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Industrial Engineering Test 2

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Industrial Engineering Test 2
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  • Question 1
    1 / -0
    In a forecasting analysis, it is required that higher weightage be given to recent demand information, the smoothening constant for this should be_________
    Solution

    Concept:

    By the exponential smoothing method:

    \({F_t} = {F_{t - 1}} + \alpha \left[ {{D_{t - 1}} - {F_{t - 1}}} \right]\)

    If we have \(\alpha = 1\)

    \({F_t} = {D_{t - 1}}\)

    Thus, the forecast equals the previous demand.
  • Question 2
    1 / -0

    Calculate the difference between 4-week and three-week moving average forecast for 2020

    Year

    2015

    2016

    2017

    2018

    2019

    Demand

    55

    60

    58

    65

    72

    Solution

    Explanation:

    Three-week moving average,

    \({x_1} = \frac{{72 + 65 + 58}}{3}\)

    x1 = 65

    Now,

    Four-week moving average,

    \({x_2} = \frac{{72 + 65 + 58 + 60}}{4}\)

    x2 = 63.75

    ⇒ x1 – x2 = 65 – 63.75

    ∴ x1 – x2 = 1.25
  • Question 3
    1 / -0
    If the previous year forecast was 110 units and actual demand was 115 units, the forecast for current year using exponential smoothening method with smoothening constant equal to 0.3 is _____
    Solution

    Concept:

    Fn+1 = Fn + α (Dn – Fn)

    Calculation:

    Given:

    D1 = 115, F1 = 110, α = 0.3

    Now,

    F2 = F1 + α (D1 – F1)

    F2 = 110 + 0.3 (115 - 110)

    ∴ F2 = 111.5
  • Question 4
    1 / -0

    Find the ratio of mean forecast error to mean absolute deviation. For the given data _____.

    Year

    Demand

    Forecast

    2013

    250

    235

    2014

    260

    265

    2015

    265

    272

    2016

    270

    252

    Solution

    Concept:

    \({\rm{Mean\;forecast\;error\;}}\left( {MFE} \right) = \frac{{\sum \left( {{D_i} - {F_i}} \right)}}{n}\)

    \({\rm{Mean\;absolute\;deviation\;}}\left( {MAD} \right) = \frac{{\sum \left| {{D_i} - {F_i}} \right|}}{n}\)

    Calculation:

    Demand

    (Di)

    Forecast

    (Fi)

    Di - Fi

    |Di - Fi|

    250

    235

    15

    15

    260

    265

    -5

    5

    265

    272

    -7

    7

    270

    252

    18

    18

     

     

    ∑ Di - Fi = 21

    ∑ = 45

     

    \(MFE = \frac{{\mathop \sum \nolimits_{i = 1}^4 \left( {{D_i} - {F_i}} \right)}}{4}\)

    \(\therefore MFE = \frac{{21}}{4}\)

    \(MAD = \frac{{\mathop \sum \nolimits_{i = 1}^4 \left| {{D_i} - {F_i}} \right|}}{4}\)

    \(\therefore MAD = \frac{{45}}{4}\)

    Now,

    \({\rm{Required\;ratio\;}} = \frac{{MFE}}{{MAD}} = \frac{{21}}{{45}}\)

    ∴ Required ratio = 0.466
  • Question 5
    1 / -0
    In a forecasting model, the linear regression technique was used for a time series forecasting method which gave the equation: F = 7 + 3t, where F is the forecast for the period. The demand for the five periods of time were 11, 14, 16, 20, 25. The sum of the absolute deviations for the above data is:
    Solution

    Concept:

    The absolute deviation is |D – F|

    The forecast for each period can be calculated from the above regression line equation.

    Calculation:

    Period(t)

    Dt

    Ft = 7 + 3t

    |Dt – Ft|

    1

    11

    10

    1

    2

    14

    13

    1

    3

    16

    16

    0

    4

    20

    19

    1

    5

    25

    22

    3


    Sum of absolute deviations is = 1 + 1 + 0 + 3 = 6

  • Question 6
    1 / -0

    For a regression model

    ∑ x = 36, ∑ y = 5432, ∑ x2 = 204, ∑ xy = 26108.8 and n = 8

    The forecast regression equation slope and constant for the data, respectively, are___
    Solution

    Concept:

    For regression model

    y = a + bx

    ∑ y = na + b ∑ x

    ∑ xy = a ∑ x + b ∑ x2

    Where, a = constant

    b = slope of equation

    ∴ From given data,

    5432 = 8a + b (36)   …1)

    Now,

    26108.8 = 36a + 204b   …2)

    On solving, equation 1) and 2)

    We get,

    a = constant = 500.628 and b = slope = 39.638
  • Question 7
    1 / -0

    Five architectural rendering jobs are waiting to be assigned at OM architects. Their work (processing) times & due dates are given in following table.

    Job

    Job work

    time (days)

    Job due

    date (days)

    A

    B

    C

    D

    E

    5

    2

    7

    3

    8

    8

    6

    17

    14

    21

     

    The jobs are sequenced according to SPT rule, then find ratio between average job completion time & total job lateness time?

    Solution

    Job sequence

    Processing time

    Flow time

    Due date

    Job lateness

    In

    Out

    B

    D

    A

    C

    E

    2

    3

    5

    7

    8

    0

    2

    5

    10

    17

    2

    5

    10

    17

    25

    6

    14

    8

    17

    21

    0

    0

    2

    0

    4

     

    Job flow time = 59

     

    Total lateness = 6

     

    Average job completion time \( = \frac{{59}}{5}\) days.

    Total lateness = 6 days

    Required ratio \( = \frac{{\left( {\frac{{59}}{5}} \right)}}{6} = 1.97\)

  • Question 8
    1 / -0

    Two operations facing, and threading are to be performed on 6 jobs. The time required for performing the jobs are shown below.

    Job

    Time for facing in minutes

    Time for threading in minutes

    1

    3

    8

    2

    12

    10

    3

    5

    9

    4

    2

    6

    5

    9

    3

    6

    11

    1

     

    The total idle time for the operations is________minutes

    Solution

    Concept:

    The jobs will be sequenced using the Johnson rule and then the idle time will be calculated.

    Steps:

    • Find out the minimum of Ai and Bi.
    • If the minimum is for a job on machine A then perform that job at the start.
    • If the minimum is for a job on machine B then perform that job at the last.
    • A job should be assigned only once.

    Calculation:

    Using the Johnson rule, sequence is:

    4

    1

    3

    2

    5

    6

    We have the table:

    Job

    Time for facing in minutes

    Time for threading in minutes

    4

    2

    6

    1

    3

    8

    3

    5

    9

    2

    12

    10

    5

    9

    3

    6

    11

    1

    Total

    42

    37

     

    Job

    IN-Facing

    OUT-Facing

    IN-Threading

    OUT-Threading

    4

    0

    2

    2

    8

    1

    2

    5

    8

    16

    3

    5

    10

    16

    25

    2

    10

    22

    25

    35

    5

    22

    31

    35

    38

    6

    31

    42

    42

    43

     

    Idle time = make span time – working time =

    Idle time (Facing) = 43 – 42 = 1 min

    Idle time (Threading) = 43 – 37 = 6 min

    Total idle time is =1 + 6 = 7 minutes
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