Business Analytics · Quiz 2 · Sep 2024

**Answer: B, C** **Topic:** Probability and Statistics — *Zero Correlation vs. Statistical Independence* ### Step-by-Step Solution **Step 1:** Recall that Pearson's correlation coefficient $\rho(X, Y) = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y}$ measures only the **linear** relationship between two random variables $X$ and $Y$, where $\text{Cov}(X, Y) = E[XY] - E[X]E[Y]$. **Step 2:** When $E[XY] = E[X]E[Y]$ (so $\text{Cov}(X, Y) = 0$) for two dependent random variables (such as $Y = X^2$ with symmetric $X$): - $\rho(X, Y) = 0$ (**B is correct**). - $X$ and $Y$ are **not linearly related** (**C is correct**). - However, zero correlation does *not* imply statistical independence (**A is false**), and $E[XY] = E[X]E[Y]$ (**D is false**). **Step 3:** Thus, **B and C** are correct. **Key Takeaway / Tip:** Zero correlation ($\rho(X, Y) = 0 \iff E[XY] = E[X]E[Y]$) means the variables have no *linear* relationship, but they may still be non-linearly dependent. A: **Incorrect:** Option A (

) does not satisfy the conditions derived in the step-by-step analysis for Zero Correlation vs. Statistical Independence (correct choice: B, C). B: **Correct:** Option B (

) matches our derived result under Zero Correlation vs. Statistical Independence. Zero correlation ($\rho(X, Y) = 0 \iff E[XY] = E[X]E[Y]$) means the variables have no *linear* relationship, but they may still be non-linearly dependent. C: **Correct:** Option C (

) matches our derived result under Zero Correlation vs. Statistical Independence. Zero correlation ($\rho(X, Y) = 0 \iff E[XY] = E[X]E[Y]$) means the variables have no *linear* relationship, but they may still be non-linearly dependent. D: **Incorrect:** Option D (

      **Business Analytics** **Section Id :** 64065373869 **Section Number :** 15 **Section type :** Online **Mandatory or Optional :** Mandat...) does not satisfy the conditions derived in the step-by-step analysis for Zero Correlation vs. Statistical Independence (correct choice: B, C).

**Answer: A** **Topic:** Demand Modeling — *Latent Demand in a Demand-Response Curve* ### Step-by-Step Solution **Step 1:** In pricing and demand-response analysis, once an optimal price $P^*$ is chosen, some potential customers who value the product below $P^*$ do not purchase it at $P^*$, representing unrealized or **latent demand**. **Step 2:** This latent demand is the additional demand area under the demand-response curve that gets unlocked when **the price is reduced below the identified optimal price**. **Step 3:** Therefore, **A** is correct. **Key Takeaway / Tip:** Reducing the price below the optimal price $P^*$ moves downward along the demand curve, capturing the latent demand region. A: **Correct:** Option A (Price is reduced below the identified optimal price) matches our derived result under Latent Demand in a Demand-Response Curve. Reducing the price below the optimal price $P^*$ moves downward along the demand curve, capturing the latent demand region. B: **Incorrect:** Option B (Price is increased beyond the identified optimal price) does not satisfy the conditions derived in the step-by-step analysis for Latent Demand in a Demand-Response Curve (correct choice: A). C: **Incorrect:** Option C (The optimal price is increased beyond the maximum available price) does not satisfy the conditions derived in the step-by-step analysis for Latent Demand in a Demand-Response Curve (correct choice: A). D: **Incorrect:** Option D (Quantity is reduced below the identified optimal quantity) does not satisfy the conditions derived in the step-by-step analysis for Latent Demand in a Demand-Response Curve (correct choice: A).

**Answer: B** **Topic:** Logistic Regression — *Objective Function of Logistic Regression* ### Step-by-Step Solution **Step 1:** Unlike Ordinary Least Squares (OLS) linear regression—which minimizes the sum of squared residuals—Logistic Regression models Bernoulli/categorical probabilities $P(Y = y_i \mid X_i)$ via the sigmoid function. **Step 2:** Its parameters $\beta$ are estimated using Maximum Likelihood Estimation (MLE), which corresponds to the **maximization of the log-likelihood** function (or equivalently, minimization of binary cross-entropy / negative log-likelihood). **Step 3:** Therefore, **B** is correct. **Key Takeaway / Tip:** Linear regression uses minimization of squared errors (OLS), whereas logistic regression uses maximization of log-likelihood (MLE). A: **Incorrect:** Option A (Minimization of squared errors) does not satisfy the conditions derived in the step-by-step analysis for Objective Function of Logistic Regression (correct choice: B). B: **Correct:** Option B (Maximization of log-likelihood) matches our derived result under Objective Function of Logistic Regression. Linear regression uses minimization of squared errors (OLS), whereas logistic regression uses maximization of log-likelihood (MLE). C: **Incorrect:** Option C (Minimization of residuals) does not satisfy the conditions derived in the step-by-step analysis for Objective Function of Logistic Regression (correct choice: B). D: **Incorrect:** Option D (None of these) does not satisfy the conditions derived in the step-by-step analysis for Objective Function of Logistic Regression (correct choice: B).

**Answer: 82** **Topic:** Pricing Optimization — *Profit-Maximizing Price for Linear Demand* ### Step-by-Step Solution **Step 1:** Write the profit function $\Pi(P) = (P - c) \cdot D(P)$ given the linear demand equation $D(P) = 720 - 5P$ and unit manufacturing cost $c = 20$: $$\Pi(P) = (P - 20)(720 - 5P) = 720P - 5P^2 - 14400 + 100P = -5P^2 + 820P - 14400$$ **Step 2:** Differentiate $\Pi(P)$ with respect to $P$ and set the derivative equal to zero to find the profit-maximizing price $P^*$: $$\frac{d\Pi}{dP} = -10P + 820 = 0 \implies 10P = 820 \implies P^* = 82$$ **Step 3:** Check the second derivative: $\frac{d^2\Pi}{dP^2} = -10 < 0$, confirming a unique global maximum at $P^* = 82$. **Key Takeaway / Tip:** Shortcut: for linear demand $D(P) = a - bP$ with unit cost $c$, the optimal price is always the average of the choke price $a/b$ and the unit cost $c$: $P^* = \frac{720/5 + 20}{2} = \frac{144 + 20}{2} = 82$.

**Answer: 0.68** **Topic:** Demand Modeling — *Elasticity of a Constant Elasticity Demand Curve* ### Step-by-Step Solution **Step 1:** Write the constant elasticity demand relationship $Q = a P^{-b}$ (where $b$ is the elasticity magnitude). **Step 2:** Take the ratio of the two given points $(P_1 = 100, Q_1 = 2400)$ and $(P_2 = 200, Q_2 = 1500)$: $$\frac{Q_2}{Q_1} = \left(\frac{P_2}{P_1}\right)^{-b} \implies \frac{1500}{2400} = \left(\frac{200}{100}\right)^{-b} \implies 2^b = \frac{2400}{1500} = 1.6$$ **Step 3:** Solve for $b$ using natural logarithms: $$b = \frac{\ln(1.6)}{\ln(2)} \approx \frac{0.4700036}{0.6931472} \approx 0.6781 \approx 0.68$$ **Key Takeaway / Tip:** Always use the log-ratio formula $b = \left|\frac{\ln(Q_2/Q_1)}{\ln(P_2/P_1)}\right|$ when the curve is explicitly specified as a constant elasticity curve.

**Answer: D** **Topic:** Linear Regression — *Definition of Multicollinearity* ### Step-by-Step Solution **Step 1:** Recall the statistical definition of **Multicollinearity** in multiple regression: it occurs when two or more **independent (explanatory) variables** are highly linearly correlated with *each other*. **Step 2:** Inspect the given options: - **A**: 'The dependent and independent variables are not-related' — False. - **B**: 'The dependent and independent variables are linearly related' — False (that is just the regression relationship itself, not multicollinearity). - **C**: 'The dependent variable is linearly related to another dependent variable' — False. - **D**: 'None of these' — **Correct**. **Step 3:** Thus, **D** is the correct choice. **Key Takeaway / Tip:** Multicollinearity is strictly a relationship among the **independent** variables ($X_j$ with $X_k$), not between the dependent variable $Y$ and the independent variables. A: **Incorrect:** Option A (The dependent and independent variables are not-related) does not satisfy the conditions derived in the step-by-step analysis for Definition of Multicollinearity (correct choice: D). B: **Incorrect:** Option B (The dependent and independent variables are linearly related) does not satisfy the conditions derived in the step-by-step analysis for Definition of Multicollinearity (correct choice: D). C: **Incorrect:** Option C (The dependent variable is linearly related to another dependent variable) does not satisfy the conditions derived in the step-by-step analysis for Definition of Multicollinearity (correct choice: D). D: **Correct:** Option D (None of these) matches our derived result under Definition of Multicollinearity. Multicollinearity is strictly a relationship among the **independent** variables ($X_j$ with $X_k$), not between the dependent variable $Y$ and the independent variables.

**Answer: A, B, C** **Topic:** Logistic Regression — *Applications of Logistic Regression* ### Step-by-Step Solution **Step 1:** Review the core capabilities and applications of Logistic Regression: - **A is true**: Binary logistic regression is the standard model for predicting binary ($0/1$ or Yes/No) outcomes. - **B is true**: Multinomial (softmax) and one-vs-rest logistic regression extend logistic regression to predict multi-class categorical outputs. - **C is true**: Because $\ln\left(\frac{p}{1-p}\right) = \beta_0 + \beta^T x$, exponentiating gives the **odds** $\frac{p}{1-p} = e^{\beta_0 + \beta^T x}$ of the occurrence of a specific event. **Step 2:** Therefore, **A, B, and C** are all valid applications. **Key Takeaway / Tip:** Logistic regression directly models the log-odds (logit) of an event and supports both binary and multi-class classification. A: **Correct:** Option A (Predicting binary outcomes) matches our derived result under Applications of Logistic Regression. Logistic regression directly models the log-odds (logit) of an event and supports both binary and multi-class classification. B: **Correct:** Option B (Predicting the multi-class output) matches our derived result under Applications of Logistic Regression. Logistic regression directly models the log-odds (logit) of an event and supports both binary and multi-class classification. C: **Correct:** Option C (Predicting the odds of the occurrence of a specific event) matches our derived result under Applications of Logistic Regression. Logistic regression directly models the log-odds (logit) of an event and supports both binary and multi-class classification. D: **Incorrect:** Option D (None of these) does not satisfy the conditions derived in the step-by-step analysis for Applications of Logistic Regression (correct choice: A, B, C).

**Answer: 314** **Topic:** Linear Regression — *Residual Degrees of Freedom in ANOVA* ### Step-by-Step Solution **Step 1:** Identify the sample size $n = 320$ data points and the number of predictors $k = 5$ ($X_1, X_2, X_3, X_4, X_5$) in the given MLR model: $$Y = 2.1 + 1.4 X_1 - 4.2 X_2 + 0.5 X_3 + 7 X_4 - 6 X_5 + \varepsilon$$ **Step 2:** Compute the Residual degrees of freedom using $df_{\text{res}} = n - k - 1$: $$df_{\text{res}} = 320 - 5 - 1 = 314$$ **Key Takeaway / Tip:** With 5 slope coefficients and 1 intercept, 6 degrees of freedom are used by the model parameters, leaving $320 - 6 = 314$ for the residuals.

**Answer: 319** **Topic:** Linear Regression — *Total Degrees of Freedom in ANOVA* ### Step-by-Step Solution **Step 1:** Recall that the **Total** degrees of freedom in an ANOVA table for a regression model with $n$ observations is $df_{\text{total}} = n - 1$. **Step 2:** Substitute $n = 320$: $$df_{\text{total}} = 320 - 1 = 319$$ **Key Takeaway / Tip:** Verify that $df_{\text{reg}} + df_{\text{res}} = 5 + 314 = 319 = df_{\text{total}}$.

**Answer: 5** **Topic:** Linear Regression — *Number of Input Features in Multiple Linear Regression* ### Step-by-Step Solution **Step 1:** Inspect the regression equation $Y = 2.1 + 1.4 X_1 - 4.2 X_2 + 0.5 X_3 + 7 X_4 - 6 X_5 + \varepsilon$. **Step 2:** There are 5 explanatory variables ($X_1, X_2, X_3, X_4, X_5$). Since no feature engineering was performed, the dataset contained **5** features. **Key Takeaway / Tip:** Each explanatory variable $X_1, \dots, X_5$ in the raw MLR equation corresponds to one feature in the dataset.

**Answer: 0.425** **Topic:** Classification Metrics — *Accuracy from a 2x2 Confusion Matrix* ### Step-by-Step Solution **Step 1:** Read the 4 entries of the given confusion matrix (Rows = Predicted class, Columns = Actual class): - True Negatives ($\text{Predicted Negative, Actual Negative}$): $\text{TN} = 145$ - False Negatives ($\text{Predicted Negative, Actual Positive}$): $\text{FN} = 155$ - False Positives ($\text{Predicted Positive, Actual Negative}$): $\text{FP} = 75$ - True Positives ($\text{Predicted Positive, Actual Positive}$): $\text{TP} = 25$ **Step 2:** Sum all four cells to find the total number of observations $N$: $$N = 145 + 155 + 75 + 25 = 400$$ **Step 3:** Compute the model **Accuracy** (in decimal): $$\text{Accuracy} = \frac{\text{TP} + \text{TN}}{N} = \frac{25 + 145}{400} = \frac{170}{400} = 0.425\text{ (or }0.42\text{ / }0.43\text{)}$$ **Key Takeaway / Tip:** Pay attention to the formatting instruction: enter the answer as a **decimal** (`0.425`, inside `0.40 to 0.44`), not as a percentage.

**Answer: 0.48** **Topic:** Classification Metrics — *Precision for the Negative Class* ### Step-by-Step Solution **Step 1:** Recall that **Precision for predicting the Negative class** is the fraction of all instances predicted as Negative that are actually Negative: $$\text{Precision}_{\text{Negative}} = \frac{\text{TN}}{\text{TN} + \text{FN}}$$ **Step 2:** From the first row (`Predicted = Negative`) of the confusion matrix, $\text{TN} = 145$ and $\text{FN} = 155$, so the total predicted negative is $145 + 155 = 300$. **Step 3:** Divide $\text{TN}$ by $300$ and round to two decimal places: $$\text{Precision}_{\text{Negative}} = \frac{145}{300} \approx 0.4833 \approx 0.48$$ **Key Takeaway / Tip:** For negative-class precision, look along the `Predicted = Negative` row ($145 + 155 = 300$).

**Answer: 0.14** **Topic:** Classification Metrics — *Recall for the Positive Class* ### Step-by-Step Solution **Step 1:** Recall that **Recall for predicting the Positive class** is the fraction of all actual Positive instances that are correctly predicted as Positive: $$\text{Recall}_{\text{Positive}} = \frac{\text{TP}}{\text{TP} + \text{FN}}$$ **Step 2:** From the second column (`Actual = Positive`) of the confusion matrix, $\text{FN} = 155$ and $\text{TP} = 25$, so the total actual positive is $155 + 25 = 180$. **Step 3:** Divide $\text{TP}$ by $180$ and round to two decimal places: $$\text{Recall}_{\text{Positive}} = \frac{25}{180} \approx 0.1389 \approx 0.14$$ **Key Takeaway / Tip:** For positive-class recall, look down the `Actual = Positive` column ($155 + 25 = 180$).

**Answer: 0.79** **Topic:** Linear Regression — *Calculating New Adjusted R-Squared After Adding a Predictor* ### Step-by-Step Solution **Step 1:** Recall the formula relating Adjusted $R^2$ to $R^2$, sample size $n$, and the number of explanatory variables $k$: $$R^2_{\text{adj}} = 1 - \frac{(1 - R^2)(n - 1)}{n - k - 1}$$ **Step 2:** Substitute the new model's values ($n = 150$ observations, $k = 9$ explanatory variables, and $R^2_{\text{new}} = 0.80$): $$R^2_{\text{adj, new}} = 1 - \frac{(1 - 0.80)(150 - 1)}{150 - 9 - 1} = 1 - \frac{0.20 \times 149}{140}$$ **Step 3:** Evaluate and round to two decimal places: $$R^2_{\text{adj, new}} = 1 - \frac{29.8}{140} \approx 1 - 0.212857 = 0.787143 \approx 0.79$$ **Key Takeaway / Tip:** Use the new predictor count $k = 9$ (so $n - k - 1 = 150 - 9 - 1 = 140$) in the denominator.

**Answer: B** **Topic:** Linear Regression — *Model Selection Using Adjusted R-Squared* ### Step-by-Step Solution **Step 1:** Compare the Adjusted $R^2$ before and after adding the 9th explanatory variable: - **Old model ($k = 8, R^2 = 0.75$)**: $$R^2_{\text{adj, old}} = 1 - \frac{(1 - 0.75)(149)}{150 - 8 - 1} = 1 - \frac{37.25}{141} \approx 0.7358$$ - **New model ($k = 9, R^2 = 0.80$)**: $$R^2_{\text{adj, new}} \approx 0.7871 \approx 0.79$$ **Step 2:** Because Adjusted $R^2$ **increased** substantially (from $0.736$ to $0.787$) even after penalizing for the extra degree of freedom, the 9th variable adds genuine explanatory power and **must be kept in the model** (**B**). **Key Takeaway / Tip:** Unlike raw $R^2$ (which never decreases when a variable is added), Adjusted $R^2$ increases if and only if the new variable's $t$-statistic satisfies $|t| > 1$. A: **Incorrect:** Option A (The new variable must be dropped from the model) does not satisfy the conditions derived in the step-by-step analysis for Model Selection Using Adjusted R-Squared (correct choice: B). B: **Correct:** Option B (The new variable must be kept in the model) matches our derived result under Model Selection Using Adjusted R-Squared. Unlike raw $R^2$ (which never decreases when a variable is added), Adjusted $R^2$ increases if and only if the new variable's $t$-statistic satisfies $|t| > 1$. C: **Incorrect:** Option C (The data is too small for fitting a regression model with 9 variables) does not satisfy the conditions derived in the step-by-step analysis for Model Selection Using Adjusted R-Squared (correct choice: B). D: **Incorrect:** Option D (None of these) does not satisfy the conditions derived in the step-by-step analysis for Model Selection Using Adjusted R-Squared (correct choice: B).

**Answer: A** **Topic:** Data Envelopment Analysis — *Input-Oriented DEA Efficient Frontier* ### Step-by-Step Solution **Step 1:** When $X_1$ and $X_2$ represent **inputs** for a fixed constant output across all Decision Making Units (DMUs), an efficient DMU is one that uses the **minimum inputs** (i.e., lies closest to the origin toward the bottom-left). **Step 2:** Inspect Figure-1 for the lower-left convex envelope of the 7 points: - DMU `1` has the smallest $X_1$ ($X_1 = 0, X_2 = 12$). - Moving down and right toward lower $X_2$, the lower-left boundary passes through DMU `2` ($3, 9$), DMU `5` ($13, 4$), and DMU `7` ($19, 5$). - Points `3`, `4`, and `6` lie above and to the right of this frontier (they consume more $X_1$ and/or $X_2$ for the same output). **Step 3:** Thus, the input-efficient frontier is **1-2-5-7**, which matches **A**. **Key Takeaway / Tip:** For an input-minimization DEA plot (fixed output), the efficient frontier is always the **south-west (bottom-left)** boundary closest to the origin. A: **Correct:** Option A (1-2-5-7) matches our derived result under Input-Oriented DEA Efficient Frontier. For an input-minimization DEA plot (fixed output), the efficient frontier is always the **south-west (bottom-left)** boundary closest to the origin. B: **Incorrect:** Option B (1-3-4-6-7) does not satisfy the conditions derived in the step-by-step analysis for Input-Oriented DEA Efficient Frontier (correct choice: A). C: **Incorrect:** Option C (3-2-5-6-4) does not satisfy the conditions derived in the step-by-step analysis for Input-Oriented DEA Efficient Frontier (correct choice: A). D: **Incorrect:** Option D (All of these) does not satisfy the conditions derived in the step-by-step analysis for Input-Oriented DEA Efficient Frontier (correct choice: A).

**Answer: B** **Topic:** Data Envelopment Analysis — *Output-Oriented DEA Efficient Frontier* ### Step-by-Step Solution **Step 1:** When $X_1$ and $X_2$ represent **outputs** produced from a fixed constant input across all DMUs, an efficient DMU is one that produces the **maximum outputs** (i.e., lies furthest from the origin toward the top-right). **Step 2:** Inspect Figure-1 for the upper-right envelope of the 7 points: - Starting from the top-left at DMU `1` ($0, 12$), moving along the outer top-right boundary goes through DMU `3` ($3, 14$), DMU `4` ($5, 13$), DMU `6` ($15, 11$), and DMU `7` ($19, 5$). - Points `2` and `5` lie strictly inside (below/left of) this outer boundary. **Step 3:** Thus, the output-efficient frontier is **1-3-4-6-7**, which matches **B**. **Key Takeaway / Tip:** For an output-maximization DEA plot (fixed input), the efficient frontier is always the **north-east (top-right)** outer boundary furthest from the origin. A: **Incorrect:** Option A (1-2-5-7) does not satisfy the conditions derived in the step-by-step analysis for Output-Oriented DEA Efficient Frontier (correct choice: B). B: **Correct:** Option B (1-3-4-6-7) matches our derived result under Output-Oriented DEA Efficient Frontier. For an output-maximization DEA plot (fixed input), the efficient frontier is always the **north-east (top-right)** outer boundary furthest from the origin. C: **Incorrect:** Option C (3-2-5-6-4) does not satisfy the conditions derived in the step-by-step analysis for Output-Oriented DEA Efficient Frontier (correct choice: B). D: **Incorrect:** Option D (All of these) does not satisfy the conditions derived in the step-by-step analysis for Output-Oriented DEA Efficient Frontier (correct choice: B).

**Answer: 6** **Topic:** Linear Programming — *Number of Constraints in the Dual Formulation* ### Step-by-Step Solution **Step 1:** Recall the fundamental rule of LP duality: each decision variable in the primal problem corresponds to one constraint in the dual problem. **Step 2:** Inspect the given Primal LP: there are $6$ non-negative decision variables ($Y_1, Y_2, Y_3, Y_4, Y_5, Y_6$). **Step 3:** Therefore, there will be **6** constraints in the dual formulation. **Key Takeaway / Tip:** Count the primal decision variables ($Y_1$ through $Y_6$) to immediately get the number of dual constraints ($6$).

**Answer: 7** **Topic:** Linear Programming — *Number of Dual Decision Variables from Standard Primal Form* ### Step-by-Step Solution **Step 1:** Look at the 5 functional constraints in the given Primal LP: - Constraint-1 ($\le 200$): 1 inequality constraint - Constraint-2 ($\le 50$): 1 inequality constraint - Constraint-3 ($= 65$): 1 equality constraint $\implies$ splits into **2** $\le$ constraints in standard form - Constraint-4 ($\le 20$): 1 inequality constraint - Constraint-5 ($= 16$): 1 equality constraint $\implies$ splits into **2** $\le$ constraints in standard form **Step 2:** Count the total number of functional constraints in the **standard form** of the primal: $$1 + 1 + 2 + 1 + 2 = 7$$ **Step 3:** Since each standard-form primal constraint corresponds to one dual decision variable, **7** decision variables are required. **Key Takeaway / Tip:** Because the prompt specifically says *'based on the standard form of the primal'*, each equality constraint ($=$) must be replaced by two $\le$ inequalities, adding $2$ extra dual variables ($5 + 2 = 7$).

**Answer: 20** **Topic:** Linear Programming — *Dual Objective Function Coefficient* ### Step-by-Step Solution **Step 1:** In Linear Programming duality, the right-hand side (RHS) constants $b_i$ of the primal constraints become the objective function coefficients of the corresponding dual decision variables $U_i$. **Step 2:** Inspect **Constraint-4** in the primal problem: $$Y_1 + Y_3 \le 20$$ **Step 3:** Its right-hand side constant is $20$, so the objective function coefficient for the dual decision variable corresponding to Constraint-4 is **20**. **Key Takeaway / Tip:** Primal RHS values map directly to Dual objective coefficients, and Primal objective coefficients map directly to Dual RHS values.

**Answer: 4** **Topic:** Linear Programming — *Complementary Slackness and Non-Zero Dual Variables* ### Step-by-Step Solution **Step 1:** Substitute the given primal solution $(Y_1, Y_2, Y_3, Y_4, Y_5, Y_6) = (4, 5, 2, 6, 10, 5)$ into each of the 5 functional constraints to check which are binding (zero slack) vs. non-binding (positive slack): - **Constraint-1**: $4 + 5 + 2 + 6 + 10 + 5 = 32 < 200$ (non-binding, slack $= 168 > 0 \implies$ dual variable $= 0$) - **Constraint-2**: $2(4) + 3(2) = 14 < 50$ (non-binding, slack $= 36 > 0 \implies$ dual variable $= 0$) - **Constraint-3**: $4(5) + 5(6) + 3(5) = 20 + 30 + 15 = 65$ (**binding equality**) - **Constraint-4**: $4 + 2 = 6 < 20$ (non-binding, slack $= 14 > 0 \implies$ dual variable $= 0$) - **Constraint-5**: $5 + 6 + 5 = 16$ (**binding equality**) **Step 2:** By Complementary Slackness, the 3 strict inequality constraints (`Constraint-1`, `Constraint-2`, `Constraint-4`) have dual variables equal to $0$. **Step 3:** The 2 binding equality constraints (`Constraint-3` and `Constraint-5`) correspond to $2 \times 2 = 4$ tight inequalities in the 7-variable standard dual formulation, yielding **4** non-zero dual variables. **Key Takeaway / Tip:** Of the 7 standard-form constraints, the 3 $\le$ constraints have positive slack (giving 0 dual values), leaving the 4 standard-form constraints coming from the 2 equalities.