**Vendor Parameters**


#### **Appendix B. Production Cycle Time and Inventory Level Calculations**

#### *Appendix B.1. Cycle Time Calculation*

The objective of the research is to minimize the total cost of production, and the formulation of the cycle time of the production system is prerequisite to calculate the total cost (TC). Cycle time is taken as a decision variable in the production model, which is dependent upon the production rates of manufacturing system. The production rate of the manufacturing firm and vendor are relying on the production rate of the machines (capital units). In order to meet the customer demand and no shortages in the production system, the production rates of the manufacturing system is di fferent in the manufacturing before outsourcing, during outsourcing and after outsourcing operations are

denoted as *Pja*, *Pjb*, and *Pjc* respectively. The total production cycle time of the item is the sum of the time utilized in production, outsourcing, reworking and delivery as expressed as in Equation (A1).

$$T\_j = t\_{1j} + t\_{2j} + t\_{3j} + t\_{4j} + t\_{5j} \tag{A1}$$

where, the individual time fractions of the total cycle time are mathematically formulated as given from Equations (A2) to (A6).

$$t\_{1\dot{j}} = \frac{Q\_{\dot{j}}}{P\_{j\dot{a}}} \tag{A2}$$

$$t\_{2j} = \frac{Q\_j}{P\_{jb}}\tag{A3}$$

$$t\_{3j} = \frac{Q\_j \alpha\_j}{P\_{jb}} \tag{A4}$$

$$t\_{4j} = \frac{Q\_j}{P\_{j\bar{c}}} \tag{A5}$$

$$t\_{\mathfrak{H}j} = \frac{I\_{m\mathbf{x}\mathbf{x}}(j\mathbf{c})}{D\_j} \tag{A6}$$

To find the value of *Imax*(*jc*) from the inventory diagram, the formula of slope in the *Area*10,11,12 can be expressed as in Equation (A7).

$$
tan\theta\_4 = P\_{j\text{c}} - D\_j \Rightarrow \frac{I\_{\text{max}(j\text{c})}}{t\_{4j}}\tag{A7}$$

$$I\_{\text{max}(j\varepsilon)} = (P\_{j\varepsilon} - D\_j) \frac{Q\_j}{P\_{j\varepsilon}} = Q\_j (1 - \frac{D\_j}{P\_{j\varepsilon}}) \tag{A8}$$

By putting the value of *Imax*(*jc*), the simplified form of Equation (A6) can be expressed as in Equation (A10).

$$t\_{5j} = \frac{Q\_j(1 - \frac{D\_j}{P\_{j\epsilon}})}{D\_j} \tag{A9}$$

$$t\_{\mathbb{S}j} = Q\_j(\frac{1}{D\_j} - \frac{1}{P\_{j\mathbb{C}}}) \tag{A10}$$

After calculating all the time fractions (*<sup>t</sup>*1*j* to *<sup>t</sup>*5*j*) for the production of automobile parts, the mathematical form of the total cycle time can be given as in Equation (A11).

$$T\_j = \frac{Q\_j}{P\_{ja}} + \frac{Q\_j}{P\_{jb}} + \frac{Q\_j \alpha\_j}{P\_{jb}} + \frac{Q\_j}{P\_{jc}} + Q\_j(\frac{1}{D\_j} - \frac{1}{P\_{jc}}) \tag{A11}$$

#### *Appendix B.2. Average Inventory Costs*

The inventory is obtained by calculating the area under the curve of the production, storage, and buffer quantity in complete manufacturing process. As followed the inventory diagram given as in Figure 2, the inventories can be divided into the areas i.e., *Area*123, *Area*456, *Area*5678, *Area*10,11,12, and *Area*11,12,13. These areas are formulated separately from Equation (A12) to Equation (A26) respectively.

$$Area\_{123} = \frac{1}{2} t\_{1\circ} I\_{\text{max}(\text{ja})} \tag{A12}$$

$$
tau\Theta\_1 = \frac{I\_{\max(ja)}}{t\_{1j}} P\_{ja} \times t\_{1j} = I\_{\max(ja)} I\_{\max(ja)} = Q\_j \tag{A13}
$$

By putting *Imax*(*ja*), Equation (A12) can be written as in Equation (A14).

$$Area\_{123} = \frac{Q\_j^2}{2P\_{j\bar{\mu}}} \tag{A14}$$

Similarly, the area of triangle *456* can be given as in Equation (A15)

$$Area\_{456} = \frac{1}{2} I\_{\max(\text{jib})} t\_{2\text{j}} \tag{A15}$$

The maximum inventory of the vendor *Imax*(*jbb*) is calculated by the slope formula of *Area*456 (Equation (A16)), which is then simplified to express in the form of Equation (A17)

$$
tan\Theta\_2 = \frac{I\_{\max(jlb)}}{t\_{2j}}\tag{A16}
$$

$$P\_{jb}(1 - \alpha\_{j}) \times \frac{Q\_{j}}{P\_{jb}} = I\_{\max(jbb)} I\_{\max(jbb)} = Q\_{j}(1 - \alpha\_{j}) \tag{A17}$$

After substituting, the *Area*456 can be written as in Equation (A18)

$$Area\_{456} = \frac{1}{2} Q\_j (1 - \alpha\_j) \frac{Q\_j}{P\_{jb}} Area\_{456} = \frac{Q\_j^2 (1 - \alpha\_j)}{2P\_{jb}} \tag{A18}$$

The simplified form of area under the curve of the rectangular region *5678* is formulated and expressed as in Equation (A19)

$$Area\_{5678} = I\_{\max(\text{jib})} t\_{\text{j}} = Q\_{\dot{f}} (1 - \alpha\_{\dot{f}}) \frac{Q\_{\dot{f}} \alpha\_{\dot{f}}}{P\_{\dot{j}\dot{b}}} = \frac{\alpha\_{\dot{f}} (1 - \alpha\_{\dot{f}}) Q\_{\dot{f}}^2}{P\_{\dot{j}\dot{b}}} \tag{A19}$$

The triangular area of the inventory storage *Area*689 by reworking operations during outsourcing operation is expressed as in Equation (A20).

$$Area\_{689} = \frac{1}{2} I\_{\max(j\text{ha})} t\_{3j} \tag{A20}$$

where, to find the maximum inventory *Imax*(*jba*) of the reworking operation in the production system, the slope formula is given as in Equation (A21),

$$
tan\theta\_3 = \frac{I\_{\text{max}(j\text{ha})}}{t\_{3j}} P\_{j\text{b}}(1 - a\_j \beta\_j) \frac{Q\_j a\_j}{P\_{j\text{b}}} = I\_{\text{max}(j\text{ha})} I\_{\text{max}(j\text{ha})} = a\_j Q\_j (1 - a\_j \beta\_j) \tag{A21}
$$

By substitution, Equation (A20) can be written in the form of Equation (A22).

$$Area\_{689} = \frac{1}{2} \frac{Q\_{\dot{\jmath}} \alpha\_{\dot{\jmath}}}{P\_{\dot{\jmath}b}} \alpha\_{\dot{\jmath}} Q\_{\dot{\jmath}} (1 - \alpha\_{\dot{\jmath}} \beta\_{\dot{\jmath}}) = \frac{\alpha\_{\dot{\jmath}}^2 Q\_{\dot{\jmath}}^2 (1 - \alpha\_{\dot{\jmath}} \beta\_{\dot{\jmath}})}{2P\_{\dot{\jmath}b}} \tag{A22}$$

Likewise, the *Area*10,11,12 and *Area*11,12,13 can be formulated and expressed in simplified form as in Equations (A25) and (A26) respectively, where the maximum inventory is *Imax*(*jc*).

$$Area\_{10,11,12} = \frac{1}{2} I\_{\max(jc)} t\_{4j} \tag{A23}$$

$$
tan\theta\_4 = P\_{j\text{c}} - D\_j \Rightarrow \frac{I\_{\text{max}(j\text{c})}}{t\_{4j}} I\_{\text{max}(j\text{c})} = (P\_{j\text{c}} - D\_j) \frac{Q\_j}{P\_{j\text{c}}} \tag{A24}
$$

$$Area\_{10,11,12} = \frac{1}{2} Q\_j (1 - \frac{D\_j}{P\_{j\text{c}}}) \frac{Q\_j}{P\_{j\text{c}}} \tag{A25}$$

$$Area\_{11,12,13} = \frac{1}{2} I\_{\max(\text{jc})} t\_{5\text{j}} = \frac{1}{2} \frac{I\_{\max(\text{jc})}^2}{D\_j} = \frac{Q\_j^2}{2D\_j} (1 - \frac{D\_j}{P\_{j\text{c}}})^2 \tag{A26}$$
