We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more   

AtoZmath.com
Support us
(New) All problem can be solved using search box
I want to sell my website www.AtoZmath.com with complete code
Home What's new College Algebra Games Feedback About us
Algebra Matrix & Vector Numerical Methods Statistical Methods Operation Research Word Problems Calculus Geometry Pre-Algebra


Home > Operation Research calculators > Data envelopment analysis (DEA method) example
Data envelopment analysis (DEA method) example ( Enter your problem)
( Enter your problem)
Algorithm and examples
  1. Example-1
  2. Example-2
2. Example-2
(Next example)

1. Example-1


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


This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


2. Example-2
(Next example)


Share this solution or page with your friends.


Home What's new College Algebra Games Feedback About us
 
Copyright © 2025. All rights reserved. Terms, Privacy
 
 

.