Integer Linear Programming - Binary (0-1) Variables 1, Fixed Cost

1. Formula
   Observation

   $X_i = 1$ if project i is selected, 0 otherwise

$X_i$

Binary decision variable indicating whether project i is selected

$i \in$ \{1, 2, 3, 4\}

1. Formula
   Observation

   Maximize $217X_1 + 125X_2 + 88X_3 + 109X_4$

$217X_1 + 125X_2 + 88X_3 + 109X_4$

Total net return from selected projects

$X_i \in$ \{0, 1\}

1. Formula
   Observation

   $58X_1 + 44X_2 + 26X_3 + 23X_4 \le 120$

$58X_1 + 44X_2 + 26X_3 + 23X_4 \le 120$

January budget constraint

$X_i \in$ \{0, 1\}

1. Formula
   Observation

   $25X_1 + 29X_2 + 13X_3 + 17X_4 \le 80$

$25X_1 + 29X_2 + 13X_3 + 17X_4 \le 80$

February budget constraint

$X_i \in$ \{0, 1\}

1. Formula
   Observation

   $43X_1 + 25X_2 + 23X_3 + 29X_4 \le 95$

$43X_1 + 25X_2 + 23X_3 + 29X_4 \le 95$

March budget constraint

$X_i \in$ \{0, 1\}

1. Formula
   Observation

   $X_1=1$, $X_2=0$, $X_3=1$, $X_4=1$

$X_1=1$, $X_2=0$, $X_3=1$, $X_4=1$

Optimal solution to the project selection problem

$X_i \in$ \{0, 1\}

1. Formula
   Observation

   Optimal Return = 414

414

Maximum net return from the optimal solution

$X_i \in$ \{0, 1\}

1. Formula
   Observation

   On-screen definition: $X_i$ = number of units produced on machine $i (i = 1, 2, 3)$.

2. Audio
   Observation

   The narrator says, "So for the units produced we write $X_i$ equals number of units produced on machine i."

$X_i$

Number of units produced on machine i.

Continuous nonnegative production quantity; specific instances are $X_1$, $X_2$, $X_3$.

1. Formula
   Observation

   On-screen definition: $Y_i = 1$ if machine i is used, 0 otherwise ($i = 1$, 2, 3).

2. Audio
   Observation

   The narrator says, "For machine one we can write $Y_1$ equals one if machine one is used, zero otherwise."

$Y_i$

Binary indicator showing whether machine i is used.

Binary variable taking values in {0,1}; specific instances are $Y_1$, $Y_2$, $Y_3$.

1. Formula
   Observation

   On-screen objective: Min $2.39X_1 + 1.99X_2 + 2.99X_3 + 300Y_1 + 250Y_2 + 400Y_3$.

2. Audio
   Observation

   The narrator reads the same expression and explains that variable costs are multiplied by units produced and fixed costs by the corresponding 0-1 variables.

$2.39X_1 + 1.99X_2 + 2.99X_3 + 300Y_1 + 250Y_2 + 400Y_3$

Total cost to be minimized, combining variable production costs and fixed setup costs.

Linear objective function in the decision variables $X_i$ and $Y_i$.

1. Formula
   Observation

   On-screen constraint: $X_1 + X_2 + X_3 = 1000$.

2. Audio
   Observation

   The narrator says, "we have an order here to supply 1000 units. So we write $X_1$ plus $X_2$ plus $X_3$ equals 1000."

$X_1 + X_2 + X_3 = 1000$

Demand satisfaction constraint requiring exactly 1000 units to be produced across all machines.

Equality constraint on total production.

1. Formula
   Observation

   On-screen constraints: $X_1$ <= 400, $X_2$ <= 550, $X_3$ <= 600.

2. Audio
   Observation

   The narrator says, "normally we just write $X_1$ less than or equal to 400, $X_2$ less than or equal to 550, and $X_3$ less than or equal to 600."

$X_i$ <= capacity_i

Machine capacity limits on production quantities before linking to binary usage variables.

Inequality constraints on individual machine outputs.

1. Audio
   Observation

   Binary variables are employed when there's a yes or no situation, that is to indicate whether a selection is made or not.

Binary variables are used to model situations where a decision is either 'yes' or 'no', such as selecting a project or not.

1. Used in integer linear programming

2. Applicable to selection problems

1. Audio
   Observation

   For example, suppose we have four different projects to consider. We can either select a project or not select it.

The project selection problem involves deciding which projects to undertake based on binary variables and constraints.

1. Projects have associated costs and returns

2. Budget constraints apply

1. Binary Variable Definition

1. Audio
   Observation

   Let's now model a fixed cost problem. Suppose a small company receives an order to supply 1000 units of a product.

The fixed cost problem involves minimizing total costs while meeting production requirements, considering both fixed and variable costs.

1. Production capacity constraints

2. Fixed and variable costs per machine

1. Binary Variable Definition

1. Formula
   Observation

   On-screen: $Y_i = 1$ if machine i is used, 0 otherwise ($i = 1$, 2, 3).

2. Audio
   Observation

   Narrator explains that because fixed costs are incurred only if a machine is used, another set of decision variables is defined.

A binary variable $Y_i$ is introduced to indicate whether machine i is used at all. It takes value 1 when the machine is active and 0 when inactive. This allows fixed costs to be included conditionally in the objective function.

1. Used when fixed costs are associated with activating a resource.

2. Applies per machine or facility in production planning models.

1. Formula
   Observation

   On-screen: Min $2.39X_1 + 1.99X_2 + 2.99X_3 + 300Y_1 + 250Y_2 + 400Y_3$.

2. Audio
   Observation

   Narrator states this is the objective function to minimize total costs.

The objective minimizes total cost as the sum of variable production costs and fixed setup costs. Variable cost coefficients multiply the number of units produced on each machine, while fixed cost coefficients multiply the corresponding binary usage indicators.

1. Requires both continuous production variables $X_i$ and binary activation variables $Y_i$.

2. Fixed costs are only counted when $Y_i = 1$.

1. Binary (0-1) Decision Variable

1. Formula
   Observation

   On-screen: $X_1 + X_2 + X_3 = 1000$.

2. Audio
   Observation

   Narrator explains equality is used because the order must be met exactly.

The total production across all machines must equal the customer order of 1000 units. An equality constraint is used rather than an inequality because the problem requires meeting the exact demand.

1. Applies when the order quantity must be fulfilled exactly.

2. Assumes no overproduction or underproduction is allowed.

1. Formula
   Observation

   On-screen: $X_1$ <= 400, $X_2$ <= 550, $X_3$ <= 600.

2. Audio
   Observation

   Narrator introduces these as normal capacity constraints before discussing their interaction with fixed costs.

Each machine has a maximum production capacity that limits how many units can be produced on it. These constraints ensure that no machine exceeds its physical or operational limit.

1. Applies to continuous production variables $X_i$.

2. Does not yet account for whether the machine is actually activated.

1. Formula
   Observation

   The slide states $Y_i = 1$ if machine i is used, 0 otherwise.

2. Formula
   Observation

   Later the slide adds $Y_i = 0-1$.

3. Audio
   Observation

   The narrator explains the two cases $Y_1=1$ and $Y_1=0$.

The model uses a separate 0-1 variable for each machine to record whether that machine is activated. The value 1 means the machine is used, and 0 means it is not used.

1. One indicator variable is associated with each machine.

2. The variable is restricted to binary values.

1. Formula
   Observation

   The displayed constraints are $X_1 \le 400Y_1$, $X_2 \le 550Y_2$, $X_3 \le 600Y_3$.

2. Audio
   Observation

   The narrator says that when $Y_1$ equals 1, up to 400 can be produced on machine 1, and when $Y_1$ equals 0, zero units should be produced.

3. Animation
   Observation

   The term $Y_1$ is highlighted, then substituted by 1 and 0 to show the two resulting bounds.

Each production quantity is bounded above by its machine capacity multiplied by the corresponding binary indicator. This single inequality simultaneously enforces the capacity limit when the machine is used and forces production to zero when the machine is not used.

1. $C_i$ is the capacity of machine i.

2. $Y_i$ is binary.

3. $X_i$ is nonnegative.

1. Binary indicator variable for machine usage

1. Audio
   Observation

   The narrator says, "We can then move the variables to the left side of the constraints".

2. Formula
   Observation

   The constraints are rewritten as $X_1 - 400Y_1 \le 0$, $X_2 - 550Y_2 \le 0$, $X_3 - 600Y_3 \le 0$.

The linking inequalities are algebraically rearranged so that all decision variables appear on the left-hand side and the right-hand side becomes 0. This puts the model into a standard linear-programming display form.

1. Equivalent to $X_i \le C$_iY_i by subtracting C_iY_i from both sides.

1. Linking production to machine activation

1. Formula
   Observation

   The slide adds $X_i \ge 0$ and $Y_i = 0-1$.

2. Audio
   Observation

   The narrator says to state that the $X_i$'s should be nonnegative and the $Y_i$'s binary.

The production variables are constrained to be nonnegative, while the activation variables are constrained to be binary 0-1 variables.

1. Applies to every machine index i in the model.

1. Binary indicator variable for machine usage

1. Formula
   Observation

   The slide displays $X_1=0$, $X_2=550$, $X_3=450$ and $Y_1=0$, $Y_2=1$, $Y_3=1$.

2. Audio
   Observation

   The narrator reads the same solution values and says only machines 2 and 3 should be used.

After solving the integer linear program, the model returns production amounts and machine-use indicators. The reported solution activates machines 2 and 3, leaves machine 1 unused, and allocates 550 units to machine 2 and 450 units to machine 3.

1. This is the solution shown for the specific numerical instance on the slide.

2. The video does not show the solver steps that produced it.

1. Linking production to machine activation
2. Domain restrictions for the decision variables

1. Audio
   Observation

   Narrator says: "whenever we have fixed costs associated with minimum requirements or maximum capacities, we usually multiply the capacity by the corresponding binary variable."

2. Formula
   Observation

   On-screen shows standard capacity constraints being replaced conceptually by linked versions involving $Y_i$.

When fixed costs are present, capacity constraints must be linked to binary activation variables so that production is only allowed if the machine is used.

1. There are fixed costs associated with using a machine.

2. Production variables $X_i$ represent units produced on each machine.

3. Binary variables $Y_i$ indicate machine usage.

For each machine i in {1,2,3}.

1. Audio
   Observation

   The narrator explicitly interprets the constraint for machine 1 in the two binary cases.

2. Formula
   Observation

   The displayed inequality is $X_1 \le 400Y_1$.

3. Animation
   Observation

   The substitution of $Y_1=1$ and $Y_1=0$ is shown visually.

For machine 1, the constraint $X_1 \le 400Y_1$ allows production up to 400 when $Y_1=1$ and forces $X_1=0$ when $Y_1=0$, assuming $X_1 \ge 0$.

1. $Y_1 \in$ \{0,1\}.

2. $X_1 \ge 0$.

3. The capacity constant is 400.

Case analysis over the binary value of $Y_1$.

1. Formula
   Observation

   The displayed solution has $Y_1=0$, $Y_2=1$, $Y_3=1$.

2. Audio
   Observation

   The narrator says this shows that only machines 2 and 3 should be used.

The reported solution uses machines 2 and 3 and does not use machine 1.

1. The displayed solution values are taken as the result of solving the model.

Specific to the numerical example shown.

1. Audio
   Observation

   In this case, our objective is to maximize return, which is $217X_1 + 125X_2 + 88X_3 + 109X_4$.

1. Expression

   $$\text{Maximize } 217X_1 + 125X_2 + 88X_3 + 109X_4$$
   Explanation

   Objective function to maximize net return

   Justification

   Based on the net returns of each project

   Shown in the video
2. Expression

   $$58X_1 + 44X_2 + 26X_3 + 23X_4 \le 120$$
   Explanation

   January budget constraint

   Justification

   Total outlays must not exceed available funds

   Shown in the video
3. Expression

   $$25X_1 + 29X_2 + 13X_3 + 17X_4 \le 80$$
   Explanation

   February budget constraint

   Justification

   Total outlays must not exceed available funds

   Shown in the video
4. Expression

   $$43X_1 + 25X_2 + 23X_3 + 29X_4 \le 95$$
   Explanation

   March budget constraint

   Justification

   Total outlays must not exceed available funds

   Shown in the video
5. Expression

   $$X_i \in \{0, 1\}$$
   Explanation

   Binary constraint

   Justification

   Each project is either selected or not

   Shown in the video

The optimal solution is $X_1=1$, $X_2=0$, $X_3=1$, $X_4=1$ with a maximum net return of 414.

1. Formula
   Observation

   On-screen objective function appears term by term with underlines labeling 'variable costs' and 'fixed costs'.

2. Audio
   Observation

   Narrator explains multiplication of variable costs by units and fixed costs by 0-1 variables.

1. Expression

   $$\text{Variable cost part: } 2.39X_1 + 1.99X_2 + 2.99X_3$$
   Explanation

   Multiply each machine's unit variable cost by the number of units produced on that machine.

   Justification

   Definition of variable cost as cost per unit times quantity.

   Shown in the video
2. Expression

   $$\text{Fixed cost part: } 300Y_1 + 250Y_2 + 400Y_3$$
   Explanation

   Multiply each machine's fixed cost by its binary usage indicator.

   Justification

   Fixed cost is incurred only if the machine is activated ($Y_i = 1$).

   Shown in the video
3. Expression

   $$\min\; (2.39X_1 + 1.99X_2 + 2.99X_3) + (300Y_1 + 250Y_2 + 400Y_3)$$
   Explanation

   Sum the variable and fixed cost components to form the total cost to minimize.

   Justification

   Total cost equals variable plus fixed costs.

   Shown in the video

The objective function correctly combines conditional fixed costs and proportional variable costs using binary and continuous variables.

1. Audio
   Observation

   Narrator contrasts simple capacity constraints with the need to link them to binary variables due to fixed costs.

2. Formula
   Observation

   Standard constraints $X_i$ <= capacity_i are shown before the explanation of modification.

1. Final modified constraints not fully displayed before clip ends.

1. Expression

   $$X_1 \leq 400,\; X_2 \leq 550,\; X_3 \leq 600$$
   Explanation

   Start with basic capacity limits assuming machines can produce up to their maximum.

   Justification

   Physical capacity of each machine.

   Shown in the video
2. Expression

   $$\text{But if } Y_i = 0, \text{ then } X_i \text{ must be } 0$$
   Explanation

   Recognize that unused machines should not produce anything, even though the simple constraint allows it.

   Justification

   Fixed cost implies activation decision; inactive machine cannot generate output.

   Derived from the video
3. Expression

   $$X_i \leq C_i Y_i \quad \text{for } i=1,2,3$$
   Explanation

   Multiply each capacity bound by the corresponding binary variable to enforce logical linkage.

   Justification

   If $Y_i=0$, RHS becomes 0 forcing $X_i=0$; if $Y_i=1$, RHS restores original capacity $C_i$.

   Derived from the video

Capacity constraints must be reformulated as $X_i \le C_i Y_i$ to respect the logical dependency between production and machine activation.

1. Formula
   Observation

   The constraint $X_1 \le 400Y_1$ is shown on screen.

2. Animation
   Observation

   $Y_1$ is replaced first by 1 and then by 0 in the highlighted expression.

3. Audio
   Observation

   The narrator explains the meaning of each case.

1. Expression

   $$X_1 \le 400Y_1$$
   Explanation

   Start from the linking constraint for machine 1.

   Justification

   Displayed formula on the slide.

   Shown in the video
2. Expression

   $$Y_1 = 1 \Rightarrow X_1 \le 400(1)=400$$
   Explanation

   If machine 1 is used, the bound becomes the full capacity 400.

   Justification

   Substitute the binary value $Y_1=1$ into the inequality.

   Shown in the video
3. Expression

   $$Y_1 = 0 \Rightarrow X_1 \le 400(0)=0$$
   Explanation

   If machine 1 is not used, the bound becomes 0.

   Justification

   Substitute the binary value $Y_1=0$ into the inequality.

   Shown in the video
4. Expression

   $$X_1 \ge 0 \text{ and } X_1 \le 0 \Rightarrow X_1=0$$
   Explanation

   Together with nonnegativity, the second case forces production on machine 1 to be exactly zero.

   Justification

   Combination of the displayed domain restriction $X_i \ge 0$ with the substituted inequality.

   Derived from the video

The single inequality $X_1 \le 400Y_1$ encodes both the capacity limit when the machine is active and the shutdown condition when it is inactive.

1. Formula
   Observation

   The constraints change from $X_i \le C$_iY_i to $X_i - C$_iY_i $\le 0$.

2. Audio
   Observation

   The narrator says to move the variables to the left side of the constraints.

1. Expression

   $$X_1 \le 400Y_1,\; X_2 \le 550Y_2,\; X_3 \le 600Y_3$$
   Explanation

   Begin with the capacity-linking inequalities in product form.

   Justification

   Displayed before the rearrangement.

   Shown in the video
2. Expression

   $$X_1 - 400Y_1 \le 0,\; X_2 - 550Y_2 \le 0,\; X_3 - 600Y_3 \le 0$$
   Explanation

   Subtract the right-hand term from both sides of each inequality.

   Justification

   Elementary algebra preserves the inequality direction when moving a term across.

   Shown in the video

The model is rewritten in a standard form with all decision variables on the left and 0 on the right.

1. Audio
   Observation

   For example, suppose we have four different projects to consider.

Select projects to maximize net return under budget constraints.

1. Four projects with associated costs and returns

2. Monthly budget constraints

Maximize net return

1. Expression

   $$\text{Maximize } 217X_1 + 125X_2 + 88X_3 + 109X_4$$
   Explanation

   Objective function

   Justification

   Based on net returns

   Shown in the video
2. Expression

   $$58X_1 + 44X_2 + 26X_3 + 23X_4 \le 120$$
   Explanation

   January constraint

   Justification

   Budget limit

   Shown in the video
3. Expression

   $$25X_1 + 29X_2 + 13X_3 + 17X_4 \le 80$$
   Explanation

   February constraint

   Justification

   Budget limit

   Shown in the video
4. Expression

   $$43X_1 + 25X_2 + 23X_3 + 29X_4 \le 95$$
   Explanation

   March constraint

   Justification

   Budget limit

   Shown in the video
5. Expression

   $$X_i \in \{0, 1\}$$
   Explanation

   Binary constraint

   Justification

   Selection decision

   Shown in the video

$X_1=1$, $X_2=0$, $X_3=1$, $X_4=1$ with a maximum net return of 414.

Solution satisfies all constraints and maximizes the objective function.

1. Diagram
   Observation

   Table lists three machines with variable costs, fixed costs, and capacities.

2. Audio
   Observation

   Narrator walks through defining variables, objective, and constraints for this specific case.

A company receives an order to supply 1000 units and must decide how much to produce on each of three machines while minimizing total cost including fixed setup costs.

1. Machine 1: variable cost 2.39/unit, fixed cost 300, capacity 400

2. Machine 2: variable cost 1.99/unit, fixed cost 250, capacity 550

3. Machine 3: variable cost 2.99/unit, fixed cost 400, capacity 600

4. Order requirement: exactly 1000 units

Formulate an integer linear program to minimize total cost subject to demand and capacity constraints.

1. Expression

   $$X_i = \text{units produced on machine } i$$
   Explanation

   Define continuous production variables for each machine.

   Justification

   Needed to model variable costs and capacity usage.

   Shown in the video
2. Expression

   $$Y_i = \begin{cases}1 & \text{if machine } i \text{ is used}\\0 & \text{otherwise}\end{cases}$$
   Explanation

   Define binary activation variables for each machine.

   Justification

   Required to conditionally include fixed costs.

   Shown in the video
3. Expression

   $$\min\; 2.39X_1 + 1.99X_2 + 2.99X_3 + 300Y_1 + 250Y_2 + 400Y_3$$
   Explanation

   Write the total cost objective combining variable and fixed components.

   Justification

   Cost structure given in table and explained verbally.

   Shown in the video
4. Expression

   $$X_1 + X_2 + X_3 = 1000$$
   Explanation

   Impose exact demand fulfillment.

   Justification

   Customer order requires precisely 1000 units.

   Shown in the video
5. Expression

   $$X_i \leq C_i Y_i \quad (i=1,2,3)$$
   Explanation

   Link production capacity to machine activation status.

   Justification

   Prevents production on inactive machines while respecting fixed cost logic.

   Derived from the video

The complete formulation includes the objective function, demand equality, and linked capacity constraints $X_i \le C_i Y_i$ with $Y_i \in {0,1}$.

Check that if $Y_i=0$ then $X_i=0$, and if $Y_i=1$ then $X_i\le C_i$; also verify total production sums to 1000.

1. Formula
   Observation

   The slide presents a complete numerical production model with three machines, an order of 1000 units, costs, capacities, and a displayed solution.

2. Audio
   Observation

   The narrator explains the linking constraints and then reports the solution values.

1. The video does not show the solver algorithm or intermediate branch-and-bound steps.

A company must supply 1000 units using three machines, each with a variable cost per unit, a fixed cost if used, and a capacity limit. Formulate and solve the resulting binary integer linear program.

1. Order requirement: 1000 units total.

2. Machine 1: variable cost 2.39, fixed cost 300, capacity 400.

3. Machine 2: variable cost 1.99, fixed cost 250, capacity 550.

4. Machine 3: variable cost 2.99, fixed cost 400, capacity 600.

5. $X_i$ is production on machine i.

6. $Y_i$ is 1 if machine i is used and 0 otherwise.

Minimize total cost subject to meeting demand, respecting capacities, and enforcing the fixed-cost logic with binary variables.

1. Expression

   $$\text{Min } 2.39X_1 + 1.99X_2 + 2.99X_3 + 300Y_1 + 250Y_2 + 400Y_3$$
   Explanation

   Write the objective as total variable cost plus total fixed cost.

   Justification

   Directly read from the displayed model.

   Shown in the video
2. Expression

   $$X_1 + X_2 + X_3 = 1000$$
   Explanation

   Impose the demand constraint that total production equals 1000 units.

   Justification

   Displayed equality constraint.

   Shown in the video
3. Expression

   $$X_1 \le 400Y_1,\; X_2 \le 550Y_2,\; X_3 \le 600Y_3$$
   Explanation

   Link each production amount to its machine-use indicator and capacity.

   Justification

   Displayed constraints and narrated interpretation.

   Shown in the video
4. Expression

   $$X_i \ge 0,\; Y_i \in \{0,1\}$$
   Explanation

   State the variable domains.

   Justification

   Added on the slide after the rearranged constraints.

   Shown in the video
5. Expression

   $$X_1=0,\; X_2=550,\; X_3=450;\quad Y_1=0,\; Y_2=1,\; Y_3=1$$
   Explanation

   Report the solution shown after solving the model.

   Justification

   Displayed solution values and narrated interpretation.

   Shown in the video

Use machines 2 and 3 only; produce 550 units on machine 2 and 450 units on machine 3, with machine 1 unused.

The displayed solution satisfies the demand equation because $0+550+450=1000$, and it respects the capacities because $550\le550$ and $450\le600$ while machine 1 is shut down.

1. Diagram
   Observation

   Table showing project costs and returns

1. Table with project data

1. Display of costs and returns for each project

1. Structure of the table

The table provides the necessary data to formulate the optimization problem.

1. Diagram
   Observation

   Persistent table showing Machine 1, 2, 3 with rows for Variable Cost/unit, Fixed Cost, and Capacity.

1. Machine columns

2. Variable cost row

3. Fixed cost row

4. Capacity row

1. Values remain static throughout the clip

1. Numerical data does not change during the segment

Provides all parameters needed to construct the optimization model.

1. Animation
   Observation

   Text definitions for $X_i$ and $Y_i$ appear sequentially below the table.

1. $X_i$ definition line

2. $Y_i$ definition line

1. Definitions fade in one after another

1. Underlying table remains visible

Visually separates the two types of decision variables required for the model.

1. Animation
   Observation

   Objective function terms are underlined and labeled as 'variable costs' and 'fixed costs'; circles highlight $Y_1=1$ and $Y_1=0$ cases.

1. Objective expression

2. Underlines

3. Circular annotations around $Y_1$ terms

1. Annotations appear progressively to emphasize structure

1. Mathematical expression itself remains unchanged

Clarifies how each part of the objective corresponds to different cost types and activation states.

1. Animation
   Observation

   Demand constraint appears first, followed by capacity constraints listed vertically beside a red separator line.

1. Demand equation

2. Capacity inequalities

3. Red vertical divider

1. Constraints appear in sequence from top to bottom

1. No numerical values change once displayed

Organizes the constraint section visually to distinguish global demand from individual machine limits.

1. Diagram
   Observation

   A static slide shows the title, a three-machine cost/capacity table, variable definitions, the objective, and the initial constraints.

2. Animation
   Observation

   The term $Y_1$ in $X_1 \le 400Y_1$ is highlighted, then replaced by (1) and later by (0).

1. Title text about supplying 1000 units.

2. Table with Machine 1, Machine 2, Machine 3.

3. Definitions of $X_i$ and $Y_i$.

4. Objective function.

5. Constraint $X_1 \le 400Y_1$.

1. $Y_1$ is highlighted.

2. $Y_1$ is temporarily replaced by 1 to show the active-machine case.

3. $Y_1$ is then replaced by 0 to show the inactive-machine case.

1. The rest of the model remains visible while the substitution is demonstrated.

2. The capacity constant 400 stays fixed during the case demonstration.

The animation visually demonstrates how one binary variable switches a capacity constraint between a positive upper bound and zero.

1. Animation
   Observation

   The three linking constraints are rewritten so that the Y terms move to the left side.

2. Formula
   Observation

   The slide then adds $X_i \ge 0$ and $Y_i = 0-1$ on the right.

1. Rearranged inequalities $X_i - C$_iY_i $\le 0$.

2. Domain labels $X_i \ge 0$ and $Y_i = 0-1$.

1. Each constraint changes from product-on-right form to all-variables-on-left form.

2. Variable-domain statements appear after the rearrangement.

1. The objective function remains unchanged.

2. The demand equation $X_1+X_2+X_3=1000$ remains unchanged.

The slide converts the logical capacity links into a more standard mathematical-programming layout and makes the variable types explicit.

1. Animation
   Observation

   Solution values appear sequentially at the top of the slide.

2. Diagram
   Observation

   A red box is drawn around the columns for machines 2 and 3 in the solution area.

1. $X_1=0$, $X_2=550$, $X_3=450$.

2. $Y_1=0$, $Y_2=1$, $Y_3=1$.

3. Red rectangular highlight around machines 2 and 3.

1. The X-values appear first, then the Y-values.

2. A red box emphasizes the solution columns for machines 2 and 3.

1. The underlying model remains visible below the solution.

2. The numerical values themselves do not change once displayed.

The visual emphasis communicates which machines are selected in the optimal plan.

1. Animation
   Observation

   The solution disappears and a new heading 'Next Video' appears.

2. Caption evidence
   Observation

   Four topic lines appear one after another: multiple choice constraints, mutually exclusive constraints, co-requisite constraints, conditional constraints.

3. Audio
   Observation

   The narrator says the next video will formulate relational constraints using binary variables, including those four types.

1. Heading 'Next Video'.

2. List of four constraint categories.

1. The current example is cleared from the top portion of the slide.

2. The four upcoming topics are revealed sequentially.

1. The lower part of the slide still shows the completed fixed-cost model.

The ending situates fixed-cost linking constraints within a broader family of binary-variable modeling techniques.

1. Audio
   Observation

   Narrator warns that writing only $X_i$ ≤ capacity_i ignores the fact that unused machines shouldn't produce.

2. Formula
   Observation

   Standard capacity constraints shown before correction discussion.

It is sufficient to write $X_i$ ≤ capacity_i without considering whether the machine is activated.

Without linking to $Y_i$, the model could allow production on a machine whose fixed cost was not paid ($Y_i=0$), violating logical consistency.

1. Audio
   Observation

   The narrator explains both cases $Y_1=1$ and $Y_1=0$ for the same inequality.

2. Formula
   Observation

   The constraint $X_1 \le 400Y_1$ contains both the capacity constant and the binary variable.

One might think $Y_i$ only records whether a machine is used, without affecting the production bound.

In this model $Y_i$ also scales the capacity limit, so setting $Y_i=0$ actively forces $X_i=0$ rather than just labeling the machine as unused.

1. Formula
   Observation

   The fixed cost appears as $300Y_1 + 250Y_2 + 400Y_3$ in the objective.

2. Formula
   Observation

   The linking constraints $X_i \le C$_iY_i are added separately.

One might think adding fixed-cost terms to the objective is enough to model machine usage correctly.

The example shows that the binary variables must also appear in linking constraints; otherwise the model would not force production to zero when a machine is not used or enforce the capacity cap in the intended way.

1. Audio
   Observation

   Binary variables are employed when there's a yes or no situation.

Binary variables are used to model the project selection problem.

1. Audio
   Observation

   Let's now model a fixed cost problem.

The fixed cost problem introduces additional considerations like fixed and variable costs.

1. Audio
   Observation

   Narrator connects definition of $Y_i$ directly to inclusion of fixed costs in objective.

2. Formula
   Observation

   $Y_i$ terms appear in objective function immediately after their definition.

Binary variables enable conditional inclusion of fixed costs in the objective function.

1. Audio
   Observation

   Narrator transitions from basic capacity constraints to the need for modification due to fixed costs.

2. Formula
   Observation

   Basic constraints shown before conceptual adjustment.

Simple capacity bounds fail to capture activation logic; they must be contrasted with linked constraints $X_i \le C_i Y_i$.

1. Formula
   Observation

   Demand constraint uses $X_1$, $X_2$, $X_3$ defined earlier.

2. Audio
   Observation

   Narrator refers back to unit production variables when stating the order requirement.

Definition of production variables is necessary before formulating the aggregate demand constraint.

1. Formula
   Observation

   The definition of $Y_i$ appears immediately above the constraints that multiply capacities by $Y_i$.

2. Audio
   Observation

   The narrator interprets the constraint using the two values of $Y_1$.

The linking constraint only makes sense after introducing the binary indicator that distinguishes used from unused machines.

1. Formula
   Observation

   The slide rewrites $X_i \le C$_iY_i as $X_i - C$_iY_i $\le 0$.

2. Audio
   Observation

   The narrator explicitly says to move the variables to the left side.

The rearranged inequalities are algebraically equivalent to the original linking constraints.

1. Formula
   Observation

   The whole slide is a worked numerical production model using $X_i$ and $Y_i$.

2. Audio
   Observation

   The narrator concludes that this completes the fixed-cost problem.

The worked example applies the binary linking-constraint method to a concrete three-machine production problem.

1. Formula
   Observation

   The solution values are displayed directly under the completed model.

2. Audio
   Observation

   The narrator says, "On solving this model, we obtain ...".

The reported variable values are part of the worked example's outcome.

1. Audio
   Observation

   The narrator says the next video will show how to formulate relational constraints using binary variables.

2. Caption evidence
   Observation

   The slide lists multiple choice, mutually exclusive, co-requisite, and conditional constraints.

The fixed-cost use of binary variables is presented as one instance within a larger family of binary-variable constraint formulations.

1. Audio
   Observation

   Binary variables are employed when there's a yes or no situation.

1. Binary Variable Definition

1. Audio
   Observation

   For example, suppose we have four different projects to consider.

1. Project Selection Problem

1. Audio
   Observation

   Let's now model a fixed cost problem.

1. Fixed Cost Problem

1. Audio
   Observation

   Narrator explicitly justifies introducing $Y_i$ due to fixed costs.

1. Binary (0-1) Decision Variable

1. Formula
   Observation

   Objective function clearly separates variable and fixed cost terms.

1. Total Cost Objective Function

1. Audio
   Observation

   Narrator explains why simple capacity constraints are insufficient with fixed costs.

1. Machine Capacity Constraints
2. Linking Capacity to Binary Usage
3. Ignoring Binary Linkage in Capacity Constraints

1. Audio
   Observation

   Narrator specifies equality because the order must be met exactly.

1. Demand Satisfaction Constraint

1. Audio
   Observation

   The narrator explains the effect of $Y_1=1$ and $Y_1=0$ on production.

2. Formula
   Observation

   The constraint $X_1 \le 400Y_1$ is shown.