1`

`(169u1+157u2)/(44v1+35v2)<=1`


A fraction with decision variables in the numerator and denominator is nonlinear. Since we are using a linear programming technique, we need to linearize the formulation, such that the denominator of the objective function is 1, then maximize the numerator.
The new formulation would be:
Max z `= 169u1+119u2`

Denominator of nonlinear ` 51v1+38v2=1`

`(169u1+119u2)-(51v1+38v2)<=0`

`(243u1+167u2)-(60v1+45v2)<=0`

`(173u1+158u2)-(43v1+33v2)<=0`

`(216u1+138u2)-(53v1+43v2)<=0`

`(155u1+161u2)-(43v1+38v2)<=0`

`(169u1+157u2)-(44v1+35v2)<=0`

and `u,v>=0`

DMU-1
solution using simplex method
Problem is
Max `Z``=````169``u_1`` + ``119``u_2`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2``0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2``0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2``0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2``0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2``0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2``0`
```51``v_1`` + ``38``v_2`=`1`
and `v_1,v_2,u_1,u_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 add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`

4. As the constraint-4 is of type '`<=`' we should add slack variable `S_4`

5. As the constraint-5 is of type '`<=`' we should add slack variable `S_5`

6. As the constraint-6 is of type '`<=`' we should add slack variable `S_6`

7. As the constraint-7 is of type '`=`' we should add artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``169``u_1`` + ``119``u_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` + ``0``S_5`` + ``0``S_6`` - ``M``A_1`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2`` + ````S_1`=`0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2`` + ````S_2`=`0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2`` + ````S_3`=`0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2`` + ````S_4`=`0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2`` + ````S_5`=`0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2`` + ````S_6`=`0`
```51``v_1`` + ``38``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,u_2,S_1,S_2,S_3,S_4,S_5,S_6,A_1 >= 0`


Iteration-1 `C_j``0``0``169``119``0``0``0``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-51``-38``169``119``1``0``0``0``0``0``0`---
`S_2``0``0``-60``-45``243``167``0``1``0``0``0``0``0`---
`S_3``0``0``-43``-33``173``158``0``0``1``0``0``0``0`---
`S_4``0``0``-53``-43``216``138``0``0``0``1``0``0``0`---
`S_5``0``0``-43``-38``155``161``0``0``0``0``1``0``0`---
`S_6``0``0``-44``-35``169``157``0``0``0``0``0``1``0`---
`A_1``-M``1``(51)``38``0``0``0``0``0``0``0``0``1``(1)/(51)=0.0196``->`
`Z=-M` `Z_j``-51M``-38M``0``0``0``0``0``0``0``0``-M`
`C_j-Z_j``51M``uarr``38M``169``119``0``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `51M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.0196` and its row index is `7`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `51`.

Entering `=v_1`, Departing `=A_1`, Key Element `=51`

`R_7`(new)`= R_7`(old)` -: 51`
`R_7`(old) = `1``51``38``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)` -: 51``0.0196``1``0.7451``0``0``0``0``0``0``0``0`


`R_1`(new)`= R_1`(old) + `51 R_7`(new)
`R_1`(old) = `0``-51``-38``169``119``1``0``0``0``0``0`
`R_7`(new) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`51 xx R_7`(new) = `1``51``38``0``0``0``0``0``0``0``0`
`R_1`(new)`= R_1`(old) + `51 R_7`(new)`1``0``0``169``119``1``0``0``0``0``0`


`R_2`(new)`= R_2`(old) + `60 R_7`(new)
`R_2`(old) = `0``-60``-45``243``167``0``1``0``0``0``0`
`R_7`(new) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`60 xx R_7`(new) = `1.1765``60``44.7059``0``0``0``0``0``0``0``0`
`R_2`(new)`= R_2`(old) + `60 R_7`(new)`1.1765``0``-0.2941``243``167``0``1``0``0``0``0`


`R_3`(new)`= R_3`(old) + `43 R_7`(new)
`R_3`(old) = `0``-43``-33``173``158``0``0``1``0``0``0`
`R_7`(new) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `0.8431``43``32.0392``0``0``0``0``0``0``0``0`
`R_3`(new)`= R_3`(old) + `43 R_7`(new)`0.8431``0``-0.9608``173``158``0``0``1``0``0``0`


`R_4`(new)`= R_4`(old) + `53 R_7`(new)
`R_4`(old) = `0``-53``-43``216``138``0``0``0``1``0``0`
`R_7`(new) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`53 xx R_7`(new) = `1.0392``53``39.4902``0``0``0``0``0``0``0``0`
`R_4`(new)`= R_4`(old) + `53 R_7`(new)`1.0392``0``-3.5098``216``138``0``0``0``1``0``0`


`R_5`(new)`= R_5`(old) + `43 R_7`(new)
`R_5`(old) = `0``-43``-38``155``161``0``0``0``0``1``0`
`R_7`(new) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `0.8431``43``32.0392``0``0``0``0``0``0``0``0`
`R_5`(new)`= R_5`(old) + `43 R_7`(new)`0.8431``0``-5.9608``155``161``0``0``0``0``1``0`


`R_6`(new)`= R_6`(old) + `44 R_7`(new)
`R_6`(old) = `0``-44``-35``169``157``0``0``0``0``0``1`
`R_7`(new) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`44 xx R_7`(new) = `0.8627``44``32.7843``0``0``0``0``0``0``0``0`
`R_6`(new)`= R_6`(old) + `44 R_7`(new)`0.8627``0``-2.2157``169``157``0``0``0``0``0``1`


Iteration-2 `C_j``0``0``169``119``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_1)`
`S_1``0``1``0``0``169``119``1``0``0``0``0``0``(1)/(169)=0.0059`
`S_2``0``1.1765``0``-0.2941``243``167``0``1``0``0``0``0``(1.1765)/(243)=0.0048`
`S_3``0``0.8431``0``-0.9608``173``158``0``0``1``0``0``0``(0.8431)/(173)=0.0049`
`S_4``0``1.0392``0``-3.5098``(216)``138``0``0``0``1``0``0``(1.0392)/(216)=0.0048``->`
`S_5``0``0.8431``0``-5.9608``155``161``0``0``0``0``1``0``(0.8431)/(155)=0.0054`
`S_6``0``0.8627``0``-2.2157``169``157``0``0``0``0``0``1``(0.8627)/(169)=0.0051`
`v_1``0``0.0196``1``0.7451``0``0``0``0``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0``0``0``0``0`
`C_j-Z_j``0``0``169``uarr``119``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `169` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.0048` and its row index is `4`. So, the leaving basis variable is `S_4`.

`:.` The pivot element is `216`.

Entering `=u_1`, Departing `=S_4`, Key Element `=216`

`R_4`(new)`= R_4`(old)` -: 216`
`R_4`(old) = `1.0392``0``-3.5098``216``138``0``0``0``1``0``0`
`R_4`(new)`= R_4`(old)` -: 216``0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0`


`R_1`(new)`= R_1`(old) - `169 R_4`(new)
`R_1`(old) = `1``0``0``169``119``1``0``0``0``0``0`
`R_4`(new) = `0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.8131``0``-2.7461``169``107.9722``0``0``0``0.7824``0``0`
`R_1`(new)`= R_1`(old) - `169 R_4`(new)`0.1869``0``2.7461``0``11.0278``1``0``0``-0.7824``0``0`


`R_2`(new)`= R_2`(old) - `243 R_4`(new)
`R_2`(old) = `1.1765``0``-0.2941``243``167``0``1``0``0``0``0`
`R_4`(new) = `0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0`
`243 xx R_4`(new) = `1.1691``0``-3.9485``243``155.25``0``0``0``1.125``0``0`
`R_2`(new)`= R_2`(old) - `243 R_4`(new)`0.0074``0``3.6544``0``11.75``0``1``0``-1.125``0``0`


`R_3`(new)`= R_3`(old) - `173 R_4`(new)
`R_3`(old) = `0.8431``0``-0.9608``173``158``0``0``1``0``0``0`
`R_4`(new) = `0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0`
`173 xx R_4`(new) = `0.8323``0``-2.8111``173``110.5278``0``0``0``0.8009``0``0`
`R_3`(new)`= R_3`(old) - `173 R_4`(new)`0.0108``0``1.8503``0``47.4722``0``0``1``-0.8009``0``0`


`R_5`(new)`= R_5`(old) - `155 R_4`(new)
`R_5`(old) = `0.8431``0``-5.9608``155``161``0``0``0``0``1``0`
`R_4`(new) = `0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0`
`155 xx R_4`(new) = `0.7457``0``-2.5186``155``99.0278``0``0``0``0.7176``0``0`
`R_5`(new)`= R_5`(old) - `155 R_4`(new)`0.0974``0``-3.4422``0``61.9722``0``0``0``-0.7176``1``0`


`R_6`(new)`= R_6`(old) - `169 R_4`(new)
`R_6`(old) = `0.8627``0``-2.2157``169``157``0``0``0``0``0``1`
`R_4`(new) = `0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.8131``0``-2.7461``169``107.9722``0``0``0``0.7824``0``0`
`R_6`(new)`= R_6`(old) - `169 R_4`(new)`0.0497``0``0.5304``0``49.0278``0``0``0``-0.7824``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0196``1``0.7451``0``0``0``0``0``0``0``0`


Iteration-3 `C_j``0``0``169``119``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_2)`
`S_1``0``0.1869``0``2.7461``0``11.0278``1``0``0``-0.7824``0``0``(0.1869)/(11.0278)=0.0169`
`S_2``0``0.0074``0``3.6544``0``11.75``0``1``0``-1.125``0``0``(0.0074)/(11.75)=0.0006`
`S_3``0``0.0108``0``1.8503``0``(47.4722)``0``0``1``-0.8009``0``0``(0.0108)/(47.4722)=0.0002``->`
`u_1``169``0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0``(0.0048)/(0.6389)=0.0075`
`S_5``0``0.0974``0``-3.4422``0``61.9722``0``0``0``-0.7176``1``0``(0.0974)/(61.9722)=0.0016`
`S_6``0``0.0497``0``0.5304``0``49.0278``0``0``0``-0.7824``0``1``(0.0497)/(49.0278)=0.001`
`v_1``0``0.0196``1``0.7451``0``0``0``0``0``0``0``0`---
`Z=0.8131` `Z_j``0``-2.7461``169``107.9722``0``0``0``0.7824``0``0`
`C_j-Z_j``0``2.7461``0``11.0278``uarr``0``0``0``-0.7824``0``0`


Positive maximum `C_j-Z_j` is `11.0278` and its column index is `4`. So, the entering variable is `u_2`.

Minimum ratio is `0.0002` and its row index is `3`. So, the leaving basis variable is `S_3`.

`:.` The pivot element is `47.4722`.

Entering `=u_2`, Departing `=S_3`, Key Element `=47.4722`

`R_3`(new)`= R_3`(old)` -: 47.4722`
`R_3`(old) = `0.0108``0``1.8503``0``47.4722``0``0``1``-0.8009``0``0`
`R_3`(new)`= R_3`(old)` -: 47.4722``0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`


`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)
`R_1`(old) = `0.1869``0``2.7461``0``11.0278``1``0``0``-0.7824``0``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`11.0278 xx R_3`(new) = `0.0025``0``0.4298``0``11.0278``0``0``0.2323``-0.1861``0``0`
`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)`0.1844``0``2.3163``0``0``1``0``-0.2323``-0.5964``0``0`


`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)
`R_2`(old) = `0.0074``0``3.6544``0``11.75``0``1``0``-1.125``0``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`11.75 xx R_3`(new) = `0.0027``0``0.458``0``11.75``0``0``0.2475``-0.1982``0``0`
`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)`0.0047``0``3.1964``0``0``0``1``-0.2475``-0.9268``0``0`


`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)
`R_4`(old) = `0.0048``0``-0.0162``1``0.6389``0``0``0``0.0046``0``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`0.6389 xx R_3`(new) = `0.0001``0``0.0249``0``0.6389``0``0``0.0135``-0.0108``0``0`
`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)`0.0047``0``-0.0412``1``0``0``0``-0.0135``0.0154``0``0`


`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)
`R_5`(old) = `0.0974``0``-3.4422``0``61.9722``0``0``0``-0.7176``1``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`61.9722 xx R_3`(new) = `0.0141``0``2.4155``0``61.9722``0``0``1.3054``-1.0456``0``0`
`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)`0.0833``0``-5.8576``0``0``0``0``-1.3054``0.328``1``0`


`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)
`R_6`(old) = `0.0497``0``0.5304``0``49.0278``0``0``0``-0.7824``0``1`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`49.0278 xx R_3`(new) = `0.0112``0``1.9109``0``49.0278``0``0``1.0328``-0.8272``0``0`
`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)`0.0385``0``-1.3805``0``0``0``0``-1.0328``0.0448``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0196``1``0.7451``0``0``0``0``0``0``0``0`


Iteration-4 `C_j``0``0``169``119``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(v_2)`
`S_1``0``0.1844``0``2.3163``0``0``1``0``-0.2323``-0.5964``0``0``(0.1844)/(2.3163)=0.0796`
`S_2``0``0.0047``0``(3.1964)``0``0``0``1``-0.2475``-0.9268``0``0``(0.0047)/(3.1964)=0.0015``->`
`u_2``119``0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0``(0.0002)/(0.039)=0.0058`
`u_1``169``0.0047``0``-0.0412``1``0``0``0``-0.0135``0.0154``0``0`---
`S_5``0``0.0833``0``-5.8576``0``0``0``0``-1.3054``0.328``1``0`---
`S_6``0``0.0385``0``-1.3805``0``0``0``0``-1.0328``0.0448``0``1`---
`v_1``0``0.0196``1``0.7451``0``0``0``0``0``0``0``0``(0.0196)/(0.7451)=0.0263`
`Z=0.8156` `Z_j``0``-2.3163``169``119``0``0``0.2323``0.5964``0``0`
`C_j-Z_j``0``2.3163``uarr``0``0``0``0``-0.2323``-0.5964``0``0`


Positive maximum `C_j-Z_j` is `2.3163` and its column index is `2`. So, the entering variable is `v_2`.

Minimum ratio is `0.0015` and its row index is `2`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `3.1964`.

Entering `=v_2`, Departing `=S_2`, Key Element `=3.1964`

`R_2`(new)`= R_2`(old)` -: 3.1964`
`R_2`(old) = `0.0047``0``3.1964``0``0``0``1``-0.2475``-0.9268``0``0`
`R_2`(new)`= R_2`(old)` -: 3.1964``0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`


`R_1`(new)`= R_1`(old) - `2.3163 R_2`(new)
`R_1`(old) = `0.1844``0``2.3163``0``0``1``0``-0.2323``-0.5964``0``0`
`R_2`(new) = `0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`
`2.3163 xx R_2`(new) = `0.0034``0``2.3163``0``0``0``0.7246``-0.1794``-0.6716``0``0`
`R_1`(new)`= R_1`(old) - `2.3163 R_2`(new)`0.181``0``0``0``0``1``-0.7246``-0.0529``0.0752``0``0`


`R_3`(new)`= R_3`(old) - `0.039 R_2`(new)
`R_3`(old) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`R_2`(new) = `0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`
`0.039 xx R_2`(new) = `0.0001``0``0.039``0``0``0``0.0122``-0.003``-0.0113``0``0`
`R_3`(new)`= R_3`(old) - `0.039 R_2`(new)`0.0002``0``0``0``1``0``-0.0122``0.0241``-0.0056``0``0`


`R_4`(new)`= R_4`(old) + `0.0412 R_2`(new)
`R_4`(old) = `0.0047``0``-0.0412``1``0``0``0``-0.0135``0.0154``0``0`
`R_2`(new) = `0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`
`0.0412 xx R_2`(new) = `0.0001``0``0.0412``0``0``0``0.0129``-0.0032``-0.0119``0``0`
`R_4`(new)`= R_4`(old) + `0.0412 R_2`(new)`0.0047``0``0``1``0``0``0.0129``-0.0166``0.0035``0``0`


`R_5`(new)`= R_5`(old) + `5.8576 R_2`(new)
`R_5`(old) = `0.0833``0``-5.8576``0``0``0``0``-1.3054``0.328``1``0`
`R_2`(new) = `0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`
`5.8576 xx R_2`(new) = `0.0086``0``5.8576``0``0``0``1.8326``-0.4536``-1.6983``0``0`
`R_5`(new)`= R_5`(old) + `5.8576 R_2`(new)`0.0919``0``0``0``0``0``1.8326``-1.759``-1.3704``1``0`


`R_6`(new)`= R_6`(old) + `1.3805 R_2`(new)
`R_6`(old) = `0.0385``0``-1.3805``0``0``0``0``-1.0328``0.0448``0``1`
`R_2`(new) = `0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`
`1.3805 xx R_2`(new) = `0.002``0``1.3805``0``0``0``0.4319``-0.1069``-0.4003``0``0`
`R_6`(new)`= R_6`(old) + `1.3805 R_2`(new)`0.0405``0``0``0``0``0``0.4319``-1.1397``-0.3555``0``1`


`R_7`(new)`= R_7`(old) - `0.7451 R_2`(new)
`R_7`(old) = `0.0196``1``0.7451``0``0``0``0``0``0``0``0`
`R_2`(new) = `0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`
`0.7451 xx R_2`(new) = `0.0011``0``0.7451``0``0``0``0.2331``-0.0577``-0.216``0``0`
`R_7`(new)`= R_7`(old) - `0.7451 R_2`(new)`0.0185``1``0``0``0``0``-0.2331``0.0577``0.216``0``0`


Iteration-5 `C_j``0``0``169``119``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(S_4)`
`S_1``0``0.181``0``0``0``0``1``-0.7246``-0.0529``0.0752``0``0``(0.181)/(0.0752)=2.4065`
`v_2``0``0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`---
`u_2``119``0.0002``0``0``0``1``0``-0.0122``0.0241``-0.0056``0``0`---
`u_1``169``0.0047``0``0``1``0``0``0.0129``-0.0166``0.0035``0``0``(0.0047)/(0.0035)=1.359`
`S_5``0``0.0919``0``0``0``0``0``1.8326``-1.759``-1.3704``1``0`---
`S_6``0``0.0405``0``0``0``0``0``0.4319``-1.1397``-0.3555``0``1`---
`v_1``0``0.0185``1``0``0``0``0``-0.2331``0.0577``(0.216)``0``0``(0.0185)/(0.216)=0.0857``->`
`Z=0.819` `Z_j``0``0``169``119``0``0.7246``0.0529``-0.0752``0``0`
`C_j-Z_j``0``0``0``0``0``-0.7246``-0.0529``0.0752``uarr``0``0`


Positive maximum `C_j-Z_j` is `0.0752` and its column index is `8`. So, the entering variable is `S_4`.

Minimum ratio is `0.0857` and its row index is `7`. So, the leaving basis variable is `v_1`.

`:.` The pivot element is `0.216`.

Entering `=S_4`, Departing `=v_1`, Key Element `=0.216`

`R_7`(new)`= R_7`(old)` -: 0.216`
`R_7`(old) = `0.0185``1``0``0``0``0``-0.2331``0.0577``0.216``0``0`
`R_7`(new)`= R_7`(old)` -: 0.216``0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`


`R_1`(new)`= R_1`(old) - `0.0752 R_7`(new)
`R_1`(old) = `0.181``0``0``0``0``1``-0.7246``-0.0529``0.0752``0``0`
`R_7`(new) = `0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`
`0.0752 xx R_7`(new) = `0.0064``0.3482``0``0``0``0``-0.0812``0.0201``0.0752``0``0`
`R_1`(new)`= R_1`(old) - `0.0752 R_7`(new)`0.1746``-0.3482``0``0``0``1``-0.6435``-0.073``0``0``0`


`R_2`(new)`= R_2`(old) + `0.2899 R_7`(new)
`R_2`(old) = `0.0015``0``1``0``0``0``0.3128``-0.0774``-0.2899``0``0`
`R_7`(new) = `0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`
`0.2899 xx R_7`(new) = `0.0249``1.3421``0``0``0``0``-0.3128``0.0774``0.2899``0``0`
`R_2`(new)`= R_2`(old) + `0.2899 R_7`(new)`0.0263``1.3421``1``0``0``0``0``0``0``0``0`


`R_3`(new)`= R_3`(old) + `0.0056 R_7`(new)
`R_3`(old) = `0.0002``0``0``0``1``0``-0.0122``0.0241``-0.0056``0``0`
`R_7`(new) = `0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`
`0.0056 xx R_7`(new) = `0.0005``0.0258``0``0``0``0``-0.006``0.0015``0.0056``0``0`
`R_3`(new)`= R_3`(old) + `0.0056 R_7`(new)`0.0006``0.0258``0``0``1``0``-0.0182``0.0256``0``0``0`


`R_4`(new)`= R_4`(old) - `0.0035 R_7`(new)
`R_4`(old) = `0.0047``0``0``1``0``0``0.0129``-0.0166``0.0035``0``0`
`R_7`(new) = `0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`
`0.0035 xx R_7`(new) = `0.0003``0.0161``0``0``0``0``-0.0038``0.0009``0.0035``0``0`
`R_4`(new)`= R_4`(old) - `0.0035 R_7`(new)`0.0044``-0.0161``0``1``0``0``0.0166``-0.0176``0``0``0`


`R_5`(new)`= R_5`(old) + `1.3704 R_7`(new)
`R_5`(old) = `0.0919``0``0``0``0``0``1.8326``-1.759``-1.3704``1``0`
`R_7`(new) = `0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`
`1.3704 xx R_7`(new) = `0.1175``6.3434``0``0``0``0``-1.4787``0.366``1.3704``0``0`
`R_5`(new)`= R_5`(old) + `1.3704 R_7`(new)`0.2093``6.3434``0``0``0``0``0.3539``-1.393``0``1``0`


`R_6`(new)`= R_6`(old) + `0.3555 R_7`(new)
`R_6`(old) = `0.0405``0``0``0``0``0``0.4319``-1.1397``-0.3555``0``1`
`R_7`(new) = `0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`
`0.3555 xx R_7`(new) = `0.0305``1.6456``0``0``0``0``-0.3836``0.0949``0.3555``0``0`
`R_6`(new)`= R_6`(old) + `0.3555 R_7`(new)`0.071``1.6456``0``0``0``0``0.0483``-1.0447``0``0``1`


Iteration-6 `C_j``0``0``169``119``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`S_1``0``0.1746``-0.3482``0``0``0``1``-0.6435``-0.073``0``0``0`
`v_2``0``0.0263``1.3421``1``0``0``0``0``0``0``0``0`
`u_2``119``0.0006``0.0258``0``0``1``0``-0.0182``0.0256``0``0``0`
`u_1``169``0.0044``-0.0161``0``1``0``0``0.0166``-0.0176``0``0``0`
`S_5``0``0.2093``6.3434``0``0``0``0``0.3539``-1.393``0``1``0`
`S_6``0``0.071``1.6456``0``0``0``0``0.0483``-1.0447``0``0``1`
`S_4``0``0.0857``4.629``0``0``0``0``-1.079``0.2671``1``0``0`
`Z=0.8254` `Z_j``0.3482``0``169``119``0``0.6435``0.073``0``0``0`
`C_j-Z_j``-0.3482``0``0``0``0``-0.6435``-0.073``0``0``0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`v_1=0,v_2=0.0263,u_1=0.0044,u_2=0.0006`

Max `Z = 0.8254`


Solution steps by BigM method

Answer is
`v_1=0.0,v_2=0.026315789473684209,u_1=0.0044279645762833912,u_2=6.4799481604146936E-4,`



DMU-2
Max z `= (243u1+167u2)/(60v1+45v2)`

`(169u1+119u2)/(51v1+38v2)<=1`

`(243u1+167u2)/(60v1+45v2)<=1`

`(173u1+158u2)/(43v1+33v2)<=1`

`(216u1+138u2)/(53v1+43v2)<=1`

`(155u1+161u2)/(43v1+38v2)<=1`

`(169u1+157u2)/(44v1+35v2)<=1`


A fraction with decision variables in the numerator and denominator is nonlinear. Since we are using a linear programming technique, we need to linearize the formulation, such that the denominator of the objective function is 1, then maximize the numerator.
The new formulation would be:
Max z `= 243u1+167u2`

Denominator of nonlinear ` 60v1+45v2=1`

`(169u1+119u2)-(51v1+38v2)<=0`

`(243u1+167u2)-(60v1+45v2)<=0`

`(173u1+158u2)-(43v1+33v2)<=0`

`(216u1+138u2)-(53v1+43v2)<=0`

`(155u1+161u2)-(43v1+38v2)<=0`

`(169u1+157u2)-(44v1+35v2)<=0`

and `u,v>=0`

DMU-2
solution using simplex method
Problem is
Max `Z``=````243``u_1`` + ``167``u_2`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2``0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2``0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2``0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2``0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2``0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2``0`
```60``v_1`` + ``45``v_2`=`1`
and `v_1,v_2,u_1,u_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 add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`

4. As the constraint-4 is of type '`<=`' we should add slack variable `S_4`

5. As the constraint-5 is of type '`<=`' we should add slack variable `S_5`

6. As the constraint-6 is of type '`<=`' we should add slack variable `S_6`

7. As the constraint-7 is of type '`=`' we should add artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``243``u_1`` + ``167``u_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` + ``0``S_5`` + ``0``S_6`` - ``M``A_1`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2`` + ````S_1`=`0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2`` + ````S_2`=`0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2`` + ````S_3`=`0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2`` + ````S_4`=`0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2`` + ````S_5`=`0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2`` + ````S_6`=`0`
```60``v_1`` + ``45``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,u_2,S_1,S_2,S_3,S_4,S_5,S_6,A_1 >= 0`


Iteration-1 `C_j``0``0``243``167``0``0``0``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-51``-38``169``119``1``0``0``0``0``0``0`---
`S_2``0``0``-60``-45``243``167``0``1``0``0``0``0``0`---
`S_3``0``0``-43``-33``173``158``0``0``1``0``0``0``0`---
`S_4``0``0``-53``-43``216``138``0``0``0``1``0``0``0`---
`S_5``0``0``-43``-38``155``161``0``0``0``0``1``0``0`---
`S_6``0``0``-44``-35``169``157``0``0``0``0``0``1``0`---
`A_1``-M``1``(60)``45``0``0``0``0``0``0``0``0``1``(1)/(60)=0.0167``->`
`Z=-M` `Z_j``-60M``-45M``0``0``0``0``0``0``0``0``-M`
`C_j-Z_j``60M``uarr``45M``243``167``0``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `60M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.0167` and its row index is `7`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `60`.

Entering `=v_1`, Departing `=A_1`, Key Element `=60`

`R_7`(new)`= R_7`(old)` -: 60`
`R_7`(old) = `1``60``45``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)` -: 60``0.0167``1``0.75``0``0``0``0``0``0``0``0`


`R_1`(new)`= R_1`(old) + `51 R_7`(new)
`R_1`(old) = `0``-51``-38``169``119``1``0``0``0``0``0`
`R_7`(new) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`51 xx R_7`(new) = `0.85``51``38.25``0``0``0``0``0``0``0``0`
`R_1`(new)`= R_1`(old) + `51 R_7`(new)`0.85``0``0.25``169``119``1``0``0``0``0``0`


`R_2`(new)`= R_2`(old) + `60 R_7`(new)
`R_2`(old) = `0``-60``-45``243``167``0``1``0``0``0``0`
`R_7`(new) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`60 xx R_7`(new) = `1``60``45``0``0``0``0``0``0``0``0`
`R_2`(new)`= R_2`(old) + `60 R_7`(new)`1``0``0``243``167``0``1``0``0``0``0`


`R_3`(new)`= R_3`(old) + `43 R_7`(new)
`R_3`(old) = `0``-43``-33``173``158``0``0``1``0``0``0`
`R_7`(new) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `0.7167``43``32.25``0``0``0``0``0``0``0``0`
`R_3`(new)`= R_3`(old) + `43 R_7`(new)`0.7167``0``-0.75``173``158``0``0``1``0``0``0`


`R_4`(new)`= R_4`(old) + `53 R_7`(new)
`R_4`(old) = `0``-53``-43``216``138``0``0``0``1``0``0`
`R_7`(new) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`53 xx R_7`(new) = `0.8833``53``39.75``0``0``0``0``0``0``0``0`
`R_4`(new)`= R_4`(old) + `53 R_7`(new)`0.8833``0``-3.25``216``138``0``0``0``1``0``0`


`R_5`(new)`= R_5`(old) + `43 R_7`(new)
`R_5`(old) = `0``-43``-38``155``161``0``0``0``0``1``0`
`R_7`(new) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `0.7167``43``32.25``0``0``0``0``0``0``0``0`
`R_5`(new)`= R_5`(old) + `43 R_7`(new)`0.7167``0``-5.75``155``161``0``0``0``0``1``0`


`R_6`(new)`= R_6`(old) + `44 R_7`(new)
`R_6`(old) = `0``-44``-35``169``157``0``0``0``0``0``1`
`R_7`(new) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`44 xx R_7`(new) = `0.7333``44``33``0``0``0``0``0``0``0``0`
`R_6`(new)`= R_6`(old) + `44 R_7`(new)`0.7333``0``-2``169``157``0``0``0``0``0``1`


Iteration-2 `C_j``0``0``243``167``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_1)`
`S_1``0``0.85``0``0.25``169``119``1``0``0``0``0``0``(0.85)/(169)=0.005`
`S_2``0``1``0``0``243``167``0``1``0``0``0``0``(1)/(243)=0.0041`
`S_3``0``0.7167``0``-0.75``173``158``0``0``1``0``0``0``(0.7167)/(173)=0.0041`
`S_4``0``0.8833``0``-3.25``(216)``138``0``0``0``1``0``0``(0.8833)/(216)=0.0041``->`
`S_5``0``0.7167``0``-5.75``155``161``0``0``0``0``1``0``(0.7167)/(155)=0.0046`
`S_6``0``0.7333``0``-2``169``157``0``0``0``0``0``1``(0.7333)/(169)=0.0043`
`v_1``0``0.0167``1``0.75``0``0``0``0``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0``0``0``0``0`
`C_j-Z_j``0``0``243``uarr``167``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `243` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.0041` and its row index is `4`. So, the leaving basis variable is `S_4`.

`:.` The pivot element is `216`.

Entering `=u_1`, Departing `=S_4`, Key Element `=216`

`R_4`(new)`= R_4`(old)` -: 216`
`R_4`(old) = `0.8833``0``-3.25``216``138``0``0``0``1``0``0`
`R_4`(new)`= R_4`(old)` -: 216``0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0`


`R_1`(new)`= R_1`(old) - `169 R_4`(new)
`R_1`(old) = `0.85``0``0.25``169``119``1``0``0``0``0``0`
`R_4`(new) = `0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.6911``0``-2.5428``169``107.9722``0``0``0``0.7824``0``0`
`R_1`(new)`= R_1`(old) - `169 R_4`(new)`0.1589``0``2.7928``0``11.0278``1``0``0``-0.7824``0``0`


`R_2`(new)`= R_2`(old) - `243 R_4`(new)
`R_2`(old) = `1``0``0``243``167``0``1``0``0``0``0`
`R_4`(new) = `0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0`
`243 xx R_4`(new) = `0.9938``0``-3.6562``243``155.25``0``0``0``1.125``0``0`
`R_2`(new)`= R_2`(old) - `243 R_4`(new)`0.0062``0``3.6562``0``11.75``0``1``0``-1.125``0``0`


`R_3`(new)`= R_3`(old) - `173 R_4`(new)
`R_3`(old) = `0.7167``0``-0.75``173``158``0``0``1``0``0``0`
`R_4`(new) = `0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0`
`173 xx R_4`(new) = `0.7075``0``-2.603``173``110.5278``0``0``0``0.8009``0``0`
`R_3`(new)`= R_3`(old) - `173 R_4`(new)`0.0092``0``1.853``0``47.4722``0``0``1``-0.8009``0``0`


`R_5`(new)`= R_5`(old) - `155 R_4`(new)
`R_5`(old) = `0.7167``0``-5.75``155``161``0``0``0``0``1``0`
`R_4`(new) = `0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0`
`155 xx R_4`(new) = `0.6339``0``-2.3322``155``99.0278``0``0``0``0.7176``0``0`
`R_5`(new)`= R_5`(old) - `155 R_4`(new)`0.0828``0``-3.4178``0``61.9722``0``0``0``-0.7176``1``0`


`R_6`(new)`= R_6`(old) - `169 R_4`(new)
`R_6`(old) = `0.7333``0``-2``169``157``0``0``0``0``0``1`
`R_4`(new) = `0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.6911``0``-2.5428``169``107.9722``0``0``0``0.7824``0``0`
`R_6`(new)`= R_6`(old) - `169 R_4`(new)`0.0422``0``0.5428``0``49.0278``0``0``0``-0.7824``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0167``1``0.75``0``0``0``0``0``0``0``0`


Iteration-3 `C_j``0``0``243``167``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_2)`
`S_1``0``0.1589``0``2.7928``0``11.0278``1``0``0``-0.7824``0``0``(0.1589)/(11.0278)=0.0144`
`S_2``0``0.0062``0``3.6562``0``11.75``0``1``0``-1.125``0``0``(0.0062)/(11.75)=0.0005`
`S_3``0``0.0092``0``1.853``0``(47.4722)``0``0``1``-0.8009``0``0``(0.0092)/(47.4722)=0.0002``->`
`u_1``243``0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0``(0.0041)/(0.6389)=0.0064`
`S_5``0``0.0828``0``-3.4178``0``61.9722``0``0``0``-0.7176``1``0``(0.0828)/(61.9722)=0.0013`
`S_6``0``0.0422``0``0.5428``0``49.0278``0``0``0``-0.7824``0``1``(0.0422)/(49.0278)=0.0009`
`v_1``0``0.0167``1``0.75``0``0``0``0``0``0``0``0`---
`Z=0.9938` `Z_j``0``-3.6562``243``155.25``0``0``0``1.125``0``0`
`C_j-Z_j``0``3.6562``0``11.75``uarr``0``0``0``-1.125``0``0`


Positive maximum `C_j-Z_j` is `11.75` and its column index is `4`. So, the entering variable is `u_2`.

Minimum ratio is `0.0002` and its row index is `3`. So, the leaving basis variable is `S_3`.

`:.` The pivot element is `47.4722`.

Entering `=u_2`, Departing `=S_3`, Key Element `=47.4722`

`R_3`(new)`= R_3`(old)` -: 47.4722`
`R_3`(old) = `0.0092``0``1.853``0``47.4722``0``0``1``-0.8009``0``0`
`R_3`(new)`= R_3`(old)` -: 47.4722``0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`


`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)
`R_1`(old) = `0.1589``0``2.7928``0``11.0278``1``0``0``-0.7824``0``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`11.0278 xx R_3`(new) = `0.0021``0``0.4305``0``11.0278``0``0``0.2323``-0.1861``0``0`
`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)`0.1567``0``2.3624``0``0``1``0``-0.2323``-0.5964``0``0`


`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)
`R_2`(old) = `0.0062``0``3.6562``0``11.75``0``1``0``-1.125``0``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`11.75 xx R_3`(new) = `0.0023``0``0.4586``0``11.75``0``0``0.2475``-0.1982``0``0`
`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)`0.004``0``3.1976``0``0``0``1``-0.2475``-0.9268``0``0`


`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)
`R_4`(old) = `0.0041``0``-0.015``1``0.6389``0``0``0``0.0046``0``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`0.6389 xx R_3`(new) = `0.0001``0``0.0249``0``0.6389``0``0``0.0135``-0.0108``0``0`
`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)`0.004``0``-0.04``1``0``0``0``-0.0135``0.0154``0``0`


`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)
`R_5`(old) = `0.0828``0``-3.4178``0``61.9722``0``0``0``-0.7176``1``0`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`61.9722 xx R_3`(new) = `0.012``0``2.419``0``61.9722``0``0``1.3054``-1.0456``0``0`
`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)`0.0708``0``-5.8368``0``0``0``0``-1.3054``0.328``1``0`


`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)
`R_6`(old) = `0.0422``0``0.5428``0``49.0278``0``0``0``-0.7824``0``1`
`R_3`(new) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`49.0278 xx R_3`(new) = `0.0095``0``1.9137``0``49.0278``0``0``1.0328``-0.8272``0``0`
`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)`0.0327``0``-1.3709``0``0``0``0``-1.0328``0.0448``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0167``1``0.75``0``0``0``0``0``0``0``0`


Iteration-4 `C_j``0``0``243``167``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(v_2)`
`S_1``0``0.1567``0``2.3624``0``0``1``0``-0.2323``-0.5964``0``0``(0.1567)/(2.3624)=0.0663`
`S_2``0``0.004``0``(3.1976)``0``0``0``1``-0.2475``-0.9268``0``0``(0.004)/(3.1976)=0.0012``->`
`u_2``167``0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0``(0.0002)/(0.039)=0.005`
`u_1``243``0.004``0``-0.04``1``0``0``0``-0.0135``0.0154``0``0`---
`S_5``0``0.0708``0``-5.8368``0``0``0``0``-1.3054``0.328``1``0`---
`S_6``0``0.0327``0``-1.3709``0``0``0``0``-1.0328``0.0448``0``1`---
`v_1``0``0.0167``1``0.75``0``0``0``0``0``0``0``0``(0.0167)/(0.75)=0.0222`
`Z=0.996` `Z_j``0``-3.1976``243``167``0``0``0.2475``0.9268``0``0`
`C_j-Z_j``0``3.1976``uarr``0``0``0``0``-0.2475``-0.9268``0``0`


Positive maximum `C_j-Z_j` is `3.1976` and its column index is `2`. So, the entering variable is `v_2`.

Minimum ratio is `0.0012` and its row index is `2`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `3.1976`.

Entering `=v_2`, Departing `=S_2`, Key Element `=3.1976`

`R_2`(new)`= R_2`(old)` -: 3.1976`
`R_2`(old) = `0.004``0``3.1976``0``0``0``1``-0.2475``-0.9268``0``0`
`R_2`(new)`= R_2`(old)` -: 3.1976``0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`


`R_1`(new)`= R_1`(old) - `2.3624 R_2`(new)
`R_1`(old) = `0.1567``0``2.3624``0``0``1``0``-0.2323``-0.5964``0``0`
`R_2`(new) = `0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`
`2.3624 xx R_2`(new) = `0.0029``0``2.3624``0``0``0``0.7388``-0.1829``-0.6847``0``0`
`R_1`(new)`= R_1`(old) - `2.3624 R_2`(new)`0.1538``0``0``0``0``1``-0.7388``-0.0494``0.0883``0``0`


`R_3`(new)`= R_3`(old) - `0.039 R_2`(new)
`R_3`(old) = `0.0002``0``0.039``0``1``0``0``0.0211``-0.0169``0``0`
`R_2`(new) = `0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`
`0.039 xx R_2`(new) = `0``0``0.039``0``0``0``0.0122``-0.003``-0.0113``0``0`
`R_3`(new)`= R_3`(old) - `0.039 R_2`(new)`0.0001``0``0``0``1``0``-0.0122``0.0241``-0.0056``0``0`


`R_4`(new)`= R_4`(old) + `0.04 R_2`(new)
`R_4`(old) = `0.004``0``-0.04``1``0``0``0``-0.0135``0.0154``0``0`
`R_2`(new) = `0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`
`0.04 xx R_2`(new) = `0``0``0.04``0``0``0``0.0125``-0.0031``-0.0116``0``0`
`R_4`(new)`= R_4`(old) + `0.04 R_2`(new)`0.004``0``0``1``0``0``0.0125``-0.0166``0.0038``0``0`


`R_5`(new)`= R_5`(old) + `5.8368 R_2`(new)
`R_5`(old) = `0.0708``0``-5.8368``0``0``0``0``-1.3054``0.328``1``0`
`R_2`(new) = `0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`
`5.8368 xx R_2`(new) = `0.0073``0``5.8368``0``0``0``1.8254``-0.4518``-1.6917``0``0`
`R_5`(new)`= R_5`(old) + `5.8368 R_2`(new)`0.0781``0``0``0``0``0``1.8254``-1.7572``-1.3637``1``0`


`R_6`(new)`= R_6`(old) + `1.3709 R_2`(new)
`R_6`(old) = `0.0327``0``-1.3709``0``0``0``0``-1.0328``0.0448``0``1`
`R_2`(new) = `0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`
`1.3709 xx R_2`(new) = `0.0017``0``1.3709``0``0``0``0.4287``-0.1061``-0.3973``0``0`
`R_6`(new)`= R_6`(old) + `1.3709 R_2`(new)`0.0344``0``0``0``0``0``0.4287``-1.1389``-0.3526``0``1`


`R_7`(new)`= R_7`(old) - `0.75 R_2`(new)
`R_7`(old) = `0.0167``1``0.75``0``0``0``0``0``0``0``0`
`R_2`(new) = `0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`
`0.75 xx R_2`(new) = `0.0009``0``0.75``0``0``0``0.2346``-0.0581``-0.2174``0``0`
`R_7`(new)`= R_7`(old) - `0.75 R_2`(new)`0.0157``1``0``0``0``0``-0.2346``0.0581``0.2174``0``0`


Iteration-5 `C_j``0``0``243``167``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(S_4)`
`S_1``0``0.1538``0``0``0``0``1``-0.7388``-0.0494``0.0883``0``0``(0.1538)/(0.0883)=1.7412`
`v_2``0``0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`---
`u_2``167``0.0001``0``0``0``1``0``-0.0122``0.0241``-0.0056``0``0`---
`u_1``243``0.004``0``0``1``0``0``0.0125``-0.0166``0.0038``0``0``(0.004)/(0.0038)=1.0512`
`S_5``0``0.0781``0``0``0``0``0``1.8254``-1.7572``-1.3637``1``0`---
`S_6``0``0.0344``0``0``0``0``0``0.4287``-1.1389``-0.3526``0``1`---
`v_1``0``0.0157``1``0``0``0``0``-0.2346``0.0581``(0.2174)``0``0``(0.0157)/(0.2174)=0.0724``->`
`Z=1` `Z_j``0``0``243``167``0``1``0``0``0``0`
`C_j-Z_j``0``0``0``0``0``-1``0``0``uarr``0``0`


Positive maximum `C_j-Z_j` is `0` and its column index is `8`. So, the entering variable is `S_4`.

Minimum ratio is `0.0724` and its row index is `7`. So, the leaving basis variable is `v_1`.

`:.` The pivot element is `0.2174`.

Entering `=S_4`, Departing `=v_1`, Key Element `=0.2174`

`R_7`(new)`= R_7`(old)` -: 0.2174`
`R_7`(old) = `0.0157``1``0``0``0``0``-0.2346``0.0581``0.2174``0``0`
`R_7`(new)`= R_7`(old)` -: 0.2174``0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`


`R_1`(new)`= R_1`(old) - `0.0883 R_7`(new)
`R_1`(old) = `0.1538``0``0``0``0``1``-0.7388``-0.0494``0.0883``0``0`
`R_7`(new) = `0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`
`0.0883 xx R_7`(new) = `0.0064``0.4064``0``0``0``0``-0.0953``0.0236``0.0883``0``0`
`R_1`(new)`= R_1`(old) - `0.0883 R_7`(new)`0.1474``-0.4064``0``0``0``1``-0.6435``-0.073``0``0``0`


`R_2`(new)`= R_2`(old) + `0.2898 R_7`(new)
`R_2`(old) = `0.0012``0``1``0``0``0``0.3127``-0.0774``-0.2898``0``0`
`R_7`(new) = `0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`
`0.2898 xx R_7`(new) = `0.021``1.3333``0``0``0``0``-0.3127``0.0774``0.2898``0``0`
`R_2`(new)`= R_2`(old) + `0.2898 R_7`(new)`0.0222``1.3333``1``0``0``0``0``0``0``0``0`


`R_3`(new)`= R_3`(old) + `0.0056 R_7`(new)
`R_3`(old) = `0.0001``0``0``0``1``0``-0.0122``0.0241``-0.0056``0``0`
`R_7`(new) = `0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`
`0.0056 xx R_7`(new) = `0.0004``0.0256``0``0``0``0``-0.006``0.0015``0.0056``0``0`
`R_3`(new)`= R_3`(old) + `0.0056 R_7`(new)`0.0005``0.0256``0``0``1``0``-0.0182``0.0256``0``0``0`


`R_4`(new)`= R_4`(old) - `0.0038 R_7`(new)
`R_4`(old) = `0.004``0``0``1``0``0``0.0125``-0.0166``0.0038``0``0`
`R_7`(new) = `0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`
`0.0038 xx R_7`(new) = `0.0003``0.0176``0``0``0``0``-0.0041``0.001``0.0038``0``0`
`R_4`(new)`= R_4`(old) - `0.0038 R_7`(new)`0.0037``-0.0176``0``1``0``0``0.0166``-0.0176``0``0``0`


`R_5`(new)`= R_5`(old) + `1.3637 R_7`(new)
`R_5`(old) = `0.0781``0``0``0``0``0``1.8254``-1.7572``-1.3637``1``0`
`R_7`(new) = `0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`
`1.3637 xx R_7`(new) = `0.0987``6.2736``0``0``0``0``-1.4715``0.3642``1.3637``0``0`
`R_5`(new)`= R_5`(old) + `1.3637 R_7`(new)`0.1768``6.2736``0``0``0``0``0.3539``-1.393``0``1``0`


`R_6`(new)`= R_6`(old) + `0.3526 R_7`(new)
`R_6`(old) = `0.0344``0``0``0``0``0``0.4287``-1.1389``-0.3526``0``1`
`R_7`(new) = `0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`
`0.3526 xx R_7`(new) = `0.0255``1.6219``0``0``0``0``-0.3804``0.0942``0.3526``0``0`
`R_6`(new)`= R_6`(old) + `0.3526 R_7`(new)`0.0599``1.6219``0``0``0``0``0.0483``-1.0447``0``0``1`


Iteration-6 `C_j``0``0``243``167``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`S_1``0``0.1474``-0.4064``0``0``0``1``-0.6435``-0.073``0``0``0`
`v_2``0``0.0222``1.3333``1``0``0``0``0``0``0``0``0`
`u_2``167``0.0005``0.0256``0``0``1``0``-0.0182``0.0256``0``0``0`
`u_1``243``0.0037``-0.0176``0``1``0``0``0.0166``-0.0176``0``0``0`
`S_5``0``0.1768``6.2736``0``0``0``0``0.3539``-1.393``0``1``0`
`S_6``0``0.0599``1.6219``0``0``0``0``0.0483``-1.0447``0``0``1`
`S_4``0``0.0724``4.6004``0``0``0``0``-1.079``0.2671``1``0``0`
`Z=1` `Z_j``0``0``243``167``0``1``0``0``0``0`
`C_j-Z_j``0``0``0``0``0``-1``0``0``0``0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`v_1=0,v_2=0.0222,u_1=0.0037,u_2=0.0005`

Max `Z = 1`


Solution steps by BigM method

Answer is
`v_1=0.0,v_2=0.022222222222222223,u_1=0.0037391700866393073,u_2=5.4719562243502051E-4,`



DMU-3
Max z `= (173u1+158u2)/(43v1+33v2)`

`(169u1+119u2)/(51v1+38v2)<=1`

`(243u1+167u2)/(60v1+45v2)<=1`

`(173u1+158u2)/(43v1+33v2)<=1`

`(216u1+138u2)/(53v1+43v2)<=1`

`(155u1+161u2)/(43v1+38v2)<=1`

`(169u1+157u2)/(44v1+35v2)<=1`


A fraction with decision variables in the numerator and denominator is nonlinear. Since we are using a linear programming technique, we need to linearize the formulation, such that the denominator of the objective function is 1, then maximize the numerator.
The new formulation would be:
Max z `= 173u1+158u2`

Denominator of nonlinear ` 43v1+33v2=1`

`(169u1+119u2)-(51v1+38v2)<=0`

`(243u1+167u2)-(60v1+45v2)<=0`

`(173u1+158u2)-(43v1+33v2)<=0`

`(216u1+138u2)-(53v1+43v2)<=0`

`(155u1+161u2)-(43v1+38v2)<=0`

`(169u1+157u2)-(44v1+35v2)<=0`

and `u,v>=0`

DMU-3
solution using simplex method
Problem is
Max `Z``=````173``u_1`` + ``158``u_2`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2``0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2``0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2``0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2``0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2``0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2``0`
```43``v_1`` + ``33``v_2`=`1`
and `v_1,v_2,u_1,u_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 add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`

4. As the constraint-4 is of type '`<=`' we should add slack variable `S_4`

5. As the constraint-5 is of type '`<=`' we should add slack variable `S_5`

6. As the constraint-6 is of type '`<=`' we should add slack variable `S_6`

7. As the constraint-7 is of type '`=`' we should add artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``173``u_1`` + ``158``u_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` + ``0``S_5`` + ``0``S_6`` - ``M``A_1`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2`` + ````S_1`=`0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2`` + ````S_2`=`0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2`` + ````S_3`=`0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2`` + ````S_4`=`0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2`` + ````S_5`=`0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2`` + ````S_6`=`0`
```43``v_1`` + ``33``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,u_2,S_1,S_2,S_3,S_4,S_5,S_6,A_1 >= 0`


Iteration-1 `C_j``0``0``173``158``0``0``0``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-51``-38``169``119``1``0``0``0``0``0``0`---
`S_2``0``0``-60``-45``243``167``0``1``0``0``0``0``0`---
`S_3``0``0``-43``-33``173``158``0``0``1``0``0``0``0`---
`S_4``0``0``-53``-43``216``138``0``0``0``1``0``0``0`---
`S_5``0``0``-43``-38``155``161``0``0``0``0``1``0``0`---
`S_6``0``0``-44``-35``169``157``0``0``0``0``0``1``0`---
`A_1``-M``1``(43)``33``0``0``0``0``0``0``0``0``1``(1)/(43)=0.0233``->`
`Z=-M` `Z_j``-43M``-33M``0``0``0``0``0``0``0``0``-M`
`C_j-Z_j``43M``uarr``33M``173``158``0``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `43M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.0233` and its row index is `7`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `43`.

Entering `=v_1`, Departing `=A_1`, Key Element `=43`

`R_7`(new)`= R_7`(old)` -: 43`
`R_7`(old) = `1``43``33``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)` -: 43``1/43``1``33/43``0``0``0``0``0``0``0``0`


`R_1`(new)`= R_1`(old) + `51 R_7`(new)
`R_1`(old) = `0``-51``-38``169``119``1``0``0``0``0``0`
`R_7`(new) = `1/43``1``33/43``0``0``0``0``0``0``0``0`
`51 xx R_7`(new) = `51/43``51``1683/43``0``0``0``0``0``0``0``0`
`R_1`(new)`= R_1`(old) + `51 R_7`(new)`51/43``0``49/43``169``119``1``0``0``0``0``0`


`R_2`(new)`= R_2`(old) + `60 R_7`(new)
`R_2`(old) = `0``-60``-45``243``167``0``1``0``0``0``0`
`R_7`(new) = `1/43``1``33/43``0``0``0``0``0``0``0``0`
`60 xx R_7`(new) = `60/43``60``1980/43``0``0``0``0``0``0``0``0`
`R_2`(new)`= R_2`(old) + `60 R_7`(new)`60/43``0``45/43``243``167``0``1``0``0``0``0`


`R_3`(new)`= R_3`(old) + `43 R_7`(new)
`R_3`(old) = `0``-43``-33``173``158``0``0``1``0``0``0`
`R_7`(new) = `1/43``1``33/43``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `1``43``33``0``0``0``0``0``0``0``0`
`R_3`(new)`= R_3`(old) + `43 R_7`(new)`1``0``0``173``158``0``0``1``0``0``0`


`R_4`(new)`= R_4`(old) + `53 R_7`(new)
`R_4`(old) = `0``-53``-43``216``138``0``0``0``1``0``0`
`R_7`(new) = `1/43``1``33/43``0``0``0``0``0``0``0``0`
`53 xx R_7`(new) = `53/43``53``1749/43``0``0``0``0``0``0``0``0`
`R_4`(new)`= R_4`(old) + `53 R_7`(new)`53/43``0``-100/43``216``138``0``0``0``1``0``0`


`R_5`(new)`= R_5`(old) + `43 R_7`(new)
`R_5`(old) = `0``-43``-38``155``161``0``0``0``0``1``0`
`R_7`(new) = `1/43``1``33/43``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `1``43``33``0``0``0``0``0``0``0``0`
`R_5`(new)`= R_5`(old) + `43 R_7`(new)`1``0``-5``155``161``0``0``0``0``1``0`


`R_6`(new)`= R_6`(old) + `44 R_7`(new)
`R_6`(old) = `0``-44``-35``169``157``0``0``0``0``0``1`
`R_7`(new) = `1/43``1``33/43``0``0``0``0``0``0``0``0`
`44 xx R_7`(new) = `44/43``44``1452/43``0``0``0``0``0``0``0``0`
`R_6`(new)`= R_6`(old) + `44 R_7`(new)`44/43``0``-53/43``169``157``0``0``0``0``0``1`


Iteration-2 `C_j``0``0``173``158``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_1)`
`S_1``0``51/43``0``49/43``169``119``1``0``0``0``0``0``(51/43)/(169)=0.007=0.007`
`S_2``0``60/43``0``45/43``243``167``0``1``0``0``0``0``(60/43)/(243)=0.0057=0.0057`
`S_3``0``1``0``0``173``158``0``0``1``0``0``0``(1)/(173)=0.0058`
`S_4``0``53/43``0``-100/43``(216)``138``0``0``0``1``0``0``(53/43)/(216)=0.0057=0.0057``->`
`S_5``0``1``0``-5``155``161``0``0``0``0``1``0``(1)/(155)=0.0065`
`S_6``0``44/43``0``-53/43``169``157``0``0``0``0``0``1``(44/43)/(169)=0.0061=0.0061`
`v_1``0``1/43``1``33/43``0``0``0``0``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0``0``0``0``0`
`C_j-Z_j``0``0``173``uarr``158``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `173` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.0057` and its row index is `4`. So, the leaving basis variable is `S_4`.

`:.` The pivot element is `216`.

Entering `=u_1`, Departing `=S_4`, Key Element `=216`

`R_4`(new)`= R_4`(old)` -: 216`
`R_4`(old) = `1.2326``0``-2.3256``216``138``0``0``0``1``0``0`
`R_4`(new)`= R_4`(old)` -: 216``0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0`


`R_1`(new)`= R_1`(old) - `169 R_4`(new)
`R_1`(old) = `1.186``0``1.1395``169``119``1``0``0``0``0``0`
`R_4`(new) = `0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.9644``0``-1.8196``169``107.9722``0``0``0``0.7824``0``0`
`R_1`(new)`= R_1`(old) - `169 R_4`(new)`0.2217``0``2.9591``0``11.0278``1``0``0``-0.7824``0``0`


`R_2`(new)`= R_2`(old) - `243 R_4`(new)
`R_2`(old) = `1.3953``0``1.0465``243``167``0``1``0``0``0``0`
`R_4`(new) = `0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0`
`243 xx R_4`(new) = `1.3866``0``-2.6163``243``155.25``0``0``0``1.125``0``0`
`R_2`(new)`= R_2`(old) - `243 R_4`(new)`0.0087``0``3.6628``0``11.75``0``1``0``-1.125``0``0`


`R_3`(new)`= R_3`(old) - `173 R_4`(new)
`R_3`(old) = `1``0``0``173``158``0``0``1``0``0``0`
`R_4`(new) = `0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0`
`173 xx R_4`(new) = `0.9872``0``-1.8626``173``110.5278``0``0``0``0.8009``0``0`
`R_3`(new)`= R_3`(old) - `173 R_4`(new)`0.0128``0``1.8626``0``47.4722``0``0``1``-0.8009``0``0`


`R_5`(new)`= R_5`(old) - `155 R_4`(new)
`R_5`(old) = `1``0``-5``155``161``0``0``0``0``1``0`
`R_4`(new) = `0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0`
`155 xx R_4`(new) = `0.8845``0``-1.6688``155``99.0278``0``0``0``0.7176``0``0`
`R_5`(new)`= R_5`(old) - `155 R_4`(new)`0.1155``0``-3.3312``0``61.9722``0``0``0``-0.7176``1``0`


`R_6`(new)`= R_6`(old) - `169 R_4`(new)
`R_6`(old) = `1.0233``0``-1.2326``169``157``0``0``0``0``0``1`
`R_4`(new) = `0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.9644``0``-1.8196``169``107.9722``0``0``0``0.7824``0``0`
`R_6`(new)`= R_6`(old) - `169 R_4`(new)`0.0589``0``0.587``0``49.0278``0``0``0``-0.7824``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0233``1``0.7674``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0233``1``0.7674``0``0``0``0``0``0``0``0`


Iteration-3 `C_j``0``0``173``158``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_2)`
`S_1``0``0.2217``0``2.9591``0``11.0278``1``0``0``-0.7824``0``0``(0.2217)/(11.0278)=0.0201`
`S_2``0``0.0087``0``3.6628``0``11.75``0``1``0``-1.125``0``0``(0.0087)/(11.75)=0.0007`
`S_3``0``0.0128``0``1.8626``0``(47.4722)``0``0``1``-0.8009``0``0``(0.0128)/(47.4722)=0.0003``->`
`u_1``173``0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0``(0.0057)/(0.6389)=0.0089`
`S_5``0``0.1155``0``-3.3312``0``61.9722``0``0``0``-0.7176``1``0``(0.1155)/(61.9722)=0.0019`
`S_6``0``0.0589``0``0.587``0``49.0278``0``0``0``-0.7824``0``1``(0.0589)/(49.0278)=0.0012`
`v_1``0``0.0233``1``0.7674``0``0``0``0``0``0``0``0`---
`Z=0.9872` `Z_j``0``-1.8626``173``110.5278``0``0``0``0.8009``0``0`
`C_j-Z_j``0``1.8626``0``47.4722``uarr``0``0``0``-0.8009``0``0`


Positive maximum `C_j-Z_j` is `47.4722` and its column index is `4`. So, the entering variable is `u_2`.

Minimum ratio is `0.0003` and its row index is `3`. So, the leaving basis variable is `S_3`.

`:.` The pivot element is `47.4722`.

Entering `=u_2`, Departing `=S_3`, Key Element `=47.4722`

`R_3`(new)`= R_3`(old)` -: 47.4722`
`R_3`(old) = `0.0128``0``1.8626``0``47.4722``0``0``1``-0.8009``0``0`
`R_3`(new)`= R_3`(old)` -: 47.4722``0.0003``0``0.0392``0``1``0``0``0.0211``-0.0169``0``0`


`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)
`R_1`(old) = `0.2217``0``2.9591``0``11.0278``1``0``0``-0.7824``0``0`
`R_3`(new) = `0.0003``0``0.0392``0``1``0``0``0.0211``-0.0169``0``0`
`11.0278 xx R_3`(new) = `0.003``0``0.4327``0``11.0278``0``0``0.2323``-0.1861``0``0`
`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)`0.2187``0``2.5264``0``0``1``0``-0.2323``-0.5964``0``0`


`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)
`R_2`(old) = `0.0087``0``3.6628``0``11.75``0``1``0``-1.125``0``0`
`R_3`(new) = `0.0003``0``0.0392``0``1``0``0``0.0211``-0.0169``0``0`
`11.75 xx R_3`(new) = `0.0032``0``0.461``0``11.75``0``0``0.2475``-0.1982``0``0`
`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)`0.0055``0``3.2018``0``0``0``1``-0.2475``-0.9268``0``0`


`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)
`R_4`(old) = `0.0057``0``-0.0108``1``0.6389``0``0``0``0.0046``0``0`
`R_3`(new) = `0.0003``0``0.0392``0``1``0``0``0.0211``-0.0169``0``0`
`0.6389 xx R_3`(new) = `0.0002``0``0.0251``0``0.6389``0``0``0.0135``-0.0108``0``0`
`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)`0.0055``0``-0.0358``1``0``0``0``-0.0135``0.0154``0``0`


`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)
`R_5`(old) = `0.1155``0``-3.3312``0``61.9722``0``0``0``-0.7176``1``0`
`R_3`(new) = `0.0003``0``0.0392``0``1``0``0``0.0211``-0.0169``0``0`
`61.9722 xx R_3`(new) = `0.0167``0``2.4315``0``61.9722``0``0``1.3054``-1.0456``0``0`
`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)`0.0988``0``-5.7627``0``0``0``0``-1.3054``0.328``1``0`


`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)
`R_6`(old) = `0.0589``0``0.587``0``49.0278``0``0``0``-0.7824``0``1`
`R_3`(new) = `0.0003``0``0.0392``0``1``0``0``0.0211``-0.0169``0``0`
`49.0278 xx R_3`(new) = `0.0132``0``1.9237``0``49.0278``0``0``1.0328``-0.8272``0``0`
`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)`0.0457``0``-1.3367``0``0``0``0``-1.0328``0.0448``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0233``1``0.7674``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0233``1``0.7674``0``0``0``0``0``0``0``0`


Iteration-4 `C_j``0``0``173``158``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`S_1``0``0.2187``0``2.5264``0``0``1``0``-0.2323``-0.5964``0``0`
`S_2``0``0.0055``0``3.2018``0``0``0``1``-0.2475``-0.9268``0``0`
`u_2``158``0.0003``0``0.0392``0``1``0``0``0.0211``-0.0169``0``0`
`u_1``173``0.0055``0``-0.0358``1``0``0``0``-0.0135``0.0154``0``0`
`S_5``0``0.0988``0``-5.7627``0``0``0``0``-1.3054``0.328``1``0`
`S_6``0``0.0457``0``-1.3367``0``0``0``0``-1.0328``0.0448``0``1`
`v_1``0``0.0233``1``0.7674``0``0``0``0``0``0``0``0`
`Z=1` `Z_j``0``0``173``158``0``0``1``0``0``0`
`C_j-Z_j``0``0``0``0``0``0``-1``0``0``0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`v_1=0.0233,v_2=0,u_1=0.0055,u_2=0.0003`

Max `Z = 1`


Solution steps by BigM method

Answer is
`v_1=0.023255813953488372,v_2=0.0,u_1=0.0055338585963050149,u_2=2.6988900531159724E-4,`



DMU-4
Max z `= (216u1+138u2)/(53v1+43v2)`

`(169u1+119u2)/(51v1+38v2)<=1`

`(243u1+167u2)/(60v1+45v2)<=1`

`(173u1+158u2)/(43v1+33v2)<=1`

`(216u1+138u2)/(53v1+43v2)<=1`

`(155u1+161u2)/(43v1+38v2)<=1`

`(169u1+157u2)/(44v1+35v2)<=1`


A fraction with decision variables in the numerator and denominator is nonlinear. Since we are using a linear programming technique, we need to linearize the formulation, such that the denominator of the objective function is 1, then maximize the numerator.
The new formulation would be:
Max z `= 216u1+138u2`

Denominator of nonlinear ` 53v1+43v2=1`

`(169u1+119u2)-(51v1+38v2)<=0`

`(243u1+167u2)-(60v1+45v2)<=0`

`(173u1+158u2)-(43v1+33v2)<=0`

`(216u1+138u2)-(53v1+43v2)<=0`

`(155u1+161u2)-(43v1+38v2)<=0`

`(169u1+157u2)-(44v1+35v2)<=0`

and `u,v>=0`

DMU-4
solution using simplex method
Problem is
Max `Z``=````216``u_1`` + ``138``u_2`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2``0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2``0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2``0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2``0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2``0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2``0`
```53``v_1`` + ``43``v_2`=`1`
and `v_1,v_2,u_1,u_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 add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`

4. As the constraint-4 is of type '`<=`' we should add slack variable `S_4`

5. As the constraint-5 is of type '`<=`' we should add slack variable `S_5`

6. As the constraint-6 is of type '`<=`' we should add slack variable `S_6`

7. As the constraint-7 is of type '`=`' we should add artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``216``u_1`` + ``138``u_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` + ``0``S_5`` + ``0``S_6`` - ``M``A_1`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2`` + ````S_1`=`0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2`` + ````S_2`=`0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2`` + ````S_3`=`0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2`` + ````S_4`=`0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2`` + ````S_5`=`0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2`` + ````S_6`=`0`
```53``v_1`` + ``43``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,u_2,S_1,S_2,S_3,S_4,S_5,S_6,A_1 >= 0`


Iteration-1 `C_j``0``0``216``138``0``0``0``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-51``-38``169``119``1``0``0``0``0``0``0`---
`S_2``0``0``-60``-45``243``167``0``1``0``0``0``0``0`---
`S_3``0``0``-43``-33``173``158``0``0``1``0``0``0``0`---
`S_4``0``0``-53``-43``216``138``0``0``0``1``0``0``0`---
`S_5``0``0``-43``-38``155``161``0``0``0``0``1``0``0`---
`S_6``0``0``-44``-35``169``157``0``0``0``0``0``1``0`---
`A_1``-M``1``(53)``43``0``0``0``0``0``0``0``0``1``(1)/(53)=0.0189``->`
`Z=-M` `Z_j``-53M``-43M``0``0``0``0``0``0``0``0``-M`
`C_j-Z_j``53M``uarr``43M``216``138``0``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `53M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.0189` and its row index is `7`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `53`.

Entering `=v_1`, Departing `=A_1`, Key Element `=53`

`R_7`(new)`= R_7`(old)` -: 53`
`R_7`(old) = `1``53``43``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)` -: 53``0.0189``1``0.8113``0``0``0``0``0``0``0``0`


`R_1`(new)`= R_1`(old) + `51 R_7`(new)
`R_1`(old) = `0``-51``-38``169``119``1``0``0``0``0``0`
`R_7`(new) = `0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`51 xx R_7`(new) = `0.9623``51``41.3774``0``0``0``0``0``0``0``0`
`R_1`(new)`= R_1`(old) + `51 R_7`(new)`0.9623``0``3.3774``169``119``1``0``0``0``0``0`


`R_2`(new)`= R_2`(old) + `60 R_7`(new)
`R_2`(old) = `0``-60``-45``243``167``0``1``0``0``0``0`
`R_7`(new) = `0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`60 xx R_7`(new) = `1.1321``60``48.6792``0``0``0``0``0``0``0``0`
`R_2`(new)`= R_2`(old) + `60 R_7`(new)`1.1321``0``3.6792``243``167``0``1``0``0``0``0`


`R_3`(new)`= R_3`(old) + `43 R_7`(new)
`R_3`(old) = `0``-43``-33``173``158``0``0``1``0``0``0`
`R_7`(new) = `0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `0.8113``43``34.8868``0``0``0``0``0``0``0``0`
`R_3`(new)`= R_3`(old) + `43 R_7`(new)`0.8113``0``1.8868``173``158``0``0``1``0``0``0`


`R_4`(new)`= R_4`(old) + `53 R_7`(new)
`R_4`(old) = `0``-53``-43``216``138``0``0``0``1``0``0`
`R_7`(new) = `0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`53 xx R_7`(new) = `1``53``43``0``0``0``0``0``0``0``0`
`R_4`(new)`= R_4`(old) + `53 R_7`(new)`1``0``0``216``138``0``0``0``1``0``0`


`R_5`(new)`= R_5`(old) + `43 R_7`(new)
`R_5`(old) = `0``-43``-38``155``161``0``0``0``0``1``0`
`R_7`(new) = `0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `0.8113``43``34.8868``0``0``0``0``0``0``0``0`
`R_5`(new)`= R_5`(old) + `43 R_7`(new)`0.8113``0``-3.1132``155``161``0``0``0``0``1``0`


`R_6`(new)`= R_6`(old) + `44 R_7`(new)
`R_6`(old) = `0``-44``-35``169``157``0``0``0``0``0``1`
`R_7`(new) = `0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`44 xx R_7`(new) = `0.8302``44``35.6981``0``0``0``0``0``0``0``0`
`R_6`(new)`= R_6`(old) + `44 R_7`(new)`0.8302``0``0.6981``169``157``0``0``0``0``0``1`


Iteration-2 `C_j``0``0``216``138``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_1)`
`S_1``0``0.9623``0``3.3774``169``119``1``0``0``0``0``0``(0.9623)/(169)=0.0057`
`S_2``0``1.1321``0``3.6792``243``167``0``1``0``0``0``0``(1.1321)/(243)=0.0047`
`S_3``0``0.8113``0``1.8868``173``158``0``0``1``0``0``0``(0.8113)/(173)=0.0047`
`S_4``0``1``0``0``(216)``138``0``0``0``1``0``0``(1)/(216)=0.0046``->`
`S_5``0``0.8113``0``-3.1132``155``161``0``0``0``0``1``0``(0.8113)/(155)=0.0052`
`S_6``0``0.8302``0``0.6981``169``157``0``0``0``0``0``1``(0.8302)/(169)=0.0049`
`v_1``0``0.0189``1``0.8113``0``0``0``0``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0``0``0``0``0`
`C_j-Z_j``0``0``216``uarr``138``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `216` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.0046` and its row index is `4`. So, the leaving basis variable is `S_4`.

`:.` The pivot element is `216`.

Entering `=u_1`, Departing `=S_4`, Key Element `=216`

`R_4`(new)`= R_4`(old)` -: 216`
`R_4`(old) = `1``0``0``216``138``0``0``0``1``0``0`
`R_4`(new)`= R_4`(old)` -: 216``0.0046``0``0``1``0.6389``0``0``0``0.0046``0``0`


`R_1`(new)`= R_1`(old) - `169 R_4`(new)
`R_1`(old) = `0.9623``0``3.3774``169``119``1``0``0``0``0``0`
`R_4`(new) = `0.0046``0``0``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.7824``0``0``169``107.9722``0``0``0``0.7824``0``0`
`R_1`(new)`= R_1`(old) - `169 R_4`(new)`0.1799``0``3.3774``0``11.0278``1``0``0``-0.7824``0``0`


`R_2`(new)`= R_2`(old) - `243 R_4`(new)
`R_2`(old) = `1.1321``0``3.6792``243``167``0``1``0``0``0``0`
`R_4`(new) = `0.0046``0``0``1``0.6389``0``0``0``0.0046``0``0`
`243 xx R_4`(new) = `1.125``0``0``243``155.25``0``0``0``1.125``0``0`
`R_2`(new)`= R_2`(old) - `243 R_4`(new)`0.0071``0``3.6792``0``11.75``0``1``0``-1.125``0``0`


`R_3`(new)`= R_3`(old) - `173 R_4`(new)
`R_3`(old) = `0.8113``0``1.8868``173``158``0``0``1``0``0``0`
`R_4`(new) = `0.0046``0``0``1``0.6389``0``0``0``0.0046``0``0`
`173 xx R_4`(new) = `0.8009``0``0``173``110.5278``0``0``0``0.8009``0``0`
`R_3`(new)`= R_3`(old) - `173 R_4`(new)`0.0104``0``1.8868``0``47.4722``0``0``1``-0.8009``0``0`


`R_5`(new)`= R_5`(old) - `155 R_4`(new)
`R_5`(old) = `0.8113``0``-3.1132``155``161``0``0``0``0``1``0`
`R_4`(new) = `0.0046``0``0``1``0.6389``0``0``0``0.0046``0``0`
`155 xx R_4`(new) = `0.7176``0``0``155``99.0278``0``0``0``0.7176``0``0`
`R_5`(new)`= R_5`(old) - `155 R_4`(new)`0.0937``0``-3.1132``0``61.9722``0``0``0``-0.7176``1``0`


`R_6`(new)`= R_6`(old) - `169 R_4`(new)
`R_6`(old) = `0.8302``0``0.6981``169``157``0``0``0``0``0``1`
`R_4`(new) = `0.0046``0``0``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.7824``0``0``169``107.9722``0``0``0``0.7824``0``0`
`R_6`(new)`= R_6`(old) - `169 R_4`(new)`0.0478``0``0.6981``0``49.0278``0``0``0``-0.7824``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0189``1``0.8113``0``0``0``0``0``0``0``0`


Iteration-3 `C_j``0``0``216``138``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`S_1``0``0.1799``0``3.3774``0``11.0278``1``0``0``-0.7824``0``0`
`S_2``0``0.0071``0``3.6792``0``11.75``0``1``0``-1.125``0``0`
`S_3``0``0.0104``0``1.8868``0``47.4722``0``0``1``-0.8009``0``0`
`u_1``216``0.0046``0``0``1``0.6389``0``0``0``0.0046``0``0`
`S_5``0``0.0937``0``-3.1132``0``61.9722``0``0``0``-0.7176``1``0`
`S_6``0``0.0478``0``0.6981``0``49.0278``0``0``0``-0.7824``0``1`
`v_1``0``0.0189``1``0.8113``0``0``0``0``0``0``0``0`
`Z=1` `Z_j``0``0``216``138``0``0``0``1``0``0`
`C_j-Z_j``0``0``0``0``0``0``0``-1``0``0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`v_1=0.0189,v_2=0,u_1=0.0046,u_2=0`

Max `Z = 1`


Solution steps by BigM method

Answer is
`v_1=0.018867924528301886,v_2=0.0,u_1=0.0046296296296296294,u_2=0.0,`



DMU-5
Max z `= (155u1+161u2)/(43v1+38v2)`

`(169u1+119u2)/(51v1+38v2)<=1`

`(243u1+167u2)/(60v1+45v2)<=1`

`(173u1+158u2)/(43v1+33v2)<=1`

`(216u1+138u2)/(53v1+43v2)<=1`

`(155u1+161u2)/(43v1+38v2)<=1`

`(169u1+157u2)/(44v1+35v2)<=1`


A fraction with decision variables in the numerator and denominator is nonlinear. Since we are using a linear programming technique, we need to linearize the formulation, such that the denominator of the objective function is 1, then maximize the numerator.
The new formulation would be:
Max z `= 155u1+161u2`

Denominator of nonlinear ` 43v1+38v2=1`

`(169u1+119u2)-(51v1+38v2)<=0`

`(243u1+167u2)-(60v1+45v2)<=0`

`(173u1+158u2)-(43v1+33v2)<=0`

`(216u1+138u2)-(53v1+43v2)<=0`

`(155u1+161u2)-(43v1+38v2)<=0`

`(169u1+157u2)-(44v1+35v2)<=0`

and `u,v>=0`

DMU-5
solution using simplex method
Problem is
Max `Z``=````155``u_1`` + ``161``u_2`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2``0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2``0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2``0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2``0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2``0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2``0`
```43``v_1`` + ``38``v_2`=`1`
and `v_1,v_2,u_1,u_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 add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`

4. As the constraint-4 is of type '`<=`' we should add slack variable `S_4`

5. As the constraint-5 is of type '`<=`' we should add slack variable `S_5`

6. As the constraint-6 is of type '`<=`' we should add slack variable `S_6`

7. As the constraint-7 is of type '`=`' we should add artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``155``u_1`` + ``161``u_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` + ``0``S_5`` + ``0``S_6`` - ``M``A_1`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2`` + ````S_1`=`0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2`` + ````S_2`=`0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2`` + ````S_3`=`0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2`` + ````S_4`=`0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2`` + ````S_5`=`0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2`` + ````S_6`=`0`
```43``v_1`` + ``38``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,u_2,S_1,S_2,S_3,S_4,S_5,S_6,A_1 >= 0`


Iteration-1 `C_j``0``0``155``161``0``0``0``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-51``-38``169``119``1``0``0``0``0``0``0`---
`S_2``0``0``-60``-45``243``167``0``1``0``0``0``0``0`---
`S_3``0``0``-43``-33``173``158``0``0``1``0``0``0``0`---
`S_4``0``0``-53``-43``216``138``0``0``0``1``0``0``0`---
`S_5``0``0``-43``-38``155``161``0``0``0``0``1``0``0`---
`S_6``0``0``-44``-35``169``157``0``0``0``0``0``1``0`---
`A_1``-M``1``(43)``38``0``0``0``0``0``0``0``0``1``(1)/(43)=0.0233``->`
`Z=-M` `Z_j``-43M``-38M``0``0``0``0``0``0``0``0``-M`
`C_j-Z_j``43M``uarr``38M``155``161``0``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `43M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.0233` and its row index is `7`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `43`.

Entering `=v_1`, Departing `=A_1`, Key Element `=43`

`R_7`(new)`= R_7`(old)` -: 43`
`R_7`(old) = `1``43``38``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)` -: 43``1/43``1``38/43``0``0``0``0``0``0``0``0`


`R_1`(new)`= R_1`(old) + `51 R_7`(new)
`R_1`(old) = `0``-51``-38``169``119``1``0``0``0``0``0`
`R_7`(new) = `1/43``1``38/43``0``0``0``0``0``0``0``0`
`51 xx R_7`(new) = `51/43``51``1938/43``0``0``0``0``0``0``0``0`
`R_1`(new)`= R_1`(old) + `51 R_7`(new)`51/43``0``304/43``169``119``1``0``0``0``0``0`


`R_2`(new)`= R_2`(old) + `60 R_7`(new)
`R_2`(old) = `0``-60``-45``243``167``0``1``0``0``0``0`
`R_7`(new) = `1/43``1``38/43``0``0``0``0``0``0``0``0`
`60 xx R_7`(new) = `60/43``60``2280/43``0``0``0``0``0``0``0``0`
`R_2`(new)`= R_2`(old) + `60 R_7`(new)`60/43``0``345/43``243``167``0``1``0``0``0``0`


`R_3`(new)`= R_3`(old) + `43 R_7`(new)
`R_3`(old) = `0``-43``-33``173``158``0``0``1``0``0``0`
`R_7`(new) = `1/43``1``38/43``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `1``43``38``0``0``0``0``0``0``0``0`
`R_3`(new)`= R_3`(old) + `43 R_7`(new)`1``0``5``173``158``0``0``1``0``0``0`


`R_4`(new)`= R_4`(old) + `53 R_7`(new)
`R_4`(old) = `0``-53``-43``216``138``0``0``0``1``0``0`
`R_7`(new) = `1/43``1``38/43``0``0``0``0``0``0``0``0`
`53 xx R_7`(new) = `53/43``53``2014/43``0``0``0``0``0``0``0``0`
`R_4`(new)`= R_4`(old) + `53 R_7`(new)`53/43``0``165/43``216``138``0``0``0``1``0``0`


`R_5`(new)`= R_5`(old) + `43 R_7`(new)
`R_5`(old) = `0``-43``-38``155``161``0``0``0``0``1``0`
`R_7`(new) = `1/43``1``38/43``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `1``43``38``0``0``0``0``0``0``0``0`
`R_5`(new)`= R_5`(old) + `43 R_7`(new)`1``0``0``155``161``0``0``0``0``1``0`


`R_6`(new)`= R_6`(old) + `44 R_7`(new)
`R_6`(old) = `0``-44``-35``169``157``0``0``0``0``0``1`
`R_7`(new) = `1/43``1``38/43``0``0``0``0``0``0``0``0`
`44 xx R_7`(new) = `44/43``44``1672/43``0``0``0``0``0``0``0``0`
`R_6`(new)`= R_6`(old) + `44 R_7`(new)`44/43``0``167/43``169``157``0``0``0``0``0``1`


Iteration-2 `C_j``0``0``155``161``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_2)`
`S_1``0``51/43``0``304/43``169``119``1``0``0``0``0``0``(51/43)/(119)=3/301=0.01`
`S_2``0``60/43``0``345/43``243``167``0``1``0``0``0``0``(60/43)/(167)=0.0084=0.0084`
`S_3``0``1``0``5``173``158``0``0``1``0``0``0``(1)/(158)=0.0063`
`S_4``0``53/43``0``165/43``216``138``0``0``0``1``0``0``(53/43)/(138)=0.0089=0.0089`
`S_5``0``1``0``0``155``(161)``0``0``0``0``1``0``(1)/(161)=0.0062``->`
`S_6``0``44/43``0``167/43``169``157``0``0``0``0``0``1``(44/43)/(157)=0.0065=0.0065`
`v_1``0``1/43``1``38/43``0``0``0``0``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0``0``0``0``0`
`C_j-Z_j``0``0``155``161``uarr``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `161` and its column index is `4`. So, the entering variable is `u_2`.

Minimum ratio is `0.0062` and its row index is `5`. So, the leaving basis variable is `S_5`.

`:.` The pivot element is `161`.

Entering `=u_2`, Departing `=S_5`, Key Element `=161`

`R_5`(new)`= R_5`(old)` -: 161`
`R_5`(old) = `1``0``0``155``161``0``0``0``0``1``0`
`R_5`(new)`= R_5`(old)` -: 161``0.0062``0``0``0.9627``1``0``0``0``0``0.0062``0`


`R_1`(new)`= R_1`(old) - `119 R_5`(new)
`R_1`(old) = `1.186``0``7.0698``169``119``1``0``0``0``0``0`
`R_5`(new) = `0.0062``0``0``0.9627``1``0``0``0``0``0.0062``0`
`119 xx R_5`(new) = `0.7391``0``0``114.5652``119``0``0``0``0``0.7391``0`
`R_1`(new)`= R_1`(old) - `119 R_5`(new)`0.4469``0``7.0698``54.4348``0``1``0``0``0``-0.7391``0`


`R_2`(new)`= R_2`(old) - `167 R_5`(new)
`R_2`(old) = `1.3953``0``8.0233``243``167``0``1``0``0``0``0`
`R_5`(new) = `0.0062``0``0``0.9627``1``0``0``0``0``0.0062``0`
`167 xx R_5`(new) = `1.0373``0``0``160.7764``167``0``0``0``0``1.0373``0`
`R_2`(new)`= R_2`(old) - `167 R_5`(new)`0.3581``0``8.0233``82.2236``0``0``1``0``0``-1.0373``0`


`R_3`(new)`= R_3`(old) - `158 R_5`(new)
`R_3`(old) = `1``0``5``173``158``0``0``1``0``0``0`
`R_5`(new) = `0.0062``0``0``0.9627``1``0``0``0``0``0.0062``0`
`158 xx R_5`(new) = `0.9814``0``0``152.1118``158``0``0``0``0``0.9814``0`
`R_3`(new)`= R_3`(old) - `158 R_5`(new)`0.0186``0``5``20.8882``0``0``0``1``0``-0.9814``0`


`R_4`(new)`= R_4`(old) - `138 R_5`(new)
`R_4`(old) = `1.2326``0``3.8372``216``138``0``0``0``1``0``0`
`R_5`(new) = `0.0062``0``0``0.9627``1``0``0``0``0``0.0062``0`
`138 xx R_5`(new) = `0.8571``0``0``132.8571``138``0``0``0``0``0.8571``0`
`R_4`(new)`= R_4`(old) - `138 R_5`(new)`0.3754``0``3.8372``83.1429``0``0``0``0``1``-0.8571``0`


`R_6`(new)`= R_6`(old) - `157 R_5`(new)
`R_6`(old) = `1.0233``0``3.8837``169``157``0``0``0``0``0``1`
`R_5`(new) = `0.0062``0``0``0.9627``1``0``0``0``0``0.0062``0`
`157 xx R_5`(new) = `0.9752``0``0``151.1491``157``0``0``0``0``0.9752``0`
`R_6`(new)`= R_6`(old) - `157 R_5`(new)`0.0481``0``3.8837``17.8509``0``0``0``0``0``-0.9752``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0233``1``0.8837``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0233``1``0.8837``0``0``0``0``0``0``0``0`


Iteration-3 `C_j``0``0``155``161``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`S_1``0``0.4469``0``7.0698``54.4348``0``1``0``0``0``-0.7391``0`
`S_2``0``0.3581``0``8.0233``82.2236``0``0``1``0``0``-1.0373``0`
`S_3``0``0.0186``0``5``20.8882``0``0``0``1``0``-0.9814``0`
`S_4``0``0.3754``0``3.8372``83.1429``0``0``0``0``1``-0.8571``0`
`u_2``161``0.0062``0``0``0.9627``1``0``0``0``0``0.0062``0`
`S_6``0``0.0481``0``3.8837``17.8509``0``0``0``0``0``-0.9752``1`
`v_1``0``0.0233``1``0.8837``0``0``0``0``0``0``0``0`
`Z=1` `Z_j``0``0``155``161``0``0``0``0``1``0`
`C_j-Z_j``0``0``0``0``0``0``0``0``-1``0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`v_1=0.0233,v_2=0,u_1=0,u_2=0.0062`

Max `Z = 1`


Solution steps by BigM method

Answer is
`v_1=0.023255813953488372,v_2=0.0,u_1=0.0,u_2=0.0062111801242236021,`



DMU-6
Max z `= (169u1+157u2)/(44v1+35v2)`

`(169u1+119u2)/(51v1+38v2)<=1`

`(243u1+167u2)/(60v1+45v2)<=1`

`(173u1+158u2)/(43v1+33v2)<=1`

`(216u1+138u2)/(53v1+43v2)<=1`

`(155u1+161u2)/(43v1+38v2)<=1`

`(169u1+157u2)/(44v1+35v2)<=1`


A fraction with decision variables in the numerator and denominator is nonlinear. Since we are using a linear programming technique, we need to linearize the formulation, such that the denominator of the objective function is 1, then maximize the numerator.
The new formulation would be:
Max z `= 169u1+157u2`

Denominator of nonlinear ` 44v1+35v2=1`

`(169u1+119u2)-(51v1+38v2)<=0`

`(243u1+167u2)-(60v1+45v2)<=0`

`(173u1+158u2)-(43v1+33v2)<=0`

`(216u1+138u2)-(53v1+43v2)<=0`

`(155u1+161u2)-(43v1+38v2)<=0`

`(169u1+157u2)-(44v1+35v2)<=0`

and `u,v>=0`

DMU-6
solution using simplex method
Problem is
Max `Z``=````169``u_1`` + ``157``u_2`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2``0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2``0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2``0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2``0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2``0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2``0`
```44``v_1`` + ``35``v_2`=`1`
and `v_1,v_2,u_1,u_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 add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`

4. As the constraint-4 is of type '`<=`' we should add slack variable `S_4`

5. As the constraint-5 is of type '`<=`' we should add slack variable `S_5`

6. As the constraint-6 is of type '`<=`' we should add slack variable `S_6`

7. As the constraint-7 is of type '`=`' we should add artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``169``u_1`` + ``157``u_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` + ``0``S_5`` + ``0``S_6`` - ``M``A_1`
subject to
` - ``51``v_1`` - ``38``v_2`` + ``169``u_1`` + ``119``u_2`` + ````S_1`=`0`
` - ``60``v_1`` - ``45``v_2`` + ``243``u_1`` + ``167``u_2`` + ````S_2`=`0`
` - ``43``v_1`` - ``33``v_2`` + ``173``u_1`` + ``158``u_2`` + ````S_3`=`0`
` - ``53``v_1`` - ``43``v_2`` + ``216``u_1`` + ``138``u_2`` + ````S_4`=`0`
` - ``43``v_1`` - ``38``v_2`` + ``155``u_1`` + ``161``u_2`` + ````S_5`=`0`
` - ``44``v_1`` - ``35``v_2`` + ``169``u_1`` + ``157``u_2`` + ````S_6`=`0`
```44``v_1`` + ``35``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,u_2,S_1,S_2,S_3,S_4,S_5,S_6,A_1 >= 0`


Iteration-1 `C_j``0``0``169``157``0``0``0``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-51``-38``169``119``1``0``0``0``0``0``0`---
`S_2``0``0``-60``-45``243``167``0``1``0``0``0``0``0`---
`S_3``0``0``-43``-33``173``158``0``0``1``0``0``0``0`---
`S_4``0``0``-53``-43``216``138``0``0``0``1``0``0``0`---
`S_5``0``0``-43``-38``155``161``0``0``0``0``1``0``0`---
`S_6``0``0``-44``-35``169``157``0``0``0``0``0``1``0`---
`A_1``-M``1``(44)``35``0``0``0``0``0``0``0``0``1``(1)/(44)=0.0227``->`
`Z=-M` `Z_j``-44M``-35M``0``0``0``0``0``0``0``0``-M`
`C_j-Z_j``44M``uarr``35M``169``157``0``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `44M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.0227` and its row index is `7`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `44`.

Entering `=v_1`, Departing `=A_1`, Key Element `=44`

`R_7`(new)`= R_7`(old)` -: 44`
`R_7`(old) = `1``44``35``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)` -: 44``1/44``1``35/44``0``0``0``0``0``0``0``0`


`R_1`(new)`= R_1`(old) + `51 R_7`(new)
`R_1`(old) = `0``-51``-38``169``119``1``0``0``0``0``0`
`R_7`(new) = `1/44``1``35/44``0``0``0``0``0``0``0``0`
`51 xx R_7`(new) = `51/44``51``1785/44``0``0``0``0``0``0``0``0`
`R_1`(new)`= R_1`(old) + `51 R_7`(new)`51/44``0``113/44``169``119``1``0``0``0``0``0`


`R_2`(new)`= R_2`(old) + `60 R_7`(new)
`R_2`(old) = `0``-60``-45``243``167``0``1``0``0``0``0`
`R_7`(new) = `1/44``1``35/44``0``0``0``0``0``0``0``0`
`60 xx R_7`(new) = `15/11``60``525/11``0``0``0``0``0``0``0``0`
`R_2`(new)`= R_2`(old) + `60 R_7`(new)`15/11``0``30/11``243``167``0``1``0``0``0``0`


`R_3`(new)`= R_3`(old) + `43 R_7`(new)
`R_3`(old) = `0``-43``-33``173``158``0``0``1``0``0``0`
`R_7`(new) = `1/44``1``35/44``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `43/44``43``1505/44``0``0``0``0``0``0``0``0`
`R_3`(new)`= R_3`(old) + `43 R_7`(new)`43/44``0``53/44``173``158``0``0``1``0``0``0`


`R_4`(new)`= R_4`(old) + `53 R_7`(new)
`R_4`(old) = `0``-53``-43``216``138``0``0``0``1``0``0`
`R_7`(new) = `1/44``1``35/44``0``0``0``0``0``0``0``0`
`53 xx R_7`(new) = `53/44``53``1855/44``0``0``0``0``0``0``0``0`
`R_4`(new)`= R_4`(old) + `53 R_7`(new)`53/44``0``-37/44``216``138``0``0``0``1``0``0`


`R_5`(new)`= R_5`(old) + `43 R_7`(new)
`R_5`(old) = `0``-43``-38``155``161``0``0``0``0``1``0`
`R_7`(new) = `1/44``1``35/44``0``0``0``0``0``0``0``0`
`43 xx R_7`(new) = `43/44``43``1505/44``0``0``0``0``0``0``0``0`
`R_5`(new)`= R_5`(old) + `43 R_7`(new)`43/44``0``-167/44``155``161``0``0``0``0``1``0`


`R_6`(new)`= R_6`(old) + `44 R_7`(new)
`R_6`(old) = `0``-44``-35``169``157``0``0``0``0``0``1`
`R_7`(new) = `1/44``1``35/44``0``0``0``0``0``0``0``0`
`44 xx R_7`(new) = `1``44``35``0``0``0``0``0``0``0``0`
`R_6`(new)`= R_6`(old) + `44 R_7`(new)`1``0``0``169``157``0``0``0``0``0``1`


Iteration-2 `C_j``0``0``169``157``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_1)`
`S_1``0``51/44``0``113/44``169``119``1``0``0``0``0``0``(51/44)/(169)=0.0069=0.0069`
`S_2``0``15/11``0``30/11``243``167``0``1``0``0``0``0``(15/11)/(243)=5/891=0.0056`
`S_3``0``43/44``0``53/44``173``158``0``0``1``0``0``0``(43/44)/(173)=0.0056=0.0056`
`S_4``0``53/44``0``-37/44``(216)``138``0``0``0``1``0``0``(53/44)/(216)=0.0056=0.0056``->`
`S_5``0``43/44``0``-167/44``155``161``0``0``0``0``1``0``(43/44)/(155)=0.0063=0.0063`
`S_6``0``1``0``0``169``157``0``0``0``0``0``1``(1)/(169)=0.0059`
`v_1``0``1/44``1``35/44``0``0``0``0``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0``0``0``0``0`
`C_j-Z_j``0``0``169``uarr``157``0``0``0``0``0``0`


Positive maximum `C_j-Z_j` is `169` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.0056` and its row index is `4`. So, the leaving basis variable is `S_4`.

`:.` The pivot element is `216`.

Entering `=u_1`, Departing `=S_4`, Key Element `=216`

`R_4`(new)`= R_4`(old)` -: 216`
`R_4`(old) = `1.2045``0``-0.8409``216``138``0``0``0``1``0``0`
`R_4`(new)`= R_4`(old)` -: 216``0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0`


`R_1`(new)`= R_1`(old) - `169 R_4`(new)
`R_1`(old) = `1.1591``0``2.5682``169``119``1``0``0``0``0``0`
`R_4`(new) = `0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.9424``0``-0.6579``169``107.9722``0``0``0``0.7824``0``0`
`R_1`(new)`= R_1`(old) - `169 R_4`(new)`0.2166``0``3.2261``0``11.0278``1``0``0``-0.7824``0``0`


`R_2`(new)`= R_2`(old) - `243 R_4`(new)
`R_2`(old) = `1.3636``0``2.7273``243``167``0``1``0``0``0``0`
`R_4`(new) = `0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0`
`243 xx R_4`(new) = `1.3551``0``-0.946``243``155.25``0``0``0``1.125``0``0`
`R_2`(new)`= R_2`(old) - `243 R_4`(new)`0.0085``0``3.6733``0``11.75``0``1``0``-1.125``0``0`


`R_3`(new)`= R_3`(old) - `173 R_4`(new)
`R_3`(old) = `0.9773``0``1.2045``173``158``0``0``1``0``0``0`
`R_4`(new) = `0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0`
`173 xx R_4`(new) = `0.9648``0``-0.6735``173``110.5278``0``0``0``0.8009``0``0`
`R_3`(new)`= R_3`(old) - `173 R_4`(new)`0.0125``0``1.8781``0``47.4722``0``0``1``-0.8009``0``0`


`R_5`(new)`= R_5`(old) - `155 R_4`(new)
`R_5`(old) = `0.9773``0``-3.7955``155``161``0``0``0``0``1``0`
`R_4`(new) = `0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0`
`155 xx R_4`(new) = `0.8644``0``-0.6034``155``99.0278``0``0``0``0.7176``0``0`
`R_5`(new)`= R_5`(old) - `155 R_4`(new)`0.1129``0``-3.192``0``61.9722``0``0``0``-0.7176``1``0`


`R_6`(new)`= R_6`(old) - `169 R_4`(new)
`R_6`(old) = `1``0``0``169``157``0``0``0``0``0``1`
`R_4`(new) = `0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0`
`169 xx R_4`(new) = `0.9424``0``-0.6579``169``107.9722``0``0``0``0.7824``0``0`
`R_6`(new)`= R_6`(old) - `169 R_4`(new)`0.0576``0``0.6579``0``49.0278``0``0``0``-0.7824``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0227``1``0.7955``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0227``1``0.7955``0``0``0``0``0``0``0``0`


Iteration-3 `C_j``0``0``169``157``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(u_2)`
`S_1``0``0.2166``0``3.2261``0``11.0278``1``0``0``-0.7824``0``0``(0.2166)/(11.0278)=0.0196`
`S_2``0``0.0085``0``3.6733``0``11.75``0``1``0``-1.125``0``0``(0.0085)/(11.75)=0.0007`
`S_3``0``0.0125``0``1.8781``0``(47.4722)``0``0``1``-0.8009``0``0``(0.0125)/(47.4722)=0.0003``->`
`u_1``169``0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0``(0.0056)/(0.6389)=0.0087`
`S_5``0``0.1129``0``-3.192``0``61.9722``0``0``0``-0.7176``1``0``(0.1129)/(61.9722)=0.0018`
`S_6``0``0.0576``0``0.6579``0``49.0278``0``0``0``-0.7824``0``1``(0.0576)/(49.0278)=0.0012`
`v_1``0``0.0227``1``0.7955``0``0``0``0``0``0``0``0`---
`Z=0.9424` `Z_j``0``-0.6579``169``107.9722``0``0``0``0.7824``0``0`
`C_j-Z_j``0``0.6579``0``49.0278``uarr``0``0``0``-0.7824``0``0`


Positive maximum `C_j-Z_j` is `49.0278` and its column index is `4`. So, the entering variable is `u_2`.

Minimum ratio is `0.0003` and its row index is `3`. So, the leaving basis variable is `S_3`.

`:.` The pivot element is `47.4722`.

Entering `=u_2`, Departing `=S_3`, Key Element `=47.4722`

`R_3`(new)`= R_3`(old)` -: 47.4722`
`R_3`(old) = `0.0125``0``1.8781``0``47.4722``0``0``1``-0.8009``0``0`
`R_3`(new)`= R_3`(old)` -: 47.4722``0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`


`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)
`R_1`(old) = `0.2166``0``3.2261``0``11.0278``1``0``0``-0.7824``0``0`
`R_3`(new) = `0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`
`11.0278 xx R_3`(new) = `0.0029``0``0.4363``0``11.0278``0``0``0.2323``-0.1861``0``0`
`R_1`(new)`= R_1`(old) - `11.0278 R_3`(new)`0.2137``0``2.7898``0``0``1``0``-0.2323``-0.5964``0``0`


`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)
`R_2`(old) = `0.0085``0``3.6733``0``11.75``0``1``0``-1.125``0``0`
`R_3`(new) = `0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`
`11.75 xx R_3`(new) = `0.0031``0``0.4648``0``11.75``0``0``0.2475``-0.1982``0``0`
`R_2`(new)`= R_2`(old) - `11.75 R_3`(new)`0.0054``0``3.2085``0``0``0``1``-0.2475``-0.9268``0``0`


`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)
`R_4`(old) = `0.0056``0``-0.0039``1``0.6389``0``0``0``0.0046``0``0`
`R_3`(new) = `0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`
`0.6389 xx R_3`(new) = `0.0002``0``0.0253``0``0.6389``0``0``0.0135``-0.0108``0``0`
`R_4`(new)`= R_4`(old) - `0.6389 R_3`(new)`0.0054``0``-0.0292``1``0``0``0``-0.0135``0.0154``0``0`


`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)
`R_5`(old) = `0.1129``0``-3.192``0``61.9722``0``0``0``-0.7176``1``0`
`R_3`(new) = `0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`
`61.9722 xx R_3`(new) = `0.0163``0``2.4517``0``61.9722``0``0``1.3054``-1.0456``0``0`
`R_5`(new)`= R_5`(old) - `61.9722 R_3`(new)`0.0966``0``-5.6437``0``0``0``0``-1.3054``0.328``1``0`


`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)
`R_6`(old) = `0.0576``0``0.6579``0``49.0278``0``0``0``-0.7824``0``1`
`R_3`(new) = `0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`
`49.0278 xx R_3`(new) = `0.0129``0``1.9396``0``49.0278``0``0``1.0328``-0.8272``0``0`
`R_6`(new)`= R_6`(old) - `49.0278 R_3`(new)`0.0446``0``-1.2817``0``0``0``0``-1.0328``0.0448``0``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0227``1``0.7955``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0227``1``0.7955``0``0``0``0``0``0``0``0`


Iteration-4 `C_j``0``0``169``157``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`(X_B)/(S_4)`
`S_1``0``0.2137``0``2.7898``0``0``1``0``-0.2323``-0.5964``0``0`---
`S_2``0``0.0054``0``3.2085``0``0``0``1``-0.2475``-0.9268``0``0`---
`u_2``157``0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`---
`u_1``169``0.0054``0``-0.0292``1``0``0``0``-0.0135``0.0154``0``0``(0.0054)/(0.0154)=0.351`
`S_5``0``0.0966``0``-5.6437``0``0``0``0``-1.3054``(0.328)``1``0``(0.0966)/(0.328)=0.2944``->`
`S_6``0``0.0446``0``-1.2817``0``0``0``0``-1.0328``0.0448``0``1``(0.0446)/(0.0448)=0.9969`
`v_1``0``0.0227``1``0.7955``0``0``0``0``0``0``0``0`---
`Z=0.9554` `Z_j``0``1.2817``169``157``0``0``1.0328``-0.0448``0``0`
`C_j-Z_j``0``-1.2817``0``0``0``0``-1.0328``0.0448``uarr``0``0`


Positive maximum `C_j-Z_j` is `0.0448` and its column index is `8`. So, the entering variable is `S_4`.

Minimum ratio is `0.2944` and its row index is `5`. So, the leaving basis variable is `S_5`.

`:.` The pivot element is `0.328`.

Entering `=S_4`, Departing `=S_5`, Key Element `=0.328`

`R_5`(new)`= R_5`(old)` -: 0.328`
`R_5`(old) = `0.0966``0``-5.6437``0``0``0``0``-1.3054``0.328``1``0`
`R_5`(new)`= R_5`(old)` -: 0.328``0.2944``0``-17.208``0``0``0``0``-3.9804``1``3.0491``0`


`R_1`(new)`= R_1`(old) + `0.5964 R_5`(new)
`R_1`(old) = `0.2137``0``2.7898``0``0``1``0``-0.2323``-0.5964``0``0`
`R_5`(new) = `0.2944``0``-17.208``0``0``0``0``-3.9804``1``3.0491``0`
`0.5964 xx R_5`(new) = `0.1756``0``-10.2621``0``0``0``0``-2.3737``0.5964``1.8183``0`
`R_1`(new)`= R_1`(old) + `0.5964 R_5`(new)`0.3893``0``-7.4722``0``0``1``0``-2.606``0``1.8183``0`


`R_2`(new)`= R_2`(old) + `0.9268 R_5`(new)
`R_2`(old) = `0.0054``0``3.2085``0``0``0``1``-0.2475``-0.9268``0``0`
`R_5`(new) = `0.2944``0``-17.208``0``0``0``0``-3.9804``1``3.0491``0`
`0.9268 xx R_5`(new) = `0.2728``0``-15.9477``0``0``0``0``-3.6889``0.9268``2.8258``0`
`R_2`(new)`= R_2`(old) + `0.9268 R_5`(new)`0.2783``0``-12.7393``0``0``0``1``-3.9364``0``2.8258``0`


`R_3`(new)`= R_3`(old) + `0.0169 R_5`(new)
`R_3`(old) = `0.0003``0``0.0396``0``1``0``0``0.0211``-0.0169``0``0`
`R_5`(new) = `0.2944``0``-17.208``0``0``0``0``-3.9804``1``3.0491``0`
`0.0169 xx R_5`(new) = `0.005``0``-0.2903``0``0``0``0``-0.0672``0.0169``0.0514``0`
`R_3`(new)`= R_3`(old) + `0.0169 R_5`(new)`0.0052``0``-0.2508``0``1``0``0``-0.0461``0``0.0514``0`


`R_4`(new)`= R_4`(old) - `0.0154 R_5`(new)
`R_4`(old) = `0.0054``0``-0.0292``1``0``0``0``-0.0135``0.0154``0``0`
`R_5`(new) = `0.2944``0``-17.208``0``0``0``0``-3.9804``1``3.0491``0`
`0.0154 xx R_5`(new) = `0.0045``0``-0.2652``0``0``0``0``-0.0613``0.0154``0.047``0`
`R_4`(new)`= R_4`(old) - `0.0154 R_5`(new)`0.0009``0``0.236``1``0``0``0``0.0479``0``-0.047``0`


`R_6`(new)`= R_6`(old) - `0.0448 R_5`(new)
`R_6`(old) = `0.0446``0``-1.2817``0``0``0``0``-1.0328``0.0448``0``1`
`R_5`(new) = `0.2944``0``-17.208``0``0``0``0``-3.9804``1``3.0491``0`
`0.0448 xx R_5`(new) = `0.0132``0``-0.7703``0``0``0``0``-0.1782``0.0448``0.1365``0`
`R_6`(new)`= R_6`(old) - `0.0448 R_5`(new)`0.0314``0``-0.5114``0``0``0``0``-0.8546``0``-0.1365``1`


`R_7`(new)`= R_7`(old)
`R_7`(old) = `0.0227``1``0.7955``0``0``0``0``0``0``0``0`
`R_7`(new)`= R_7`(old)`0.0227``1``0.7955``0``0``0``0``0``0``0``0`


Iteration-5 `C_j``0``0``169``157``0``0``0``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``u_2``S_1``S_2``S_3``S_4``S_5``S_6`MinRatio
`S_1``0``0.3893``0``-7.4722``0``0``1``0``-2.606``0``1.8183``0`
`S_2``0``0.2783``0``-12.7393``0``0``0``1``-3.9364``0``2.8258``0`
`u_2``157``0.0052``0``-0.2508``0``1``0``0``-0.0461``0``0.0514``0`
`u_1``169``0.0009``0``0.236``1``0``0``0``0.0479``0``-0.047``0`
`S_4``0``0.2944``0``-17.208``0``0``0``0``-3.9804``1``3.0491``0`
`S_6``0``0.0314``0``-0.5114``0``0``0``0``-0.8546``0``-0.1365``1`
`v_1``0``0.0227``1``0.7955``0``0``0``0``0``0``0``0`
`Z=0.9686` `Z_j``0``0.5114``169``157``0``0``0.8546``0``0.1365``0`
`C_j-Z_j``0``-0.5114``0``0``0``0``-0.8546``0``-0.1365``0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`v_1=0.0227,v_2=0,u_1=0.0009,u_2=0.0052`

Max `Z = 0.9686`


Solution steps by BigM method

Answer is
`v_1=0.022727272727272728,v_2=0.0,u_1=8.71786554212957E-4,u_2=0.0052307193252777587,`

Final Score, weight and WeightedData table is
NoScoreRankv1v2u1u2(v1)(v2)(u1)(u2)v1 * (v1)v2 * (v2)u1 * (u1)u2 * (u2)`sum v_i * (v_i)`
10.82543746513816911900.026315790.004427960.00064799010.748326010.077111381
211604524316700.022222220.003739170.0005472010.908618330.091381671
31143331731580.0232558100.005533860.00026989100.957357540.042642461
41153432161380.0188679200.00462963010101
51143381551610.02325581000.0062111810011
60.96855486544351691570.0227272700.000871790.00523072100.147331930.821222931


Final Projection table is
NoScoreRankv1Projection =
v1 * Score
Diff (%) =
(Projection - v1)/v1 * 100
v2Projection =
v2 * Score
Diff (%) =
(Projection - v2)/v2 * 100
u1Projection =
u1 * `sum v_i * (v_i)`
Diff (%) =
(Projection - u1)/u1 * 100
u2Projection =
u2 * `sum v_i * (v_i)`
Diff (%) =
(Projection - u2)/u2 * 100
10.825437465142.0973074-17.463831.3666212-17.4616916901191190
211606004545024324301671670
311434303333017317301581580
411535304343021621601381380
511434303838015515501611610
60.9685548654442.61641384-3.143533.8994201-3.1416916901571570


Final Slack table is
NoScoreRankslack v1slack v2slack u1slack u2
10.825437460.34817537000
2110000
3110000
4110000
5110000
60.96855486500.5113737700



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