7.Ano Backlogs Are Allowed

OM HW Chapter 11

7.aNo backlogs are allowed

Period / Mar. / Apr. / May / Jun. / July / Aug. / Sep. / Total
Forecast / 50 / 44 / 55 / 60 / 50 / 40 / 51 / 350
Output
Regular / 40 / 40 / 40 / 40 / 40 / 40 / 40 / 280
Overtime / 8 / 8 / 8 / 8 / 8 / 3 / 8 / 51
Subcontract / 2 / 0 / 3 / 12 / 2 / 0 / 0 / 19
Output - Forecast / 0 / 4 / –4 / 0 / 0 / 3 / –3
Inventory
Beginning / 0 / 0 / 4 / 0 / 0 / 0 / 3
Ending / 0 / 4 / 0 / 0 / 0 / 3 / 0
Average / 0 / 2 / 2 / 0 / 0 / 1.5 / 1.5 / 7
Backlog / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0
Costs:
Regular / 3,200 / 3,200 / 3,200 / 3,200 / 3,200 / 3,200 / 3,200 / 22,400
Overtime / 960 / 960 / 960 / 960 / 960 / 360 / 960 / 6,120
Subcontract / 280 / 0 / 420 / 1,680 / 280 / 0 / 0 / 2,660
Inventory / 0 / 20 / 20 / 0 / 0 / 15 / 15 / 70
Total / 4,440 / 4,180 / 4,600 / 5,840 / 4,440 / 3,575 / 4,175 / 31,250

b.Level strategy

Period / Mar. / Apr. / May / Jun. / July / Aug. / Sep. / Total
Forecast / 50 / 44 / 55 / 60 / 50 / 40 / 51 / 350
Output
Regular / 40 / 40 / 40 / 40 / 40 / 40 / 40 / 280
Overtime / 8 / 8 / 8 / 8 / 8 / 8 / 8 / 56
Subcontract / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 14
Output - Forecast / 0 / 6 / –5 / –10 / 0 / 10 / –1
Inventory
Beginning / 0 / 0 / 6 / 1 / 0 / 0 / 1
Ending / 0 / 6 / 1 / 0 / 0 / 1 / 0
Average / 0 / 3 / 3.5 / .5 / 0 / .5 / .5 / 8
Backlog / 0 / 0 / 0 / 9 / 9 / 0 / 0 / 18
Costs:
Regular / 3,200 / 3,200 / 3,200 / 3,200 / 3,200 / 3,200 / 3,200 / 22,400
Overtime / 960 / 960 / 960 / 960 / 960 / 960 / 960 / 6,720
Subcontract / 280 / 280 / 280 / 280 / 280 / 280 / 280 / 1,960
Inventory / 30 / 35 / 5 / 0 / 5 / 5 / 80
Backlog / 180 / 180 / 360
Total / 4,440 / 4,470 / 4,475 / 4,625 / 4,620 / 4,445 / 4,445 / 31,520

21.Given:

We have the following forecasts and customer orders over the next eight weeks:

1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Forecast / 50 / 50 / 50 / 50 / 50 / 50 / 50 / 50
Customer Orders / 52 / 35 / 20 / 12

Beginning inventory = 0 units

Lot size = 75 units

Use the MPS rule of ordering when project on-hand inventory would be negative without production.

The calculations for MPS and projected on-hand inventory are shown below:

Week / Inventory From Previous Week / Requirements* / Net Inventory before MPS / (75)
MPS / Projected
On-hand Inventory
1 / 0 / 52 / -52 / 75 / 23
2 / 23 / 50 / -27 / 75 / 48
3 / 48 / 50 / -2 / 75 / 73
4 / 73 / 50 / 23 / – / 23
5 / 23 / 50 / -27 / 75 / 48
6 / 48 / 50 / -2 / 75 / 73
7 / 73 / 50 / 23 / – / 23
8 / 23 / 50 / -27 / 75 / 48

*Requirements equal the larger of forecast and customer orders in each week.

Net Inventory before MPS = Inventory from previous week – Current week’s requirements.

Projected on-hand inventory = Inventory from previous week – Current week’s requirements + Current week’s MPS.

Note: We need a MPS quantity whenever Net Inventory before MPS < 0 units (i.e., when it would be negative).

Example calculations for projected on-hand inventory:

Week 1:

Net Inventory before MPS = 0 – 52 = -52. Warning: This is below the desired level of 0 units. We must plan for 75 units of MPS.

Projected on-hand inventory = -52 + 75 = 23.

Week 2:

Net Inventory before MPS = 23 – 50 = -27. Warning: This is below the desired level of 0 units. We must plan for 75 units of MPS.

Projected on-hand inventory = -27 + 75 = 48.

The final MPS is shown below:

Beg. Inv. = 0 / Week
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Forecast / 50 / 50 / 50 / 50 / 50 / 50 / 50 / 50
Customer Orders / 52 / 35 / 20 / 12
Projected on-hand inventory / 23 / 48 / 73 / 23 / 48 / 73 / 23 / 48
MPS / 75 / 75 / 75 / 75 / 75 / 75

22.Calculate the ATP for Problem 21:

ATP (first period) = Beginning inventory + MPS (first period) – sum of customer orders until (but not including) the period of the next MPS.

ATP (other periods) = MPS (current period) – sum of customer orders until (but not including) the period of the next MPS.

*We calculate ATP in the first period and in all other periods with MPS quantities.

ATP (Week 1) = 0 + 75 – (52) = 23

ATP (Week 2) = 75 – (35) = 40

ATP (Week 3) = 75 – (20 + 12) = 43

ATP (Week 5) = 75 – (0) = 75

ATP (Week 6) = 75 – (0 + 0) = 75

ATP (Week 8) = 75 – (0) = 75

OM HW Chapter 12

7. / Master
Schedule for: X / Week / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Quantity / 80 / 30
Week
Part X / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Gross requirements / 80 / 30
Lot size: LFL / Scheduled receipts
On hand
Net requirements / 80 / 30
Planned order receipt / 80 / 30
Planned order release / 80 / 30
Week
Part B / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Gross requirements / 160 / 60
Scheduled receipts
Lot size: LFL / On hand / 30 / 30 / 30 / 30 / 30 / 30 / 0 / 0 / 0
Net requirements / 130 / 60
Planned order receipt / 130 / 60
Planned order release / 130 / 60
Week
Part D / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Gross requirements / 260 / 240 / 120 / 90
Lot size = 50 / Scheduled receipts / 50 / 50 / 50 / 50
On hand / 20 / 70 / 70 / 120 / 10 / 60 / 20 / 0 / 10
Net requirements / 140 / 180 / 100 / 40
Planned order receipt / 150 / 200 / 100 / 50
Planned order release / 150 / 200 / 100 / 50

*Note that the lead time is a function of the lot size.

8. / Master
Schedule / Week / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6
Quantity / 40A / 60B / 30C
ALT = 1 wk. / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6
Gross requirements / 40
Scheduled receipts
Projected on hand
Net requirements / 40
Planned order receipt / 40
Planned order release / 40
BLT = 1 wk. / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6
Gross requirements / 60
Scheduled receipts
Projected on hand / 10 / 10 / 10 / 10 / 10 / 10
Net requirements / 50
Planned order receipt / 50
Planned order release / 50
CLT = 2 wks. / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6
Gross requirements / 30
Scheduled receipts
Projected on hand / 10 / 10 / 10 / 10 / 10 / 10 / 10
Net requirements / 20
Planned order receipt / 20
Planned order release / 20
DLT = 1 wk. / Beg. Inv. / 1 / 2 / 3 / 4 / 5 / 6
Gross requirements / 40 / 180
Scheduled receipts / 100
Projected on hand / 25 / 125 / 125 / 125 / 85 / 5
Net requirements / 95
Planned order receipt / 100
Planned order release / 100

OM HW Chapter 13

3.D = 1,215 bags/yr.

S = $10

H = $75

b.Q/2 = 18/2 = 9 bags

d.

e.Assuming that holding cost per bag increases by $9/bag/year

Q = 17 bags

Increase by [$1,428.71 – $1,350] = $78.71

5.D = 750 pots/mo. x 12 mo./yr. = 9,000 pots/yr.

Price = $2/pot S = $20 H = ($2)(.30) = $.60/unit/year

TC = 232.35 + 232.36

= 464.71

If Q = 1500

TC = 450 + 120 = $570

Therefore the additional cost of staying with the order size of 1,500 is:

$570 – $464.71 = $105.29

b. Only about one half of the storage space would be needed.

9.p = 5,000 hotdogs/day

u = 250 hotdogs/day

300 days per year

S = $66

H = $.45/hotdog per yr.

b.D/Qo = 75,000/4,812 = 15.59, or about 16 runs/yr.

c.run length: Qo/p = 4,812/5,000 = .96 days, or approximately 1 day

10.p = 50/ton/day

u = 20 tons/day

200 days/yr.

S = $100

H = $5/ton per yr.

Average is tons [approx. 3,098 bags]

c.Run length =

d.Runs per year:

e.Q = 258.2

TC =

TCorig.= $1,549.00

TCrev. = $ 774.50

Savings would be $774.50

18. Daily usage = 800 ft./day Lead time = 6 days

Service level desired: 95 percent. Hence, risk should be 1.00 – .95 = .05

This requires a safety stock of 1,800 feet.

ROP = expected usage + safety stock

= 800 ft./day x 6 days + 1,800 ft. = 6,600 ft.

20. LT demand = 600 lb.

LT = 52 lb.

risk = 4%  Z = 1.75

a. ss = ZLT = 1.75 (52 lbs.) = 91 lbs.

b.ROP = Average demand during lead time + safety stock

ROP = 600 + 91 = 691 lbs.

With no safety stock risk is 50%.

OM HW Chapter 14

3. / N = ? / N = / DT(1 + X)
D = 200 lb. per day / C
T = 2/8 = .250 day / = / 200(.250) (1.08) / = 2.7 [round to 3]
C = 120 lb. / 20
X = .08

a. Cycle 1234
A 6655
B 3333
C 1111
D 4455
E 2222 / b. Cycle 1 2
A 1111
B 6 6
C 2 2
D 9 9
E 4 4

c. 4 cycles = lower inventory, more flexibility

2 cycles = fewer changeovers

6.The smallest daily quantity is 5. None of the other daily quantities are divisible by 5, however. Using five cycles and producing one unit of each product per cycle would leave some products, as shown in the following table. Hence, if five cycles are used, some extra units will have to be made in some cycles.

Product / Daily quantity / Nearest multiple of 4 without exceeding daily quantity / Units short
F / 9 / 1 / 4
G / 8 / 1 / 3
H / 5 / 1 / –
K / 6 / 1 / 1

One possibility would be:

Cycle / 1 / 2 / 3 / 4 / 5
Pattern / F(2)–G–H–K / F(2)–G(2)–H–K / F(2)–G(2)–H–K / F(2)–G(2)–H–K / F–G–H–K(2)
Extra units(s) / F / F G / F G / F G / K

HW Chapter 15

Solutions

1.H = $10 per item per year (2,000 item) = $20,000 per year.

D = 3 days

Incremental holding cost = $20,000 (3/365) = $164.38

This exceeds the savings that would result by using 5-day freight. Therefore, use 2-day freight.

2.H = .30(80 boxes)($200 per box) = $4,800

2-day analysis:

d = 1 day

Savings = $40

Incremental holding cost = H

= $4,800

= $13.15

Net savings = $40 – $13.15 = $26.85

6-day analysis

d = 5 days

Savings = $120

Incremental holding cost = H

= $4,800

= $65.75

Net savings = $120 – $67.75 = $54.25

Thus, the 6-day alternative would yield the greater savings.

3.H = .35 (300 boxes) ($140 per box) (1/365 days) = $40.274

Cost = FC + dH

A / B
Option / Cost / Option / Cost
2 days / $580.55 / 2 days / $605.55
3 days / 580.82 / 4 days / 611.10
9 days / 762.46 / 7 days / 691.92

Ship 2-day using A.

OM HW Chapter 16

6. / a. / FCFS: A–B–C–D
SPT: D–C–B–A
EDD: C–B–D–A
CR: A–C–D–B
FCFS: / Job time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
A / 14 / 14 / 20 / 0
B / 10 / 24 / 16 / 8
C / 7 / 31 / 15 / 16
D / 6 / 37 / 17 / 20
37 / 106 / 44
SPT: / Job time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
D / 6 / 6 / 17 / 0
C / 7 / 13 / 15 / 0
B / 10 / 23 / 16 / 7
A / 14 / 37 / 20 / 17
37 / 79 / 24
EDD: / Job time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
C / 7 / 7 / 15 / 0
B / 10 / 17 / 16 / 1
D / 6 / 23 / 17 / 6
A / 14 / 37 / 20 / 17
84 / 24

Critical Ratio

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / 14 / 20 / (20 – 0) / 14 = 1.43
B / 10 / 16 / (16 – 0) /10 = 1.60
C / 7 / 15 / (15 – 0) / 7 = 2.14
D / 6 / 17 / (17 – 0) / 6 = 2.83

Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After the completion of Job A, the revised critical ratios are:

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 10 / 16 / (16 – 14) /10 = 0.20
C / 7 / 15 / (15 – 14) / 7 = 0.14
D / 6 / 17 / (17 – 14) / 6 = 0.50

Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After the completion of Job C, the revised critical ratios are:

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 10 / 16 / (16 – 21) /10 = –0.50
C / – / – / –
D / 6 / 17 / (17 – 21) / 6 = –0.67

Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.

The critical ratio sequence is A–C–D–B and the makespan is 37 days.

Critical Ratio sequence / Processing Time (Days) / Flow time / Due Date / Tardiness
A / 14 / 14 / 20 / 0
C / 7 / 21 / 15 / 6
D / 6 / 27 / 17 / 10
B / 10 / 37 / 16 / 21
 / 99 / 37
b. / / FCFS / SPT / EDD / CR
26.50 / 19.75 / 21.00 / 24.75
11.0 / 6.00 / 6.00 / 9.25
2.86 / 2.14 / 2.27 / 2.67

c. SPT is superior.

10. / a. / Job / Machine A / Machine B
/ a / 16 / 5 / 7
/ b / 3 / 2 / 13 / Thus, the sequence is e–b–g–h–d–c–a–f.
/ c / 9 / 6 / 6
/ d / 8 / 7 / 5
/ e / 2 / 1 / 14
/ f / 12 / 4 / 8
/ g / 18 / 14 / 3
/ h / 20 / 11 / 4
b.

025234351607688

e / b / g / h / d / c / a / f
/ e / b / g / h / d / c / a / f

0216294354616776818892

c. Original idle time for B: 2 + 9 + 7 = 18 hrs., and original makespan is 92.

The last two tasks in the sequence are a and f. Splitting both of their completion times evenly, we get the following results.

Machine 1Machine 2

a182.5

a282.5

f162

f262

After splitting, we get the following Gantt chart:

02523435160 68 76 82 88

e / b / g / h / d / c / a1 / a2 / f1 / f2
/ e / b / g / h / d / c / / a1 / / a2 / / f1 / / f2

021629435461676870.57678.58490

After splitting, idle time is: 2 + 1 + 5.5 + 3.5 + 4 = 16 hours, and the new makespan = 90.

There is a savings of 2 hr.