give answer in 2 step with explanation at the end of each step and final answer at the end:

Required information Consider the following problem. Maximize Z
=
-
5
x
_
(
1
)
+
5
x
_
(
2

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:

Required information Consider the following problem. Maximize Z
=
-
5
x
_
(
1
)
+
5
x
_
(
2
)
+
1
3
x
_
(
3
)
,
 ﻿subject to 
-
x
_
(
1
)
+
x
_
(
2
)
+
3
x
_
(
3
)
<
=
2
0
 
1
2
x
_
(
1
)
+
4
x
_
(
2
)
+
1
0
x
_
(
3
)
<
=
9
0
 ﻿and x
_
(
j
)
>
=
0
(
j
=
1
,
2
,
3
)
.
 ﻿If we let x
_
(
4
)
 ﻿and x
_
(
5
)
 ﻿be the slack variables for the respective constraints, the simplex method yields the following final set of equations: 
(
0
)
 ﻿Z
,
+
2
x
_
(
3
)
+
5
x
_
(
4
)
=
1
0
0
 
(
1
)
 
-
x
_
(
1
)
+
x
_
(
2
)
+
3
x
_
(
3
)
+
x
_
(
4
)
=
2
0
 
(
2
)
 
1
6
x
_
(
1
)
-
2
x
_
(
3
)
-
4
x
_
(
4
)
+
x
_
(
5
)
=
1
0
.
 ﻿Now you are to conduct sensitivity analysis by independently investigating the following change in the original model. For the following change, use the sensitivity analysis procedure to revise this set of equations 
(
in tableau form
)
 ﻿and convert it to proper form from Gaussian elimi

VIPSolutions · Accepted Answer

Let's answer this sensitivity analysis question **step by step**.

---

## **Step 1: Revise the Simplex Tableau for the Change**

**Change:** The coefficient of $ x_3 $ in the objective function changes from 13 to **7**.

### **Adjust Row 0 (Z-row) for the Change**

Original Z-row in the tableau:
$$ Z + 2x_3 + 5x_4 = 100 $$

This means the coefficients in the tableau are:
- $ x_1 = 0 $
- $ x_2 = 0 $
- $ x_3 = 2 $
- $ x_4 = 5 $
- $ x_5 = 0 $
- Right side = 100

#### **How to Update Z-row?**

The change in the objective coefficient of $ x_3 $ (from 13 to 7) means:
- The difference, $ \Delta c_3 $, is $ 7 - 13 = -6 $.
- To revise the Z-row:  
  Add $ \Delta c_3 $ times the $ x_3 $ column of the tableau to the Z-row.

From the tableau, the $ x_3 $ column for each row is:
- Z-row: 2
- $ x_2 $-row: 3
- $ x_5 $-row: -2

So, update the Z-row using:
$$ \text{New Z-row} = \text{Old Z-row} + (-6) \times \text{Row coefficients of } x_3 $$

So, the new Z-row coefficients:
- $ x_1: 0 + (-6) \times (-1) = 6 $
- $ x_2: 0 + (-6) \times 1 = -6 $
- $ x_3: 2 + (-6) \times 3 = 2 - 18 = -16 $
- $ x_4: 5 + (-6) \times 1 = 5 - 6 = -1 $
- $ x_5: 0 + (-6) \times 0 = 0 $
- Right side: $ 100 + (-6) \times 20 = 100 - 120 = -20 $

**New Z-row:**  
$ Z + 6x_1 - 6x_2 - 16x_3 - x_4 = -20 $

### **Tableau after the change:**

| Bas var | Eq No | Z | $x_1$ | $x_2$ | $x_3$ | $x_4$ | $x_5$ | Right side |
|---------|-------|---|---------|---------|---------|---------|---------|------------|
| Z       | 0     | 1 |   6     |  -6     |  -16    | -1      |  0      |   -20      |
| $x_2$ | 1     | 0 |  -1     |   1     |   3     |  1      |  0      |    20      |
| $x_5$ | 2     | 0 |  16     |   0     |  -2     | -4      |  1      |    10      |

**End of Step 1 Explanation:**  
We have revised the tableau to reflect the new coefficient of $ x_3 $ in the objective function by updating the Z-row using the sensitivity analysis procedure.

---

## **Step 2: Test the Current Basic Solution for Feasibility and Optimality**

### **Feasibility**

The basic variables are $ x_2 $ and $ x_5 $:

- $ x_2 = 20 \geq 0 $
- $ x_5 = 10 \geq 0 $

All other variables ($ x_1, x_3, x_4 $) are non-basic and set to 0.

**Conclusion:** The current basic solution is **feasible**.

---

### **Optimality**

For maximization, the solution is optimal if the coefficients of non-basic variables ($ x_1, x_3, x_4 $) in the Z-row are **≤ 0**.

From the Z-row after the change:
- $ x_1: 6 $ (**not optimal, since 6 > 0**)
- $ x_3: -16 $ (ok)
- $ x_4: -1 $ (ok)

**Conclusion:** The basic solution is **not optimal** after the change, because the coefficient of $ x_1 $ in the Z-row is positive.

---

## **Final Answer**

### **Step 1:**  
We revised the simplex tableau after the change in $ c_3 $ by updating the Z-row. The new Z-row is:  
$ Z + 6x_1 - 6x_2 - 16x_3 - x_4 = -20 $.

### **Step 2:**  
The current basic solution ($ x_2 = 20, x_5 = 10 $) is **feasible** but **not optimal** since the Z-row coefficient for $ x_1 $ is positive (6), indicating that further improvement is possible.

---

**Final Answer:**  
After changing $ c_3 $ to 7, the current solution remains feasible but is no longer optimal.

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