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# ENGR Fundamentals of Logistics Engineering

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```									ENGR 3210 Fundamentals of
Logistics Engineering

Class 33
November 11, 2002
How do you size a warehouse?
What are the important variables?
   Throughput or demand for goods moving through
   Inventory turnover (defines how much inventory
is stored in the warehouse)
   Storage size requirements for the units to be stored
   The fraction of the total warehouse that is actually
used for storage (vs. aisle and other space)
 Material handling equipment and storage
systems used helps to define this fraction

2
A simple example
   The warehouse will handle 2,400,000 lbs of
widgets per month
   Monthly turnover of inventory is 3 turns per
month (turnover is monthly sales divided by
average inventory)
   The storage system to be used requires that 50%
of the floor space would be devoted to aisles
   Only 70% of the warehouse is really used for
storage. The rest will be used for office facilities
   One pound of widgets is packaged in boxes that
are 0.5 cubic feet in size
   The boxes of widgets can be stacked 16 feet high
on pallets                                             3
How many square feet do we
need in the warehouse?
   How many pounds of widgets are going to be stored in the
warehouse?
 2,400,000 lbs / 3 turns per month =800,000 lbs

   How many cubic feet do the widgets take up?
 800,000 lbs (0.5 cu. ft. per lb.) = 400,000 cu. ft.

   How many square feet do the widgets occupy?
 400,000 cu. Ft. / 16 ft. = 25,000 sq ft.

   How much aisle space do we need?
 An equal amount to the widget space = 25,000 sq. ft.

   What about office space ?
 50,000 sq. ft. / 70% = 71,428 sq, ft in the entire
warehouse
4
The real question here is not size but cost
(If we are to minimize Total Logistics
Costs)
   Construction of the warehouse is assumed to cost
\$30 / sq. ft.
   The construction cost is amortized over 20 years
with an interest rate that breaks the construction
cost down to \$1.50 per sq. ft. each year.
   The annual fixed cost to operate a warehouse of
this type is \$3.00 per sq. ft.
   The materials handling charge for this material is
\$0.05 per lb. of throughput.

5
What is the annual cost of
operating the warehouse?
   What is needed to pay off the construction loan?
 71,428 sq. ft. (\$1.50 per sq. ft.) = \$107,142

   What is needed to operate the warehouse for a
year?
 71,428 sq. ft. (\$3.00 per sq. ft.) = \$214,284

   What is needed to handle the material for a year?
 2,400,000 lbs. (12months) (\$0.05) =
\$1,440,000
   Total cost = \$1,761,426 per year
6
A simplified computation for the
example
 Space (sq. ft.)=Monthly demand(lbs.) x
(1/3)(0.5/16)(1/0.5)(1/.7)
= Monthly demand x .03
 Annual Cost = Space (sq. ft.) (\$4.50) +
Monthly demand (\$0.60)

7
   What is the cost of the warehouse if the aisle
requirements are:
 35% of the total space?

 40% of the total space?

 45% of the total space?

   What is the cost of the warehouse if the stacking
height is:
 17 feet?

 18 feet?

 19 feet?

   What could you afford to spend on equipment to
achieve each of these improvements?               8
Throughput    2,400,000 lbs/mo                                     Construction Cost/sf/yr                  \$1.50
Turns                 3 turns/mo                                   Operation Cost/sf/yr                     \$3.00
Handling Cost/lb                         \$0.05
Cubic feet/lb                             0.50

SQUARE FEET REQUIRED

Stack height
Aisle %             14             15           16           17             18           19          20
35.00%      62,794         58,608       54,945       51,713         48,840       46,270      43,956
40.00%      68,027         63,492       59,524       56,022         52,910       50,125      47,619
45.00%      74,212         69,264       64,935       61,115         57,720       54,682      51,948
50.00%      81,633         76,190       71,429       67,227         63,492       60,150      57,143
55.00%      90,703         84,656       79,365       74,697         70,547       66,834      63,492
60.00%     102,041         95,238       89,286       84,034         79,365       75,188      71,429

ANNUAL COST

Stack height
Aisle %              14            15            16           17            18           19           20 AVG \$/ft
35.00%   \$1,722,575    \$1,703,736    \$1,687,253   \$1,672,708    \$1,659,780   \$1,648,213   \$1,637,802  \$14,129
40.00%   \$1,746,122    \$1,725,714    \$1,707,857   \$1,692,101    \$1,678,095   \$1,665,564   \$1,654,286  \$15,306
45.00%   \$1,773,952    \$1,751,688    \$1,732,208   \$1,715,019    \$1,699,740   \$1,686,070   \$1,673,766  \$16,698
50.00%   \$1,807,347    \$1,782,857    \$1,761,429   \$1,742,521    \$1,725,714   \$1,710,677   \$1,697,143  \$18,367
55.00%   \$1,848,163    \$1,820,952    \$1,797,143   \$1,776,134    \$1,757,460   \$1,740,752   \$1,725,714  \$20,408
60.00%   \$1,899,184    \$1,868,571    \$1,841,786   \$1,818,151    \$1,797,143   \$1,778,346   \$1,761,429  \$22,959
Avg \$/5%        \$35,322       \$32,967       \$30,907      \$29,089       \$27,473      \$26,027      \$24,725        9
The results graphed

\$1,900,000

\$1,850,000

\$1,800,000

\$1,750,000                                                     35%
40%
\$1,700,000                                                     45%
50%
\$1,650,000
55%
\$1,600,000                                                    60%
60%
\$1,550,000                                            55%
50%
\$1,500,000                                          45%
14                                   40%
15   16   17                  35%
18
19   20

10
Which variable promises the biggest
Throughput    2,400,000 lbs/mo                                     Construction Cost/sf/yr                  \$1.50
Turns                 3 turns/mo                                   Operation Cost/sf/yr                     \$3.00
Handling Cost/lb                         \$0.05
Cubic feet/lb                             0.50

SQUARE FEET REQUIRED

Stack height
Aisle %             14             15           16           17             18           19          20
35.00%      62,794         58,608       54,945       51,713         48,840       46,270      43,956
40.00%      68,027         63,492       59,524       56,022         52,910       50,125      47,619
45.00%      74,212         69,264       64,935       61,115         57,720       54,682      51,948
50.00%      81,633         76,190       71,429       67,227         63,492       60,150      57,143
55.00%      90,703         84,656       79,365       74,697         70,547       66,834      63,492
60.00%     102,041         95,238       89,286       84,034         79,365       75,188      71,429

ANNUAL COST

Stack height
Aisle %              14            15            16           17            18           19           20 AVG \$/ft
35.00%   \$1,722,575    \$1,703,736    \$1,687,253   \$1,672,708    \$1,659,780   \$1,648,213   \$1,637,802  \$14,129
40.00%   \$1,746,122    \$1,725,714    \$1,707,857   \$1,692,101    \$1,678,095   \$1,665,564   \$1,654,286  \$15,306
45.00%   \$1,773,952    \$1,751,688    \$1,732,208   \$1,715,019    \$1,699,740   \$1,686,070   \$1,673,766  \$16,698
50.00%   \$1,807,347    \$1,782,857    \$1,761,429   \$1,742,521    \$1,725,714   \$1,710,677   \$1,697,143  \$18,367
55.00%   \$1,848,163    \$1,820,952    \$1,797,143   \$1,776,134    \$1,757,460   \$1,740,752   \$1,725,714  \$20,408
60.00%   \$1,899,184    \$1,868,571    \$1,841,786   \$1,818,151    \$1,797,143   \$1,778,346   \$1,761,429  \$22,959
Avg \$/5%        \$35,322       \$32,967       \$30,907      \$29,089       \$27,473      \$26,027      \$24,725      11
Which has a steeper slope?

\$1,900,000

\$1,850,000

\$1,800,000

\$1,750,000                                                     35%
40%
\$1,700,000                                                     45%
50%
\$1,650,000
55%
\$1,600,000                                                    60%
60%
\$1,550,000                                            55%
50%
\$1,500,000                                          45%
14                                   40%
15   16   17                  35%
18
19   20

12
Implications for design
   For these economic assumptions the total costs are
more sensitive to reducing aisle width than
increasing stacking height
   Therefore we would favor materials handling and
storage systems which minimized aisle width
(over increasing stacking height)
   Reducing aisle width by 5% will save about
\$30,000 annually
   How much could I spend on equipment to achieve
this?
13
Engineering Economics greatly
simplified
   A dollar in your pocket today is worth more than
the promise of a dollar one year from now
   If you had a dollar today you could put it into a
savings account earning interest and at the end of
one year you could extract the dollar plus the
interest
   There is a “time value of money”
   Comparing equipment investment strategies that
have different initial costs and varying annual
costs and benefits require comparing the costs at a
single point in time                               14
Engineering Economics generally refers
to the process of making economic
comparisons of Engineering investments
like the Equipment in the warehouse
 Forecast or estimate the “streams” of costs
(or cost savings over time)
 Convert those streams to a common point in
time using interest rate concepts
 Compare the alternative designs at a
common point in time in order to make the
most economical choice
15
A very simple example of the
time value of money
   Would you prefer \$3000 now or \$3100 a year
from now?
 If I had \$3000 now and could invest it in a bank
earning 3% interest, the bank would give me
\$3090 at the end of the year. Take the \$3100
next year.
 If I could invest it in another bank which
offered 5% interest, that bank would give me
\$3150 at the end of the year. Take the \$3000
now.
16
Translating present worth to
future sums using interest
i = annual rate of interest
n = number of interest periods (usually years)
P = the principal or present value of money
F= the future amount of money

17
Relating F to P is what it is all
F=P(1+i) at the end of the first year
F= (P(1+i))(1+i)=P(1+i)2 at the end of the
second year

F=P(1+i)n at the end of the nth year

Or P=F/(1+i)n to bring a future value back to
a present value
18
This enables comparison of
alternatives with different cash
flows
Alternative A:
Costs \$10,000 now but returns \$4,000 each
of the next 3 years
Alternative B:
Costs \$8,000 now but returns \$3,000 each
of the next 3 years

19
Graphic Cash Flow Diagrams
 Interest periods on horizontal axis
 Upward arrows represent receipts (positive
cash flow)
 Downward arrows represent costs (negative
cash flows)
 Flows assumed to be at end of period

20
The two alternatives depicted
graphically:

\$4,000 \$4,000 \$4,000
Alternative A
0      1      2      3

\$10,000
\$3,000 \$3,000 \$3,000
Alternative B
0     1      2    3
\$8,000

21
How do you compare them using
Net Present Worth?
 Assume some time value of money or
interest rate
 Convert the cash flows to a common point
in time
 Select the alternative with the largest net
present worth

22
To convert to the present use the
formula: P=F/(1+i)n

\$4,000 \$4,000 \$4,000
Alternative A
0      1      2      3

\$10,000
\$3,000 \$3,000 \$3,000
Alternative B
0     1      2    3
\$8,000

23
Choosing an appropriate interest
rate
   One way is to think about what it will cost to
borrow the money from a bank
   Another way to acquire capital is to sell shares of
stock
   A third method is to think about what could be
earned if the money were invested in some other
“project”
   Often a blending of the three
   Seek guidance from your accounting or finance
department
24
To convert to the present use the
formula: P=F/(1+i)n (Choose 8% for i)

\$4,000 \$4,000 \$4,000 Each future
value must
Alternative A
0      1                   be converted
2      3
back to the
\$10,000                        present time, 0
\$3,000 \$3,000 \$3,000
Alternative B
0     1       2     3 PA,2=4000/(1+.08)2
\$8,000                     PA,2= 3429

25
To convert to the present use the
formula: P=F/(1+i)n

\$3429 \$4,000        \$4,000 Each future
value must
Alternative A
0      1                  be converted
2      3
back to the
\$10,000                       present time, 0
\$3,000 \$3,000 \$3,000
Alternative B
0     1      2     3 PA,2=4000/(1+.08)2
\$8,000                     PA,2= 3429

26
Convert all the future values to
present values using P=F/(1+i)n
\$3704        Net Present Worth
= \$10,308 - \$10,000
\$3429
Alternative A         \$3175        = \$308
0     1    2   3
\$10,000
\$2778         Net Present Worth
\$2572         = \$7,731 - \$8,000
\$2381        = - \$269
Alternative B       0      1   2    3
\$8,000

27
Lets go back to the equipment
expense question
 How much can I afford to spend on
equipment to reduce the aisle percentage by
5%?
 Reducing the aisle percentage by 5% saves
us \$30,000 annually
\$30k \$30k              \$30k

??
28
Some key questions needed to
find the appropriate investment in
the equipment
 What is the useful life of the equipment?
 Are there operating charges that should be
estimated for each year?
 Is there a salvage value at the end of the
useful life?
 What is an appropriate interest rate?

29
Including \$5k operating costs and a
salvage value of \$5k at the end of 8 years

\$30k \$30k          \$30k     \$5k

??    \$5k                      \$5k

30
We could invest 146,000 dollars
Interest=       0.08   Future        Present
Value         Value
1                   25    23.14815
2                   25    21.43347
3                   25    19.84581
4                   25    18.37575
5                   25    17.01458
6                   25    15.75424
7                   25    14.58726
8                   30    16.20807
Total          146.3673

31

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