Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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}
Question 5 |
The forecast for the monthly demand of a product is given in the table below.
The forecast is made by using the exponential smoothing method. The exponential smoothing coefficient used in forecasting the demand is
The forecast is made by using the exponential smoothing method. The exponential smoothing coefficient used in forecasting the demand is
0.1 | |
0.4 | |
0.5 | |
1 |
Question 5 Explanation:
\begin{aligned} F_{t}&=F_{t-1}+\alpha\left(D_{t-1}-F_{t-1}\right)\\ \text{For 2nd month }F_{t}&=31.8, \text{ for }1^{\text {st }} \text{month}\\ F_{t-1} &=32 \text { and } D_{t-1}=30 \\ 31.8 &=32+\alpha(30-32) \\ 2 \alpha &=32-31.8 \\ \alpha &=0.1 \end{aligned}
Question 6 |
The table presents the demand of a product. By simple three-months moving average method, the demand-forecast of the product for the month of September is
490 | |
510 | |
530 | |
536.67 |
Question 6 Explanation:
3 month moving average method :
\begin{aligned} \mathrm{F}_{\mathrm{sep}} &=\frac{\mathrm{D}_{\mathrm{Aug}}+\mathrm{D}_{\mathrm{July}}+\mathrm{D}_{\text {June }}}{3} \\ &=\frac{560+475+495}{3}=510 \end{aligned}
\begin{aligned} \mathrm{F}_{\mathrm{sep}} &=\frac{\mathrm{D}_{\mathrm{Aug}}+\mathrm{D}_{\mathrm{July}}+\mathrm{D}_{\text {June }}}{3} \\ &=\frac{560+475+495}{3}=510 \end{aligned}
Question 7 |
The time series forecasting method that gives equal weightage to each of the m most recent observations is
Moving average method | |
Exponential smoothing with linear trend | |
Triple Exponential smoothing | |
Kalman Filter |
Question 7 Explanation:
It gives equal weightage to all data points.
Question 8 |
The demand for a two-wheeler was 900 units and 1030 units in April 2015 and May 2015, respectively. The forecast for the month of April 2015 was 850 units. Considering a smoothing constant of 0.6, the forecast for the month of June 2015 is
850 units | |
927 units | |
965 units | |
970 units |
Question 8 Explanation:
\begin{aligned} F_{May} &=F_{April} +0.6(D_{April}-F_{April})\\ &=580+0.6(900-850) \\ &= 850+30\\ &= 880\; unit\\ F_{June} &=F_{May} +0.6(D_{May}-F_{May})\\ &=880+0.6(1030-880) \\ &= 880+90\\ &= 970\; unit\\ \end{aligned}
Question 9 |
Sales data of a product is given in the following table:
Regarding forecast for the month of June, which one of the following statements is TRUE?
Regarding forecast for the month of June, which one of the following statements is TRUE?
Moving average will forecast a higher value compared to regression. | |
Higher the value of order N, the greater will be the forecast value by moving average. | |
Exponential smoothing will forecast a higher value compared to regression. | |
Regression will forecast a higher value compared to moving average. |
Question 9 Explanation:
As regression follow a pattern
\text{(forecast)}_{\text {June }} according to regression \gt 25
but according to moving average,
T_{\text {June }} \lt 25
\text{(forecast)}_{\text {June }} according to regression \gt 25
but according to moving average,
T_{\text {June }} \lt 25
Question 10 |
For a canteen, the actual demand for disposable cups was 500 units in January and 600 units in February. The forecast for the month of January was 400 units. The forecast for the month of March considering smoothing coefficient as 0.75 is ________
569 | |
986 | |
451 | |
258 |
Question 10 Explanation:
\begin{aligned} F_{\mathrm{Feb}} &=F_{\mathrm{Jan}}+\alpha\left[D_{\mathrm{Jan}}-F_{\mathrm{Jan}}\right] \\ &=400+0.75[500-400] \\ &=475 \\ F_{\mathrm{March}} &=F_{\mathrm{Feb}}+\alpha\left[D_{\mathrm{Feb}}-F_{\mathrm{Feb}}\right] \\ &=475+0.75[600-475] \\ &=568.75 \approx 569 \end{aligned}
There are 10 questions to complete.
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