# Supply Chain Management - 4th edition

Document Sample

```					Chapter 12: Determining Optimal Level of Product Availability

Exercise Solutions

1.

*
  C  50  0.2941
u
CSL
C  C 50  120
u       o

Optimal lot-size = O  NORMINV(CSL ,  ,  ) = NORMINV(0.2941,100,40) = 78.34
*                           *

Given that p = \$200, s = \$30, c = \$150:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$2,657

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 7.41

Expected understock =

( – O)[1 – NORMDIST((O – )/, 0, 1, 1)] +  NORMDIST((O – )/, 0, 1, 0) = 29.07

EXCEL worksheet 12-1 illustrates these computations

2.

With revised forecasting:

*
     C   u

50
 0.2941
CSL
Cu  Co               50  120
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.2941,100,15) = 91.88

Given that p = \$200, s = \$30, c = \$150:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$4,121

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

1
= 2.78

Expected understock =

( – O)[1 – NORMDIST((O – )/, 0, 1, 1)] +  NORMDIST((O – )/, 0, 1, 0) = 10.9

EXCEL worksheet 12-2 illustrates these computations

3.

Mean demand during lead time =DL= (2000)(2) = 4000

Standard deviation of demand during lead time = L =  D L = 500 2 = 707

Safety inventory = ROP – DL = 6000 – 4000 = 2000

CSL = NORMDIST (6000, 4000, 707, 1) = 0.9977

Cost of overstocking = (0.25)(40) = \$10

HQ                10  10000
Justifying cost of understocking: Cu =                                                    \$411
(1  CSL) D year (1  0.9977)  2000  52

Optimal CSL =          C       u

80
 0.8889
Cu  Co               80  10

Optimal safety stock = (NORMSINV (0.8889)) (707) = 863 units

EXCEL worksheet 12-3 illustrates these computations

4.

Using the current policy:

*
  C  30  0.75
u
CSL
C  C 30  10
u       o

Optimal lot-size = O  NORMINV(CSL ,  ,  ) = NORMINV(0.75,20000,10000) = 26,745
*                          *

Given that p = \$60, s = \$20, c = \$30:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

2
+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$472,889

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 8,236

Using South America option:

*
     C   u

30
 0.857
CSL
C C u        o
30  5
 NORMINV(CSL ,  ,  ) = NORMINV(0.857,20000,10000)
*                           *
Optimal lot-size = O
= 30,676

Given that p = \$60, s = \$25, c = \$30:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$521,024

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 11,407

So, it is evident that using South America option results in increased expected profits, but also

increases the production capacity requirements needed at Champion.

EXCEL worksheet 12-4 illustrates these computations

5.

Current sourcing (one line):

Reguplo:

*
     C   u

100
 0.8333
CSL
Cu  Co               100  20
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.8333,10000,1000) =
= 10,967

Given that p = \$200, s = \$80, c = \$100:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

3
– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$970,018

Each of the other models:

*
     C   u

110
 0.7857
CSL
Cu  Co               110  30
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.7857,1000,700) =
= 1,554

Given that p = \$220, s = \$80, c = \$110:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$81,421

Total expected profits = \$970,018 + 3(\$81,421) = \$1,214,280

Tailored sourcing policy:

The computations are exactly the same with revised data for Reguplo (c = \$90) and for each of
the other three models ( c= \$120)

Total expected profits = \$1,281,670

Thus, it is benefical to utilize the tailored sourcing option due to increased expected profits. This
option increases the optimal production lot size for Reguplo and decreases the lot sizes for each
of the other three options.

EXCEL worksheet 12-5 illustrates these computations

6.
IBM:

*
     C   u

35
 0.7447
CSL
C C  u       o
35  12
 NORMINV(CSL ,  ,  ) = NORMINV(0.7447,5000,2000) = 6,316
*                           *
Optimal lot-size = O

Given that p = \$50, s = \$3, c = \$15:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

4
+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$144,796

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 1,622

Similarly, the other three are evaluated and the results are summarized below:

Outputs                                                 AT&T          HP             Cisco
Optimal cycle service level                             0.7447       0.7447          0.7447
Optimal production size                                  8,645        5,316           5,447

Expected profits                                    \$207,245      \$109,796        \$106,776

Expected overstock                                      2,028        1,622           1,785

Total production lot size = 6316 + 8,645 + 5,316 + 5,447 = 25,723
Total expected profits = \$144,796 + \$207,245 + \$109,796 + \$106,776 = \$568,612
Total expected overstock = 1,622 + 2,028 + 1,622 + 1,785 = 7,057 (= amount donated to charity
on average)

EXCEL worksheet 12-6 illustrates these computations

7.

With aggregation:

Anticipated demand = 5,000 + 7,000 + 4,000 + 4,000 = 20,000
Standard deviation =           20002  25002  20002  22002  4369

*
   C   u

32
 0.8889
CSL
C Cu       o
32  4
 NORMINV(CSL ,  ,  ) = NORMINV(0.8889,20000,4369)
*                         *
Optimal lot-size = O
= 25,333

Given that p = \$50, s = \$14, c = \$18:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$610,210

5
Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 5,568

As can be seen from the results above, postponement increases the expected profit and decreases
the amount of overstock.

EXCEL worksheet 12-7 illustrates these computations

8.
(a)

Cost of overstocking, CO =              \$   0.50

Cost of understocking, CU =             \$   1.00

Mean demand                             50,000
Standard deviation of demand
=                                       15,000

Optimal CSL =       C   u

1
 0.67
Cu  Co             1  0.5

Optimal order quantity = (NORMSINV (0.67))(15,000) + 50,000 = 56,461

(b)

Cost of overstocking, CO =              \$   0.50

Cost of understocking, CU =             \$   5.00

Mean demand                             50,000
Standard deviation of demand
=                                       15,000

Optimal CSL =     C     u

5
 0.91
C C
u        o
5  0.5

Optimal order quantity = (NORMSINV (0.91))(15,000) + 50,000 = 70,028

EXCEL worksheet 12-8 illustrates these computations

6
9.

(a)

Mean demand =                         5,000

Standard deviation of demand =        2,000

Cost of overstocking, CO              \$ 40.00

Order size =                          6,000

CSL (implied by the order size) = NORMDIST (6000-5000/2000) = 0.691

Implied cost of understocking, CU = (CO)(CSL)/(1-CSL) = (40)(0.691)/(1-0.691) = \$89.64

(b)

Mean demand =                         5,000

Standard deviation of demand =        2,000

Cost of overstocking, CO              \$ 40.00

Order size =                          8,000

CSL (implied by the order size) = NORMDIST (8000-5000/2000) = 0.933

Implied cost of understocking, CU = (CO)(CSL)/(1-CSL) = (40)(0.933)/(1-0.933) = \$558.74

EXCEL worksheet 12-9 illustrates these computations

10.

Current policy:

*
     C   u

45
 0.6923
CSL
Cu  Co       45  20
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.6923,4000,1750) = 4879

Given that p = \$125, s = \$60, c = \$80:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

7
+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$140,001

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 1,224

Southern Hemisphere option:

*
   C   u

45
 0.90
CSL
C Cu       o
45  5
 NORMINV(CSL ,  ,  ) = NORMINV(0.9,4000,1750) = 6243
*                            *
Optimal lot-size = O

Given that p = \$125, s = \$75, c = \$80:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$164,644

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 2,326

EXCEL worksheet 12-10 illustrates these computations

11.

(a)

Mean demand during lead time =DL= (40)(1) = 40

Standard deviation of demand during lead time = L =  D L = 5 1 = 5

Safety inventory = ROP – DL = 45 – 40 = 5

CSL = NORMDIST (45, 40, 5, 1) = 0.8413

Cost of holding one unit for one year = (0.25)(4) = \$1

HQ                 1 200
Justifying cost of understocking: Cu =                                                      \$0.086
(1  CSL) D year (1  0.8413)  40  365

8
(b)

HQ(CSL)           1 200  0.8413
Justifying cost of understocking: Cu =                                                      \$0.073
(1  CSL) D year (1  0.8413)  40  365

(c)

HQ                   1  200
Desired CSL = 1                         = 1                  = 0.9909
Cu   D   year
1.5  40  365

Desired safety stock = (NORMSINV(0.9909))(5) = 11.8

Desired reorder point = 40 + 11.8 = 51.8

EXCEL worksheet 12-11 illustrates these computations

12.

Without postponement:

For each box:

*
     C   u

10
 0.7692
CSL
Cu  Co        10  3
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.7692,20000,8000)
= 25,891

Given that p = \$20, s = \$7, c = \$10:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$168,362

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 6,965

Total expected profits = 4(168,362) = \$673,446

Total expected overstock = 4(6,965) = 27,860

Total production quantity = 4(25,891) = 103,564

9
With postponement:

Anticipated demand = 20,000 + 20,000 + 20,000 + 20,000 = 80,000
Standard deviation =           80002  80002  80002  80002  16000

*
  C  8  0.6154
u
CSL
C C 85
u       o

Optimal lot-size = O  NORMINV(CSL ,  ,  ) = NORMINV(0.6154,80000,16000)
*                     *

= 84,694

Given that p = \$20, s = \$7, c = \$12:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$560,515

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 9,003

Indifferent:

At a unit cost of \$10.7 the two options, i.e., postponement and no postponement would be
indifferent. This unit cost is obtained by using the solver option in EXCEL by considering cell 21
as the changing cell while cell 35 is utilized as the target cell with a value of \$673,446.

EXCEL worksheet 12-12 illustrates these computations

13.

The with and without postponement calculations are similar to problem 12 (EXCEL worksheet
12-13 illustrates these computations), but what is new in this problem is the tailored
postponement which is discussed below:

Tailored postponement:

Popular style without postponement:
C           15
CSL   u  15  7  0.6818
*

Cu Co
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.6818,30000,5000)
= 32,364

Given that p = \$35, s = \$13, c = \$20:

10
Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$410,757

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 3,396

Other three styles with postponement:

Aggregated expected demand = 8,000 + 8,000 + 8,000 = 24,000
Standard deviation =     40002  40002  40002  6928

*
     C   u

14
 0.6182
CSL
Cu  Co       14  8
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.6182,24000,6928)
= 26,083

Given that p = \$35, s = \$13, c = \$21.4:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$268,281

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 18,083

Total expected profit = \$410,757 + \$268,281 = \$679,038

Total expected overstock = 3,396 + 18,083 = 21,479

EXCEL worksheet 12-13 illustrates these computations

14.

Without discount:

*
     C   u

65
 0.6842
CSL
Cu  Co       65  30
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.6842,20000,8000)

11
= 23,836

Given that p = \$95, s = \$0, c = \$30:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$1,029,731

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 5,470

With discount:

Optimal lot-size = O*  25,000

Given that p = \$95, s = \$0, c = \$28:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$1,076,941

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 6,295

Expected profits increase with discount.

EXCEL worksheet 12-14 illustrates these computations

15.

Without discount:

*
     C   u

7
 0.7
CSL
Cu  Co       73
Optimal lot-size = O*  NORMINV(CSL* ,  ,  ) = NORMINV(0.7,70000,25000)
= 83,110

Given that p = \$10, s = \$0, c = \$3:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

12
+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$403,077

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 17,869

With discount:

Optimal lot-size = O*  100,000

Given that p = \$10, s = \$0, c = \$2.75:

Expected profits = (p – s) NORMDIST((O – )/, 0, 1, 1)

– (p – s) NORMDIST((O – )/, 0, 1, 0) – O (c – s) NORMDIST(O, , , 1)

+ O (p – c) [1 – NORMDIST(O, , , 1)] = \$410,974

Expected overstock = (O – )NORMDIST((O – )/, 0, 1, 1) +  NORMDIST((O – )/, 0, 1, 0)

= 31,403

Expected profits increase with discount.

EXCEL worksheet 12-15 illustrates these computations

16.

a. the manufacturer should order :

40-Gb                20-Gb                 6-Gb
26,772               47,419               84,054

b. The expected profits for the units are:

40-Gb               20-Gb                 6-Gb
\$1,664,888          \$2,048,931           \$2,080,846

c. If the available capacity is limited to 140,000 units the manufacturer should order:

40-Gb               20-Gb                 6-Gb
26,772               41,300               72,028

13
Formatted: Right: -2.12"

expected profits would be:
40-Gb                20-Gb             6-Gb
\$1,790,125          \$2,072,482         \$2,002,170

Formatted: Justified
EXCEL worksheet 12-16 illustrates these computations.        Formatted: Justified

14

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