CH07s

98

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

S U P P L E M E N T

Capacity Planning

DISCUSSION QUESTIONS 1. Design   capacity   is   the   theoretical   maximum   output   of   a system in a given period. Effective capacity is the capacity a firm can expect to achieve given its current product mix, methods of scheduling, maintenance, and standards of quality. 2. The fundamental assumptions of break­even analysis are   

Fixed costs do not vary with volume Unit variable costs do not vary with volume Unit revenues do not vary with volume

3. The manager obtains data for use in break­even analysis from  

Cost data: industrial engineering and accounting Demand and revenue data: marketing

4. Revenue   data,   when   plotted,   do   not   fall   on   a   straight   line because of volume discounts, etc. 5. Lagging is preferred when short­term options like overtime and   subcontracting   are   relatively   low   cost   and/or   easy   to   use. Leading is preferred when a firm cannot afford to lose customers for lack of product availability, and overtime, etc., are not available. 6. NPV   determines   the   discounted   or   time   value   of   money, comparing cost and income streams over periods of time. Process decisions may incur much of their expense early in the life of the equipment, but the stream of revenues may follow for decades. NPV is the appropriate analytical tool for that situation. 7. Effective capacity is the capacity a firm can expect to achieve given   its   current   product   mix,   methods   of   scheduling, maintenance, and standards of quality. 8. Efficiency is the actual output as a percent of effective capacity. Efficiency = actual output/effective capacity 9. Expected output = effective capacity  efficiency

ACTIVE MODEL EXERCISES ACTIVE MODEL S7.1: Productivity 1. Due to an anticipated decrease in demand the firm is considering dropping one of its shifts. What will be the capacity if they do so? 134,400 2. Another option would be to maintain 3 shifts but only work on weekdays. What will be the capacity if they select this option? 144,000 3. As the effective capacity rises, how does this affect both the utilization and efficiency. Utilization is unaffected but the efficiency drops.

4. As   the   actual   output   rises,   how   does   this   affect   both   the utilization and efficiency? Both utilization and efficiency rise.

ACTIVE MODEL S7.2: Breakeven Analysis 1. Use the scrollbars to determine what happens to the breakeven point as the fixed costs increase? The variable costs increase? The selling price increases? If the fixed or variable costs increase then the breakeven point  increases,   while if  the selling  price increases  then the breakeven point decreases. 2. What is the percentage increase (over 5,714) to the breakeven point if the fixed costs increase by 10% to $11,000? If the variable costs increase by 10% to $2.48? If the price per unit increases by 10% to $4.40? If the fixed costs rise by 10% the break even point rises by 10%. In this case if the variable costs rise by 10% then the BEP rises by 15%. If the price per unit increases by 10% then the breakeven point FALLS by 19%. 3. In  order  to  cut  the  breakeven  point  in  half,   by  how  much would   the   FIXED   costs   have   to   decrease?   The   variable   costs? How much would the selling price have to increase? Fixed costs – $5,000; Variable costs by $1.75 from $2.25 to $.50; The selling price would need to increase by the same $1.75.

END-OF-CHAPTER PROBLEMS S7.1    Utilization =

actual output 6,000 = = 0.857  85.7% design capacity 7,000

S7.2    Efficiency =

actual output 4,500 = = 0.692 or 69.2% effective capacity 6,500

S7.3     Expected output = (effective capacity)  (efficiency)   (6,500)(.88)  5,720 S7.4     Efficiency =

actual (expected) output 800 = = 88.9% effective capacity 900

S7.5 Efficiency =

actual output 400   or  0.80 = effective capacity effective capacity

Thus, effective capacity =

400  500 0.80

99

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

S7.6 Expected output = (effective capacity)  (efficiency) = 90  0.90 = 81 chairs

S7.10 In 2006, Eye Associates has 8 machines @ 2500. In year 2011 it needs capacity of 20,100.

S7.7 Actual (expected) output = hours  efficiency = 8 hr  5 days  2 shifts  4 machines  0.95 = 320  0.95 = 304 hrs S7.8 Design: 93,600  0.95 = 88,920 Fabrication: 156,000  1.03 = 160,680 Finishing: 62,400  1.05 = 65,520

(a) Therefore, if it adds 3,000 to capacity in 2006, total capacity   in   2011   will   be   23,000   lenses,   more   than adequate. Exceeds by 2,900. (b) If it buys the standard machine in 2006, its capacity in 2011 will be 22,500 lenses, still more than adequate; the smaller machine will suffice. Exceeds by 2,400. S7.11 Where:

S7.9 Year

X

1996 1997 1998 1999 2000 2001 2002 2003 2004 2005

1 2 3 4 5 6 7 8 9 10 X 5 = 5

X2

Y 15.00 15.50 16.25 16.75 16.90 17.24 17.50 17.30 17.75 18.10 Y 168.2 = 9

X 2 =

1 4 9 16 25 36 49 64 81 100 38 5

XY

XY =

15.00 31.00 48.75 67.00 84.50 103.44 122.50 138.40 159.75 181.00 951.3 4

Regression Output Constant X Coefficient Std err of Y est R squared No. of observations Degrees of freedom Std err of coef.

15.11 0.312 0.301 0.916 8 10 8 0.033

x  5.5 n y Y  16.829 n 25.745 b  0.312  82 a  16.829  0.312  5.5  15.11 X

Year 2006 = a + bx11, therefore 15.11 + 0.312  11 = 15.11 + 3.43 = 18.54, or 18,540 lenses Year 2008 = a + bx13, therefore 15.11 + 0.312  13 = 15.11 + 4.056 = 19.17, or 19,160 lenses Year 2010 = a + bx15, therefore 15.11 + 0.312  15 = 15.11 + 4.68 = 19.79, or 19,790 lenses (a) 2006 capacity needs = 18.54 thousand 2008 capacity needs = 19.17 thousand 2010 capacity needs = 19.79 thousand (b) Requirements in 2010 are for 19.79(  1000) lenses. Therefore,   Eye   Associates   will   need   8   machines (19,790/2,500 = 7.9, round up to 8).

Design Capacity = 2,000 students Effective Capacity = 1,500 students Actual Output = 1,450 students Therefore: Utilization 

actual output 1,450   72.5% design capacity 2,000

Efficiency 

actual output 1,450   96.7% effective capacity 1,500

S7.12 (a) Proposal A breakeven in units is: Fixed cost $50,000 $50,000    6,250 units SP  VC 20  12 8 (b) Proposal B breakeven in units is: Fixed cost $70,000 $70,000    7,000 units SP  VC 20  10 10 S7.13 (a) Proposal A breakeven in dollars is: Fixed cost 1  VC SP



$50,000  $10,000 1  12 20



$60,000  $150,000 0.40

(b) Proposal B breakeven in dollars is: BEP$ =

Fixed cost $70,000  $10,000 = 1  VC 1  10 SP 20



$80,000  $160,000  0.50

S7.14 Set Proposal A = Proposal B (SPA  VC A ) X A  FA  (SPB  VCB ) X B  FB (20  12)X  50,000  (20  10)X  70,000 (8)X  50,000  (10)X  70,000 (8)X + 20,000  (10)X 20,000  10 X  8 X 20,000  2 X 10,000  X 20,000  1,667 pizzas; 14  2 30,000                   BEPB   2,353 pizzas 14  1.25

S7.15   (a)   BEPA 

(b   &   c)   For   both   quantities,   oven   A   is   slightly   more profitable (but oven B is catching up).

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

100

S7.15  (cont’d) Oven A Fixed cost Revenue Variable cost

$20,000.00 $14.00 $2.00

Oven B

Unit Sales of

$30,000.00 $14.00 $1.25

$9,000  $12,000 

(d)  20,000 + 2Xa  = 30,000 + 1.25Xb                                  .75X = 10,000                                       X = 13,333 pizzas S7.16 Given:  Price ( P) = $8 unit Variable cost (V ) = $4 unit  Fixed cost (F ) = $50,000   (a)  Breakeven in units is given by: BEPx 

Profit A –  $88,000 $124,000

  Breakeven is given by: (a)  BEPx 

F 325,000 325,000    32,500 units P  V 30  20 10

(b)  BEP$ 

F 325,000 325,000    $975,000 1  0.667 1  VP 1  20 30

S7.20 Option A: Stay as is   Option B: add new equipment Units   Price  VC   FC = Profit

Profit A = 30,000   1.00  0.50   14,000

F 50,000 50,000    12,500 units PV 84 4

= $1,000

Profit B = 50,000   1.00  0.60   20,000

  (b)  Breakeven in dollars is given by: BEP$ 

F 50,000 50,000    $100,000  1  0.50 1  48 1  VP

  (c)  Profit is given by: Profit  Volume  Contribution  Fixed cost   100,000   8  4   50,000

= $0 Therefore, the company should stay with the present equipment. S7.21 Option A: Stay as is   Option B: Add new equipment, raise selling price Units   Price  VC   FC = Profit

Profit A = 30,000   1.00  0.50   14,000

 400,000  50,000  $350,000

= $1,000

S7.17 Given: Price ( P ) = $0.05 unit Variable cost (V ) = $0.01 unit  Fixed cost ( F ) = $15,000   Breakeven is given by: BEP$ 

Profit B – $84,750 $123,000

F 15,000 15,000    $18,750 0.01 1  0.2 1  VP 1  0.05

F 15,000 15,000   P  V 0.05  0.01 0.04  375,000 copies

BEPx 

S7.18 Given:  Price ( P ) = $30 unit Variable cost (V ) = $20 unit  Fixed cost ( F ) = $250,000   Breakeven is given by: BEP$ 

F 250,000 250,000    $750,000 20 1  0.667 1  VP 1  30

BEPx 

F 250,000 250,000    25,000 units P V 30  20 10

S7.19 Given:  Price (P)  $30 unit Variable cost (V )  $20 unit Fixed cost (F )  $250,000  $75,000  $325,000

Profit B = 45,000   1.10  0.60   20,000 = $2,500

Therefore, the company should choose option B: add the new equipment and raise the selling price. S7.22 Where:   FC = $37,500 VC = $1.75     P = 2.50   (a)  Break­even quantity for the manual process in units: 

F 37,500   50,000 bags P  V 2.50  1.75

  (b) Revenue at the break­even quantity: 50,000  2.50  $125,000 and  37,500 BEP$  V 1 P 37,500   $125,000 1.75 1 2.50 (c) Break­even   quantity   for   the   mechanized   process: where: F = 75,000     P = 1.25 75,000 BEPu   60,000 bags 2.50  1.25 (d) Revenue   at   the   break­even   quantity   for   the   mecha­ nized process: 60,000 × 2.50 = $150,000  75,000 and  = $150,000 1.25 1 2.50

101

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

(e) Monthly profit or loss of the manual process if they expect to sell 60,000 bags of lettuce per month: Profit   =   2.50(60,000)   –   37,500   –   1.75(60,000)  = $7,500

S7.24 (a)  Break­even volume: Total fixed cost =     1800 rent, utilities, etc. + 2000  entertainment =    3800

(f) Monthly profit or loss of the mechanized process if they expect to sell 60,000 bags of lettuce per month 2.50(60,000) – 75,000 – 1.25(60,000) = 0.0 (breakeven) (g) They should be indifferent to the process selected at 75,000 bags. 37,500  1.75 X  75,000  1.25 X 37,500  .5 X 75,000  X (h) The manual process be preferred over the mechanized process below 75,000 bags. The mechanized process be preferred over the manual process above 75,000 bags.

Selling Price Volume Revenue Drinks Meals Desserts/etc. Sandwiches

1.50 10.00 2.50 6.25

P Drinks 1.50 Meals 10.00 Desserts 2.50 Lunch 6.25

BEP$ 

30000 10000 10000 20000

Percent of Total Revenue

45000 100000 25000 125000 295000

0.153 0.339 0.085 0.423 1.00 0

V

V/P

1–V/P

Wi

1– (V/P)Wi

0.75 5.00 1.00 3.25

0.50 0.50 0.40 0.52

0.50 0.50 0.60 0.48

0.153 0.339 0.085 0.423 1.00 0

0.077 0.170 0.051 0.203 0.501

F 





i

  1  VPi

 Wi

3800  $7,584.83 0.501

(b)  Number of meals per day at breakeven = 9 S7.23 (a) Yes, Total profit now: [40,000  (2.00 – 0.75)] – $20,000 = $30,000 Total profit with new machine: [50,000  (2.00 – .50)] – $25,000 = $50,000 (b) The equipment choice changes at 20,000 units. .75x  20,000  .50 x  25,000 .25x  5,000 x  20,000 units

Fraction BE Units BE Selling of Total Dollar per Units Price Revenue Volume Month per Day Drinks 1.50 Meals 10.00 Desserts/et 2.50 c. Sandwiches 6.25

0.153 0.339 0.085

1160.48 2571.26 644.71

774 258 258

26 9 8

0.424

3208.28

514

18

S7.25 (a)  Break­even volume: Total fixed cost =     1800  rent, utilities, etc. + 2000 entertainment =      3800 Selling Price Drinks Meals Desserts/et c. Sandwiches

Volume Revenue

1.50 10.00 2.50

30000 10000 10000

45000 100000 25000

0.153 0.339 0.085

6.25

20000

125000 29500 0

0.424 1.00 0

Total Variable Food Cost Cost Factor*

(c) At   a   volume   of   15,000   units,   the   current   process should be used. Drinks

Percent of Total Revenue

0.75

1.10

Total Variable Cost 0.83

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

Meals Desserts/etc. Lunch/Sandwich es

5.00 1.00 3.25

1.43 1.10 1.43

7.15 1.10 4.65

* The total variable cost factor for meals and sandwiches is developed as:  1.00 food cost  0.33 labor, at two­thirds of food cost 0.10 variable expenses at 10% of food costs 1.43

102

Total sales at breakeven × 25% of sales Price of wine 986.19 × 0.25 = = 140.9 servings $1.75

(b)  No. of wine servings  = at breakeven           103

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

The total variable cost factor for drinks and desserts/wines is developed as: 1.00 food cost 0.10 variable expenses at 10% of costs 1.10 total variable expense

Drinks Meals Dessert s Lunch

BEP$ 

P

V

V/P

1–V/P

Wi

[1– (V/P)Wi

 1.50 10.00  2.50

0.83 7.15 1.10

0.55 0.72 0.44

0.45 0.28 0.56

0.153 0.339 0.085

0.069 0.095 0.048

 6.25

4.65

0.74

0.26

0.423 1.000

0.110 0.322

F

  1  Vi Pi



 Wi

3800  $11,801.24 0.322

Selling Price

Fraction of Total Revenue

1.50 10.00 2.50

0.153 0.339 0.085

1805.59 4000.62 1003.11

1204   401   402

6.25

0.424

5003.73

 810

Drinks Meals Desserts/Win e Lunch/ Sandwiche s

S7.27 (a)

Dollar Volume BEP Units

(b) Branch B which represents Option B­Modernize 2nd floor, has the highest expected value, $74,000. S7.28

(b) Monthly breakeven, to include a profit of $35,000 per year Total  Fixed Cost =     1800 rent, utilities, etc.    + 2000 entertainment + 2917 (35,000 /12) profit =    6717 BEP$ 

F 

V



i

  1  Pi

 Wi

6717  $20,860.25 0.322

Selling Price

Fraction of Total Revenue

1.50 10.00 2.50

0.153 0.339 0.085

3191.62 7071.62 1773.12

2128 708 710

6.25

0.424

8844.75

1516

Drinks Meals Desserts/Win e Lunch/ Sandwiche s

Dollar Volume BEP Units

Prefer   to   build   a   large   line.   Large   line   has   a   payoff   of $100,000. Small line has a payoff of $66,666 + 0 = $66,666.

S7.26 (a)   Break­even volume, where total fixed cost = labor   (at $250) + booth rental (at 5  $50) = $500.

Item Soft Drinks Wine Coffee Candy Totals

Selling Price 1.00 1.75 1.00 1.00

Variable Var. Cost Total Cost Factor (%) Var. Cost 0.65 0.95 0.30 0.30

1.1 1.1 1.1 1.1

Breakeven = TFC/wt. contribution = 500/0.507 = $986.19

0.715 1.045 0.330 0.330

Estimated Contributio Percent n 1 Revenue Weighted (vc/sp) Revenue 0.285 0.403 0.670 0.670

0.25 0.25 0.30 0.20 1.0 0

0.071 0.101 0.201 0.134 0.50 7

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

S7.29

* NPV factor from Table S7.1.

Initial investment = $75,000 Salvage value = $45,000 Five­year return = $15,000 Cost of capital = 12% NPV annuity factor  5 years @ 12% = 3.605 Present value = 3.605  15,000 = $54,075 Present value of salvage: 0.567  45,000 = $25,515 Net present value = 54,075 + 25,515 – 75,000 = $4,590 S7.30 Initial investment = $65,000 Eight­year return = $16,000 per year Cost of capital = 10% NPV annuity factor  8 years @ 10% = 5.33 Present value = 5.33  $16000 = $85,280 Net present value = $85,280 – $65,000 = $20,280

P

F (1  i) N



2000 (1  0.09)3



NPV   for   Machine   A   is   –$22,988;   NPV   for   Machine   B   is – $27,026. Therefore, Machine A should be recommended. S7.34 Expense

2000  $1,544.40 1.2950

or from Table S7.1 NPV = F  PVF9%, 3 = 2000 = 0.772 = $1,544

Two Large Ovens

3750 750

5000 0

750

400

750

1000

Three Small Ovens NPV Factor*

Now 1 2 3 4 5

Expense Expense Expense Expense Expense Expense

3750 1500 1500 1500 1500 1500

1.000 0.877 0.769 0.675 0.592 0.519

5

Salvage revenue

750

0.519

S7.32

NPV –3750 –1316 –1154 –1013 –888 –779 –8900 +389 –8511

P

F (1  i)

N



5600 15

(1.08)



* NPV factor from Table S7.1.

5600  $1,765.35 3.17

Two Large Ovens NPV Factor*

or from Table S7.1

Year

NPV = F  PVF8%, 15 = 5600  0.315 = $1,764

Now 1 2 3 4 5

Expense Expense Expense Expense Expense Expense

5000 400 400 400 400 400

1.000 0.877 0.769 0.675 0.592 0.519

5

Salvage revenue

1000

0.519

S7.33 Expense Original cost Labor per year Maintenance per year Salvage value

Machine A

Machine B

10000 2000 4000 2000

20000 4000 1000 7000

Year

NPV Factor*

NPV –10000 –5358 –4782 –4272 –24412 +1424

Now 1 2 3

Expense Expense Expense Expense

10000 6000 6000 6000

1.000 0.893 0.797 0.712

3

Salvage revenue

2000

0.712

–22988 * NPV factor from Table S7.1. Machine B Year

NPV –5000 –351 –308 –270 –237 –208 –6374 +519 –5855

Machine A

3

Three Small Ovens

Original cost Excess labor per year Maintenance per year Salvage value

Year

S7.31

Now 1 2 3

104

NPV Factor*

NPV

Expense Expense Expense Expense

20000 5000 5000 5000

1.000 0.893 0.797 0.712

Salvage revenue

7000

0.712

–20000 –4465 –3985 –3560 –32010 +4984 –27026

* NPV factor from Table S7.1.

(a) NPV of the three small ovens = –$8,511; NPV of the two   large   ovens   =   –$5,855.   Therefore,   you   should recommend that the firm purchase the two large ovens. (b) The basic assumptions made with regard to the ovens are:  The ovens are of equal quality  The ovens are of equivalent production capacity (c) The   basic   assumptions   made   with   regard   to methodology are:  Future interest rates are known  Payments are made at the end of each time period

105

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

S7.35

F 300   100 crepes (a)  BEPx1  P –V 4–1 0 F2 0   0 (b)  BEPx 2  P – V2 4–(1+1.5) 1.5

S7.39 (a) Proposal A: Profit at 8,500 units Profit = (SP  VC )X  F   @ 8,500 for Proposal A:  (20  12)8,500  50,000 = 18,000  @ 8,500 for Proposal B:  (20  10)8,500  70,000 = 15,000 

If fixed costs are zero, and V < P, then profitable from start (c,d) Stand: Profit1 = P(BEP)  F + x (BEP) = 4(350)  (300) + (1  350) = $750  (better option) Portable: Profit2 = P(BEP)  F + x (BEP) = 4(350)  (0) + (2.5  350) = $525

Proposal A is best. (b) Proposal B: Profit at 15,000 units @ 15,000 units for Proposal A:  (20  12)15,000  50,000 = $70,000 

F –F 300 – 0  200 (e)  BEP1vs 2  1 2  V2 – V1 2.5 – 1  So if BEPx < 200, Portable  If BEPx > 200, Stand

@ 15,000 units for Proposal B:  (20  10)15,000  70,000 = $80,000  Proposal B is best.

Demand would have to be different by 150 (i.e. demand would  have to drop from 350 to 200).

INTERNET HOMEWORK PROBLEMS Solutions   to   problems   on   our   companion   web   site   home   page (www.prenhall.com/heizer). S7.36 (a) Proposal A breakeven in units is: Fixed cost 50,000 50,000    6,250 units SP  VC 20  12 8 (b) Proposal B breakeven in units is: Fixed cost 70,000 70,000    7,000 units SP  VC 20  10 10 S7.37 (a) Proposal A breakeven in dollars is: Fixed cost 50,000 50,000    $125,000 0.40 1  VC 1  12

S7.40 Investment A net income, using Table S7.2 19000   PVF9%, 6  – 61,000 = 19,000  4.486 – 61,000 = $24,234 Investment B Net Income NPV Factor*

Year Now 1

Expense Revenue

74,000 19,000

1.000 0.917

2

Revenue

20,000

0.842

3

Revenue

21,000

0.772

4

Revenue

22,000

0.708

5

Revenue

21,000

0.650

6

Revenue

20,000

0.596

7

Revenue

11,000

0.547

20

SP

(b) Proposal B breakeven in dollars is: Fixed cost 70,000 70,000 BEP$     $140,000 0.50 1  VC 1  10 SP

20

S7.38 Set Proposal A = Proposal B; Solve for units (SPA  VC A ) X A  FA  (SPB  VCB ) X B  FB (20  12)X  50,000  (20  10)X  70,000 (8)X  50,000  (10)X  70,000 (8)X  20,000  (10)X 20,000  10 X  8 X 20,000  2 X 10,000  X

NPV –74,000 +17,42 3 +16,84 0 +16,21 2 +15,57 6 +13,65 0 +11,92 0 +6,017 23,638

* From Table S7.1

Therefore, Investment A, with a payoff of $24,234, would be pre­ ferred over Investment B, with a payoff of $23,638.

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

S7.41 Initial investment = $20,000 NPV Factor*

Year 1 2 3 4 5 6 7 8 9

0.909 0.826 0.751 0.683 0.621 0.564 0.513 0.467 0.424

Cash Flow 1 Cash P $1,000 1,000 3,000 15,000 3,000 1,000 — 1,000 —

Cash Flows Cash Flow 2 Cash P

$909 826 2,253 10,245 1,863 564 — 467 — $17,12 7

$7,000 6,000 5,000 4,000 4,000 4,000 4,000 2,000 —

Cash Flow 3 Cash P

$6,363 4,956 3,755 2,732 2,484 2,256 2,052 934 — $25,532

$10,000 5,000 3,000 2,000 1,000 1,000 1,000 — 1,000

$9,090 4,130 2,253 1,366 621 564 513 — 425 $18,96 2

*The NPV from investment 2 is highest, at $5,532 (after initial investment of $20,000 is subtracted).

S7.42 At 11 percent, the net present value is –$7,677.89. At 4 percent, the net present value is $5,378.54. They should purchase at 4 percent but not at 11 percent. Vinyl Siding Machine Solution at 11%

Period Period Period Period Period Period Period Total

0 1 2 3 4 5 6

Inflow

Outflow

0 20,000 15,000 15,000 15,000 10,000 10,000 85,00 0

70,000 0 0 0 0 0 0 70,00 0

PV Factor at 11% 1 0.9009 0.8116 0.7312 0.6587 0.5935 0.5346

PV Inflow

PV Outflow

0 18,018.02 12,174.34 10,967.87 9,880.965 5,934.513 5,346.409 62,322.11

70,000 0 0 0 0 0 0 70,00 0

PV (Inflow–Outflow) –70,000 18,018.02 12,174.34 10,967.87 9,880.965 5,934.513   5,346.409 – 7,677.89

Vinyl Siding Machine Solution at 4% Inflow Period Period Period Period Period Period Period Total

0 1 2 3 4 5 6

Outflow

PV Factor at 4%

PV Inflow

0 20,000 15,000 15,000 15,000 10,000 10,000 85,000

70,000 1 0 0 0.9615 19,230.77 0 0.9246 13,868.34 0 0.889 13,334.95 0 0.8548 12,822.06 0 0.8219   8,219.271 0 0.7903   7,903.146 70,00 75,378.54 0 S7.43 The net present value of the receipts is $89,711.58. Inflow Outflow Period Period Period Period Period Total

0 1 2 3 4

0 50,000 30,000 0 20,000 100,00 0

1 0 0 0 0 0

PV Factor at 6% Present value 0 0.9434 0.89 0.8396 0.7921

0 47,169.81 26,699.89 0 15,841.87 89,711.5 8

PV Outflow 70,000 0 0 0 0 0 0 70,00 0

PV (Inflow–Outflow) –70,000 19,230.77 13,868.34 13,334.95 12,822.06   8,219.271   7,903.146 5,378.53 9

106

107

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

S7.44 Machine A’s NPV is $81,323.16; machine B’s NPV is $85,982.66. Machine B has the higher NPV. The lower annual returns are more than offset by a lower initial cost and by the   salvage value. Milling Machine A

PV Factor at 7%

Outflow

0 80,000 80,000 80,000 80,000 80,000 80,000 480,00 0

300,000 0 0 0 0 0 0 300,00 0

Milling Machine B

Inflow

Outflow

PV Factor at 7%

PV Inflow

Period Period Period Period Period Period Period

0 60,000 60,000 60,000 60,000 60,000 90,000

220,000 0 0 0 0 0 0

1 0.9346 0.8734 0.8163 0.7629 0.713 0.6663

0 56,074.77 52,406.32 48,977.88 45,773.71 42,779.17 59,970.80

220,000 0 0 0 0 0 0

–220,000 56,074.77 52,406.32 48,977.88 45,773.71 42,779.17 59,970.8

305,982.70

220,00

85,982.66

Period Period Period Period Period Period Period Total

0 1 2 3 4 5 6

0 1 2 3 4 5 6

Total

390,00

S7.45

1 0.9346 0.8734 0.8163 0.7629 0.713 0.6663

220,000

PV Inflow

PV Outflow

0 74,766.35 69,875.09 65,303.83 61,031.62 57,038.89 53,307.38 381,323.2 0

300,000 0 0 0 0 0 0 300,00 0

PV (Inflow–Outflow)

Inflow

PV Factor Inflow Outflow at 6% Present value Period Period Period Period Period Total

0 1 2 3 4

0 20,000 0 30,000 50,000

1 0 0 0 0

100,00 0

0

0 0.9434 0.89 0.8396 0.7921

0 18,867.92 0 25,188.58 39,604.68 83,661.1 9

PV Outflow

–300,000 74,766.35 69,875.09 65,303.83 61,031.62 57,038.89 53,307.38 81,323.1 6 PV (Inflow–Outflow)

2. What kind of major changes can take place in APH’s demand forecast   that   would   leave   the   hospital   with   an   underutilized   facility (namely what are the risks connected with this capacity decision)? a) Demand   will   not   continue   to   grow   dramatically.   The hospital believes that the new building will attract new OB/GYN   doctors   to   deliver   there.   The   other   major hospital chain in Orlando, Florida Hospital, has also just announced   a   major   expansion.   This   may   flood   the hospital bed market in the short run.

VIDEO CASE STUDY

b) The population boom in Central Florida could abate with rising housing prices that are discouraging future growth.

CAPACITY PLANNING AT ARNOLD PALMER  HOSPITAL

c) There are always unforeseen disasters in medicine that could   damage   the   hospital’s   sterling   reputation   (e.g., lawsuits, drop in quality).

The Arnold Palmer Hospital video for this case is available on the video cassette/DVD from Prentice Hall that accompanies this text. (Its   running   time   is   9   minutes.)   Also   note   that   the   Global Company Profile in Chapter 6 highlights this hospital. 1. Given the discussion in the text, what approach is being taken by APH toward matching capacity to demand? Referring   to   text   Figure   S7.4,   Arnold   Palmer   Hospital’s capacity first lagged demand (part c), and now is leading demand with   incremental   expansion   (part   a).   The   new   building   will provide sufficient capacity for several years. The top 2 floors (left unfinished for additional beds) and operating rooms (on the 4th floor, available for horizontal expansion) will be built out when needed.

d) There is a nursing shortage that could create a staffing bottleneck   if   not   corrected.   Recently,   the   two   major hospital chains in Central Florida got into a bidding war in attempts to recruit each other’s nurses.

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

3. Use   regression   analysis   to   forecast   the   point   at   which Swanson   needs   to   “build   out”   the   top   two   floors   of   the   new building. Regression analysis on the birth data in Table S7.3 yields: Y = projected births = 5067 + 569x (where x = time in years. x = 1 is 1995, x = 2 is 1996,… x = 10 in 2004.) 2

The   R   =   .988,   so   R   =   .98,   a   very   high   coefficient   of determination



To forecast the point at which the top 2 floors will need to be built out; we examine 2005, x = 11 and get y = 11,326; for 2006, x = 12 gives y = 11, 895; for 2007, x = 13 gives 12,464; for 2008, x = 14 gives 13,033; for 2009, x = 15 gives 13,602. So the top two floors need to be built out before 2009. 

INTERNET CASE STUDIES* 

1

CAPACITY PLANNING AT SHOULDICE  1 HOSPITAL

Class discussion: Although this case is structured as a capacity case, it does lend itself to a much broader discussion of service operations.   Consistent   with   our   approach   of   focusing   on   the 10 strategic decisions of operations, note for instance: 





Design   and   selection   of   the   product.   Patients  are  selected— only patients with uncomplicated external hernias and who are in good health are admitted. Quality  is  much  easier  with  a  standardized  low­risk  product and   a   state­of­the­art   process   such   as   the   one   that   exists   at Shouldice. Documented in the case by a 1% recurrence rate vs 10% for general hospitals. Process   design,   although   for   a   service,   is   nearly   ideal.   The system is process oriented around reception, dining, operation rooms, recovery, etc. But because patients move consistently in batches between processes, it has some of the attributes of a repetitive   process   (i.e.,   an   assembly   line).   The   process effectively   supports   the   desired   results   by   providing   early ambulation,   a   standardized   medical   procedure   with   medical staff   who   desire  

Check In Monday Tuesday Wednesday Thursday Friday Saturday Sunday Total





to do only this procedure, and a country club atmosphere that is pleasant to both staff and patient. A lot of self­care with early ambulation from operating table and to meals reduces costs and improves recovery. The short three­day process also supports a good schedule from the point of view of the staff—allowing low staffing on the weekends. The one hole in the system is that the facility does not appear to meet the demand, and some of the facility is underutilized on the weekend. Scheduling: Aggregate scheduling is easy with one product and constant   demand.   Short   term   scheduling,   with   surgery   on weekdays, and five days of work tapering off on weekends is convenient   for   both   staff   and   patients.   Reduced   staffing   on Saturday   and   Sunday   is   popular.   Moreover,   the   schedule provides   full   utilization   of   facilities   on   weekdays   with underutilization on weekends. Location:   suburban   Toronto,   Canada,   makes   it   reasonably convenient for its worldwide customer base. Human   resource   management   appears   to   be   well   handled. Although the case is not explicit, the turnover of staff is low and the working conditions pleasant. Supply chain and inventory have little impact on the case, but the   steady   consistent   flow   of   patients   does   make   the development of good relations with suppliers easy and aid in the establishment of efficient inventory procedures. Maintenance and reliability, because of the consistent nature of the   process,   should   be   known   and   easily   scheduled,   with weekends available for major work.

Discussion Questions for Shouldice Case  1. Shouldice beds are only fully utilized three days per week, but doctors operate five days: so the question of utilization has at least these two perspectives. Utilization = actual/design = 450/(90  7) = .71 = 71%  2. Per the table below, if surgery is added on Saturday and held at   30   per   day,   then   added   beds   are   wasted   capacity.   Beds   are available on Saturday under current operating policies.

Beds Used under Current Operations (150 per week) Monday Tuesday Wednesday Thursday Friday Saturday 30

30 30

30 30 30

30 30 30

30 30 none

Sunday

30 none

30 60

30 90

90

90

1

Source:   Adopted   from   R.   D.   Chase,   F.   R.   Jacobs.   and   N.   J.   Acquilano. Operation Management for Competitive Edge  10/e (Boston: Irwin McGraw Hill, 2004), 404–5. *These   case   studies   are   found   at   our   companion   web   site www.prenhall.com/heizer

108

60

30

30 30

109

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

happy  with  life,   then  the  decision  might  well  be  made  on  other Check In Monday Tuesday Wednesday Thursday Friday Saturday Sunday Total

Beds Used with Saturday Surgery (180 criteria.) per week) Monday Tuesday Wednesday Thursday Friday Saturday 30

30 60

30 30

30 90

30 30 30

90

30 30 30

90

30 30 30

90

30 30 none 60

Sunday

30 30 60

Utilization = 540/(90 )  .857 85.7% 3.   If beds are increased by 50% (to 135) but surgery is held at 30 per day, the added capacity is wasted.  With the added beds, surgeries could move from 30 per day to 45 per day.  A 50% increase in bed capacity needs to be matched with 45 surgeries Monday, Tuesday, Wednesday, and Thursday to fully utilize facilities 4 days per week.  If 30 surgeries are performed each day in 5 rooms, then 6 are performed in each room. To perform 45 per day, the rooms   will   need   to   be   occupied   9   hrs   per   day   or   more rooms   will   need   to   be   added.   Extending  the  hours  may complicate the smooth recovery process used at Shouldice. More operating rooms are recommended. Beds Used with 50% Increase in Beds (225 per week if surgery keeps up) Check In Monday Tuesday Wednesday Thursday Friday Saturday Sunday Monday Tuesday Wednesday Thursday Friday Saturday Sunday Total

45

45 45

45 45 45

45 45 45

45 45 none

45 none

45 90

45 13 5

135

135

90

45

45 45

Several perspectives on cost exist, just as do capacity options. Here are some hypothetial costs for one capacity  option: 15 more surgeries per day. Income for 15 more surgeries at $2,400 each  Less surgery cost at $800 per surgery  Less other costs (room/meals/supplies; assume $500/day) [Not given in case] Daily income after surgery costs  Five surgeries per day  5 days  50 weeks

$ $ $

36,000 12,000 15,000

$ 9,000 $ 11,250,000

 1,250 surgeries Bed cost over 5 years = ($100,000  45 beds)/5 years  $ 900,000 Net income per year $

If the expansion decision is made on the basis of ROI, this is a very good investment. (NPV is ignored, but can be added—no interest rate is given in the case.) The   expansion   could   be   made   on   other   basis   (i.e.,   Shouldice investors and personnel decide they are doing a good job and are

Daily receipts Less surgery cost Less est. of other costs Daily income Yearly income (9,000  1,250)

Yearly write-off of bed cost Yearly income

SUPPLEMENT 7 C A PA C I T Y P L A N N I N G

2

SOUTHWESTERN UNIVERSITY: D 1. Determine weighted contribution.

Items

Selling Price/ea

Var. Cost/ea

VC/SP

1(VC/SP)

$1.50 $2.00 $2.00 $2.50 $1.00

$0.75 $0.50 $0.80 $1.00 $0.40

0.50 0.25 0.40 0.40 0.40

0.50 0.75 0.60 0.60 0.60

Soft Drinks Coffee Hot Dogs Hamburgers Misc. Snacks

Percent of Revenues

Determine fixed cost per game: Prorated salaries/5 games = 2,400 sq. ft  $2 6 people  6 booths  $7  5 hrs.

25% 25% 20% 20% 10% 100 %

0.125 0.1875 0.120 0.120 0.060 0.612 5

$ 20,000.00 $ 4,800.00 $ 1,260.00 $ 26,060.00 $ 42,546.94

Breakeven = $26,060/0.6125 =

    Total Sales at BE $42,546.94

2.  At 70,000 attendees, each spend the following: Percent of Revenues Soft Drinks Coffee Hot Dogs Hamburgers Misc. Snacks

Percent Contribution

= Percent of Sales =

25% 25% 20% 20% 10% Total sales at BE

$ $ $ $ $

10,636.73 10,636.73 8,509.39 8,509.39 4,254.69 $ 42,546.94

=

Sales per person 70,000 $ 0.152 $ 0.152 $ 0.122 $ 0.122 $ 0.061 $ 0.609

Average per person food sale

At 27,000 attendees, each spend the following: Percent of Revenues Soft Drinks Coffee Hot Dogs Hamburgers Misc. Snacks

25% 25% 20% 20% 10% Total Sales at BE =

= Percent of Sales =

Sales per person (/27,000)

$ 10,636.73 $ 10,636.73 $ 8,509.39 $ 8,509.39 $ 4,254.69 $ 42,546.94

$ 0.394 $ 0.394 $ 0.315 $ 0.315 $ 0.158 $ 1.576

Average per person food sale

Units sales at breakeven Percent of Revenues Soft Drinks Coffee Hot Dogs Hamburgers Misc. Snacks

25% 25% 20% 20% 10% Total Sales at BE =

= Percent of Sales = $10,636.73 $10,636.73 $8,509.39 $8,509.39 $4,254.69 $42,546.9 4

This data indicates that drinks (soft drinks and coffee) total This data indicates that hot dogs and hamburgers total This data indicates that misc. snacks total Maddux thinks his forecast is safe and that sales will exceed his

Selling Price

Number of Units Sold of Break-even (/SP)

$1.50 $2.00 $2.00 $2.50 $1.00

7,091.2 5,318.4 4,254.7 3,403.8 4,254.7

12,409.6   7,658.5   4,254.7

110