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Page 1
Forecasting
• Simple Moving Average-
?? ?? =
1
?? (?? ?? + ?? ?? -1
+ ?? ?? -2
+ ?? ?? -3
+ ? )
• Moving Weight Average-
Weight Moving Average =
? W
i
D
i
n
t=1
? W
i
n
i=1
• Single (Simple) Exponential Smoothing-
?? ?? = ?? ?? -1
+ ?? (?? ?? -1
- ?? ?? -1
)
???? ?? ?? = (1 - ?? )?? ?? -1
+ ?? ?? ?? -1
if previous forecasting is not given
?? ?? = ?? ?? ?? + ?? (1 - ?? )?? ?? -1
+ ?? (1 - ?? )
2
?? ?? -2
… ….
Where ?? ?? = Smoothed average forecast for period t
?? ?? -1
=Previous period forecast
?? =Smoothing constant
• Linear Regression- ?? = ?? + ????
? ?? = ???? + ?? ? ??
? ???? = ?? ? ?? + ?? ? ?? 2
• Forecasting Error-
?? ?? = (?? ?? - ?? ?? )
• Bias-
???????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? =??
• Mean Absolute Deviation-
?????? =
?? ?? ? |?? ?? - ?? ?? |
?? ?? =??
• Mean Square Error-
?????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? ?? =??
• Mean Absolute Percentage Error-
???????? =
?? ?? ?
|?? ?? - ?? ?? |
?? ?? ?? ?? =?? × 100
Inventory
If D= demand/year, ?? ?? =Order cost, ?? ?? = Carrying cost, P= Purchase price/unit
?? *
=Economic Order Quantity, K= Production Rate and ?? ?? =Shortage Cost/unit/period
• Case-1 Purchase Model With Instantaneous Replenishment and Without Shortage-
1. ?????? ?? *
= v
2?? ?? ?? ?? ?? at EOQ ?????????????????? ???????? = ?????????? ????????
2. ???? . ???? ?????????? =
?? ?? *
3. ???????? ?????????? ?????? ?????????? =
?? *
??
4. ?????????? ???????? = ???????? ???????? + ?????????????????? ???????? + ?????????? ????????
= (?? × ?? ) + (
?? 2
× ?? ?? ) + (
?? ?? × ?? ?? )= (?? × ?? ) + v2?? ?? ?? ?? ??
• Case-2 Manufacturing Model Without Shortage-
Page 2
Forecasting
• Simple Moving Average-
?? ?? =
1
?? (?? ?? + ?? ?? -1
+ ?? ?? -2
+ ?? ?? -3
+ ? )
• Moving Weight Average-
Weight Moving Average =
? W
i
D
i
n
t=1
? W
i
n
i=1
• Single (Simple) Exponential Smoothing-
?? ?? = ?? ?? -1
+ ?? (?? ?? -1
- ?? ?? -1
)
???? ?? ?? = (1 - ?? )?? ?? -1
+ ?? ?? ?? -1
if previous forecasting is not given
?? ?? = ?? ?? ?? + ?? (1 - ?? )?? ?? -1
+ ?? (1 - ?? )
2
?? ?? -2
… ….
Where ?? ?? = Smoothed average forecast for period t
?? ?? -1
=Previous period forecast
?? =Smoothing constant
• Linear Regression- ?? = ?? + ????
? ?? = ???? + ?? ? ??
? ???? = ?? ? ?? + ?? ? ?? 2
• Forecasting Error-
?? ?? = (?? ?? - ?? ?? )
• Bias-
???????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? =??
• Mean Absolute Deviation-
?????? =
?? ?? ? |?? ?? - ?? ?? |
?? ?? =??
• Mean Square Error-
?????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? ?? =??
• Mean Absolute Percentage Error-
???????? =
?? ?? ?
|?? ?? - ?? ?? |
?? ?? ?? ?? =?? × 100
Inventory
If D= demand/year, ?? ?? =Order cost, ?? ?? = Carrying cost, P= Purchase price/unit
?? *
=Economic Order Quantity, K= Production Rate and ?? ?? =Shortage Cost/unit/period
• Case-1 Purchase Model With Instantaneous Replenishment and Without Shortage-
1. ?????? ?? *
= v
2?? ?? ?? ?? ?? at EOQ ?????????????????? ???????? = ?????????? ????????
2. ???? . ???? ?????????? =
?? ?? *
3. ???????? ?????????? ?????? ?????????? =
?? *
??
4. ?????????? ???????? = ???????? ???????? + ?????????????????? ???????? + ?????????? ????????
= (?? × ?? ) + (
?? 2
× ?? ?? ) + (
?? ?? × ?? ?? )= (?? × ?? ) + v2?? ?? ?? ?? ??
• Case-2 Manufacturing Model Without Shortage-
1. ?????? ?? *
=
v
2?? ?? ?? ?? ?? (1-
?? ?? )
2. ?? 1
=
?? *
??
3. ?? 2
=
?? *
(1-
?? ?? )
??
4. Total optimum cost = v2?? ?? ?? ?? ?? (1 -
?? ?? )
• Case 3 Purchase Model With Shortage-
1. ?? = ?????? = v
2?? ?? ?? ?? ?? (
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) , ?? 2
= ?? - ?? 1
3. ?? =
?? ?? , ?? 1
=
?? 1
?? and ?? 2
=
?? 2
??
4. Total optimum cost =v2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? )
• Case 4 Manufacturing Model With Shortfall
1. ?? = ?????? =
v
2?? ?? ?? ?? ?? ×(1-
?? ?? )
(
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) × (1 -
?? ?? )
Page 3
Forecasting
• Simple Moving Average-
?? ?? =
1
?? (?? ?? + ?? ?? -1
+ ?? ?? -2
+ ?? ?? -3
+ ? )
• Moving Weight Average-
Weight Moving Average =
? W
i
D
i
n
t=1
? W
i
n
i=1
• Single (Simple) Exponential Smoothing-
?? ?? = ?? ?? -1
+ ?? (?? ?? -1
- ?? ?? -1
)
???? ?? ?? = (1 - ?? )?? ?? -1
+ ?? ?? ?? -1
if previous forecasting is not given
?? ?? = ?? ?? ?? + ?? (1 - ?? )?? ?? -1
+ ?? (1 - ?? )
2
?? ?? -2
… ….
Where ?? ?? = Smoothed average forecast for period t
?? ?? -1
=Previous period forecast
?? =Smoothing constant
• Linear Regression- ?? = ?? + ????
? ?? = ???? + ?? ? ??
? ???? = ?? ? ?? + ?? ? ?? 2
• Forecasting Error-
?? ?? = (?? ?? - ?? ?? )
• Bias-
???????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? =??
• Mean Absolute Deviation-
?????? =
?? ?? ? |?? ?? - ?? ?? |
?? ?? =??
• Mean Square Error-
?????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? ?? =??
• Mean Absolute Percentage Error-
???????? =
?? ?? ?
|?? ?? - ?? ?? |
?? ?? ?? ?? =?? × 100
Inventory
If D= demand/year, ?? ?? =Order cost, ?? ?? = Carrying cost, P= Purchase price/unit
?? *
=Economic Order Quantity, K= Production Rate and ?? ?? =Shortage Cost/unit/period
• Case-1 Purchase Model With Instantaneous Replenishment and Without Shortage-
1. ?????? ?? *
= v
2?? ?? ?? ?? ?? at EOQ ?????????????????? ???????? = ?????????? ????????
2. ???? . ???? ?????????? =
?? ?? *
3. ???????? ?????????? ?????? ?????????? =
?? *
??
4. ?????????? ???????? = ???????? ???????? + ?????????????????? ???????? + ?????????? ????????
= (?? × ?? ) + (
?? 2
× ?? ?? ) + (
?? ?? × ?? ?? )= (?? × ?? ) + v2?? ?? ?? ?? ??
• Case-2 Manufacturing Model Without Shortage-
1. ?????? ?? *
=
v
2?? ?? ?? ?? ?? (1-
?? ?? )
2. ?? 1
=
?? *
??
3. ?? 2
=
?? *
(1-
?? ?? )
??
4. Total optimum cost = v2?? ?? ?? ?? ?? (1 -
?? ?? )
• Case 3 Purchase Model With Shortage-
1. ?? = ?????? = v
2?? ?? ?? ?? ?? (
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) , ?? 2
= ?? - ?? 1
3. ?? =
?? ?? , ?? 1
=
?? 1
?? and ?? 2
=
?? 2
??
4. Total optimum cost =v2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? )
• Case 4 Manufacturing Model With Shortfall
1. ?? = ?????? =
v
2?? ?? ?? ?? ?? ×(1-
?? ?? )
(
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) × (1 -
?? ?? )
3. ?? 1
= (1 -
?? ?? ) ?? - ?? 2
4. ?? =
?? ?? , ?? 1
=
?? 1
?? -?? and ?? 2
=
?? 1
??
1. ?? 3
=
?? 3
?? and ?? 4
=
?? 2
?? -??
• Lead Time Demand + Safety Stock = Reorder Point
PERT and CPM
• EFT = EST + activity time
• LFT = LST + Duration of activity
• Total Float-
• Free Float- FFo= (Ej-Ei)-Tij
• Independent Float -
Example-
1. Total float = L2 – (E1 + t12) = 57 – (20 + 19) = 18
2. Free float = E2 – E1 – t12 = 0
3. Independent float = E2 – (L1 + t12) = -18
• PERT Expected time- ?? ?? =
?? 0
+4?? ?? +?? ?? 6
1. t0 = Optimistic time i.e., shortest possible time to complete the activity
if all goes well.
2. tp = Pessimistic time i.e., longest time that an activity could take if
everything goes wrong.
3. tm = Most likely time i.e., normal time of an activity would take.
• Standard deviation-
• Variance -
• Crashing-
• Standard Normal Variation (SNV)-
Page 4
Forecasting
• Simple Moving Average-
?? ?? =
1
?? (?? ?? + ?? ?? -1
+ ?? ?? -2
+ ?? ?? -3
+ ? )
• Moving Weight Average-
Weight Moving Average =
? W
i
D
i
n
t=1
? W
i
n
i=1
• Single (Simple) Exponential Smoothing-
?? ?? = ?? ?? -1
+ ?? (?? ?? -1
- ?? ?? -1
)
???? ?? ?? = (1 - ?? )?? ?? -1
+ ?? ?? ?? -1
if previous forecasting is not given
?? ?? = ?? ?? ?? + ?? (1 - ?? )?? ?? -1
+ ?? (1 - ?? )
2
?? ?? -2
… ….
Where ?? ?? = Smoothed average forecast for period t
?? ?? -1
=Previous period forecast
?? =Smoothing constant
• Linear Regression- ?? = ?? + ????
? ?? = ???? + ?? ? ??
? ???? = ?? ? ?? + ?? ? ?? 2
• Forecasting Error-
?? ?? = (?? ?? - ?? ?? )
• Bias-
???????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? =??
• Mean Absolute Deviation-
?????? =
?? ?? ? |?? ?? - ?? ?? |
?? ?? =??
• Mean Square Error-
?????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? ?? =??
• Mean Absolute Percentage Error-
???????? =
?? ?? ?
|?? ?? - ?? ?? |
?? ?? ?? ?? =?? × 100
Inventory
If D= demand/year, ?? ?? =Order cost, ?? ?? = Carrying cost, P= Purchase price/unit
?? *
=Economic Order Quantity, K= Production Rate and ?? ?? =Shortage Cost/unit/period
• Case-1 Purchase Model With Instantaneous Replenishment and Without Shortage-
1. ?????? ?? *
= v
2?? ?? ?? ?? ?? at EOQ ?????????????????? ???????? = ?????????? ????????
2. ???? . ???? ?????????? =
?? ?? *
3. ???????? ?????????? ?????? ?????????? =
?? *
??
4. ?????????? ???????? = ???????? ???????? + ?????????????????? ???????? + ?????????? ????????
= (?? × ?? ) + (
?? 2
× ?? ?? ) + (
?? ?? × ?? ?? )= (?? × ?? ) + v2?? ?? ?? ?? ??
• Case-2 Manufacturing Model Without Shortage-
1. ?????? ?? *
=
v
2?? ?? ?? ?? ?? (1-
?? ?? )
2. ?? 1
=
?? *
??
3. ?? 2
=
?? *
(1-
?? ?? )
??
4. Total optimum cost = v2?? ?? ?? ?? ?? (1 -
?? ?? )
• Case 3 Purchase Model With Shortage-
1. ?? = ?????? = v
2?? ?? ?? ?? ?? (
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) , ?? 2
= ?? - ?? 1
3. ?? =
?? ?? , ?? 1
=
?? 1
?? and ?? 2
=
?? 2
??
4. Total optimum cost =v2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? )
• Case 4 Manufacturing Model With Shortfall
1. ?? = ?????? =
v
2?? ?? ?? ?? ?? ×(1-
?? ?? )
(
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) × (1 -
?? ?? )
3. ?? 1
= (1 -
?? ?? ) ?? - ?? 2
4. ?? =
?? ?? , ?? 1
=
?? 1
?? -?? and ?? 2
=
?? 1
??
1. ?? 3
=
?? 3
?? and ?? 4
=
?? 2
?? -??
• Lead Time Demand + Safety Stock = Reorder Point
PERT and CPM
• EFT = EST + activity time
• LFT = LST + Duration of activity
• Total Float-
• Free Float- FFo= (Ej-Ei)-Tij
• Independent Float -
Example-
1. Total float = L2 – (E1 + t12) = 57 – (20 + 19) = 18
2. Free float = E2 – E1 – t12 = 0
3. Independent float = E2 – (L1 + t12) = -18
• PERT Expected time- ?? ?? =
?? 0
+4?? ?? +?? ?? 6
1. t0 = Optimistic time i.e., shortest possible time to complete the activity
if all goes well.
2. tp = Pessimistic time i.e., longest time that an activity could take if
everything goes wrong.
3. tm = Most likely time i.e., normal time of an activity would take.
• Standard deviation-
• Variance -
• Crashing-
• Standard Normal Variation (SNV)-
Linear Programming
Simplex Method Case 1. Maximization Problem
3 1 5 2
/ 3 1 2 2 18 ( )
=+
+ ? -
MaxZ x x
s t x x I
1 4 ( )
2 6 ( )
1, 2 0
?-
?-
?
x II
x III
xx
Standard Form:
Max Z = 3x1 + 5x2 + 0w1 + 0w2 + 0w3
3x1 + 2x2 + w1 + 0w2 + 0w3 = 18
x1 + 0x2 + 0w1 + w2 + 0w3 = 4
0x1 + x2 + 0w1 + 0w2 + w3 = 6
To prepare initial Table:
Table - I
• Ij = (Zj-cj) = ( ?aij.ci)-cj
Interpretation of Simplex Table
Table - I
• Key Column ? Min Ij [ Most Negative ]
• Key Row ? Min positive ratio.
How to get next table ?
• Leaving variable : w3
• Entering variable : x2
• Key no. = 1
• For old key row : New No.= Old No./key No.
• For other rows:
( .).
( .)
..
.
=-
CorrespondingKeyRowNo
CorrespondingKeyColumnNo
NewNo Old No
KeyNo
• 18 ? 18 - (6*2)/1 = 6
• I(w3) = 0 ? 0 - [1*(-5)]/1 = 5
Page 5
Forecasting
• Simple Moving Average-
?? ?? =
1
?? (?? ?? + ?? ?? -1
+ ?? ?? -2
+ ?? ?? -3
+ ? )
• Moving Weight Average-
Weight Moving Average =
? W
i
D
i
n
t=1
? W
i
n
i=1
• Single (Simple) Exponential Smoothing-
?? ?? = ?? ?? -1
+ ?? (?? ?? -1
- ?? ?? -1
)
???? ?? ?? = (1 - ?? )?? ?? -1
+ ?? ?? ?? -1
if previous forecasting is not given
?? ?? = ?? ?? ?? + ?? (1 - ?? )?? ?? -1
+ ?? (1 - ?? )
2
?? ?? -2
… ….
Where ?? ?? = Smoothed average forecast for period t
?? ?? -1
=Previous period forecast
?? =Smoothing constant
• Linear Regression- ?? = ?? + ????
? ?? = ???? + ?? ? ??
? ???? = ?? ? ?? + ?? ? ?? 2
• Forecasting Error-
?? ?? = (?? ?? - ?? ?? )
• Bias-
???????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? =??
• Mean Absolute Deviation-
?????? =
?? ?? ? |?? ?? - ?? ?? |
?? ?? =??
• Mean Square Error-
?????? =
?? ?? ?(?? ?? - ?? ?? )
?? ?? ?? =??
• Mean Absolute Percentage Error-
???????? =
?? ?? ?
|?? ?? - ?? ?? |
?? ?? ?? ?? =?? × 100
Inventory
If D= demand/year, ?? ?? =Order cost, ?? ?? = Carrying cost, P= Purchase price/unit
?? *
=Economic Order Quantity, K= Production Rate and ?? ?? =Shortage Cost/unit/period
• Case-1 Purchase Model With Instantaneous Replenishment and Without Shortage-
1. ?????? ?? *
= v
2?? ?? ?? ?? ?? at EOQ ?????????????????? ???????? = ?????????? ????????
2. ???? . ???? ?????????? =
?? ?? *
3. ???????? ?????????? ?????? ?????????? =
?? *
??
4. ?????????? ???????? = ???????? ???????? + ?????????????????? ???????? + ?????????? ????????
= (?? × ?? ) + (
?? 2
× ?? ?? ) + (
?? ?? × ?? ?? )= (?? × ?? ) + v2?? ?? ?? ?? ??
• Case-2 Manufacturing Model Without Shortage-
1. ?????? ?? *
=
v
2?? ?? ?? ?? ?? (1-
?? ?? )
2. ?? 1
=
?? *
??
3. ?? 2
=
?? *
(1-
?? ?? )
??
4. Total optimum cost = v2?? ?? ?? ?? ?? (1 -
?? ?? )
• Case 3 Purchase Model With Shortage-
1. ?? = ?????? = v
2?? ?? ?? ?? ?? (
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) , ?? 2
= ?? - ?? 1
3. ?? =
?? ?? , ?? 1
=
?? 1
?? and ?? 2
=
?? 2
??
4. Total optimum cost =v2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? )
• Case 4 Manufacturing Model With Shortfall
1. ?? = ?????? =
v
2?? ?? ?? ?? ?? ×(1-
?? ?? )
(
?? ?? +?? ?? ?? ?? )
2. ?? 1
= v
2?? ?? ?? ?? ?? (
?? ?? ?? ?? +?? ?? ) × (1 -
?? ?? )
3. ?? 1
= (1 -
?? ?? ) ?? - ?? 2
4. ?? =
?? ?? , ?? 1
=
?? 1
?? -?? and ?? 2
=
?? 1
??
1. ?? 3
=
?? 3
?? and ?? 4
=
?? 2
?? -??
• Lead Time Demand + Safety Stock = Reorder Point
PERT and CPM
• EFT = EST + activity time
• LFT = LST + Duration of activity
• Total Float-
• Free Float- FFo= (Ej-Ei)-Tij
• Independent Float -
Example-
1. Total float = L2 – (E1 + t12) = 57 – (20 + 19) = 18
2. Free float = E2 – E1 – t12 = 0
3. Independent float = E2 – (L1 + t12) = -18
• PERT Expected time- ?? ?? =
?? 0
+4?? ?? +?? ?? 6
1. t0 = Optimistic time i.e., shortest possible time to complete the activity
if all goes well.
2. tp = Pessimistic time i.e., longest time that an activity could take if
everything goes wrong.
3. tm = Most likely time i.e., normal time of an activity would take.
• Standard deviation-
• Variance -
• Crashing-
• Standard Normal Variation (SNV)-
Linear Programming
Simplex Method Case 1. Maximization Problem
3 1 5 2
/ 3 1 2 2 18 ( )
=+
+ ? -
MaxZ x x
s t x x I
1 4 ( )
2 6 ( )
1, 2 0
?-
?-
?
x II
x III
xx
Standard Form:
Max Z = 3x1 + 5x2 + 0w1 + 0w2 + 0w3
3x1 + 2x2 + w1 + 0w2 + 0w3 = 18
x1 + 0x2 + 0w1 + w2 + 0w3 = 4
0x1 + x2 + 0w1 + 0w2 + w3 = 6
To prepare initial Table:
Table - I
• Ij = (Zj-cj) = ( ?aij.ci)-cj
Interpretation of Simplex Table
Table - I
• Key Column ? Min Ij [ Most Negative ]
• Key Row ? Min positive ratio.
How to get next table ?
• Leaving variable : w3
• Entering variable : x2
• Key no. = 1
• For old key row : New No.= Old No./key No.
• For other rows:
( .).
( .)
..
.
=-
CorrespondingKeyRowNo
CorrespondingKeyColumnNo
NewNo Old No
KeyNo
• 18 ? 18 - (6*2)/1 = 6
• I(w3) = 0 ? 0 - [1*(-5)]/1 = 5
Table - II
• Key Column ? Min Ij
• Key Row ? Min positive ratio
Table - III
• This is the final Table
The Optimal Solution is x1 = 2, x2 = 6
giving Z = 36
Type of Solutions : Basic, Feasible/Infeasible, Optimal/ Non-Optimal,
Unique/Alternative Optimal, Bounded/Unbounded, Degenerate/Non-Degenerate
• Analysis of Solution
1. This is a Basic solution, as values of basic variables are Positive
2. This is a feasible solution, as values of basic variables, not containing
Artificial Variable, are Positive and all constraints are satisfied
3. This Feasible solution is an Optimal, as all values in Index Row are positive.
4. If there is an Artificial Variable, as Basic variable in final table, it is called as
Infeasible solution
5. This solution is unique Optimal, as the number of zeroes are equal to number
of basic variables in Index Row in final Table.
6. If the number of zeroes are more than number of basic variables in Index Row
in final Table, it is a case of more than one optimal solutions.
7. This is a Bounded Solution, as the values of all Basic variables in final table,
are finite positive.
8. This is a Non-degenerate Solution, as value of none of the basic variables is
Zero , in final table.
9. If value of at least one of the basic variables is Zero in Index Row in final
Table, it is a Degenerate Solution.
• Duality With Example
1. Case-1
Max.Z = x1 - x2 + 3x3
s/t x1 + x2 + x3 ? 10
2 x1 - x2 - x3 ? 2
2x1 - 2x2 - 3x3 ? 6
x1 , x2, x3 ? 0
Dual of this would be
Min Z = 10y1 +2y2 + 6y3
s/t y1 + 2 y2 + 2y3 ? 1
y1 - y2 - 2y3 ? -1
y1 - y2 - 3y3 ? 3
y1 , y3 , y3 ? 0 ,
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FAQs on Industrial Engineering Formulas for GATE ME Exam - Industrial Engineering - Mechanical Engineering
| 1. What are some important industrial engineering formulas that are frequently asked in the GATE ME Exam for Mechanical Engineering? |  |
Ans. Some important industrial engineering formulas that are frequently asked in the GATE ME Exam for Mechanical Engineering include:
1. Time Study Formula: Time Study is a technique used to determine the standard time required to perform a specific task. The formula used for time study is:
Standard Time = (Observed Time x Rating Factor) / Performance Rating
2. Work Sampling Formula: Work Sampling is a statistical technique used to determine the percentage of time spent on different activities in a work process. The formula used for work sampling is:
Percentage of Time = (Number of Occurrences of an Activity / Total Number of Observations) x 100
3. Efficiency Formula: Efficiency is a measure of how well a process or system is performing. The formula used for efficiency calculation is:
Efficiency = (Total Output / Total Input) x 100
4. Productivity Formula: Productivity is a measure of the output produced per unit of input. The formula used for productivity calculation is:
Productivity = (Total Output / Total Input)
5. Break-even Analysis Formula: Break-even analysis is used to determine the point at which the total revenue equals the total cost, resulting in neither profit nor loss. The formula used for break-even analysis is:
Break-even Point = Fixed Costs / (Selling Price per Unit - Variable Cost per Unit)
| 2. How can industrial engineering formulas help in the GATE ME Exam preparation for Mechanical Engineering? | |
Ans. Industrial engineering formulas play a crucial role in the GATE ME Exam preparation for Mechanical Engineering in the following ways:
1. Understanding Concepts: Learning and practicing industrial engineering formulas helps in understanding the underlying concepts and principles of industrial engineering.
2. Problem Solving: Industrial engineering formulas provide a structured approach to solving problems related to time study, work sampling, efficiency, productivity, break-even analysis, and other areas of industrial engineering.
3. Time Management: By knowing and applying industrial engineering formulas, students can solve numerical problems more efficiently, saving valuable time during the exam.
4. Scoring High Marks: Since industrial engineering formulas are frequently asked in the GATE ME Exam, having a strong command over these formulas increases the chances of scoring high marks in the exam.
5. Real-life Applications: Industrial engineering formulas are not only important for exams but also have practical applications in industries. Understanding these formulas helps in applying them to real-life scenarios.
| 3. Can you provide an example of how the time study formula is used in the GATE ME Exam? | |
Ans. Sure! Here's an example of how the time study formula is used in the GATE ME Exam:
Suppose a time study is conducted to determine the standard time required to assemble a product. The observed time for assembling one unit is measured as 10 minutes, and the performance rating is determined as 90%. Now, we can calculate the standard time using the time study formula:
Standard Time = (Observed Time x Rating Factor) / Performance Rating
Standard Time = (10 minutes x 100) / 90
Standard Time = 11.11 minutes
So, the standard time required to assemble one unit of the product is 11.11 minutes. This calculation helps in analyzing the efficiency and productivity of the assembly process.
| 4. How is efficiency different from productivity in the context of industrial engineering? | |
Ans. Efficiency and productivity are two important metrics used in the context of industrial engineering. Here's how they differ:
Efficiency: Efficiency is a measure of how well a process or system is performing. It represents the ratio of output achieved to the input used. It is usually expressed as a percentage. Higher efficiency indicates that a process or system is utilizing its resources effectively and producing the desired output with minimum waste.
Productivity: Productivity, on the other hand, is a measure of the output produced per unit of input. It represents the overall effectiveness of a system or process in converting inputs into outputs. It is usually measured in terms of units produced per unit of time or per unit of input. Higher productivity indicates that a system or process is capable of producing more output with the same level of input.
In summary, efficiency focuses on how well resources are utilized, while productivity focuses on the overall output achieved relative to the input used.
| 5. How can break-even analysis be applied in real-life industrial scenarios? | |
Ans. Break-even analysis is a valuable tool used in real-life industrial scenarios to make informed decisions about pricing, production volume, and cost management. Here are some applications of break-even analysis in industrial scenarios:
1. Pricing Decisions: Break-even analysis helps determine the minimum selling price required to cover all costs and achieve the desired profit margin. It helps in setting competitive prices that ensure profitability.
2. Production Planning: Break-even analysis helps in determining the production volume required to cover all costs and avoid losses. It helps in optimizing production levels, allocating resources efficiently, and managing inventory.
3. Cost Management: Break-even analysis helps identify the cost structure of a business and highlights areas where costs can be reduced or controlled. It aids in identifying cost drivers and implementing cost-saving measures.
4. New Product Introduction: Break-even analysis helps assess the feasibility of introducing a new product by estimating the sales volume required to cover all costs and generate profits. It aids in evaluating the potential profitability of new product ventures.
5. Financial Decision-making: Break-even analysis provides insights into the financial health of a business and helps in making informed decisions related to investments, expansion, diversification, or downsizing.
In conclusion, break-even analysis is a versatile tool that helps in various aspects of industrial decision-making, enabling businesses to optimize their operations and achieve profitability.
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