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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
1. Example-1
(Previous example)

2. Example-2


Data envelopment analysis (DEA method)
InuptOutupt
10,2100
8,480
12,1.5120


Solution:

DMU-1
Max z `= (100u1)/(10v1+2v2)`

`(100u1)/(10v1+2v2)<=1`

`(80u1)/(8v1+4v2)<=1`

`(120u1)/(12v1+1.5v2)<=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 `= 100u1`

Denominator of nonlinear ` 10v1+2v2=1`

`(100u1)-(10v1+2v2)<=0`

`(80u1)-(8v1+4v2)<=0`

`(120u1)-(12v1+1.5v2)<=0`

and `u,v>=0`

DMU-1
solution using simplex method
Problem is
Max `Z``=````100``u_1`
subject to
` - ``10``v_1`` - ``2``v_2`` + ``100``u_1`≤`0`
` - ``8``v_1`` - ``4``v_2`` + ``80``u_1`≤`0`
` - ``12``v_1`` - ``1.5``v_2`` + ``120``u_1`≤`0`
```10``v_1`` + ``2``v_2`=`1`
and `v_1,v_2,u_1 >= 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 artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``100``u_1`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` - ``M``A_1`
subject to
` - ``10``v_1`` - ``2``v_2`` + ``100``u_1`` + ````S_1`=`0`
` - ``8``v_1`` - ``4``v_2`` + ``80``u_1`` + ````S_2`=`0`
` - ``12``v_1`` - ``1.5``v_2`` + ``120``u_1`` + ````S_3`=`0`
```10``v_1`` + ``2``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,S_1,S_2,S_3,A_1 >= 0`


Iteration-1 `C_j``0``0``100``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-10``-2``100``1``0``0``0`---
`S_2``0``0``-8``-4``80``0``1``0``0`---
`S_3``0``0``-12``-1.5``120``0``0``1``0`---
`A_1``-M``1``(10)``2``0``0``0``0``1``(1)/(10)=0.1``->`
`Z=-M` `Z_j``-10M``-2M``0``0``0``0``-M`
`C_j-Z_j``10M``uarr``2M``100``0``0``0``0`


Positive maximum `C_j-Z_j` is `10M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.1` and its row index is `4`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `10`.

Entering `=v_1`, Departing `=A_1`, Key Element `=10`

`R_4`(new)`= R_4`(old)` -: 10`
`R_4`(old) = `1``10``2``0``0``0``0`
`R_4`(new)`= R_4`(old)` -: 10``0.1``1``0.2``0``0``0``0`


`R_1`(new)`= R_1`(old) + `10 R_4`(new)
`R_1`(old) = `0``-10``-2``100``1``0``0`
`R_4`(new) = `0.1``1``0.2``0``0``0``0`
`10 xx R_4`(new) = `1``10``2``0``0``0``0`
`R_1`(new)`= R_1`(old) + `10 R_4`(new)`1``0``0``100``1``0``0`


`R_2`(new)`= R_2`(old) + `8 R_4`(new)
`R_2`(old) = `0``-8``-4``80``0``1``0`
`R_4`(new) = `0.1``1``0.2``0``0``0``0`
`8 xx R_4`(new) = `0.8``8``1.6``0``0``0``0`
`R_2`(new)`= R_2`(old) + `8 R_4`(new)`0.8``0``-2.4``80``0``1``0`


`R_3`(new)`= R_3`(old) + `12 R_4`(new)
`R_3`(old) = `0``-12``-1.5``120``0``0``1`
`R_4`(new) = `0.1``1``0.2``0``0``0``0`
`12 xx R_4`(new) = `1.2``12``2.4``0``0``0``0`
`R_3`(new)`= R_3`(old) + `12 R_4`(new)`1.2``0``0.9``120``0``0``1`


Iteration-2 `C_j``0``0``100``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`(X_B)/(u_1)`
`S_1``0``1``0``0``100``1``0``0``(1)/(100)=0.01`
`S_2``0``0.8``0``-2.4``(80)``0``1``0``(0.8)/(80)=0.01``->`
`S_3``0``1.2``0``0.9``120``0``0``1``(1.2)/(120)=0.01`
`v_1``0``0.1``1``0.2``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0`
`C_j-Z_j``0``0``100``uarr``0``0``0`


Positive maximum `C_j-Z_j` is `100` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.01` and its row index is `2`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `80`.

Entering `=u_1`, Departing `=S_2`, Key Element `=80`

`R_2`(new)`= R_2`(old)` -: 80`
`R_2`(old) = `0.8``0``-2.4``80``0``1``0`
`R_2`(new)`= R_2`(old)` -: 80``0.01``0``-0.03``1``0``0.0125``0`


`R_1`(new)`= R_1`(old) - `100 R_2`(new)
`R_1`(old) = `1``0``0``100``1``0``0`
`R_2`(new) = `0.01``0``-0.03``1``0``0.0125``0`
`100 xx R_2`(new) = `1``0``-3``100``0``1.25``0`
`R_1`(new)`= R_1`(old) - `100 R_2`(new)`0``0``3``0``1``-1.25``0`


`R_3`(new)`= R_3`(old) - `120 R_2`(new)
`R_3`(old) = `1.2``0``0.9``120``0``0``1`
`R_2`(new) = `0.01``0``-0.03``1``0``0.0125``0`
`120 xx R_2`(new) = `1.2``0``-3.6``120``0``1.5``0`
`R_3`(new)`= R_3`(old) - `120 R_2`(new)`0``0``4.5``0``0``-1.5``1`


`R_4`(new)`= R_4`(old)
`R_4`(old) = `0.1``1``0.2``0``0``0``0`
`R_4`(new)`= R_4`(old)`0.1``1``0.2``0``0``0``0`


Iteration-3 `C_j``0``0``100``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`(X_B)/(v_2)`
`S_1``0``0``0``(3)``0``1``-1.25``0``(0)/(3)=0``->`
`u_1``100``0.01``0``-0.03``1``0``0.0125``0`---
`S_3``0``0``0``4.5``0``0``-1.5``1``(0)/(4.5)=0`
`v_1``0``0.1``1``0.2``0``0``0``0``(0.1)/(0.2)=0.5`
`Z=1` `Z_j``0``-3``100``0``1.25``0`
`C_j-Z_j``0``3``uarr``0``0``-1.25``0`


Positive maximum `C_j-Z_j` is `3` and its column index is `2`. So, the entering variable is `v_2`.

Minimum ratio is `0` and its row index is `1`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `3`.

Entering `=v_2`, Departing `=S_1`, Key Element `=3`

`R_1`(new)`= R_1`(old)` -: 3`
`R_1`(old) = `0``0``3``0``1``-1.25``0`
`R_1`(new)`= R_1`(old)` -: 3``0``0``1``0``0.3333``-0.4167``0`


`R_2`(new)`= R_2`(old) + `0.03 R_1`(new)
`R_2`(old) = `0.01``0``-0.03``1``0``0.0125``0`
`R_1`(new) = `0``0``1``0``0.3333``-0.4167``0`
`0.03 xx R_1`(new) = `0``0``0.03``0``0.01``-0.0125``0`
`R_2`(new)`= R_2`(old) + `0.03 R_1`(new)`0.01``0``0``1``0.01``0``0`


`R_3`(new)`= R_3`(old) - `4.5 R_1`(new)
`R_3`(old) = `0``0``4.5``0``0``-1.5``1`
`R_1`(new) = `0``0``1``0``0.3333``-0.4167``0`
`4.5 xx R_1`(new) = `0``0``4.5``0``1.5``-1.875``0`
`R_3`(new)`= R_3`(old) - `4.5 R_1`(new)`0``0``0``0``-1.5``0.375``1`


`R_4`(new)`= R_4`(old) - `0.2 R_1`(new)
`R_4`(old) = `0.1``1``0.2``0``0``0``0`
`R_1`(new) = `0``0``1``0``0.3333``-0.4167``0`
`0.2 xx R_1`(new) = `0``0``0.2``0``0.0667``-0.0833``0`
`R_4`(new)`= R_4`(old) - `0.2 R_1`(new)`0.1``1``0``0``-0.0667``0.0833``0`


Iteration-4 `C_j``0``0``100``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`v_2``0``0``0``1``0``0.3333``-0.4167``0`
`u_1``100``0.01``0``0``1``0.01``0``0`
`S_3``0``0``0``0``0``-1.5``0.375``1`
`v_1``0``0.1``1``0``0``-0.0667``0.0833``0`
`Z=1` `Z_j``0``0``100``1``0``0`
`C_j-Z_j``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=0.1,v_2=0,u_1=0.01`

Max `Z = 1`


Solution steps by BigM method

Answer is
`v_1=0.1,v_2=0.0,u_1=0.01,`


DMU-2
Max z `= (80u1)/(8v1+4v2)`

`(100u1)/(10v1+2v2)<=1`

`(80u1)/(8v1+4v2)<=1`

`(120u1)/(12v1+1.5v2)<=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 `= 80u1`

Denominator of nonlinear ` 8v1+4v2=1`

`(100u1)-(10v1+2v2)<=0`

`(80u1)-(8v1+4v2)<=0`

`(120u1)-(12v1+1.5v2)<=0`

and `u,v>=0`

DMU-2
solution using simplex method
Problem is
Max `Z``=````80``u_1`
subject to
` - ``10``v_1`` - ``2``v_2`` + ``100``u_1`≤`0`
` - ``8``v_1`` - ``4``v_2`` + ``80``u_1`≤`0`
` - ``12``v_1`` - ``1.5``v_2`` + ``120``u_1`≤`0`
```8``v_1`` + ``4``v_2`=`1`
and `v_1,v_2,u_1 >= 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 artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``80``u_1`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` - ``M``A_1`
subject to
` - ``10``v_1`` - ``2``v_2`` + ``100``u_1`` + ````S_1`=`0`
` - ``8``v_1`` - ``4``v_2`` + ``80``u_1`` + ````S_2`=`0`
` - ``12``v_1`` - ``1.5``v_2`` + ``120``u_1`` + ````S_3`=`0`
```8``v_1`` + ``4``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,S_1,S_2,S_3,A_1 >= 0`


Iteration-1 `C_j``0``0``80``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-10``-2``100``1``0``0``0`---
`S_2``0``0``-8``-4``80``0``1``0``0`---
`S_3``0``0``-12``-1.5``120``0``0``1``0`---
`A_1``-M``1``(8)``4``0``0``0``0``1``(1)/(8)=0.125``->`
`Z=-M` `Z_j``-8M``-4M``0``0``0``0``-M`
`C_j-Z_j``8M``uarr``4M``80``0``0``0``0`


Positive maximum `C_j-Z_j` is `8M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.125` and its row index is `4`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `8`.

Entering `=v_1`, Departing `=A_1`, Key Element `=8`

`R_4`(new)`= R_4`(old)` -: 8`
`R_4`(old) = `1``8``4``0``0``0``0`
`R_4`(new)`= R_4`(old)` -: 8``0.125``1``0.5``0``0``0``0`


`R_1`(new)`= R_1`(old) + `10 R_4`(new)
`R_1`(old) = `0``-10``-2``100``1``0``0`
`R_4`(new) = `0.125``1``0.5``0``0``0``0`
`10 xx R_4`(new) = `1.25``10``5``0``0``0``0`
`R_1`(new)`= R_1`(old) + `10 R_4`(new)`1.25``0``3``100``1``0``0`


`R_2`(new)`= R_2`(old) + `8 R_4`(new)
`R_2`(old) = `0``-8``-4``80``0``1``0`
`R_4`(new) = `0.125``1``0.5``0``0``0``0`
`8 xx R_4`(new) = `1``8``4``0``0``0``0`
`R_2`(new)`= R_2`(old) + `8 R_4`(new)`1``0``0``80``0``1``0`


`R_3`(new)`= R_3`(old) + `12 R_4`(new)
`R_3`(old) = `0``-12``-1.5``120``0``0``1`
`R_4`(new) = `0.125``1``0.5``0``0``0``0`
`12 xx R_4`(new) = `1.5``12``6``0``0``0``0`
`R_3`(new)`= R_3`(old) + `12 R_4`(new)`1.5``0``4.5``120``0``0``1`


Iteration-2 `C_j``0``0``80``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`(X_B)/(u_1)`
`S_1``0``1.25``0``3``100``1``0``0``(1.25)/(100)=0.0125`
`S_2``0``1``0``0``(80)``0``1``0``(1)/(80)=0.0125``->`
`S_3``0``1.5``0``4.5``120``0``0``1``(1.5)/(120)=0.0125`
`v_1``0``0.125``1``0.5``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0`
`C_j-Z_j``0``0``80``uarr``0``0``0`


Positive maximum `C_j-Z_j` is `80` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.0125` and its row index is `2`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `80`.

Entering `=u_1`, Departing `=S_2`, Key Element `=80`

`R_2`(new)`= R_2`(old)` -: 80`
`R_2`(old) = `1``0``0``80``0``1``0`
`R_2`(new)`= R_2`(old)` -: 80``0.0125``0``0``1``0``0.0125``0`


`R_1`(new)`= R_1`(old) - `100 R_2`(new)
`R_1`(old) = `1.25``0``3``100``1``0``0`
`R_2`(new) = `0.0125``0``0``1``0``0.0125``0`
`100 xx R_2`(new) = `1.25``0``0``100``0``1.25``0`
`R_1`(new)`= R_1`(old) - `100 R_2`(new)`0``0``3``0``1``-1.25``0`


`R_3`(new)`= R_3`(old) - `120 R_2`(new)
`R_3`(old) = `1.5``0``4.5``120``0``0``1`
`R_2`(new) = `0.0125``0``0``1``0``0.0125``0`
`120 xx R_2`(new) = `1.5``0``0``120``0``1.5``0`
`R_3`(new)`= R_3`(old) - `120 R_2`(new)`0``0``4.5``0``0``-1.5``1`


`R_4`(new)`= R_4`(old)
`R_4`(old) = `0.125``1``0.5``0``0``0``0`
`R_4`(new)`= R_4`(old)`0.125``1``0.5``0``0``0``0`


Iteration-3 `C_j``0``0``80``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`S_1``0``0``0``3``0``1``-1.25``0`
`u_1``80``0.0125``0``0``1``0``0.0125``0`
`S_3``0``0``0``4.5``0``0``-1.5``1`
`v_1``0``0.125``1``0.5``0``0``0``0`
`Z=1` `Z_j``0``0``80``0``1``0`
`C_j-Z_j``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=0.125,v_2=0,u_1=0.0125`

Max `Z = 1`


Solution steps by BigM method

Answer is
`v_1=0.125,v_2=0.0,u_1=0.0125,`


DMU-3
Max z `= (120u1)/(12v1+1.5v2)`

`(100u1)/(10v1+2v2)<=1`

`(80u1)/(8v1+4v2)<=1`

`(120u1)/(12v1+1.5v2)<=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 `= 120u1`

Denominator of nonlinear ` 12v1+1.5v2=1`

`(100u1)-(10v1+2v2)<=0`

`(80u1)-(8v1+4v2)<=0`

`(120u1)-(12v1+1.5v2)<=0`

and `u,v>=0`

DMU-3
solution using simplex method
Problem is
Max `Z``=````120``u_1`
subject to
` - ``10``v_1`` - ``2``v_2`` + ``100``u_1`≤`0`
` - ``8``v_1`` - ``4``v_2`` + ``80``u_1`≤`0`
` - ``12``v_1`` - ``1.5``v_2`` + ``120``u_1`≤`0`
```12``v_1`` + ``1.5``v_2`=`1`
and `v_1,v_2,u_1 >= 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 artificial variable `A_1`

After introducing slack,artificial variables
Max `Z``=````0``v_1`` + ``0``v_2`` + ``120``u_1`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` - ``M``A_1`
subject to
` - ``10``v_1`` - ``2``v_2`` + ``100``u_1`` + ````S_1`=`0`
` - ``8``v_1`` - ``4``v_2`` + ``80``u_1`` + ````S_2`=`0`
` - ``12``v_1`` - ``1.5``v_2`` + ``120``u_1`` + ````S_3`=`0`
```12``v_1`` + ``1.5``v_2`` + ````A_1`=`1`
and `v_1,v_2,u_1,S_1,S_2,S_3,A_1 >= 0`


Iteration-1 `C_j``0``0``120``0``0``0``-M`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3``A_1`MinRatio
`(X_B)/(v_1)`
`S_1``0``0``-10``-2``100``1``0``0``0`---
`S_2``0``0``-8``-4``80``0``1``0``0`---
`S_3``0``0``-12``-1.5``120``0``0``1``0`---
`A_1``-M``1``(12)``1.5``0``0``0``0``1``(1)/(12)=0.0833``->`
`Z=-M` `Z_j``-12M``-1.5M``0``0``0``0``-M`
`C_j-Z_j``12M``uarr``1.5M``120``0``0``0``0`


Positive maximum `C_j-Z_j` is `12M` and its column index is `1`. So, the entering variable is `v_1`.

Minimum ratio is `0.0833` and its row index is `4`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `12`.

Entering `=v_1`, Departing `=A_1`, Key Element `=12`

`R_4`(new)`= R_4`(old)` -: 12`
`R_4`(old) = `1``12``1.5``0``0``0``0`
`R_4`(new)`= R_4`(old)` -: 12``0.0833``1``0.125``0``0``0``0`


`R_1`(new)`= R_1`(old) + `10 R_4`(new)
`R_1`(old) = `0``-10``-2``100``1``0``0`
`R_4`(new) = `0.0833``1``0.125``0``0``0``0`
`10 xx R_4`(new) = `0.8333``10``1.25``0``0``0``0`
`R_1`(new)`= R_1`(old) + `10 R_4`(new)`0.8333``0``-0.75``100``1``0``0`


`R_2`(new)`= R_2`(old) + `8 R_4`(new)
`R_2`(old) = `0``-8``-4``80``0``1``0`
`R_4`(new) = `0.0833``1``0.125``0``0``0``0`
`8 xx R_4`(new) = `0.6667``8``1``0``0``0``0`
`R_2`(new)`= R_2`(old) + `8 R_4`(new)`0.6667``0``-3``80``0``1``0`


`R_3`(new)`= R_3`(old) + `12 R_4`(new)
`R_3`(old) = `0``-12``-1.5``120``0``0``1`
`R_4`(new) = `0.0833``1``0.125``0``0``0``0`
`12 xx R_4`(new) = `1``12``1.5``0``0``0``0`
`R_3`(new)`= R_3`(old) + `12 R_4`(new)`1``0``0``120``0``0``1`


Iteration-2 `C_j``0``0``120``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`(X_B)/(u_1)`
`S_1``0``0.8333``0``-0.75``100``1``0``0``(0.8333)/(100)=0.0083`
`S_2``0``0.6667``0``-3``(80)``0``1``0``(0.6667)/(80)=0.0083``->`
`S_3``0``1``0``0``120``0``0``1``(1)/(120)=0.0083`
`v_1``0``0.0833``1``0.125``0``0``0``0`---
`Z=0` `Z_j``0``0``0``0``0``0`
`C_j-Z_j``0``0``120``uarr``0``0``0`


Positive maximum `C_j-Z_j` is `120` and its column index is `3`. So, the entering variable is `u_1`.

Minimum ratio is `0.0083` and its row index is `2`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `80`.

Entering `=u_1`, Departing `=S_2`, Key Element `=80`

`R_2`(new)`= R_2`(old)` -: 80`
`R_2`(old) = `0.6667``0``-3``80``0``1``0`
`R_2`(new)`= R_2`(old)` -: 80``0.0083``0``-0.0375``1``0``0.0125``0`


`R_1`(new)`= R_1`(old) - `100 R_2`(new)
`R_1`(old) = `0.8333``0``-0.75``100``1``0``0`
`R_2`(new) = `0.0083``0``-0.0375``1``0``0.0125``0`
`100 xx R_2`(new) = `0.8333``0``-3.75``100``0``1.25``0`
`R_1`(new)`= R_1`(old) - `100 R_2`(new)`0``0``3``0``1``-1.25``0`


`R_3`(new)`= R_3`(old) - `120 R_2`(new)
`R_3`(old) = `1``0``0``120``0``0``1`
`R_2`(new) = `0.0083``0``-0.0375``1``0``0.0125``0`
`120 xx R_2`(new) = `1``0``-4.5``120``0``1.5``0`
`R_3`(new)`= R_3`(old) - `120 R_2`(new)`0``0``4.5``0``0``-1.5``1`


`R_4`(new)`= R_4`(old)
`R_4`(old) = `0.0833``1``0.125``0``0``0``0`
`R_4`(new)`= R_4`(old)`0.0833``1``0.125``0``0``0``0`


Iteration-3 `C_j``0``0``120``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`(X_B)/(v_2)`
`S_1``0``0``0``(3)``0``1``-1.25``0``(0)/(3)=0``->`
`u_1``120``0.0083``0``-0.0375``1``0``0.0125``0`---
`S_3``0``0``0``4.5``0``0``-1.5``1``(0)/(4.5)=0`
`v_1``0``0.0833``1``0.125``0``0``0``0``(0.0833)/(0.125)=0.6667`
`Z=1` `Z_j``0``-4.5``120``0``1.5``0`
`C_j-Z_j``0``4.5``uarr``0``0``-1.5``0`


Positive maximum `C_j-Z_j` is `4.5` and its column index is `2`. So, the entering variable is `v_2`.

Minimum ratio is `0` and its row index is `1`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `3`.

Entering `=v_2`, Departing `=S_1`, Key Element `=3`

`R_1`(new)`= R_1`(old)` -: 3`
`R_1`(old) = `0``0``3``0``1``-1.25``0`
`R_1`(new)`= R_1`(old)` -: 3``0``0``1``0``0.3333``-0.4167``0`


`R_2`(new)`= R_2`(old) + `0.0375 R_1`(new)
`R_2`(old) = `0.0083``0``-0.0375``1``0``0.0125``0`
`R_1`(new) = `0``0``1``0``0.3333``-0.4167``0`
`0.0375 xx R_1`(new) = `0``0``0.0375``0``0.0125``-0.0156``0`
`R_2`(new)`= R_2`(old) + `0.0375 R_1`(new)`0.0083``0``0``1``0.0125``-0.0031``0`


`R_3`(new)`= R_3`(old) - `4.5 R_1`(new)
`R_3`(old) = `0``0``4.5``0``0``-1.5``1`
`R_1`(new) = `0``0``1``0``0.3333``-0.4167``0`
`4.5 xx R_1`(new) = `0``0``4.5``0``1.5``-1.875``0`
`R_3`(new)`= R_3`(old) - `4.5 R_1`(new)`0``0``0``0``-1.5``0.375``1`


`R_4`(new)`= R_4`(old) - `0.125 R_1`(new)
`R_4`(old) = `0.0833``1``0.125``0``0``0``0`
`R_1`(new) = `0``0``1``0``0.3333``-0.4167``0`
`0.125 xx R_1`(new) = `0``0``0.125``0``0.0417``-0.0521``0`
`R_4`(new)`= R_4`(old) - `0.125 R_1`(new)`0.0833``1``0``0``-0.0417``0.0521``0`


Iteration-4 `C_j``0``0``120``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`(X_B)/(S_2)`
`v_2``0``0``0``1``0``0.3333``-0.4167``0`---
`u_1``120``0.0083``0``0``1``0.0125``-0.0031``0`---
`S_3``0``0``0``0``0``-1.5``(0.375)``1``(0)/(0.375)=0``->`
`v_1``0``0.0833``1``0``0``-0.0417``0.0521``0``(0.0833)/(0.0521)=1.6`
`Z=1` `Z_j``0``0``120``1.5``-0.375``0`
`C_j-Z_j``0``0``0``-1.5``0.375``uarr``0`


Positive maximum `C_j-Z_j` is `0.375` and its column index is `5`. So, the entering variable is `S_2`.

Minimum ratio is `0` and its row index is `3`. So, the leaving basis variable is `S_3`.

`:.` The pivot element is `0.375`.

Entering `=S_2`, Departing `=S_3`, Key Element `=0.375`

`R_3`(new)`= R_3`(old)` -: 0.375`
`R_3`(old) = `0``0``0``0``-1.5``0.375``1`
`R_3`(new)`= R_3`(old)` -: 0.375``0``0``0``0``-4``1``2.6667`


`R_1`(new)`= R_1`(old) + `0.4167 R_3`(new)
`R_1`(old) = `0``0``1``0``0.3333``-0.4167``0`
`R_3`(new) = `0``0``0``0``-4``1``2.6667`
`0.4167 xx R_3`(new) = `0``0``0``0``-1.6667``0.4167``1.1111`
`R_1`(new)`= R_1`(old) + `0.4167 R_3`(new)`0``0``1``0``-1.3333``0``1.1111`


`R_2`(new)`= R_2`(old) + `0.0031 R_3`(new)
`R_2`(old) = `0.0083``0``0``1``0.0125``-0.0031``0`
`R_3`(new) = `0``0``0``0``-4``1``2.6667`
`0.0031 xx R_3`(new) = `0``0``0``0``-0.0125``0.0031``0.0083`
`R_2`(new)`= R_2`(old) + `0.0031 R_3`(new)`0.0083``0``0``1``0``0``0.0083`


`R_4`(new)`= R_4`(old) - `0.0521 R_3`(new)
`R_4`(old) = `0.0833``1``0``0``-0.0417``0.0521``0`
`R_3`(new) = `0``0``0``0``-4``1``2.6667`
`0.0521 xx R_3`(new) = `0``0``0``0``-0.2083``0.0521``0.1389`
`R_4`(new)`= R_4`(old) - `0.0521 R_3`(new)`0.0833``1``0``0``0.1667``0``-0.1389`


Iteration-5 `C_j``0``0``120``0``0``0`
`B``C_B``X_B``v_1``v_2``u_1``S_1``S_2``S_3`MinRatio
`v_2``0``0``0``1``0``-1.3333``0``1.1111`
`u_1``120``0.0083``0``0``1``0``0``0.0083`
`S_2``0``0``0``0``0``-4``1``2.6667`
`v_1``0``0.0833``1``0``0``0.1667``0``-0.1389`
`Z=1` `Z_j``0``0``120``0``0``1`
`C_j-Z_j``0``0``0``0``0``-1`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`v_1=0.0833,v_2=0,u_1=0.0083`

Max `Z = 1`


Solution steps by BigM method

Answer is
`v_1=0.0833333333333333,v_2=1.4802973661668756E-16,u_1=0.0083333333333333332,`

Final Score, weight and WeightedData table is
NoScoreRankv1v2u1(v1)(v2)(u1)v1 * (v1)v2 * (v2)u1 * (u1)`sum v_i * (v_i)`
1111021000.100.011011
21184800.12500.01251011
311121.51200.0833333300.008333331011


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
111101002201001000
21188044080800
311121201.51.501201200


Final Slack table is
NoScoreRankslack v1slack v2slack u1
111000
211000
311000


This material is intended as a summary. Use your textbook for detail explanation.
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1. Example-1
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