CH07s

October 24, 2017 | Author: Petra Raditya Ivanny | Category: Net Present Value, Business Economics, Financial Accounting, Management Accounting, Economies
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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 
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