Exponential Smoothing Method MCQ Quiz - Objective Question with Answer for Exponential Smoothing Method - Download Free PDF
Last updated on Apr 3, 2025
Latest Exponential Smoothing Method MCQ Objective Questions
Exponential Smoothing Method Question 1:
For a hotel, the actual demand for disposable cup was 600 units in January and 700 units February. The forecast for the month of January was 500 units. What will be forecast for the month of March. Use simple exponential smoothening method. [Smoothening coefficient = 0.8]
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 1 Detailed Solution
Concept:
In Simple Exponential Smoothing, the forecast is updated based on the previous actual and forecasted values using the formula:
\( F_{t} = \alpha A_{t-1} + (1 - \alpha) F_{t-1} \)
Where, Ft is the forecast for the current period, \(A_{t-1}\) is the actual demand of the previous period, and \(F_{t-1}\) is the forecast for the previous period.
Calculation:
Given:
Actual demand in January = 600 units, Forecast for January = 500 units
Actual demand in February = 700 units
Smoothing coefficient, \(\alpha\) = 0.8
Forecast for February:
\( F_{Feb} = 0.8 \times 600 + 0.2 \times 500 = 480 + 100 = 580 \)
Forecast for March:
\( F_{Mar} = 0.8 \times 700 + 0.2 \times 580 = 560 + 116 = 676 \)
Exponential Smoothing Method Question 2:
The sales of a product during the last four years were 840, 860, 850, 870 units. The forecast for the fourth year was 855. If the forecast for the fifth year, using simple exponential smoothening, is equal to the forecast using the three period moving average, what will be the value of exponential smoothening constant?
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 2 Detailed Solution
Concept:
The given sales data for the last four years are: 840, 860, 850, and 870 units.
The three-period moving average is given by:
\( F_5 = \frac{A_2 + A_3 + A_4}{3} \)
\( F_5 = \frac{860 + 850 + 870}{3} \)
\( F_5 = \frac{2580}{3} = 860 \)
The simple exponential smoothing equation is:
\( F_5 = \alpha A_4 + (1 - \alpha) F_4 \)
Where:
A4 = 870 (Actual sales in Year 4)
F4 = 855 (Forecast for Year 4)
F5 = 860 (Forecast for Year 5 using both methods)
Calculation:
Substituting values in the equation:
\( 860 = \alpha (870) + (1 - \alpha)(855) \)
Expanding:
\( 860 = 870\alpha + 855 - 855\alpha \)
\( 860 - 855 = (870 - 855) \alpha \)
\( 5 = 15\alpha \)
\( \alpha = \frac{5}{15} = \frac{1}{3} \)
Exponential Smoothing Method Question 3:
The sales of water purifiers in a shop in 4 consecutive months are as given below. The magnitude of difference in the forecast for the next month using the exponential smoothening method with a smoothening constant of 0.3 and weighted moving average method with the weights as 0.1, 0.2, 0.3 and 0.4 will be ______water purifiers (in Integer)
Months | Di |
January | 1080 |
February | 950 |
March | 1050 |
April | 1120 |
Answer (Detailed Solution Below) 5.5 - 6.5
Exponential Smoothing Method Question 3 Detailed Solution
Calculation:
Forecast for the January is not given.
Assuming the forecast for January is equal to demand in January. so, the table can be generated using the forecasting formula for February, March, and April month to get the forecast for May month.
Months | Di | Fi (Forecast) |
January | 1080 | 1080 |
February | 950 | 1080 |
March | 1050 | 1041 |
April | 1120 | 1044 |
F2 = F1 + α (D1 - F1)
⇒ F2 = 1080 + 0.3 (1080 - 1080) = 1080
⇒ F3 = 1080 + 0.3 (950 - 1080) = 1041
⇒ F4 = 1041 + 0.3 (1050 - 1041) = 1044
Next month's forecast using exponential smoothening
F5 = F4 + α (D4 - F4)
F5 =1044 + 0.3 (1120-1044) = 1066.8
⇒ F5 ≃1067 water purifiers
Forecast using weighted moving average is given as,
⇒ F'5 =1120 × 0.4 + 1050 × 0.3 + 950 × 0.2 + 1080 × 0.1
⇒ F'5 = 1061 water purifiers
Therefore difference in forecast = 1067-1061 = 6
Exponential Smoothing Method Question 4:
The expression to calculate Ft (Smooth average forecast for period t) in exponential smoothing method is [Where, Ft-1 = Previous period forecast
Dt-1 = Previous period demand, α = Smoothing constant]
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 4 Detailed Solution
Concept:
Exponential smoothing method:
\({F_t} = {F_{t - 1}} + α \left[ {{D_{t - 1}} - {F_{t - 1}}} \right]\)
where α = smoothing constant and the term (Dt-1 - Ft-1) represents the error present in the forecast.
Additional Information In the exponential smoothing method of the forecast, the forecast for the next period is equal to
Ft = α Dt-1 + (1 - α) Ft-1
If we further expand the expression
Ft = α Dt-1 + (1 - α) (α Dt-2 + (1-α) Ft-2
Ft = α Dt-1 + α (1-α ) Dt-2 + (1 - α )2 Ft-2
If we put the value of α = 0.8 (Close to 1) then,
Ft = 0.8 Dt-1 + 0.16 Dt-2 + 0.04 Ft-2
It can be observed from the above expression that recent data is given more weightage compared to the previous data, therefore, the higher value of exponential smoothing constant (Close to 1) is used for changing pattern of demand or to follow recent demand.
Exponential Smoothing Method Question 5:
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 α = 0.2, then the sales forecast for the month of February of the same year is ______ units (in integer).
Answer (Detailed Solution Below) 104
Exponential Smoothing Method Question 5 Detailed Solution
Concept:
Exponential smoothing method:
This method gives weight to all the previous data and the pattern of weight assigned is exponentially decreasing in order with most recent data is given highest weight.
The forecast for next period is equal to:
Ft = α Dt-1 + (1- α) Ft-1
where, Ft-1 = old forecast, α = exponential smoothing constant,
Dt-1 = latest figure sale or latest demand, Ft = next forecast
Calculation:
Given:
α = 0.2 , Dt-1 = 120 units , error e = 20 units
e = Dt-1 - Ft-1 = 20 ⇒ Ft-1 = 100
Ft = α Dt-1 + (1- α) Ft-1
Ft = 0.2 × 120 + (1- 0.2) 100
Ft = 24 + 80 = 104 units
Top Exponential Smoothing Method MCQ Objective Questions
Which one of the following is not a casual forecasting method?
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 6 Detailed Solution
Download Solution PDFExplanation:
- Forecasting is the prediction of future sells or demand of the particular product.
- It is a projection based upon past data and art of human judgement.
Types of forecasting method
Qualitative or Subjective |
Quantitative or Objective |
Judgemental
|
Time series
Casual or Econometrics
|
Used for long-range and new product |
Used for Short-range and for old products |
In exponential smoothening method, which one of the following is true?
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 7 Detailed Solution
Download Solution PDFConcept:
The general form of forecasting is Ft = Ft-1 + α. (Dt-1 – Ft-1)
where α = smoothing constant and its range is 0 ≤ α ≤ 1
For Immediate forecast → high value of "α" is high and less for other forecasts.
Hence the high value of forecast is only chosen when the nature of demand is not reliable rather unstable.
\(α = \frac{2}{{n + 1}}\)
where, n = no. of period of moving average, Ft = recent forecast, Ft-1 = previous forecast, Dt-1 = previous demand
- If α = 0, then Ft = Ft-1 ……….…..(limit of stability)
- If α = 1 then Ft = Dt-1…………….(limit of responsiveness)
The current period forecast becomes equal to last period forecast for the value of smoothing constant equal to
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 8 Detailed Solution
Download Solution PDFExplanation:
Forecast value in Smoothing constant method is given by-
Ft = Ft-1 + α [ Dt-1 - Ft-1 ]
where Ft = Current period forecast, Ft-1 = last period forecast, Dt-1 = last period demand, α = smoothing constant
for α = 0
Ft = Ft-1
Hence for α = 0 only the current period forcast becomes equal to last period forecast.
Important Points
For α = 1 current period forecast will become equal to the last period demand.
Which of the following is a technique used for forecasting?
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 9 Detailed Solution
Download Solution PDFExplanation:
Forecasting
- Forecasting is defined as estimating the future value that a parameter will take. Most scientific forecasting methods forecast the future value using past data.
- Some simple forecasting models using time series data are simple average, moving average and simple exponential smoothing.
Moving average Method or rolling average Method:
- In this method, fresh average is calculated at the end of each period by adding the actual demand data for the most recent period and deleting the data for the order period. It gives equal weight to each of the most recent observations.
\({F_{n+1}} = \frac{{{D_1} + {D_2} + {D_3} + {D_4} + \ldots \ldots \ldots \ldots \ldots \ldots + {D_n}}}{n}\)
Weighted moving average Method:
- This method gives unequal weight to each demand data with more weight to recent data.
\({F_{n+1}} = \left[ {{w_{1}} \times {D_{1}} +{w_{2}\times {D_{2}}} +..........+ {w_{n}} \times {D_{n}}} \right]\)
Exponential Smoothing Method:
- This method gives weight to all the previous data and the pattern of weight assigned is exponentially decreasing in order with most recent data is given the highest weight.
- In exponential smoothing method of forecast, the forecast for the next period is equal to
Ft = α Dt-1 + (1 - α) Ft-1
where, Dt-1 = latest figure sale or latest demand, Ft-1 = old forecast, α = exponential smoothing constant
Additional Information
Project
- A project may be defined as a combination of interrelated activities which must be executed in a certain order before the entire task can be completed.
- The aim of planning is to develop a sequence of activities of the project so that the project completion time and cost are properly balanced.
- To meet the objective of systematic planning, the management has evolved several techniques applying network strategy.
- PERT (Programme Evaluation and Review Technique) and CPM (Critical Path Method) are network techniques which have been widely used for planning, scheduling and controlling the large and complex projects.
Difference between PERT and CPM (Critical Path Method)
PERT |
CPM |
1. Probabilistic approach |
1. Deterministic approach |
2. Three-time estimate |
2. One - time estimate |
3. Event oriented network model |
3. Activity-oriented network model |
4. The slack concept is used |
4. Float concept is used |
5. Project crashing is not possible |
5. Project crashing is possible |
6. Deals with probabilistic time estimates |
6. Deals with deterministic time estimates |
Gantt charts:
- Gantt charts are mainly used to allocate resources to activities.
- The resources allocated to activities include staff, hardware, and software.
- Gantt charts are useful for resource planning. A Gantt chart is a special type of bar chart where each bar represents an activity.
- The bars are drawn along a timeline.
- The length of each bar is proportional to the duration of time planned for the corresponding activity.
Control charts:
- Control chart is a graphical representation of the collected information.
- It indicates whether a process is in control or out of control.
- It determines process variability and detects unusual variations taking place in a process.
- It ensures product quality level.
- It provides information about the selection and setting of tolerance limits.
Which of the following forecasting methods takes a fraction of forecast error into account for the next period forecast?
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 10 Detailed Solution
Download Solution PDFExplanation:
Forecast error (ei) for a period is defined as the difference between actual and actual and forecasted demand.
ei = Actual demand - Forecast demand ⇒ Di - Fi
Exponential forecasting:
\({F_T} = {F_{T - 1}} + α ({D_{T - 1}} - {F_{T - 1}})\)
where
FT is the forecast for the next period
\(({D_{T - 1}} - {F_{T - 1}})\) is the forecast error and
α is the smoothing constant.
Thus exponential smoothing takes into account the forecast error of the previous period, for the forecast of the next period.
Additional Information
Simple Average Method:
In the simple moving average, we take the average of the past data points for future demand.
For 'n' period moving average forecast will be given by:
\({F_{n+1}} = \frac{{{D_1}\;+\;{D_2}\;+\;{D_3}\;+ \;{D_4}\;+\;\ldots \ldots \ldots \ldots \ldots \ldots \;+\;{D_n}}}{n}\)
Weighted Moving Average Method:
In weighted moving average, the highest weightage is given to recent data & it decreases for older data points.
For n period weighted moving average, weightage is as follows:
\(\frac{n}{{{\rm{\Sigma }}n}},\;\frac{{n - 1}}{{{\rm{\Sigma }}n}},\;\frac{{n - 2}}{{{\rm{\Sigma }}n}}, \;- - - - - - - ,\frac{1}{{{\rm{\Sigma }}n}}\)
\({F_{n+1}} = \left[ ({{w_{1}} \times {D_{1}})\;+\;({w_{2}\times {D_{2}}})\;+\;..........+\;({w_{n}} \times {D_{n}}})\right]\)
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 _______.
Answer (Detailed Solution Below) 568 - 570
Exponential Smoothing Method Question 11 Detailed Solution
Download Solution PDFConcept:
𝐹𝑡+1 = 𝛼𝐷𝑡 + (1 − 𝛼) 𝐹𝑡 = 𝐹𝑡 + α (Dt - Ft)
This equation can be used when simple exponential smoothing is used as a forecasting model.
Here Ft represents the forecast for period t, Dt is the known demand for period t and α is the smoothing constant.
Calculation:
Forecaster for February:
F2 = 400 + 0.75 (500 – 400) = 475
Forecaster for March:
F3 = 475 + 0.75 (600 – 475) = 568.75An 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 α = 0.2, then the sales forecast for the month of February of the same year is ______ units (in integer).
Answer (Detailed Solution Below) 104
Exponential Smoothing Method Question 12 Detailed Solution
Download Solution PDFConcept:
Exponential smoothing method:
This method gives weight to all the previous data and the pattern of weight assigned is exponentially decreasing in order with most recent data is given highest weight.
The forecast for next period is equal to:
Ft = α Dt-1 + (1- α) Ft-1
where, Ft-1 = old forecast, α = exponential smoothing constant,
Dt-1 = latest figure sale or latest demand, Ft = next forecast
Calculation:
Given:
α = 0.2 , Dt-1 = 120 units , error e = 20 units
e = Dt-1 - Ft-1 = 20 ⇒ Ft-1 = 100
Ft = α Dt-1 + (1- α) Ft-1
Ft = 0.2 × 120 + (1- 0.2) 100
Ft = 24 + 80 = 104 units
The sale of cycles in a shop in four consecutive months are given as 70, 68, 82, 95. Exponentially smoothing average method with a smoothing factor of 0.4 is used in forecasting. The expected number of sales in the next month is
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 13 Detailed Solution
Download Solution PDFConcept:
When only demand/sales are given from previous data, then forecast for the next month:
\(F_t=F_{t-1}+\alpha(D_{t-1}-F_{t-1})\)
ft = forecast of next month,
ft-1 = forecast for the previous month and so on.
Dt-1 = Demand of the previous month
Calculation:
Given:
Month |
Dt |
Ft |
1 |
70 |
70 |
2 |
68 |
|
3 |
82 |
|
4 |
95 |
|
\(F_t=F_{t-1}+\alpha(D_{t-1}-F_{t-1})\)
\(F_2=F_{1}+\alpha(D_{1}-F_{1})\)
\(F_2=70+0.4 \times (70-70)\)
F2 = 70
\(F_3=F_{2}+\alpha(D_{2}-F_{2})\)
\(F_3=70+0.4 \times (68-70)\)
F3 = 69.2
\(F_4=F_{3}+\alpha(D_{3}-F_{3})\)
\(F_4=69.2+0.4 \times (82-69.2)\)
F4 = 74.32
\(F_5=F_{4}+\alpha(D_{4}-F_{4})\)
\(F_5=74.32+0.4 \times (95-74.32)\)
F5 = 82.59
Note: In the options, 82.59 is not available so always select the higher number from the given answer.
In simple exponential smoothing forecasting, to give higher weightage to recent demand information, the smoothing constant must be close to
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 14 Detailed Solution
Download Solution PDFExplanation:
In exponential smoothing method of forecast the forecast for the next period is equal to
Ft = α Dt-1 + (1 - α) Ft-1
If we further expand the expression
Ft = α Dt-1 + (1 - α) (α Dt-2 + (1-α) Ft-2
Ft = α Dt-1 + α (1-α ) Dt-2 + (1 - α )2 Ft-2
If we put the value of α = 0.8 (Close to 1) then,
Ft = 0.8 Dt-1 + 0.16 Dt-2 + 0.04 Ft-2
It can be observed from the above expression that recent data is given more weightage compared to the previous data, therefore, the higher value of exponential smoothing constant (Close to 1) is used for changing pattern of demand or to follow recent demand.
If the actual demand of a product is 62, a previous year's forecast is 57, and the value of smoothing constant is 0.3, what would be the forecast for the current year using exponential smoothing method of forecasting?
Answer (Detailed Solution Below)
Exponential Smoothing Method Question 15 Detailed Solution
Download Solution PDFConcept:
For exponential smoothing method,
\(F_t = F_{t-1} + α (D_{t-1} – F_{t-1})\)
where Ft-1 = previous year's forecast, Ft = current year's forecast, Dt-1 = previous year's demand, α = smoothing constant
Calculation:
Given:
Ft-1 = 57, Dt-1 = 62, α = 0.3
\(F_t = F_{t-1} + α (D_{t-1} – F_{t-1})\)
\(\Rightarrow F_{t}= 57+ 0.3\times (62 – 57)\)
F t = 58.5.