Given Information:
Smoothing constant = α = 0.25
Initial forecast salary = F₀ = $55,000
Actual salaries = A = $60,000, $72,000, $84,500, and $96,000
Required Information:
Forecast salaries = F = ?
Answer:
![F_(1) = \$56,250\\F_(2) =\$ 60,187.5\\F_(3) = \$66,265.6\\F_(4) = \$73,699.2\\](https://img.qammunity.org/2021/formulas/mathematics/college/hbfmpu4m3diq2wny6v4y7kwr0gf6zshnp8.png)
Explanation:
The exponential smoothing model is given by
![F_(n) = \alpha \cdot A_(n - 1) + (1 - \alpha ) F_(n - 1)](https://img.qammunity.org/2021/formulas/mathematics/college/rqd2gzjre9l89udkrlaqmx792r8idokw5a.png)
Where
is the forecast salary for nth graduate class
α is the smoothing constant
is the actual salary of n - 1 graduate class
is the forecast salary of n - 1 graduate class
For n = 1
![F_(1) = 0.25 \cdot A_0} + (1-0.25) \cdot F_(0)\\F_(1) = 0.25 \cdot 60,000} + (0.75) \cdot 55,000\\F_(1) = 56,250](https://img.qammunity.org/2021/formulas/mathematics/college/y143xh0xdd4i7qcnzexxb7ooezfkhpeg94.png)
For n = 2
![F_(2) = 0.25 \cdot A_1} + (1-0.25) \cdot F_(1)\\F_(2) = 0.25 \cdot 72,000} + (0.75) \cdot 56,250\\F_(2) = 60,187.5](https://img.qammunity.org/2021/formulas/mathematics/college/9kfiwev978w6pc3f3ccz7a4pssze3ybf1q.png)
For n = 3
![F_(3) = 0.25 \cdot A_2} + (1-0.25) \cdot F_(2)\\F_(3) = 0.25 \cdot 84,500} + (0.75) \cdot 60,187.5\\F_(3) = 66,265.625](https://img.qammunity.org/2021/formulas/mathematics/college/2dynpuk9js7y5fuy453rglwq9yl2j0f95s.png)
For n = 4
![F_(4) = 0.25 \cdot A_3} + (1-0.25) \cdot F_(3)\\F_(4) = 0.25 \cdot 96,000} + (0.75) \cdot 66,265.625\\F_(4) = 73,699.218](https://img.qammunity.org/2021/formulas/mathematics/college/ywwofuai6q8yyguiw8x60d691j5y9age2o.png)
Therefore, the foretasted starting salaries are
![F_(1) = \$56,250\\F_(2) =\$ 60,187.5\\F_(3) = \$66,265.6\\F_(4) = \$73,699.2\\](https://img.qammunity.org/2021/formulas/mathematics/college/hbfmpu4m3diq2wny6v4y7kwr0gf6zshnp8.png)