Transcript Lecture 19.pptx

Lecture 19
Capacity Management and Planning (Revision)
Books
• Introduction to Materials Management, Sixth Edition, J. R. Tony Arnold, P.E., CFPIM, CIRM, Fleming
College, Emeritus, Stephen N. Chapman, Ph.D., CFPIM, North Carolina State University, Lloyd M.
Clive, P.E., CFPIM, Fleming College
• Operations Management for Competitive Advantage, 11th Edition, by Chase, Jacobs, and Aquilano, 2005,
N.Y.: McGraw-Hill/Irwin.
• Operations Management, 11/E, Jay Heizer, Texas Lutheran University, Barry Render, Graduate School of
Business, Rollins College, Prentice Hall
Objectives
•
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Capacity
Capacity Management
Matching Capacity and Demand
Capacity Planning process
Resource Planning
Inputs to capacity planning
Managing Demand
Tactics for managing demand
Approaches to capacity expansion
Breakeven analysis
Decision tree and capacity expansion
Net present value
Capacity planning issues
Capacity
• The throughput, or the number of units a
facility can hold, receive, store, or produce
in a period of time
• Determines
fixed costs
• Determines if
demand will
be satisfied
• Three time horizons
Capacity Management
• Capacity management is planning and controlling
resources needed to meet production objectives
– Planning - determining resources needed to meet the
priority plan; selecting methods to make that capacity
available
– Controlling - monitoring output, comparing it with the
plan, and taking corrective action
Matching Capacity and Demand
• Demand Management
– vary prices
– change lead times
– encourage/discourage business
• Capacity Management
–
–
–
–
adjust staffing
adjust equipment and processes
change methods to facilitate production
redesign the product to facilitate production
Capacity Planning Process
• Determine the capacity available at each work center.
• Translate the released and planned orders into the
capacity required at each work center in each time
period.
• Sum up the capacities required for each work center to
determine the load on each work center in each time
period.
• Resolve differences between available capacity and
required capacity.
Planning Over a Time Horizon
Long-range
planning
Add facilities
Add long lead time equipment
Intermediaterange planning
Subcontract
Add equipment
Add shifts
Short-range
planning
Add personnel
Build or use inventory
*
Modify capacity
* Limited options exist
*
Schedule jobs
Schedule personnel
Allocate machinery
Use capacity
Resource Planning
• RP involves long-range capacity resource
requirements and is directly linked to production
planning.
• RP involves changes in manpower, capital
equipment, product design, or other facilities that
take a long time to acquire and eliminate.
• The production plan and RP set the limits and
levels for production.
Rough-Cut Capacity Planning
RCCP is the process to check the feasibility of the master
production schedule (MPS) (end items), provide warnings
of any bottlenecks, ensure utilization of work centers, and
advise vendors of capacity requirements.
Capacity Requirements Planning
• CRP is directly linked to the material requirements plan
(MRP) (component items).
• CRP is the most detailed, complete, and accurate of the
capacity planning techniques and is highly important in the
immediate time periods.
Inputs for Capacity Planning
• Inputs needed are:
– open shop orders from open order file
– planned order releases from material
requirements plan
– routings and time standards from the routing file
– lead times and work center capacities from the
work center file
Inputs for Capacity Planning
• Planned order releases are determined by the computer’s
MRP logic based upon the gross requirements for a
particular part. Used to help assess the total capacity
required in future time periods.
Inputs for Capacity Planning
• Routing file is the path that work follows from work
center to work center as it is completed. It should
contain:
–
–
–
–
–
–
Operations to be performed
Sequence of operations
Work centers to be used
Possible alternate work centers
Tooling needed for each operation
Standard times: setup times and run times per piece
Inputs for Capacity Planning
• Work center file is composed of a number of machines or
workers capable of doing the same work. It contains the
following information
– Move time is the time normally taken to move material from
one workstation to another.
– Wait time is the time a job is at a work center after completion
and before being moved.
– Queue time is the time a job waits at a work center before
being handled.
– Lead time is the sum of queue, setup, run, wait, and move
times.
Determining Capacity Available
• What is Capacity available?
– The capability of a system or resource to produce a
quantity of output in a particular time period.
APICS Dictionary
Determining Capacity Available
• How do we calculate Capacity?
– To calculate available capacity, we need to know:
• Available time
• Utilization
• Efficiency
Available Time
• Available time depends on the number of
machines, the number of workers, and the hours of
operation. The number of hours a work center can
be used.
If a work center has three equivalent machines and
works eight hours a day for five days a week, what
is its weekly available time?
Available time = 3 x 8 x 5 = 120 hours per week
Utilization
• Utilization is the percentage of time that the work
center is actually active.
Hours actually worked
Utilization =
x 100%
Available hours
What is the utilization if a work center is available for
120 hours a week, but produces goods for only 100
hours?
100
Utilization =
x 100% = 83.3%
120
Efficiency
Standard Time is the time it would take a qualified
operator working at a normal pace to do the job using
the proscribed method.
Efficiency
• Efficiency is a measure (as a percentage) of the actual
output compared with the standard expected output.
Standard hours of work
Efficiency =
x 100%
Hours actually worked
What is the efficiency if a work center is used 100 hours
per week and produces 120 standard hours of work?
120
Efficiency =
x 100% = 120%
100
Utilization and Efficiency Problem
A work center produces 90 standard hours of work in one
week. The hours available are 80, and 70 are actually
worked. Calculate the utilization and efficiency of the
work center.
70
Utilization =
x 100% = 87.5%
80
90
Efficiency =
x 100% = 128.57%
70
Rated Capacity
• Rated capacity is a measure of the output that can be
expected for a work center.
Rated Capacity = available time x utilization x efficiency
A work center has 3 machines and is operated 8 hours a
day for 5 days/week. Past utilization has been 75% &
efficiency has been 110%. What is the rated capacity?
Available time = 3 x 8 x 5 = 120 hours per week
Rated capacity = 120 x 0.75 x 1.10 = 99 standard hours
Rated Capacity Problem
A work center consists of six machines that are available 16
hours per day for five days a week. Utilization is 80%, and
efficiency is 120%. What is the rated weekly capacity?
Rated Capacity Problem
Available time = 6 x 16 x 5
= 480 hours per week
Rated capacity = 480 x 0.80 x 1.20
= 460.8 standard hours
Demonstrated (Measured) Capacity
• Demonstrated capacity is proven capacity
calculated from actual performance data. The
result is average capacity.
Over the previous four weeks, a work center
produced 110, 140, 120, and 130 standard hours of
work. What is the demonstrated capacity?
Demonstrated capacity =
110 +140 +120 +130
 125 standard hours
4
Required Capacity
• Determining required capacity is a two step process:
determine the time needed for each
order at each work center
sum up the capacity required
for individual orders to obtain
a load
Required Capacity
• Time Needed for Each Order:
– The sum of the setup time and the run time
A work center is to process 150 units of gear shaft SG 123
on work order 333. The setup time is 1.5 hours, and the
run time is 0.2 hours per piece. What is the standard time
needed to run the order?
Total standard time = setup time + run time
= 1.5 + (150 x 0.2) = 31.5 std. hrs.
Scheduling Orders
• Calculate the operation time required at each work center
– Operation time = setup time + (run time per piece x number
of pieces)
• Calculate lead time
– Lead time = operation time + queue time + wait time +
move time
Scheduling Orders
• Back Scheduling is to start with the due date and,
using the lead times, to work back to find the start
date for each operation.
Resolving Differences
• Change Capacity
•
•
•
•
•
Overtime or undertime
Hiring or layoff
Shift work force
Alternate routings
Subcontract
• Alter Load
• Alter lot sizes
• Reschedule
Design and Effective Capacity
 Design capacity is the maximum
theoretical output of a system
 Normally expressed as a rate
 Effective capacity is the capacity a firm
expects to achieve given current
operating constraints
 Often lower than design capacity
Utilization and Efficiency
Utilization is the percent of design capacity achieved
Utilization = Actual output/Design capacity
Efficiency is the percent of effective capacity
achieved
Efficiency = Actual output/Effective capacity
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Efficiency = 148,000/175,000 = 84.6%
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Efficiency = 148,000/175,000 = 84.6%
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Efficiency = 84.6%
Efficiency of new line = 75%
Expected Output = (Effective Capacity)(Efficiency)
= (175,000)(.75) = 131,250 rolls
Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Efficiency = 84.6%
Efficiency of new line = 75%
Expected Output = (Effective Capacity)(Efficiency)
= (175,000)(.75) = 131,250 rolls
Average unit cost
(dollars per room per night)
Economies and Diseconomies of Scale
25 - room roadside
motel
50 - room
roadside motel
Economies of
scale
25
75 - room
roadside motel
Diseconomies of
scale
50
Number of Rooms
75
Managing Demand
 Demand exceeds capacity
 Curtail demand by raising prices, scheduling
longer lead time
 Long term solution is to increase capacity
 Capacity exceeds demand
 Stimulate market
 Product changes
 Adjusting to seasonal demands
 Produce products with complementary
demand patterns
Break-Even Analysis
 Technique for evaluating process
and equipment alternatives
 Objective is to find the point in
dollars and units at which cost
equals revenue
 Requires estimation of fixed costs,
variable costs, and revenue
Break-Even Analysis
 Fixed costs are costs that continue
even if no units are produced
 Depreciation, taxes, debt, mortgage
payments
 Variable costs are costs that vary
with the volume of units produced
 Labor, materials, portion of utilities
 Contribution is the difference between
selling price and variable cost
Break-Even Analysis
Assumptions
 Costs and revenue are linear
functions
 Generally not the case in the real
world
 We actually know these costs
 Very difficult to accomplish
 There is no time value of money
Break-Even Analysis
–
Total revenue line
900 –
800 –
Cost in dollars
700 –
Total cost line
Break-even point
Total cost = Total revenue
600 –
500 –
Variable cost
400 –
300 –
200 –
100 –
Fixed cost
|
|
|
|
|
|
|
|
|
|
|
–
0 100 200 300 400 500 600 700 800 900 1000 1100
|
Volume (units per period)
Break-Even Analysis
BEPx = break-even point in
units
BEP$ = break-even point in
dollars
P = price per unit (after all
discounts)
x = number of units produced
TR
F
V
TC
=
=
=
=
total revenue = Px
fixed costs
variable cost per unit
total costs = F + Vx
Break-even point
occurs when
TR = TC
or
Px = F + Vx
F
BEPx =
P-V
Break-Even Analysis
BEPx = break-even point in
units
BEP$ = break-even point in
dollars
P = price per unit (after all
discounts)
BEP$ = BEPx P
F
=
P
P-V
F
=
(P - V)/P
F
=
1 - V/P
x = number of units produced
TR
F
V
TC
=
=
=
=
total revenue = Px
fixed costs
variable cost per unit
total costs = F + Vx
Profit = TR - TC
= Px - (F + Vx)
= Px - F - Vx
= (P - V)x - F
Break-Even Example
Fixed costs = $10,000
Direct labor = $1.50/unit
BEP$ =
F
=
1 - (V/P)
Material = $.75/unit
Selling price = $4.00 per unit
$10,000
1 - [(1.50 + .75)/(4.00)]
Break-Even Example
Fixed costs = $10,000
Direct labor = $1.50/unit
BEP$ =
F
=
1 - (V/P)
Material = $.75/unit
Selling price = $4.00 per unit
$10,000
1 - [(1.50 + .75)/(4.00)]
$10,000
=
= $22,857.14
.4375
BEPx =
F
=
P-V
$10,000
= 5,714
4.00 - (1.50 + .75)
Break-Even Example
50,000 –
Revenue
40,000 –
Break-even
point
Dollars
30,000 –
Total
costs
20,000 –
Fixed costs
10,000 –
–|
0
|
|
|
|
|
2,000
4,000
6,000
8,000
10,000
Units
Break-Even Example
Multiproduct Case
BEP$ =
where
V
P
F
W
i
F
∑
Vi
1x (Wi)
Pi
= variable cost per unit
= price per unit
= fixed costs
= percent each product is of total dollar sales
= each product
Multiproduct Example
Fixed costs = $3,500 per month
Item
Sandwich
Soft drink
Baked potato
Tea
Salad bar
Price
$2.95
.80
1.55
.75
2.85
Cost
$1.25
.30
.47
.25
1.00
Annual Forecasted
Sales Units
7,000
7,000
5,000
5,000
3,000
Multiproduct Example
Fixed costs = $3,500 per month
Annual Forecasted
Item
Price
Cost
Sales Units
Sandwich
$2.95
$1.25
7,000
Soft drink
.80
.30
7,000
Baked potato
1.55
.47 Annual 5,000 Weighted
% of Contribution
Tea Selling Variable .75
.25Forecasted 5,000
Item (i)
Price (P) Cost (V) (V/P) 1 - (V/P) Sales $
Sales (col 5 x col 7)
Salad bar
2.85
1.00
3,000
Sandwich
Soft drink
Baked
potato
Tea
Salad bar
$2.95
.80
1.55
$1.25
.30
.47
.42
.38
.30
.58
.62
.70
$20,650
5,600
7,750
.446
.121
.167
.259
.075
.117
.75
2.85
.25
1.00
.33
.35
.67
.65
3,750
8,550
$46,300
.081
.185
1.000
.054
.120
.625
F
BEPExample
=
Multiproduct
V
$
∑
1-
i
Pi
x (Wi)
Fixed costs = $3,500 per month
$3,500
x 12
Annual
Forecasted
=
= $67,200
.625
Item
Price
Cost
Sales Units
Sandwich
$2.95
$1.25
7,000
$67,200
Daily
Soft drink
.80
.30
7,000
=
= $215.38
sales Annual
312 days
Baked potato
1.55
.47
5,000 Weighted
% of Contribution
Tea Selling Variable .75
.25Forecasted 5,000
Item (i)
Price (P) Cost (V) (V/P) 1 - (V/P) Sales $
Sales (col 5 x col 7)
.446 x $215.383,000
Salad bar
2.85
1.00
= 32.6  33
Sandwich
$2.95
$1.25
.42
.58
$20,650
.446
.259
$2.95
sandwiches
Soft drink
Baked
potato
Tea
Salad bar
.80
1.55
.30
.47
.38
.30
.62
.70
5,600
7,750
.75
2.85
.25
1.00
.33
.35
.67
.65
3,750
8,550
$46,300
.121
.075
per
day
.167
.117
.081
.185
1.000
.054
.120
.625
Decision Trees and Capacity Decision
Market favorable (.4)
Market unfavorable (.6)
Market favorable (.4)
Medium plant
Market unfavorable (.6)
Market favorable (.4)
Market unfavorable (.6)
$100,000
-$90,000
$60,000
-$10,000
$40,000
-$5,000
$0
Decision Trees and Capacity Decision
Market favorable (.4)
Market unfavorable (.6)
Market favorable (.4)
Medium plant
Large Plant
Market unfavorable (.6)
EMV = (.4)($100,000)
+ (.6)(-$90,000)
Market favorable (.4)
EMV = -$14,000
Market unfavorable (.6)
$100,000
-$90,000
$60,000
-$10,000
$40,000
-$5,000
$0
Decision Trees and Capacity Decision
-$14,000
Market favorable (.4)
Market unfavorable (.6)
$100,000
-$90,000
$18,000
Market favorable (.4)
Medium plant
Market unfavorable (.6)
$60,000
-$10,000
$13,000
Market favorable (.4)
Market unfavorable (.6)
$40,000
-$5,000
$0
Strategy-Driven Investment
 Operations may be responsible for
return-on-investment (ROI)
 Analyzing capacity alternatives
should include capital investment,
variable cost, cash flows, and net
present value
Net Present Value (NPV)
F
P=
(1 + i)N
where
F
P
i
N
= future value
= present value
= interest rate
= number of years
Net Present Value (NPV)
F
P=
(1 + i)N
Whilewhere
this worksFfine,
it
= future
value
is cumbersome
P for
= present value
larger values of N
i = interest rate
N = number of years
NPV Using Factors
F
P=
= FX
N
(1 + i)
where
Portion of
Table S7.1
X = a factor from Table S7.1
defined as = 1/(1 + i)N and
F = future value
Year
1
2
3
4
5
5%
.952
.907
.864
.823
.784
6%
.943
.890
.840
.792
.747
7%
.935
.873
.816
.763
.713
…
10%
.909
.826
.751
.683
.621
Present Value of an Annuity
An annuity is an investment which generates
uniform equal payments
S = RX
where
X = factor from Table S7.2
S = present value of a series of
uniform annual receipts
R = receipts that are received every
year of the life of the investment
Present Value of an Annuity
Year
1
2
3
4
5
5%
.952
1.859
2.723
4.329
5.076
6%
.943
1.833
2.676
3.465
4.212
7%
.935
1.808
2.624
3.387
4.100
…
10%
.909
1.736
2.487
3.170
3.791
Present Value of an Annuity
$7,000 in receipts per for 5 years
Interest rate = 6%
From Table S7.2
X = 4.212
S = RX
S = $7,000(4.212) = $29,484
Present Value With Different Future
Receipts
Investment A’s
Cash Flow
Investment B’s
Cash Flow
Year
Present Value
Factor at 8%
$10,000
$9,000
1
.926
9,000
9,000
2
.857
8,000
9,000
3
.794
7,000
9,000
4
.735
Present Value With Different Future
Receipts
Investment A’s
Present Values
Investment B’s
Present Values
1
$9,260 = (.926)($10,000)
$8,334 = (.926)($9,000)
2
7,713 = (.857)($9,000)
7,713 = (.857)($9,000)
3
6,352 = (.794)($8,000)
7,146 = (.794)($9,000)
4
5,145 = (.735)($7,000)
6,615 = (.735)($9,000)
Year
Totals
Minus initial
investment
Net present
value
$28,470
$29,808
-25,000
-26,000
$3,470
$3,808
Capacity Planning Issues
• To determine the capacity required to achieve the MPS
relative to the capacity available
• To make necessary capacity adjustments before creating
crises
• To determine an the appropriate level of safety capacity
• To determine the required level of detail and the critical
machines and work centers
• To use an appropriate technique given the tradeoff between
accuracy and computational effort
Finite Capacity Planning Example
The Single Square Company summarizes the capacity
requirements for three of its key resources from each of its
three product lines. A typical report (before any action is
taken) is shown on the next slide.
Finite Capacity Planning Example (Continued)
Key resource type
Machine Number
Drill 1 1
Drill 2 2
Drill 3 1
Percent capacity by line
Shifts
A
B
3
22
13
3
36
48
1
52
24
C
24
36
21
Filer 12 3
Filer 23 3
2
12
0
20
14
8
16
40
Dryer 1 2
Dryer 2 2
Dryer 3 2
2
2
1
28
72
84
36
51
27
51
43
24
a.
b.
Total
59
120
97
Remarks
116
116
135
What actions would you recommend? (Assume each
machine type is equivalent in terms of capacity.)
What other observations would you have for management,
based on the preceding report?
Finite Capacity Planning Example:
Single Square Company
Key Resource Type
Machine Number Shifts
Drill 1
1
3
Drill 2
2
3
Drill 3
1
1
Total
3
6
1
10
% Capacity by Line
A
B
22
13
36
48
52
24
C
24
36
21
Tot.
59
120
97
Equiv.
1.77
7.20
.97
9.94
Remarks
Req'd.
Move
from
2 to 1
Filer 1
Filer 2
2
3
3
3
6
9
15
2
12
0
20
14
8
16
40
.96
3.60 _
4.56
Excess
capacity
Dryer 1
Dryer 2
Dryer 3
2
2
2
2
2
1
4
4
2
10
28
72
84
36
51
27
52
43
24
116
166
135
4.64
6.64
2.70
13.98
Increase
to three
shifts
Finite Capacity Planning Example:
Single Square Company (continued)
a. There is enough capacity now in the Drill and Filer areas. To correct
the imbalance in the Drills, it is necessary to move some of the load
from Drill 2 to Drill 1. In the case of the Dryers, it is necessary to
increase capacity. Adding shifts will provide enough for the current
load.
b. Why is there so much capacity for the Filers? They can be brought
back to one shift and are okay. Secondly, if there is any upward trend
in the loads for the Dryers, it behooves management to begin to
worry about where they can get additional dryer capacity.
Twin Disc Capacity Bill Report
73
Twin Disc Capacity Bill Example
Suppose, in the Twin Disc example, it was decided to move all
production for product lines F and I from the Maag grinder
(CEA) to the Reishauer grinder (CAB). What's the resulting
total percentage of capacity for each machine and the amount
for product lines F and I? (Note the Reishauer is three times
faster than the Maag grinder so the Maag takes three times as
many hours to complete a job.) Assume setup time is
negligible.
Twin Disc Capacity Bill Example
For Line I
Grinding Hours
On Maag
= 3044 x .08 = 243.52
On Reishauer
= 243.52/3 = 81.17
% of Reishauer
= 81.17/950 = 9
New % of Line I on Reishauer
= 9 + 1 = 10%
New % of Line I on Maag
= 0%
For Line F:
Grinding Hours
On Maag
= 3044 x .05 = 152.20
On Reishauer
= 152.20/3 = 50.73
% of Reishauer
= 50.73/950 = 5
New % of Line F on Reishauer
= 5 + 2 = 7%
New % of Line F on Maag
= 0%
New Totals:
New total for Reishauer
New total for Maag
= 69 + 9 + 5 = 83%
= 113- 8 - 5 = 100%
Applicon Capacity Status Report
76
Applicon Capacity Bill Example
Applicon has the following capacity bills for items 207 and
208:
207
Work center
ALF-A
HLT-A
MIS-A
MVX-A
PCB-A
PCB-P
208
Hours/unit
Work center Hours/unit
0.5 ALF-T
0.3
1.0 HLT-T
0.8
0.8 MIS-T
0.6
0.8 MVX-T
0.5
0.5 PCB-P
0.9
1.0 PCB-T
1.4
Applicon Capacity Bill Example
a. A customer wants to know if Applicon can deliver 100 units
each of items 207 and 208 during the next month (20
working days). Assume there are adequate materials, work
center MVX-T'S crew has been increased to one full person
(making standard capacity = 160), the increase has come by
reducing MVX-A to 6.5 persons. and no other orders have
been booked. Use the standard capacity data and conditions
in the previous slides as the basis for your analysis.
b. Suppose the customer (in part a. above) decided to delay the
order for several months. How many units of item 207
alone could be added to the MPS in the next month? How
about item 208?
Applicon
Analysis:
For Item 207
Work CenterUnits Possible
ALF-A
HLT-A
MIS-A
MVX-A
PCB-A
PCB-P
(480 - 70)/0.5 = 820 units
(800 - 438)/1.0 = 362 units
(800 - 270)/0.8 = 662 units
(1120 - 80 - 399)/0.8 = 801 units
(160 - 52)/0.5 = 216 units
(960 - 408)/1.0 = 552 units
BUT this is if only item 207 is made.
Applicon
If BOTH items are made in equal quantities, PCB-P is a shared
work center so:
PCB-P (960 - 408)/(1.0 + 0.9) = 291 of each.
For Item 208
Work Center
Units Possible
ALF-T
(80 - 5)/0.3 = 250 units
HLT-T
(160 - 85)/0.8 = 94 units
MIS-T
(80 - 14)/0.6 = 110 units
MVX-T (160 - 79)/0.5 = 162 units
PCB-P
(960 - 408)/(1.0 + 0.9) = 291 units
PCB-T
(1680 - 918)/1.4 = 544 units
Applicon
a.
The analysis says that 100 units of each cannot
be added to the MPS without something else
being done. The problem is with item 208 in
work center HLT-T. To make 100 units of item
208, work center HLT-T will have to work
overtime-unless capacity can be provided from a
less highly utilized work center or perhaps
greater productivity can be gained in some other
way.
b. The analysis shows that 216 units of item 207 or
a total of 94 units of item 208 can be made.
End of Lecture 19