Quantitative Review I
Capacity and Constraint Management
(Break-Even Analysis)
Portland Radio Company (PRC) is trying to decide
whether or not to introduce a new model. If they
introduce it, there will be additional fixed costs of
$400,000 per year. The variable costs have been
estimated to be $30 per radio. If PRC sells the new
radio model for $40 per radio, how many must they
sell to break even?
F
Q
SP  VC
$400,000
Q
 40,000units
$40  $30
Which Year to Break Even (BE)
Year
BE Quantity
(At)
Available for Sale
(Bt=At+Dt-1)
Demand
(Ct)
Left Over /Inventory
(Dt=Bt-Ct)
1
40,000
40,000
30,000
10,000
2
40,000
50,000
35,000
15,000
3
40,000
55,000
60,000
-5,000
Which Month of Year 3 to Break Even
Yearly Demand = 60,000
Monthly Demand = 60,000/12 = 5,000
BE = 5,000 x 11 months = 55,000 or Month 11 of Year 3
OR 55,000/60,000 = 0.9167
0.9167 x 12 = 11
Operations and Productivity
Productivity
# Outputs
P
# Inputs
Labor Productivity
Three workers paint twenty-four tables in eight hours
Inputs: 24 hours of labor (3 workers x 8 hours)
Outputs: 24 painted tables
Outputs 24 tables
P

 1 table/ hour
Inputs 24 hours
Multifactor Productivity
• Convert all inputs & outputs to $ value
Example:
– 200 units produced sell for $12 each
– materials cost $6 per unit
– 40 hours of labor were required at $10 an hour
200 units $12 / unit
P
200 units $6 / unit  40 hours $10 / hour
$2,400
P
 1.50
$1,600
Productivity Index
• Can be used to compare a process’
productivity at a given time (P2) to the
same process’ productivity at an earlier
time (P1)
P2  P1
Growth Rate 
P1
Productivity Growth Rate
Example:
– Last week, a company produced 150 units using 200
hours of labor.
– This week, the same company produced 170 units
using 240 hours of labor.
150 units
P1 
 0.75 units / hour
200 hours
170 units
P2 
 0.71 units / hour
240 hours
P2  P1 0.71 0.75
Growth Rate 

 0.05
P1
0.75
or a negative5% growth rate
If inputs increase by 25% and outputs decrease
by 10%, what is the percentage change in
productivity? Assume P1 = 1/1.
1
P1   1
1
0.90
P2 
 0.72
1.25
0.72  1
%Change
1
%Change .28or28%decrease
Production increased to 75 from 65 pieces per day.
Defective items have dropped from 12 to 5 pieces per
day. Production facility operates eight hours per day.
Seven people work daily in the plant. What is the
change in productivity?
Period 1
Period 2
Output = 65 Defects = 12 Net Output = 53
Output = 75 Defects = 5 Net Output = 70
Period 1
Period 2
Input = 8 hours x 7 workers = 56
Input = 8 hours x 7 workers = 56
P1 = 53/56 = 0.95
P2 = 70/56 = 1.25
Change = (1.25-0.95)/0.95 = 0.32
Revenue Management Systems
Revenue Management Systems (Hotel)
 Contribution to profit and overhead ($)
= [(Selling Price – Variable Cost) x Demand]1 +
[(Selling Price – Variable Cost) x Demand]2 +… +
[(Selling Price – Variable Cost) x Demand]n
 Hotel management effectiveness (%)
=
=
Actual hotel revenue
Maximum possible hotel revenue
Actual prices for each room night x
Actual number of room nights rented
Maximum legal price for each room night x
Maximum number of room nights available in hotel
Hotel Management
Characteristic/Variable
Business
Hotel Customers
Customers for this day
(Demand)
Average price/room night
(Selling Price)
Variable cost/room night
(Variable cost)
Maximum price/room
night
Maximum number rooms
available for sale this day
400 room nights
rented
Convention
Association
Hotel Customers
800 room nights
rented
$200
$120
$50
$50
$250
$150
500 room nights
available
900 room nights
available
Hotel Management
 Contribution to profit and overhead ($)
= [(Selling Price – Variable Cost) x Demand]
= [($200 - $50) x 400] + [($120 - $50) x 800]
= ($150 x 400) + ($70 x 800) = $116,000
 Hotel management effectiveness (%)
=
Actual prices for each room night x
Actual number of room nights rented
Maximum legal price for each room night x
Maximum number of room nights available in hotel
= ($200 x 400 rooms) + ($120 x 800 rooms)
($250 x 500 rooms) + ($150 x 900 rooms)
= ($80,000 + $96,000) / ($125,000 + $135,000) = 67.69%
Revenue Management Systems (Airlines)
A regional airline that operates a 50-seat jet prices
the ticket for one popular business flight at $250. If
the airline overbooks the reservations, overbooked
passengers receive a $300 travel voucher. The
airline is considering overbooking by up to 5 seats,
and the demand for the flight always exceeds the
number of reservations it might accept. The
probabilities of the number of passengers who
show up is given in the following table:
Number of reservations
50
51
52
53
54
55
45
0.100
0.080
0.060
0.040
0.020
0.010
46
0.150
0.130
0.125
0.050
0.040
0.030
Number of passengers showing up
47
48
49
50
51
52
53
54
55
0.150 0.200 0.300 0.100
0.180 0.150 0.250 0.110 0.100
0.175 0.200 0.250 0.100 0.050 0.040
0.070 0.200 0.250 0.150 0.100 0.080 0.060
0.050 0.090 0.120 0.210 0.180 0.140 0.100 0.050
0.040 0.060 0.100 0.120 0.200 0.180 0.150 0.090 0.020
Overbooking Strategies and Analysis
Number of reservations
50
51
52
53
54
55
45
0.100
0.080
0.060
0.040
0.020
0.010
46
0.150
0.130
0.125
0.050
0.040
0.030
Number of passengers showing up
47
48
49
50
51
52
53
54
55
0.150 0.200 0.300 0.100
0.180 0.150 0.250 0.110 0.100
0.175 0.200 0.250 0.100 0.050 0.040
0.070 0.200 0.250 0.150 0.100 0.080 0.060
0.050 0.090 0.120 0.210 0.180 0.140 0.100 0.050
0.040 0.060 0.100 0.120 0.200 0.180 0.150 0.090 0.020
Expected revenue for 50 reservations
= $250 x (45*.1 + 46*.15 + 47*.15 + 48*.2 + 49*.3 + 50*.1) = $11,937.50
Expected re venue for 51 reservations
= [$250 x (45*.08 + 46*.13 + 47*.18 + 48*.15 + 49*.25 + 50*.11)] – ($300 x .1) = 10,717.50
Expected re venue for 52 reservations
= [$250 x (45*.06 + 46*.125 + 47*.175 + 48*.2 + 49*.25 + 50*.1)] – [$300 x (.05+.04)]
= $10,854.25
Expected re venue for 53 reservations
= [$250 x (45*.04 + 46*.05 + 47*.07 + 48*.2 + 49*.25 + 50*.15)] – [$300 x (.1+.08+.06)]
= $9,113
Expected revenue for 54 reservations = $6,306.50
Expected revenue for 55 reservations = $4,180.50
Forecasting
Lauren's Beauty Boutique has experienced the
following weekly sales. Calculate a 3 period moving
average for Week 6.
Week Sales
1
2
3
4
5
6
432
396
415
478
460
451
415  478  460
Week 6Sales 
 451
3
A firm has the following order history over the last 6
months. What would be a 3-month weighted moving
average forecast for July, using weights of 60% for the
most recent month, 20% for the month preceding the
most recent month, and 20% for the month preceding
that one?
January
February
March
April
May
June
July
120
95
100
75
100
50
65
JulyOrder (0.6)(50)  (0.2)(100)  (0.2)(75)  65
Exponential Smoothing
Ft 1  At  1   Ft
– Last period’s actual value (At)
– Last period’s forecast (Ft)
– Select value of smoothing coefficient, between 0
and 1.0
Summary of Single Exponential Smoothing
Milk-Sales Forecasts with α = 0.2
F2 = .2(172) + .8(172) =172
F3 = .2(217) + .8(172) = 181
F4 = .2(190) + .8(181) = 182.8
F5 = .2(233) + .8(182.8) = 192.84
F6 = .2(179) + .8(192.84) = 190.07
F7……….
You start with past data and calculate forecasts working forward.
Determine forecast for periods 7 and 8
exponential smoothing with alpha = 0.2 and the
period 6 forecast being 375.
Period Actual
1
300
2
315
3
290
4
345
5
320
6
360
7
375
8
Exponential
Smoothing
Period 7 = 0.2(360) + 0.8(375)
= 72 + 300 = 372.0
Period 8 = 0.2(375) + 0.8(372)
= 75 + 297.6 = 372.6
375.0
372.0
372.6
Quarterly Forecasting
Expected total demand in 2012 is 3,000 units.
Given the historical sales figures below, derive a
forecast for each quarter in 2012.
750 x 0.43
Q1
Q2
Q3
Q4
Total
Quarter
2009
250
500
700
900
2350
587.5
250/587.5
0.43
0.85
1.19
1.53
2010
270
530
800
970
2570
642.5
0.42
0.82
1.25
1.51
2011
310
600
850
1000
0.45
0.87
1.23
1.45
0.43
0.85
1.22
1.50
2760
690
2012
324
636
917
1123
3000
(1.53+1.51+1.45)/3
750
The Regression Equation or
Trend Forecast
Tx  y  a  bX
Tx
= trend forecast or y variable
a
= estimate of Y-axis intercept where X = 0
b
= estimate of slope of the demand line
X
= period number or independent variable
Linear Regression
•
b
 XY  X  Y 
 X 2  X  X 
•
Identify dependent (y) and
independent (x) variables
Solve for the slope of the line
XY  n X Y

b
 X  n((X) )
2
•
2
Solve for the y intercept
a  Y  bX
•
Develop your equation for the
trend line
Tx or y = a + bX
A maker of golf shirts has been tracking the relationship between
sales and advertising dollars. Use linear regression to find out
what sales might be if the company invested $60,000 in
advertising next year.
1
Sales
$ (Y)
Advertising
$ (X)
XY
X^2
130
48
6240
2304
XY  n X Y

b
 X  n((X) )
2
2
151
52
7852
2704
3
150
50
7500
2500
4
158
55
8690
3025
a  Y  bX
Tx 
5
Total
589
205
Average
147.25
51.25
30282
10533
2633.25
2
y  a  bX
XY  n X Y

b
 X  n((X) )
2
2
30282 4(51.25)(147.25) 95.75
b

 3.579
2
10533 4(51.25)
26.75
a  Y  bX
a = 147.25-3.579(51.25) = -36.17
a  Y  bX
Y = a + bX
Y = -36.17 + 3.579X
Y = -36.17 + 3.579(60)
Y = 178.57 or $178,570 in sales
Tracking Forecast Error Over Time
• Mean Absolute Deviation (MAD)
1 n
– A good measure of the actual error in
MAD =  Ai  Fi
a forecast
n i =1
• Tracking Signal (TS)
– Exposes bias (positive or negative)
Positive TS = under-forecasting
Negative TS = over-forecasting
actual- forecast

TS 
MAD
Mean Absolute Deviation
• MAD sums only absolute values errors, both
positive and negative errors add to the sum
and the average size of the error (whether
positive or negative) is determined.
1 n
MAD =  Ai  Fi
n i =1
n = number of periods of data
F = forecast of demand in period i
A = actual demand in period i
A company is comparing the accuracy of two forecasting
methods. Forecasts using both methods are shown below
along with the actual values for January through May. The
company also uses a tracking signal with ±4 limits to decide
when a forecast should be reviewed. Which forecasting
method is best?
Method A
Method B
Month
Actual
sales
Forecast
Error
Abs. Value
Forecast
Error
Abs. Value
Jan.
30
28
2
2
27
3
3
Feb.
26
25
1
1
25
1
1
Mar.
32
32
0
0
29
3
3
Apr.
29
30
-1
1
27
2
2
May
31
30
1
1
29
2
2
MAD = 5/5 = 1
TS = 3/1 = 3
MAD = 11/5 = 2.2
TS = 11/2.2 = 5
Capacity and Constraint Management
Capacity Utilization
 Theoretical Capacity: Maximum output
rate under ideal conditions
 Effective Capacity: Maximum output
rate under normal (realistic) conditions
actualoutput rate
Utilization 
(100%)
capacity
Computing Capacity Utilization
In the bakery example, the design capacity is 60
custom cakes per day. On average, this bakery can
make 40 custom cake per day. Currently, the
bakery is producing 56 cakes per day. What is the
bakery’s capacity utilization relative to both
theoretical and effective capacity?
actualoutput
Utilization effective 
(100%)
effectivecapacity
actualoutput
Utilization theoretical 
(100%)
theoretical capacity
Computing Capacity Utilization
In the bakery example, the design capacity is 60
custom cakes per day. On average, this bakery can
make 40 custom cake per day. Currently, the
bakery is producing 56 cakes per day. What is the
bakery’s capacity utilization relative to both
theoretical and effective capacity?
Utilization effective 
Utilization
actualoutput
56
(100%) (100%) 140%
effectivecapacity
40
theoretical 
actualoutput
56
(100%) (100%) 93%
theoretical capacity
60
A clinic has been set up to give flu shots to the
elderly in a large city. The theoretical capacity is 80
seniors per hour, and the effective capacity is 55
seniors per hour. Yesterday the clinic was open for
ten hours and gave flu shots to 350 seniors.
What is the effective utilization?
What is the theoretical utilization?
Utilization effective 
actualoutput
350/10
(100%)
(100%) 63.64%
effectivecapacity
55
actualoutput
350/10
Utilization theoretical 
(100%)
(100%) 43.75%
theoretical capacity
80
Decision Trees
• Build from the present to the future:
– Distinguish between decisions (under
your control) & chance events (out of
your control, but can be estimated to a
given probability)
• Solve from the future to the present:
– Generate an expected value for each
decision point based on probable
outcomes of subsequent events
Should you expand large or small?
Low Demand (0.40)
$60,000 Exp Rev
.40 x $60,000 = $24,000
$24,000 + $60,000
= $84,000
Expand Large
.60 x $100,000 = $60,000
High Demand (0.60)
Low Demand (0.40)
$100,000 Exp Rev
$50,000 Exp Rev
.40 x $50,000 = $20,000
Expand Small
$20,000 + $42,000
= $62,000
Expand
High
Demand
(0.60)
.60 x $70,000
= $42,000
$70,000 Exp Rev
$70,000 > $45,000
Do not Expand
$45,000 Exp Rev
Capacity Analysis
 Capacity analysis determines the throughput
capacity of workstations in a system.
 A bottleneck is a limiting factor or constraint.
 A bottleneck has the lowest effective capacity in a
system or takes the most time.
 What is the bottleneck?
Station A (50 seconds)
 What is the maximum production per hour?
MaximumOut put 
AvailableT ime 3600 sec/ hour

 72Units / hour
Bottleneck
50 sec/ unit
That’s all, folks!
Thanks for today!