Question 1 |
A machine produces a defective component with a probability of 0.015. The
number of defective components in a packed box containing 200 components
produced by the machine follows a Poisson distribution. The mean and the
variance of the distribution are
3 and 3, respectively | |
\sqrt{3} and \sqrt{3} , respectively | |
0.015 and 0.015, respectively | |
3 and 9, respectively |
Question 1 Explanation:
P = 0.015
n = 200
mean =\lambda =np= 200 \times 0.015 = 3
variance =\sigma ^2=\lambda =3
n = 200
mean =\lambda =np= 200 \times 0.015 = 3
variance =\sigma ^2=\lambda =3
Question 2 |
Parts P1-P7 are machined first on a milling machine
and then polished at a separate machine. Using the
information in the following table, the minimum
total completion time required for carrying out both
the operations for all 7 parts is __________ hours.
\begin{array}{|c|c|c|} \hline \text{Part} & \text{Milling (hours)} &\text{Polishing (hours)} \\ \hline P1 & 8 &6 \\ \hline P2 & 3& 2\\ \hline P3& 3& 4\\ \hline P4& 4& 6\\ \hline P5& 5&7 \\ \hline P6& 6 & 4\\ \hline P7& 2& 1 \\ \hline \end{array}
\begin{array}{|c|c|c|} \hline \text{Part} & \text{Milling (hours)} &\text{Polishing (hours)} \\ \hline P1 & 8 &6 \\ \hline P2 & 3& 2\\ \hline P3& 3& 4\\ \hline P4& 4& 6\\ \hline P5& 5&7 \\ \hline P6& 6 & 4\\ \hline P7& 2& 1 \\ \hline \end{array}
31 | |
33 | |
30 | |
32 |
Question 2 Explanation:
Sequencing model :
\begin{array}{|c|c|c|}\hline \text{Part}&\text{Milling (hours)}& \text{Polishing (hours)} \\ \hline P_1&8&6\\ \hline P_2&3&2\\ \hline P_3&3&4\\ \hline P_4&4&6\\ \hline P_5&5&7\\ \hline P_6&6&4\\ \hline P_7&2&1\\ \hline \end{array}
Optimum job sequence (According to Johnson's algorithm)
P_3,P_4,P_5,P_1,P_6,P_2,P_7
Optimum Make-span (Minimum total completion time)
\begin{array}{|c|c|c|c|c|c|c|}\hline \text{Part}& \text{Milling}& & &\text{Polishing}&& \\ \hline & \text{In time}& \text{Processing time} &\text{Out time}&\text{In time}& \text{Processing time} &\text{Out time}\\ \hline P_3& 0 & 3 & 3 & 3 &4 &7\\ \hline P_4& 3 & 4 & 7 & 7 &6 &13\\ \hline P_5& 7 & 5 & 12 & 13 &7 &20\\ \hline P_1& 12 & 8 & 20 & 20 &6 &26\\ \hline P_6& 20 & 6 & 26 & 26 &4 &30\\ \hline P_2& 26 & 3 & 29 & 30 &2 &32\\ \hline P_7& 29 & 2 & 31 & 32 &1 &33\\ \hline \end{array}
Minimum total completion time of all parts = 33 hours
\begin{array}{|c|c|c|}\hline \text{Part}&\text{Milling (hours)}& \text{Polishing (hours)} \\ \hline P_1&8&6\\ \hline P_2&3&2\\ \hline P_3&3&4\\ \hline P_4&4&6\\ \hline P_5&5&7\\ \hline P_6&6&4\\ \hline P_7&2&1\\ \hline \end{array}
Optimum job sequence (According to Johnson's algorithm)
P_3,P_4,P_5,P_1,P_6,P_2,P_7
Optimum Make-span (Minimum total completion time)
\begin{array}{|c|c|c|c|c|c|c|}\hline \text{Part}& \text{Milling}& & &\text{Polishing}&& \\ \hline & \text{In time}& \text{Processing time} &\text{Out time}&\text{In time}& \text{Processing time} &\text{Out time}\\ \hline P_3& 0 & 3 & 3 & 3 &4 &7\\ \hline P_4& 3 & 4 & 7 & 7 &6 &13\\ \hline P_5& 7 & 5 & 12 & 13 &7 &20\\ \hline P_1& 12 & 8 & 20 & 20 &6 &26\\ \hline P_6& 20 & 6 & 26 & 26 &4 &30\\ \hline P_2& 26 & 3 & 29 & 30 &2 &32\\ \hline P_7& 29 & 2 & 31 & 32 &1 &33\\ \hline \end{array}
Minimum total completion time of all parts = 33 hours
Question 3 |
An electric car manufacturer underestimated the
January sales of car by 20 units, while the actual
sales was 120 units. If the manufacturer uses
exponential smoothing method with a smoothing
constant of \alpha = 0.2, then the sales forecast for the
month of February of the same year is _______units
(in integer).
96 | |
104 | |
120 | |
128 |
Question 3 Explanation:
Actual sales in Jan ( D_{Jan}) = 120 units
Under estimate in Jan = D_{Jan}- F_{Jan}=20
D_{Jan} = 100 units
Smoothing constant ( \alpha = 0.2)
Forecast for Feb ( D_{Jan}) = ?
F_{Feb} = F_{Jan} + \alpha (D_{Jan} - F_{Jan})= 100 + 0.2 (120 - 100) = 104 \;units
Under estimate in Jan = D_{Jan}- F_{Jan}=20
D_{Jan} = 100 units
Smoothing constant ( \alpha = 0.2)
Forecast for Feb ( D_{Jan}) = ?
F_{Feb} = F_{Jan} + \alpha (D_{Jan} - F_{Jan})= 100 + 0.2 (120 - 100) = 104 \;units
Question 4 |
The demand and forecast of an item for five months are given in the table.
\begin{array}{|c|c|c|}\hline \textbf{Month} & \textbf{Demand} & \textbf{Forecast} \\ \hline \text{April} & \text{225} & \text{200} \\ \hline \text{May} & \text{220} & \text{240}\\ \hline \text{June} & \text{285} & \text{300} \\ \hline \text{July} & \text{290} & \text{270} \\ \hline \text{August} & \text{250} & \text{230} \\ \hline \end{array}
The Mean Absolute Percent Error (MAPE) in the forecast is _______% (round off to two decimal places)
\begin{array}{|c|c|c|}\hline \textbf{Month} & \textbf{Demand} & \textbf{Forecast} \\ \hline \text{April} & \text{225} & \text{200} \\ \hline \text{May} & \text{220} & \text{240}\\ \hline \text{June} & \text{285} & \text{300} \\ \hline \text{July} & \text{290} & \text{270} \\ \hline \text{August} & \text{250} & \text{230} \\ \hline \end{array}
The Mean Absolute Percent Error (MAPE) in the forecast is _______% (round off to two decimal places)
4.45 | |
12.25 | |
18.42 | |
8.08 |
Question 4 Explanation:
\begin{array}{|c|c|c|c|c|} \hline \text { March } & \mathrm{Di} & \mathrm{Fi} & \mathrm{ei} & \left|\frac{\mathrm{ei}}{\mathrm{Di}} \times 100\right| \\ \hline \text { April } & 225 & 200 & 25 & 11.11 \% \\ \hline \text { May } & 220 & 240 & -20 & 9.09 \% \\ \hline \text { June } & 285 & 300 & -15 & 5.26 \% \\ \hline \text { July } & 290 & 270 & 20 & 6.896 \% \\ \hline \text { August } & 250 & 230 & 20 & 8.0 \% \\ \hline & & & & \sum \frac{e i}{D i} \times 100 \mid=40.356 \\ \hline \end{array}
\text{MAPE}=\frac{\sum\left|\frac{e i}{D i} \times 100\right|}{n}=8.0712 \%
\text{MAPE}=\frac{\sum\left|\frac{e i}{D i} \times 100\right|}{n}=8.0712 \%
Question 5 |
Daily production capacity of a bearing manufacturing company is 30000 bearings. The daily demand of the bearing is 15000. The holding cost per year of keeping a bearing in the inventory is Rs. 20. The setup cost for the production of a batch is Rs. 1800. Assuming 300 working days in a year, the economic batch quantity in number of bearings is ______ (in integer).
36254 | |
20145 | |
40250 | |
42145 |
Question 5 Explanation:
\begin{aligned} Q^{\star} &=\sqrt{\frac{2 D \times C_{0}}{C_{h}} \times \frac{P}{P-d}} \\ &=\sqrt{\frac{2 \times 15000 \times 300 \times 1800}{20} \times\left(\frac{30000}{30000-15000}\right)} \\ &=40249.2 \simeq 40250 \text { units } \end{aligned}
There are 5 questions to complete.
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