Sample Solution

MSTE-002 Solved Assignment 2025

1. State whether the following statements are True or False. Give reasons in support of your answers.
   (a) The solution of a transportation problem with 5 rows (supplies) and 4 columns (destinations) is feasible if number of possible allocations are 8.
   (b) The moving averages of suitable period in a time-series are free from the influences of seasonal and cyclic variations.
   (c) If the basic solutions for a system of equations are $(-2,0,1),(0,1,3),(-2,3,0)$$(-2,0,1),(0,1,3),(-2,3,0)$(-2,0,1),(0,1,3),(-2,3,0)(-2,0,1),(0,1,3),(-2,3,0)$(-2,0,1),(0,1,3),(-2,3,0)$, then only $(0,1,3)$$(0,1,3)$(0,1,3)(0,1,3)$(0,1,3)$ is feasible.
   (d) In the stepwise selection method of multiple regression model, once a variable enters in the model then it always remains in the model.
   (e) An enterprise requires 1000 units per month. The ordering cost is estimated to be 50 per order. The purchase price is 20 per unit and the carrying cost per unit is $10\mathrm{\%}$$10\mathrm{\%}$10%10 \%$10\mathrm{\%}$ of it. Then the economic lot size to be ordered is 775 .
2. Use the penalty $(\mathrm{Big}M)$$(\mathrm{Big}M)$(Big M)(\operatorname{Big} M)$(\mathrm{Big}M)$ method to solve the following LP problem:

3. A company has three production facilities $S_1,S_2$$S_1,S_2$S_(1),S_(2)S_1, S_2$S_1,S_2$ and $S_3$$S_3$S_(3)S_3$S_3$ with production capacity of 7,9 and 18 units (in 100 s ) per week of a product, respectively. These units are to be shipped to four warehouses $D_1,D_2,D_3$$D_1,D_2,D_3$D_(1),D_(2),D_(3)D_1, D_2, D_3$D_1,D_2,D_3$ and $D_4$$D_4$D_(4)D_4$D_4$ with requirement of 5, 6,7 and 14 units (in 100 s) per week, respectively. The transportation costs (in Rs) per unit between factories to warehouses are given in the table below:

7. A firm wants to know whether there is any linear relationship between the sales (X) and its yearly revenue (Y). The records for 10 years were examined and the following results were obtained:

Solution:

Question:-1

State whether the following statements are True or False. Give reasons in support of your answers.

Answer:

• False.
• Justification: In a transportation problem with $m$$m$mm$m$ rows (supplies) and $n$$n$nn$n$ columns (destinations), a basic feasible solution requires exactly $m+n-1$$m+n-1$m+n-1m + n – 1$m+n-1$ allocations (positive entries) to be feasible, assuming total supply equals total demand. Here, $m=5$$m=5$m=5m = 5$m=5$, $n=4$$n=4$n=4n = 4$n=4$, so $m+n-1=5+4-1=8$$m+n-1=5+4-1=8$m+n-1=5+4-1=8m + n – 1 = 5 + 4 – 1 = 8$m+n-1=5+4-1=8$. However, the statement says "possible allocations are 8," which is ambiguous. Typically, a transportation problem with 5 rows and 4 columns has $5\times 4=20$$5\times 4=20$5xx4=205 \times 4 = 20$5\times 4=20$ possible allocations (cells). A feasible basic solution requires exactly 8 non-zero allocations, not that the total number of possible allocations is 8. Since the problem has 20 possible allocations, the statement is false unless misinterpreted.

• True.
• Justification: Moving averages smooth out short-term fluctuations in a time series. If the period of the moving average matches the length of the seasonal cycle (e.g., 12 months for yearly seasonality), it effectively removes seasonal variations by averaging over the cycle. Similarly, moving averages can dampen cyclic variations (longer-term oscillations) if the period is appropriately chosen, as they focus on the trend component. Thus, with a suitable period, moving averages eliminate seasonal and cyclic influences, leaving the trend.

• False.
• Justification: For a solution to be feasible in a linear programming context, it must satisfy all constraints, including non-negativity (if applicable). The statement claims only $(0,1,3)$$(0,1,3)$(0,1,3)(0,1,3)$(0,1,3)$ is feasible among the given basic solutions. However, without the specific system of equations or constraints, we assume a typical linear programming problem where variables are non-negative ($x_1,x_2,x_3\ge 0$$x_1,x_2,x_3\ge 0$x_(1),x_(2),x_(3) >= 0x_1, x_2, x_3 \geq 0$x_1,x_2,x_3\ge 0$).
  • $(-2,0,1)$$(-2,0,1)$(-2,0,1)(-2,0,1)$(-2,0,1)$: Contains a negative value ($-2$$-2$-2-2$-2$), so it is not feasible if non-negativity is required.
  • $(0,1,3)$$(0,1,3)$(0,1,3)(0,1,3)$(0,1,3)$: All components are non-negative, so it is feasible if it satisfies the system’s constraints.
  • $(0,3,0)$$(0,3,0)$(0,3,0)(0,3,0)$(0,3,0)$: All components are non-negative, so it is also feasible if it satisfies the constraints.
    Since $(0,3,0)$$(0,3,0)$(0,3,0)(0,3,0)$(0,3,0)$ is non-negative and a basic solution, it could be feasible unless additional constraints exclude it. The claim that only $(0,1,3)$$(0,1,3)$(0,1,3)(0,1,3)$(0,1,3)$ is feasible is false, as $(0,3,0)$$(0,3,0)$(0,3,0)(0,3,0)$(0,3,0)$ may also be feasible without specific constraints ruling it out.

• False.
• Justification: In stepwise regression (specifically, stepwise selection or backward elimination with re-evaluation), variables can be added or removed based on statistical criteria (e.g., p-values or AIC). In forward stepwise selection, a variable may enter the model but can be removed later if it becomes insignificant after adding other variables. For example, if variable $X_1$$X_1$X_(1)X_1$X_1$ is added early but later has a high p-value due to correlation with a newly added variable, it may be dropped. Thus, variables do not always remain in the model.

• True.
• Justification: The economic order quantity (EOQ) is calculated using the formula:
  $$Q=\sqrt{\frac{2DS}{H}}$$$$Q=\sqrt{\frac{2DS}{H}}$$Q=sqrt((2DS)/(H))Q = \sqrt{\frac{2DS}{H}}$$Q=\sqrt{\frac{2DS}{H}}$$
  where:
  • $D=1000\times 12=12,000$$D=1000\times 12=12,000$D=1000 xx12=12,000D = 1000 \times 12 = 12,000$D=1000\times 12=12,000$ units/year (annual demand, assuming monthly demand is converted),
  • $S=50$$S=50$S=50S = 50$S=50$ (ordering cost per order),
  • $H=10\mathrm{\%}\times 20=2$$H=10\mathrm{\%}\times 20=2$H=10%xx20=2H = 10\% \times 20 = 2$H=10\mathrm{\%}\times 20=2$ (carrying cost per unit per year, as 10% of the purchase price).
  Plugging in:
  $$Q=\sqrt{\frac{2\times 12,000\times 50}{2}}=\sqrt{\frac{1,200,000}{2}}=\sqrt{600,000}\approx 774.6$$$$Q=\sqrt{\frac{2\times 12,000\times 50}{2}}=\sqrt{\frac{1,200,000}{2}}=\sqrt{600,000}\approx 774.6$$Q=sqrt((2xx12,000 xx50)/(2))=sqrt((1,200,000)/(2))=sqrt(600,000)~~774.6Q = \sqrt{\frac{2 \times 12,000 \times 50}{2}} = \sqrt{\frac{1,200,000}{2}} = \sqrt{600,000} \approx 774.6$$Q=\sqrt{\frac{2\times 12,000\times 50}{2}}=\sqrt{\frac{1,200,000}{2}}=\sqrt{600,000}\approx 774.6$$
  Rounding to the nearest integer, $Q\approx 775$$Q\approx 775$Q~~775Q \approx 775$Q\approx 775$.

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