calculator techniques

1. Calculator Techniques
(Calc 101)
"𝑊𝑖𝑡ℎ𝑜𝑢𝑡 𝑚𝑦 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟, 𝐼 𝑎𝑚 𝑛𝑜𝑡ℎ𝑖𝑛𝑔
𝑊𝑖𝑡ℎ𝑜𝑢𝑡 𝑚𝑒, 𝑚𝑦 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟 𝑖𝑠 𝑛𝑜𝑡ℎ𝑖𝑛𝑔"

2. Table of Contents
• BASIC FUNCTIONS (CALC AND SHIFT SOLVE)
• MODE FUNCTIONS
• EQUATION SOLVER (EQN)
• COMPLEX (CMPLX)
• STATISTICS (STAT)
• MATRIX
• SUMMATION
• Note that this lecture is made for PPT presentation

3. SHIFT SOLVE
• APPLICATIONS
• FINDING THE VALUE OF A SINGLE VARIABLE
• FINDING THE ROOTS OF A POLYNOMIAL EXPRESSION
WHETHER LINEAR OR NONLINEAR
• HOW TO USE
• ENTER THE EXPRESSION THEN PRESS SHIFT THEN PRESS
• NOTE
• CALCULATOR MUST BE IN COMP MODE (PRESS MODE -> 1)

4. EXAMPLE
1.) FIND THE VALUE OF X IN THIS EXPRESSION:
3x + 4 = 7

5. SOLUTION FOR #1
• MANUAL
3x + 4 = 7
Steps:
1. Transpose 4 to the right side
3x = 7 – 4 = 3
2. Divide both sides of the equation
by 3 to isolate the variable
1/3 * (3x) = (3) * 1/3
x = 1
• CALCULATOR
Steps:
1. Input the function in your
calculator (To input the variable,
press “ALPHA” then press the
variable of your choice)
2. Press “SHIFT” then
x = 1

6. EXAMPLE
2.) FIND THE ROOTS OF THE 3rd DEGREE POLYNOMIAL EXPRESSION:
(x3 + 4x2 - 5x – 14)

7. SOLUTION FOR #2
• MANUAL
(x3 + 4x2 - 5x - 14)
Steps:
1.) 1 4 -5 -14 2
2 12 14
1 6 7 0
x = {2, -3+ 2, -3- 2}
• CALCULATOR
Steps:
1. Input the function in your
calculator (To input the variable,
press “ALPHA” then press the
variable of your choice)
2. Press “SHIFT+Solve”

8. EXAMPLE
3.) FIND THE ROOTS OF THE FOLLOWING EXPONENTIAL/LOGARITHMIC
EXPRESSION:
0
2
3 

 x
x
e
e

9. SOLUTION FOR #3
MANUAL
0
)
1
)(
3
(
0
2
3
0
2
3
2








 
x
x
x
x
x
x
e
e
e
e
e
e
)
(
1
097
.
1
)
3
ln(
)
3
ln(
)
ln(
3
reject
e
x
e
e x
x
x






Calculator
• Do the same with the
“ShiftSolve” procedure
• In order to type “e”,
press “ALPHA”
• Or you may use
“SHIFT”

10. CALC FUNCTION
• APPLICATIONS
• WHEN USING THE SAME EQUATION BUT WITH DIFFERENT
VALUES OF THE INPUT VARIABLES (i.e. Exponential Smoothing)
• HOW TO USE
• ENTER THE EXPRESSION THEN PRESS
• NOTE
• CALCULATOR MUST BE IN COMP MODE

11. EXAMPLE
4.) The company has accumulated the demand data in the table below
for its computers for the past twelve months, from which it wants to
consider exponential smoothing forecasts with a smoothing constant
equal to 0.30. Calculate the forecast on next January (13th period).

12. SOLUTION FOR #4
MANUAL
Remembering the formula for exponential smoothing:
𝐹𝑡 = 𝐹𝑡−1 + 𝛼(𝐴𝑡−1 − 𝐹𝑡−1)
and by continuous use of the formula we get the following data:

13. SOLUTION FOR #4
CALCULATOR
Steps:
1. To have a forecast for the month of February we use the “Naive
method” thus February will have a forecast of 37.
2. Use the formula for exponential smoothing to get a forecast for
March: 𝐹𝑡+1 = 37 + 0.3(40 − 37) = 37.9
3. Next modify the formula above in the following manner:
𝐴𝑛𝑠 + 0.3(𝑋 − 𝐴𝑛𝑠)
4. Press
5. The calculator will be needing a value for “X”. The value of “X”
should be the actual demand in the CURRENT month.

14. SOLUTION FOR #4

15. PROBLEM
• Determine the forecast for period 7 using exponential smoothing with
a smoothing constant of 0.4.
Period Demand
1 60
2 65
3 55
4 58
5 64
6 62
Demand for period 7 is 61.2992 ≈ 61.30 𝑢𝑛𝑖𝑡𝑠

16. CALC 101 (MODE FUNCTIONS)
• MODE FUNCTIONS
• EQUATION SOLVER (EQN)
• COMPLEX (CMPLX)
• STATISTICS (STAT)
• MATRIX

17. EQN SOLVER
• APPLICATIONS
• 2 EQUATIONS, 2 UNKNOWNS
• 3 EQUATIONS, 3 UNKNOWNS
• FINDING THE ROOTS OF A 2nd DEGREE EQUATION (QUADRATIC FORMULA)
• FINDING THE ROOTS OF A 3rd DEGREE EQUATION
• HOW TO USE
• PRESS “MODE”-> ->

18. EXAMPLE
5. Find the roots of the following systems of linear equations
𝑥 + 3𝑦 = 5
2𝑥 + 𝑦 = 6

19. SOLUTION FOR #5
MANUAL
2[𝑥 + 3𝑦 = 5]
- 2𝑥 + 𝑦 = 6
0 + 5𝑦 = 4; 𝑦 = 4/5
𝑥 + 3(
4
5
) = 5
𝑥 + 12/5 = 5
𝑥 = 5 −
12
5
= 13/5
CALCULATOR
Press “MODE”
Choose “5:EQN”
Choose “1”
Input the values
Press “=“

20. EXAMPLE
6. Find the roots of the following systems of linear equations
0
3
3
3
0
2
8
3
3
0
1
2
4










z
y
z
x
y
y
x
38
/
5
38
/
43
38
/
31





Z
Y
X

21. COMPLEX MODE (CMPLX)
• APPLICATIONS
• MATHEMATICAL OPERATIONS OF COMPLEX NUMBERS
• FINDING THE MAGNITUDE AND DIRECTION OF RESULTANT
• HOW TO USE
• PRESS “MODE”, THEN PRESS “2”

22. EXAMPLE
7. Determine the magnitude and direction of the resultant of these
forces:
X
Y
F1=8 kN
F2=10 kN
F3=14 kN
5
12
45o
25o

23. SOLUTION FOR #7
MANUAL
X
Y
F1=8 kN
F2=10 kN
F3=14 kN
5
12
45o
25o
𝐹𝑋 = 8 cos 45 − 10 cos 25 + 14 cos tan−1
12
5
𝐹𝑋 ≈ 1.98 𝑘𝑁
𝐹𝑌 = 8 sin 45 + 10 sin 25 − 14 sin tan−1
12
5
𝐹𝑌 ≈ − 3.04 𝑘𝑁
𝑅 = ( 𝐹𝑋)2 + ( 𝐹𝑌)2 = ( 1.98)2 + (−3.04)2≈ 3.63 𝑘𝑁
𝜃 = tan−1
𝐹𝑌
𝐹𝑋
= tan−1(
−3.04
1.98
) ≈ −56.93𝑜 𝑜𝑟 56.93𝑜 𝑆 𝑜 𝐸
CALCULATOR
• Press “ MODE”
• Choose “2:CMPLX”
• Type
(To type “L”, press “SHIFT” )
• Press “=” then “SHIFT”
• Choose “3”
• Press “=”
)
83
.
67
270
(
14
)
25
180
(
10
45
8







NOTE THAT THE 0o STARTS FROM THE
RIGHT AND ROTATES COUNTER CLOCKWISE

24. EXAMPLE
8. Determine the magnitude and direction of the resultant of these
forces:
X
Y
F1=28 kN
F2=20 kN
F3=25 kN
5
12
45o
65o
F4=24 kN
R = 29.13 kN
Θ = 71.13o N of E

25. STATISTICS (STAT)
• APPLICATIONS
• REGRESSION
• ARITHMETIC/GEOMETRIC SEQUENCES
• COMPUTATION OF STATISTIC (MEAN AND STD. DEV OF 1 OR 2
GROUPS)
• NORMAL DISTRIBUTION CALCULATIONS
• LEARNING CURVES
• ANALYSIS OF VARIANCE (1-WAY, BLOCK DESIGN, 2-WAY, 3-WAY)
• DEPRECIATION
• HOW TO USE
• PRESS “MODE” THEN “3”

26. STATISTICAL FUNCTIONS
1 – Variable Simple Linear
Regression
Quadratic Regression
Exponential
Regression
AB Exponential
Regression
Logarithmic
Regression
Power Regression
Inverse Regression

27. EXAMPLE
9. (Arithmetic Progression/Sequence)
The 6th term of an arithmetic progression is 12 and the 30th term is 180.
Find the following:
a.) Common difference
b.) The 1st term
c.) Sum of the first 60 terms
d.) The term that has a value of 250
e.) The sum between the 12th & 37th term (inclusive)

28. SOLUTION FOR #9
Press “Mode” then “3” then “2” and input the ff:
a. Common Difference:
Press “AC” then “Shift” then “1”
Choose “Reg” then Choose “2: B”
X Y
6 12
30 180
b. The 1st term:
Press “AC” then Press “1” then “Shift” then “1”
Choose “Reg” then Choose “5: ŷ”
Your screen should now display 1ŷ (this means “What is the value of y,
given the value of x is 1) then press “=“
c. Sum of the 1st 60 terms
Press “AC” then Press “Shift” then to place the summation function
Now input the following “Xŷ,1,60)”
Your screen should now display (Xŷ,1,60) then press “=“
(the syntax is “function, 1st value of variable, last value of variable”)
d. The term that has a value of 250
Similar to question b, but instead of using ŷ we use x
Answer is 40th term
e. The sum between the 12th & 37th term (inclusive)
Similar to question c, but the starting value is 12 and ending is 37
Answer is 3679

29. EXAMPLE
10. (Geometric Progression/Sequence)
The value first four term of a certain sequence are 2, 6, 18, 54.
a.) Find the 12th term
b.) What term has a value of 9,565,938
c.) Find the sum of the first 10 terms
For this example the solution is the same as before, the only difference
is that you should use the Power Regression.
Press “MODE”
Choose “3:STAT”
Choose “6:AˑB^X”
Answers:
a. 354294 b. 15th term c. 59048

30. EXAMPLE
11.) Regression
Given the data below, answer the following questions:
a.) Calculate the least squares estimate of the model? What is the
equation?
b.) What is the value of the dependent variable given that the value of
the independent is 15?
c.) What is the coefficient of correlation? Coefficient of determination?
X 12 9 11 10 13 8 9 11 12
Y 51 48 52 50 55 45 47 50 53

31. SOLUTION FOR #11
MANUAL
a. Least-squares estimate and equation
  
 
  
n
n
Y
X
XY
n
Y
X
n
XY
Y
Y
X
X
X
X
X
X
X
X
b
















 

2
2
2
2
2
1
n
X
n
Y
X
Y b
b
b

 


 1
1
0
Y
Y
where
X
Y
b
b
b
b
of
value
predicted
the
=
ˆ
slope
sample
the
=
intercept
sample
the
=
:
ˆ
1
0
1
0


b.) What is the value of the dependent variable given that the value of
the independent is 15?
Substitute the value of x as 15 in the model.
c.) What is the coefficient of correlation? Coefficient of determination?
Coefficient of correlation (r):
Coefficient of determination (r2)
)
(
)
(
)
(
yy
xx
xy
S
S
S
r


2
1
1
2
)
( 











n
i
n
x
i
i
xx
n
x
x
S
2
1
1
2
)
( 











n
i
n
i
i
i
yy
n
y
y
S





















n
i
n
i
i
n
i
i
i
i
xy
n
y
x
y
x
S
1
1
1
)
(
CALCULATOR
a. Least-squares estimate and equation
• PRESS “MODE” THEN “3” THEN “2”
• INPUT THE VALUES OF X AND Y ON THE TABLE DISPLAYED
• PRESS “AC” THEN “SHIFT” THEN “1” THEN “REG”
• “A” → b0 and “B” → b1
b. What is the value…
Like in the arithmetic/geometric sequences: Input “15ŷ”
c. Coefficient of correlation and determination
PRESS “SHIFT” THEN “1” THEN “REG” THEN “R” TO GET CORRELATION
AND TO GET DETERMINATION JUST SQUARE THE VALUE OF R.
Answers:
a. y=31.375+1.775x b. 58 c. r=0.95 r2=0.91

32. PROBLEM
• Healthy Hamburgers has a chain of 12 stores in northern Illinois. Sales
figures and profits for the stores are given in the following table.
Obtain a regression line for the data, and predict profit for a store
assuming sales of $10 million. Unit Sales, (in
$ millions)
Profits, (in $
millions)
$7 $0.15
2 0.1
6 0.13
4 0.15
14 0.25
15 0.27
16 0.24
12 0.2
14 0.27
20 0.44
15 0.34
7 0.17
𝒚 = 𝟎. 𝟎𝟓𝟎𝟔 + 𝟎. 𝟎𝟏𝟓𝟗𝒙
𝒚 𝟏𝟎 ≈ 𝟎. 𝟐𝟎𝟗 (𝒊𝒏 $ 𝒎𝒊𝒍𝒍𝒊𝒐𝒏𝒔)

33. MEAN & STD. DEV
• FOR SINGLE (1) VARIABLE: PRESS “MODE” THEN “3” THEN “1”
• FOR TWO (2) VARIABLES: PRESS “MODE” THEN “3” THEN “2”

34. EXAMPLE
12. A marketer has conducted a survey for the average monthly income
of 9 families of 2 different classes. Determine the mean and standard
deviation of the 2 classes.
Group 1 46 25 28 38 37 40 41 41 49
Group 2 48 46 37 42 41 42 35 28 32

35. SOLUTION FOR #12
MANUAL
 
1
2
2






n
X
X
s
n
X
X
i
i
CALCULATOR
PRESS “MODE” THEN ‘3” THEN “2” THEN INPUT THE VALUES ON THE
TABLE DISPLAYED (NOTE: BE CONSISTENT WITH YOUR DATA)
THEN PRESS “SHIFT” THEN “1” THEN “VAR” THEN CHOOSE “4:𝑠𝑥”
(NOTE THE DIFFERENCE BETWEEN 𝜎𝑋 𝐴𝑁𝐷 𝑠𝑥, 𝜎𝑋is population std. dev.
while 𝑠𝑥 is sample std. dev.)
Answers:
𝜇1=38.33 𝜇2=39
𝜎1=7.713 𝜎2=6.54

36. EXAMPLE
12. A marketer has conducted a survey for the average monthly income
of 9 families of 2 different classes. Determine the mean and standard
deviation of the 2 classes.
Group 1 46 25 28 38 37 40 41 41 49
Group 2 48 46 37 42 41 42 35 28 32

37. NORMAL DISTRIBUTION CALCULATIONS
• PRESS “MODE” THEN “3” THEN “1” (NOTE THAT THIS FUNCTION ONLY
WORKS IN THE “1-VAR” MODE)

38. EXAMPLE
13. The diameter of a shaft in an optical storage drive is normally
distributed with mean 0.2500 inch and standard deviation 0.0005 inch.
The specifications on the shaft are 0.2500 ± 0.0015 inch. What
proportion of shafts conforms to specifications? What proportions
exceed the upper specification? What proportions exceed the lower
specification?

39. SOLUTION FOR #13
MANUAL
𝑃 𝑋 < 0.2485 = 𝑃
0.2485 − 0.25
0.0005
< 𝑍 = 𝑃 −3 < 𝑍 = 0.00135
𝑃 𝑋 > 0.2515 = 1 − 𝑃 𝑋 < 0.2515 = 1 − 𝑃
0.2515 − 0.25
0.0005
< 𝑍
= 1 − 𝑃 3 < 𝑍 = 0.00135
CALCULATOR
PRESS “MODE” THEN “3” THEN “1” THEN “AC” THEN “SHIFT” THEN “1”
THEN “DISTR”
NOTE: IN ORDER TO USE THIS FUNCTION, WE MUST KNOW THE “Z” VALUE BEFOREHAND

40. SOLUTION FOR #13
• To know the probability of conforming:
INPUT THE FF: 2xQ(3)
• To know the probability of exceeding the upper specification:
INPUT THE FF: R(3)
• To know the probability of not meeting the lower specification:
INPUT THE FF: P(-3)

41. EXAMPLE
14. A teacher has 40 students enrolled in her statistics class. The table
below displays the scores of those students in their first quiz. What is
the probability that in a 50 points quiz, a student would get a score
greater than 40?
SCORE 35 46 25 28 38 37 40 44 41 49
FREQ 15 2 9 1 2 1 1 3 4 2

42. SOLUTION FOR #14
MANUAL
SOLVE FOR THE MEAN AND STD. DEV
THEN USE THE FORMULA 𝑍 =
𝑋−𝜇
𝜎𝜇
THEN LOCATE THE Z-VALUE IN NORMAL DIST. TABLE
CALCULATOR
TO TURN ON FREQUENCY PRESS “SHIFT” THEN “MODE” THEN “DOWN”
THEN “STAT” THEN “ON” THEN “AC”
PRESS “MODE” THEN “3” THEN “1” THEN INPUT THE VALUES THEN
“AC”
INPUT “40” THEN PRESS “SHIFT” THEN “1” THEN “DISTR” THEN “▲t”
(YOUR DISPLAY SHOULD READ 40▲t) THEN “=“
THEN PRESS “SHIFT” THEN “1” THEN “DISTR” THEN “R(“ THEN “ANS”
THEN “)” THEN “=“
CAUTION: THE “>t” function uses the formula 𝑍 =
𝑋−𝜇
𝜎𝜇
which assumes
that the data come from a population and not a sample thus using the
population std. dev
Answer:
0.2593

43. PROBLEM
• The pull-off force for 40 connectors is measured in a laboratory test.
Data for the 40 test specimens follow:
241, 203, 201, 251, 236, 190, 258, 195, 195, 238, 245, 175
237, 249, 255, 210, 209, 178, 210, 220, 245, 198, 212, 175
194, 194, 235, 199, 185, 190, 225 245, 220, 183, 187,
248, 209, 249, 213, 218
What is the probability that a connector would have a pull-off force
between 211 and 220?
𝑷 𝟐𝟏𝟏 < 𝑿 < 𝟐𝟐𝟎 = 𝟎. 𝟏𝟒𝟑𝟕𝟓

44. LEARNING CURVES
• PRESS “MODE” THEN “3” THEN “7”

45. EXAMPLE
• 15. An aircraft company has an order to refurbish the interiors of 18
jet aircraft. The work has a learning curve percentage of 80. On the
basis of experience with similar jobs, the industrial engineering
department estimates that the first plane will require 300 hours to
refurbish. Estimate the amount of time needed to complete:
a. The fifth plane.
b. The first five planes.
c. All 18 planes.

46. SOLUTION FOR #15
PRESS “MODE” THEN “3” THEN “7”
THEN INPUT
a.) INPUT “5ŷ”
b.) INPUT " (𝑋ŷ, 1,5)"
c.) INPUT " (𝑋ŷ, 1,18)"
X Y
1 300
2 240
(WE GOT 240 FROM THE KNOWLEDGE
THAT WHEN THE OUTPUT DOUBLES THE
TIME IS MULTIPLIED BY THE LEARNING
PERCENTAGE)
a.178.69hrs
b.b. 1121.32hrs
c. c. 2914.85hrs

47. PROBLEM
• A contractor intends to bid on a job installing 30 airport security
systems. Because this will be a new line of work for the contractor, he
believes there will be a learning effect for the job. After reviewing
time records from a similar type of activity, the contractor is
convinced that an 85 percent curve is appropriate. He estimates that
the first job will take his crew eight days to install. How many days
should the contractor budget for:
a. The first 10 installations?
b. The second 10 installations?
c. The final 10 installations?
56.92 DAYS
42.28 DAYS
37.50 DAYS

48. PROBLEM
• An analyst has estimated that there will be an 84 percent learning
curve for an assembly operation. The first assembly takes 48 min, and
the standard time is set at 25 min. How long will it take the operator
to reach the standard time?
TOTAL TIME = 435.44 mins

49. ANOVA
• TYPES OF ANOVA
• 1-WAY (ONE FACTOR)
• RANDOMIZED COMPLETE BLOCK DESIGN (TWO FACTORS BUT ONE IS
DISREGARDED)
• 2-WAY WITH REPLICATION
• 2-WAY WITHOUT REPLICATION
• 3-WAY

50. ANOVA
• BRIEF INTRO INTO ANOVA
FACTOR
LEVELS
TREATMENT

51. EXAMPLE
• 16. (1-WAY) The tensile strength of Portland cement is being studied.
Four different mixing techniques can be used economically. A
completely randomized experiment was conducted and the following
data were collected:
• Complete the ANOVA table
Mixing
Technique
Tensile Strength (lb/in2)
1 3129 3000 2865 2890
2 3200 3300 2975 3150
3 2800 2900 2985 3050
4 2600 2700 2600 2765

52. SOLUTION FOR #16
THE 1-WAY ANOVA TABLE
• TO USE THE SHORTCUT WE MUST FIRST DO THIS:
• COMPUTE THE MEAN FOR EACH TREATMENT
• COMPUTE THE GRAND MEAN
Source Sum of Squares Degrees of Freedom Mean Square F-Comp F-Crit
TREATMENT SSTr a – 1 MSTR=SSTR/(a-1) =MSTR/MSE
ERROR SSE N – a MSE=SSE/(N-a)
TOTAL SST N – 1
Mixing
Technique
Tensile Strength (lb/in2) MEAN
GRAND
MEAN
1 3129 3000 2865 2890 2971
2931.8125
2 3200 3300 2975 3150 3156.25
3 2800 2900 2985 3050 2933.75
4 2600 2700 2600 2765 2666.25
NOW THAT WE HAVE DETERMINED THE MEANS AND THE GRAND
MEAN WE CAN NOW SOLVE FOR THE NEEDED SUM OF SQUARES:
• WE FIRST SOLVE SST: TO DO THIS PRESS “MODE” THEN “3” THEN “2”
(NOTE: MAKE SURE THAT FREQUENCY IS OFF)
• FOR THE X-COLUMN INPUT THE DATA FROM THE TABLE AND FOR THE
Y-COLUMN INPUT THE GRAND MEAN THEN PRESS “AC”
• PRESS” “SHIFT” THEN “1” THEN CHOOSE “3:SUM” THEN INPUT THE
FOLLOWING EXPRESSION “ 𝑥2 − 2 𝑥𝑦 + 𝑦2"
NOW THAT WE HAVE DETERMINED THE MEANS AND THE GRAND
MEAN WE CAN NOW SOLVE FOR THE NEEDED SUM OF SQUARES:
• NEXT WE SOLVE FOR SSTr: TO DO THIS PRESS “MODE” THEN “3” THEN
“2” (NOTE: THIS TIME MAKE SURE THAT FREQUENCY IS ON)
• FOR THE X-COLUMN INPUT THE MEAN PER TREATMENT AND FOR THE
Y-COLUMN INPUT THE GRAND MEAN
• FOR THE FREQ-COLUMN INPUT THE NUMBER OF OBSERVATIONS PER
TREATMENT
• THEN INPUT AGAIN " 𝑥2 − 2 𝑥𝑦 + 𝑦2"
NOW THAT WE HAVE DETERMINED THE MEANS
AND THE GRAND MEAN WE CAN NOW SOLVE FOR
THE NEEDED SUM OF SQUARES:
•LASTLY TO SOLVE FOR SSE WE REMEMBER THE
FOLLOWING PROPERTY: "𝑆𝑆𝑇 = 𝑆𝑆𝑇𝑟 + 𝑆𝑆𝐸"
Source of
Variation
SS df MS F-Comp F-Crit
Treatment 489740.2 3 163246.7 12.72811 3.490295
Error 153908.3 12 12825.69
Total 643648.4 15

53. PROBLEM
• An engineer would like to test the effect of 4 brands of gasoline on
the mileage of a car. The data are as follows:
Complete the ANOVA Table.
BRAND
OBSERVATION
1 2 3 4
A 21 24 23 24
B 19 21 20 22
C 20 24 23 24
D 22 26 23 21
Source of
Variation
SS df MS F P-value F crit
Treatment 17.6875 3 5.895833 1.979021 0.170983 3.490295
Error 35.75 12 2.979167
Total 53.4375 15

54. EXAMPLE
• 17. (RCBD or Blocked Design) A chemist wishes to test the effect of
four chemical agents on the strength of a particular type of cloth.
Because there might be variability from one bolt to another, the
chemist decides to use a randomized block design, with the bolts of
cloth considered as blocks. She selects five bolts and applies all four
chemicals in random order to each bolt. The resulting tensile
strengths follow.
Complete the ANOVA Table
Chemical
Bolt
1 2 3 4 5
1 73 68 74 71 67
2 73 67 75 72 70
3 75 68 78 73 68
4 73 71 75 75 69

55. SOLUTION FOR #17
THE RCBD ANOVA TABLE
• SAME PROCEDURE AS BEFORE: WE GET THE MEANS PER TREATMENT
AND THE GRAND MEAN BUT THIS TIME WE ALSO GET THE MEANS
PER BLOCK
• COMPUTATION FOR SSB IS SIMILAR TO SSTr, THE ONLY DIFFERENCE IS
THE USE OF THE MEANS PER BLOCK IN THE X-COLUMN
Source Sum of Squares Degrees of Freedom Mean Square F-Comp F-Crit
TREATMENT SSTr a – 1 MSTR=SSTR/(a - 1) =MSTR/MSE
BLOCKS SSB b – 1 MSB=SSB/(b – 1)
ERROR SSE (a – 1)*(b – 1)
𝑀𝑆𝐸 =
𝑆𝑆𝐸
(a – 1)∗(b – 1)
TOTAL SST N – 1
ANOVA
Source of
Variation
SS df MS F P-value F crit
Treatment 12.95 3 4.316667 2.376147 0.121144 3.490295
Blocks 157 4 39.25
Error 21.8 12 1.816667
Total 191.75 19

56. EXAMPLE
18. (2-WAY W/O REPLICATION) The shear strength of an adhesive is
thought to be affected by the application pressure and temperature. A
factorial experiment is performed in which both factors are assumed to
be fixed.
Complete the ANOVA table
Pressure
(lb/in2)
Temperature (°F)
250 260 270
120 9.6 11.28 9
130 9.69 10.1 9.57
140 8.43 11.01 9.03
150 9.98 10.44 9.8

57. SOLUTION FOR #18
THE 2-WAY ANOVA TABLE W/O REPLICATION
THE COMPLETION OF THE ANOVA TABLE IS SIMILAR TO THE RCBD
ANOVA TABLE
Source Sum of Squares Degrees of Freedom Mean Square F-Comp F-Crit
A SSA a – 1 MSA=SSA/(a - 1) =MSA/MSE
B SSB b – 1 MSB=SSB/(b – 1) =MSB/MSE
ERROR SSE (a – 1)*(b – 1)
𝑀𝑆𝐸 =
𝑆𝑆𝐸
(a – 1)∗(b – 1)
TOTAL SST N – 1
ANOVA
Source of
Variation
SS df MS F P-value F crit
Rows 0.580692 3 0.193564 0.539204 0.672704 4.757063
Columns 4.65765 2 2.328825 6.487329 0.031618 5.143253
Error 2.153883 6 0.358981
Total 7.392225 11

58. PROBLEM
• Seeds of 4 different types of corn are planted in 5 blocks. Each block is
divided into 4 plots, which are then randomly assigned to the 4 types.
The yields (in bushels per acre) are as follows:
We wish to determine whether the type of corn, and plot are
significant to determining the yield
Complete the ANOVA table
TYPES OF CORN
I II III IV
Blocks
A 12 15 10 14
B 15 19 12 11
C 14 18 15 12
D 11 16 12 16
E 16 17 11 14
Source of
Variation
SS df MS F P-value F crit
Rows 10 4 2.5 0.646552 0.639893 3.259167
Columns 67.6 3 22.53333 5.827586 0.010751 3.490295
Error 46.4 12 3.866667
Total 124 19

59. EXAMPLE
• 19. (2-WAY WITH REPLICATION) An article in Industrial Quality Control
describes an experiment to investigate the effect of two factors (glass
type and phosphor type) on the brightness of a television tube. The
response variable measured is the current (in microamps) necessary
to obtain a specified brightness level. The data are shown in the
following table. Complete the ANOVA table.
Glass Type
Phosphor Type
1 2 3
1
280 300 290
290 310 285
285 295 290
2
230 260 220
235 240 225
240 235 230

60. SOLUTION FOR #19
THE 2-WAY ANOVA TABLE WITH REPLICATION
THE COMPUTATIONS FOR SSA, SSB, SSE, AND SST ARE THE SAME AS
BEFORE.
THE COMPUTATION FOR SSAB IS AS FOLLOWS:
Source Sum of Squares Degrees of Freedom Mean Square F-Comp F-Crit
A SSA a – 1 MSA=SSA/(a - 1) =MSA/MSE
B SSB b – 1 MSB=SSB/(b – 1) =MSB/MSE
INTERACTION SSAB (a – 1)*(b – 1)
𝑀𝑆𝐴𝐵 =
𝑆𝑆𝐴𝐵
(a – 1)∗(b – 1)
=MSAB/MSE
ERROR SSE ab(n – 1)
𝑀𝑆𝐸 =
𝑆𝑆𝐸
ab(n−1)
TOTAL SST abn – 1
• NOW TO COMPUTE FOR SSAB WE DO THE FF:
• GET THE MEAN PER COMBINATION OF LEVELS
• PRESS “MODE” THEN “3” THEN “2” (NOTE THAT FREQUENCY MUST BE ON)
• INPUT IN THE X-COLUMN THE MEANS PER COMBINATION AND IN THE
Y-COLUMN INPUT THE GRAND MEAN
• FOR THE FREQ-COLUMN INPUT THE NUMBER OF REPLICATIONS
• THEN INPUT " 𝑥2 − 2 𝑥𝑦 + 𝑦2 − 𝑆𝑆𝐴 − 𝑆𝑆𝐵"
Source of
Variation
SS df MS F P-value F crit
Sample 14450 1 14450 273.7895 1.26E-09 4.747225
Columns 933.3333 2 466.6667 8.842105 0.004364 3.885294
Interaction 133.3333 2 66.66667 1.263158 0.317801 3.885294
Within 633.3333 12 52.77778
Total 16150 17

61. EXAMPLE
• 20. (3-WAY ANOVA) A mechanical engineer is studying the surface
roughness of a part produced in a metal-cutting operation. Three
factors, feed rate (A), depth of cut (B), and tool angle (C), are of
interest. All three factors have been assigned two levels, and two
replicates of a factorial design are run. Data are shown in the table
below.

62. SOLUTION FOR #20
THE 3-WAY ANOVA TABLE
• COMPUTATIONS FOR SSA, SSB, AND SSC ARE THE SAME AS WITH 2-WAY
ANOVA
• COMPUTATIONS FOR SSAB, SSAC, AND SSBC ARE THE SAME AS WITH 2-WAY
ANOVA (i.e. SSBC = " 𝑥2 − 2 𝑥𝑦 + 𝑦2 − 𝑆𝑆𝐵 − 𝑆𝑆𝐶")
• COMPUTATIONS FOR SSABC IS
= 𝑥2
− 2 𝑥𝑦 + 𝑦2
− 𝑆𝑆𝐴 − 𝑆𝑆𝐵 − 𝑆𝑆𝐶 − 𝑆𝑆𝐴𝐵 − 𝑆𝑆𝐴𝐶 − 𝑆𝑆𝐵𝐶

63. DEPRECIATION
• DEPRECIATION METHODS
• STRAIGHT LINE
• DECLINING BALANCE
• DOUBLE DECLINING BALANCE
• SUM OF YEARS
• SINKING FUND

64. DEPRECIATION
METHOD MODE DATA ENTRY
ANNUAL
DEPRECIATION
(𝒅)
DEPRECIATION ON THE
Mth YEAR
TOTAL
DEPRECIATION
AFTER Mth YEAR
BOOK VALUE AT
Mth YEAR
STRAIGHT LINE 3-2 B B
FC-Mŷ Mŷ
DECLINING
BALANCE
3-6
- (M-1)ŷ-Mŷ
DOUBLE
DECLINING
BALANCE
SUM OF YEARS 3-3
SINKING FUND
- - 𝐹𝐶 − 𝑆𝑉
0
𝑛−1
[(1 + 𝑖)𝑥] BV(M-1) - BVM FC-BVM
FC-
0
𝑀−1
[𝑑(1 + 𝑖)𝑥
]
X Y
0 FC
n SV
X Y
0 FC
1 1 −
2
𝑛
𝐹𝐶
X Y
0 FC
n SV
n+1 SV

65. EXAMPLE
• 21. If an asset has a first cost of $50,000 with a $10,000 estimated
salvage value after 5 years, (a) calculate the annual depreciation and
(b) calculate the book value of the asset after the 3rd year, using
straight line depreciation.

66. SOLUTION FOR #21
• PRESS “MODE” THEN “3” THEN “2” THEN INPUT THE VALUES THEN “AC”
• To compute annual depreciation: PRESS “SHIFT” THEN “1” THEN CHOOSE
“5:REG” THEN CHOOSE “2:B”
• To compute for the book value after the 3rd year: INPUT “4ŷ”
X Y
0 50,000
5 10,000

67. EXAMPLE
• 22. Underwater electroacoustic transducers were purchased for use
in SONAR applications. The equipment will be DDB depreciated over
an expected life of 12 years. There is a first cost of $25,000 and an
estimated salvage of $2500. (a) Calculate the depreciation and book
value for years 1 and 4. (b) Calculate the implied salvage value after
12 years.

68. SOLUTION FOR #22
• PRESS “MODE” THEN “3” THEN “6” THEN INPUT THE VALUES THEN “AC”
• To compute depreciation for Mth year: INPUT “(M-1)ŷ-Mŷ”
• To compute for the book value for Mth year : INPUT “Mŷ”
• To compute for salvage value after 12 years: INPUT “12ŷ”
X Y
0 25,000
1 1 −
2
12
25,000

69. EXAMPLE
• 23. Calculate the SYD depreciation charges for 2 years for
electro-optics equipment with FC=$25,000, SV=$4000, and an 8-year
recovery period.

70. SOLUTION FOR #23
• PRESS “MODE” THEN “3” THEN “3” THEN INPUT THE VALUES THEN
“AC”
• To compute depreciation for year 2: INPUT “(2-1)ŷ-2ŷ”
X Y
0 25,000
8 4000
9 4000

71. PROBLEM
• An asset for drilling was purchased ad placed in service by a
petroleum production company. Its cost basis is $60,000, and it has
an estimated market value of $12,000 at the end of an estimated
useful life of 14 years. Compute the depreciation amount in the third
year and the book value at the end of the fifth year of life by each of
these methods. (a) Straight Line (b) Declining Balance (c) Double
Declining Balance (d) Sum of Year (e) Sinking Fund (𝑖 = 8%)
(c) D3=$6,297.38 and BV6=$23,794.17
(a) D3= $3,428.57 and BV6=$39,428.57
(b) D3=$5,177.50 and BV6=$30,101.81
(d) D3=$5,485.71 and BV6=$28,457.14
(e) D3=$2,312.10 and BV6=$45,458.36

72. MATRIX
• APPLICATIONS
• MATRIX ALGEBRA
• OBTAINING THE DETERMINANT OF MATRICES OF SIZES UPTO 4x4
• STEADY-STATE PROBABILITIES (MARKOV CHAINS)
• HOW TO USE
• PRESS “MODE’’ THEN “6”

73. EXAMPLE
• 24. Determine the determinant of the ff. 3x3 matrix:
𝐴 =
1 5 4
7 6 9
2 3 8

74. SOLUTION FOR #24
• PRESS “MODE” THEN “6” THEN CHOOSE ANY MATRIX AND INPUT THE
VALUES
• PRESS “AC” THEN “SHIFT” THEN “4” THEN CHOOSE “7:det”
• PRESS “SHIFT” THEN “4” THEN CHOOSE THE MATRIX YOU HAVE
STORED THE DATA THEN “=“

75. EXAMPLE
• 25. Determine the determinant of the 4x4 matrix.
















3
3
3
1
0
2
1
0
2
1
2
1
1
1
1
1
A

76. SOLUTION FOR #25
• Choose any number from the matrix (we call
it pivot, choose “1” for convenience) (For this
example, our pivot will be the one encircled)
• Let:
Blue = MatC (1x3 matrix)
Green = MatB (3x1 matrix)
Violet = MatA (3x3 matrix)
















3
3
3
1
0
2
1
0
2
1
2
1
1
1
1
1
A
• Type the following: “det(MatA – MatB*MatC)”

77. PROBLEM
• What is the determinant of the following 4x4 matrix?
Det = -72


















3
3
3
2
0
2
1
0
2
1
2
1
1
1
1
2
A

78. EXAMPLE
• 26. (Steady-State Probabilities) Suppose that a farmland has three
states (1) good, (2) fair, and (3) poor. In the long run, what are the
probabilities that the land will be in each of the three states?
1 2 3
1
2
3
.3 .6 .1
.1 .6 .3
.05 .4 .55

79. SOLUTION FOR #26
• INPUT THE DATA IN A MATRIX
• THEN PRESS “SHIFT” THEN “4” THEN CHOOSE THE MATRIX THAT HAS
THE DATA THEN PRESS “x2” 5 times
• Your display should read “MatA22222” THEN PRESS “=“
• Note: The “press x2 5 times is subjective”, the logic is that you just press “2”
just enough so that per column, all the entries are equal.
.1016 .5254 .3728
.1016 .5254 .3728
.1016 .5254 .3728

80. SUMMATION
• APPLICATIONS
• TIME VALUE OF MONEY
• DISCRETE PROBABILTY DISTRIBUTIONS
• QUEUING THEORY

81. TIME VALUE OF MONEY
• PRESENT VALUE
• FUTURE VALUE
• ANNUITY
• UNIFORM GRADIENT
• GEOMETRIC GRADIENT

82. TIME VALUE OF MONEY
Formulas:
• Annuity:
• Uniform Gradient:
• Geometric Gradient:





 












i
i
A
F
i
i
i
A
P
n
n
n
1
)
1
(
)
1
(
1
)
1
(

















 n
n
n
i
n
i
i
i
i
G
P
)
1
(
)
1
(
1
)
1
(
1
 

















i
r
i
n
A
i
r
r
i
r
i
A
P
n
n
;
)
1
)(
(
;
)
1
(
)
1
(
1
1

83. TIME VALUE OF MONEY
Formulas for Calculator:
• Annuity:
• Uniform Gradient:
• Geometric Gradient:

 







n
x
x
n
n
x
x
i
A
F
i
A
P
1
)
(
1
)
)
1
)(
((
)
)
1
)(
((

 











n
x
x
n
n
x
x
i
X
G
A
F
i
X
G
A
P
1
)
(
1
)
)
1
))(
1
(
((
)
)
1
))(
1
(
((

 











n
x
x
n
x
n
x
x
x
i
r
A
F
i
r
A
P
1
)
(
1
1
1
)
)
1
(
)
1
(
(
)
)
1
(
)
1
(
(
“x=1” if the
first payment
was made on
first period
Fixed

84. TIME VALUE OF MONEY
Formulas: Comparison
• Annuity:
• Uniform Gradient:
• Geometric Gradient:









 n
n
i
i
i
A
P
)
1
(
1
)
1
(

















 n
n
n
i
n
i
i
i
i
G
P
)
1
(
)
1
(
1
)
1
(
1
 

















i
r
i
n
A
i
r
r
i
r
i
A
P
n
n
;
)
1
)(
(
;
)
1
(
)
1
(
1
1







n
x
x
x
i
r
A
P
1
1
)
)
1
(
)
1
(
(







n
x
x
i
X
G
A
P
1
)
)
1
))(
1
(
((





n
x
x
i
A
P
1
)
)
1
)(
((
NOTE: MAKE SURE THAT YOU ARE IN
COMP MODE

85. EXAMPLE
• 27. How much money should you be willing to pay now for a
guaranteed $600 per year for 9 years starting next year, at a rate of
return of 16% per year?

86. SOLUTION FOR #27
• TYPE 𝑥=1
9
(600(1.16)−𝑥)
1 2 3 4
$600
P=? i = 16%
……………………
……………………
9
$600
$600 $600
$600

87. EXAMPLE
28.) Ten payments are to be made to an account listed below which
bears interest at the rate of 12% compounded annually.
End of Year Payment ($)
1 500
2 1100
𝟑
⋮
1700
⋮
10 5900
How much will there be in
the account at the end of
tenth year?

88. SOLUTION FOR #28
• TYPE:
1 2 3 4
$500
$1,100
$1,700
i = 12%
……………………
……………………
$5,900
$2,300
10






10
1
10
)
)
12
.
1
))(
1
(
600
500
((
x
x
X
F
F=?

89. EXAMPLE
• 29. Determine the present worth of a geometric gradient series with
a cash flow of $50,000 in year 1 and increases of 6% each year
through year 8. The interest rate is 10% per year.

90. SOLUTION FOR #29
• Type 




8
1
1
)
)
1
.
1
(
)
06
.
1
(
000
,
50
(
x
x
x
P

91. PROBLEM
• Engineer Hector would like to have a financially independent life after
he retires. To do that he must have P1M by the age of 32. He has
started saving up for his retirement fund when he was 22y/o at a
bank that has an interest rate of 12% (Assume that he is depositing at
the end of each year). How much must he save annually in order to
achieve his goal?
A.) P57,984.16
B.) P56,489.61
C.) P56,984.16
D.) P57,849.61

92. PROBLEM
• Calculate the present equivalent at i=15% per year, using arithmetic
gradient.
PRESENT VALUE = $19,053.39
EOY CASH FLOWS ($)
1 8,000
2 7,000
3 6,000
4 5,000

93. DISCRETE PROBABILITY DISTRIBUTIONS
• BINOMIAL DISTRIBUTION
• HYPERGEOMETRIC DISTRIBUTION
• GEOMETRIC DISTRIBUTION
• NEGATIVE BINOMIAL DISTRIBUTION
• POISSON DISTRIBUTION
*Note: This method can be applied as long as
you know their Probability Mass Functions (PMF))
Also make sure that you are in COMP mode.

94. EXAMPLE
• 30. A multiple choice quiz question has 15 questions, each with 4
possible answers of which only 1 is correct. What is the probability
that the student will get at least 10 correct answers. Assuming that
the student did a shotgun method.

95. SOLUTION FOR #30
MANUAL
THE PMF OF THE BINOMIAL DISTRIBUTION IS:
𝑃 𝑋 = 𝑥 = 𝑛𝐶𝑥 ∗ 𝑝𝑥
∗ (1 − 𝑝)𝑛−𝑥
; 𝑝 = 0.25, 1 − 𝑝 = 0.75, 𝑛 = 15
SOLUTION:
𝑃 𝑋 ≥ 10 = 𝑃 𝑋 = 10 + 𝑃 𝑋 = 11 + ⋯ + 𝑃 𝑋 = 15
𝑃 𝑋 ≥ 10
= 15𝐶10 (.25)10(.75)5 + 15𝐶11 (.25)11(.75)4
+ 15𝐶12 (.25)12(.75)3 + 15𝐶13 (.25)13(.75)2
+ 15𝐶14 (.25)14(.75)1 + 15𝐶15 (.25)15(.75)0
=7.9494x10-4
CALCULATOR
TYPE THE FF: 10
15
(15𝐶𝑋 (.25)𝑋(.75)15−𝑋)
• TO INPUT THE COMBINATION OPERATOR PRESS “SHIFT” THEN “÷”

96. PROBLEM
• Contamination is a problem in the manufacture of magnetic storage
disks. Assume that the number of particles of contamination that
occur on a disk surface has a Poisson distribution, and the average
number of particles per square centimeter of media surface is 0.1.
The area of a disk under study is 100 square centimeters. Determine
the probability that 12 or fewer particles occur in the area of the disk
under study.
The PMF of the Poisson distribution is 𝑃 𝑋 = 𝑥 =
(𝑒ߣ𝑇)(ߣ𝑇𝑥)
𝑥!
P(X ≤ 12) = 0.792

97. QUEUING THEORY
• Case 1: SINGLE SERVER (M/M/1):(GD/∞/∞)
• Case 2: SINGLE SERVER FINITE SYSTEM LIMIT (M/M/1):(GD/N/∞)
• Case 3: MULTIPLE SERVER (M/M/s):(GD/∞/∞)
• Case 4: MULTIPLE SERVER FINITE SYSTEM LIMIT (M/M/s):(GD/N/∞)
• Case 5: SINGLE/MULTIPLE SERVER WITH FINITE POPULATION
(M/M/R):(GD/N/N)
• Note that the following formulas only apply if the probability
distribution is Markovian (arrival is Poisson, service is Exponential)

98. QUEUING THEORY
• MEASURES OF PERFORMANCE
• P0= probability of having zero customers in the system
• Pn= probability of having “n” customers in the system
• Lq= average number of customers waiting in queue
• L= average number of customers in the system
• Wq= average time a customer spends waiting in queue
• W= average time a customer spends in the system
• PARAMETERS
• ߣ = Arrival rate
• μ = Service rate

99. QUEUING THEORY
• TRANSITION DIAGRAMS
THE NUMBER OF
CUSTOMERS
IN THE SYSTEM
THE ARRIVAL
RATE
THE SERVICE
RATE

100. EXAMPLE
• 31. A fast-food restaurant has one drive-through window. An average
of 40 customers per hour arrive at the window. It takes an average of
1 minute to serve a customer. There is a limit of only 5 customers per
line. Assume that inter-arrival and service times are exponential.
Draw the transition diagram

101. SOLUTION FOR #31

102. QUEUING THEORY
• 𝑀𝑎𝑛𝑢𝑎𝑙: 𝑃0 =
1
1+
𝜆0
𝜇1
+
𝜆0𝜆1
𝜇1𝜇2
+
𝜆0𝜆1𝜆2
𝜇1𝜇2𝜇3
+⋯+
𝜆0𝜆1𝜆2⋯𝜆𝑛
𝜇1𝜇2𝜇3⋯𝜇𝑛+1
• 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟: 𝑃0 = ( 0
50
(
𝜆𝑥
𝜇𝑥))−1 NOTE THAT THIS FORMULA IS ONLY
APPLICABLE FOR CASES 1 AND 2.
FOR CASE 2 JUST CHANGE THE “50” TO
N.
FORMULAS FOR Pn:
𝑃1 = 𝑃0 ∗
𝜆0
𝜇1
𝑃2 = 𝑃0 ∗
𝜆0𝜆1
𝜇1𝜇2
⋮
𝑃𝑛 = 𝑃0 ∗
𝜆0𝜆1𝜆2 ⋯ 𝜆𝑛
𝜇1𝜇2𝜇3 ⋯ 𝜇𝑛+1
NOTE THAT
ρ =
𝜆
𝜇

103. QUEUING THEORY
CASE P0 Pn L Lq W Wq
1 (
𝑥=0
50
( 𝜌𝑥))−1 𝑃1 = 𝑃0 ∗ 𝜌1
𝑃2 = 𝑃0 ∗ 𝜌2
⋮
𝑃𝑛 = 𝑃0 ∗ 𝜌𝑛
𝜆𝑒𝑓𝑓 = (1 − 𝑝𝑁)𝜆
𝑥=0
50
( 𝑥 ∗ 𝑃0 ∗ 𝜌𝑥
)
𝑥=1
50
( 𝑥 − 1 ∗ 𝑃0 ∗ 𝜌𝑥
)
𝐿
𝜆𝑒𝑓𝑓
𝐿𝑞
𝜆𝑒𝑓𝑓
𝑥=0
𝑁
( 𝑥 ∗ 𝑃0 ∗ 𝜌𝑥
)
𝑥=1
𝑁
( (𝑥 − 1) ∗ 𝑃0 ∗ 𝜌𝑥)
2 (
𝑥=0
𝑁
( 𝜌𝑥))−1

104. CASE P0 Pn L Lq W Wq
3 (
𝑥=0
𝑠−1
(
𝜌𝑥
𝑥!
) +
𝜌𝑠
𝑠!
𝑥=𝑠
50
(
𝜌
𝑠
)𝑥−𝑠
)−1
𝜌𝑛
𝑠!
∗ 𝑃0
; 0 ≤ 𝑛 ≤ 𝑠
𝜌𝑛
𝑠! 𝑠𝑛−𝑠
∗ 𝑃0
; 𝑠 ≤ 𝑛 ≤ 𝑁
𝐿𝑞 + 𝜌 𝑥=𝑠
50
( (𝑥 − 𝑠) ∗ 𝑃0 ∗
𝜌𝑥
𝑠! 𝑠𝑥−𝑠)
𝐿
𝜆𝑒𝑓𝑓
𝐿𝑞
𝜆𝑒𝑓𝑓
𝐿𝑞 +
𝜆𝑒𝑓𝑓
𝜇
;
𝜆𝑒𝑓𝑓 = (1 − 𝑝𝑁)𝜆 𝑥=𝑠
𝑁
( (𝑥 − 𝑠) ∗ 𝑃0 ∗
𝜌𝑥
𝑠! 𝑠𝑥−𝑠)
4 (
𝑥=0
𝑠−1
(
𝜌𝑥
𝑥!
) +
𝜌𝑠
𝑠!
𝑥=𝑠
𝑁
(
𝜌
𝑠
)𝑥−𝑠)−1

105. CASE P0 Pn L Lq W Wq
5 (s=1) (
𝑥=0
𝑁
(
𝑁!
(𝑁 − 𝑥)!
)(𝜌)𝑥)−1 𝑃0 ∗ (
𝑁!
(𝑁 − 𝑛)!
)(𝜌)𝑛
𝐿𝑞 + 1 − 𝑃
0
𝑥=1
𝑁
( 𝑥 − 1 ∗ 𝑃0
∗ (
𝑁!
(𝑁 − 𝑥)!
)(𝜌)𝑥
)
; 𝜆𝑒𝑓𝑓 = (N − L)
𝐿
𝜆𝑒𝑓𝑓
𝐿𝑞
𝜆𝑒𝑓𝑓
5 (s>1)
(
𝑥=0
𝑠−1
(
𝑁!
𝑁 − 𝑥 ! 𝑥!
)(𝜌𝑥
)
+
𝑥=𝑠
𝑁
(
𝑁!
𝑁 − 𝑥 ! 𝑠! 𝑠𝑥−𝑠
)(𝜌𝑥
))−1
𝑃0(
𝑁!
𝑁 − 𝑛 ! 𝑛!
)(𝜌)𝑛
If n≤s
𝑃0(
𝑁!
𝑁 − 𝑥 ! 𝑠! 𝑠𝑛−𝑠
)(𝜌)𝑛
If n≥s
𝑥=0
𝑠−1
( 𝑥𝑃𝑥) + 𝐿𝑞
+ s(1 −
𝑥=0
𝑠−1
( 𝑃𝑥)) 𝑥=𝑠
𝑁
( 𝑥 − 𝑠)𝑃𝑥

106. Example

107. Problem

108. THANK YOU FOR LISTENING
• For any tutorials (mainly in OR/STAT), comments, or clarifications you
may reach me at hctrbnd@gmail.com or you may message me
directly on Facebook.
“Things never get easier, you just get better”
“Good luck to all of you future Certified Industrial Engineers”
-Hec