Example 9.1 - An EOQ Model For Bedrock's Problem: Input Cells Are Shaded

A B C D E F G H I J
1 Example 9.1 - An EOQ Model for Bedrock's Problem

2
3 Input Cells are shaded 100
4 Annual Demand 12,000 <-------- Graph Table --------->
5 Ordering Cost £50.00 Order Holding Ordering Annual
6 Unit Cost £25.00 size cost cost cost
7 Unit holding cost per year (two options) 100 £375 £6,000 £6,375
8 (i) in £s per year 200 £750 £3,000 £3,750
9 (ii) as % of unit cost 30.0% 300 £1,125 £2,000 £3,125
10 Unit holding cost per year = £7.50 400 £1,500 £1,500 £3,000
11 500 £1,875 £1,200 £3,075
12 Output 600 £2,250 £1,000 £3,250
13 EOQ 400.00 700 £2,625 £857 £3,482
14 No. of Orders/Year 30.0 800 £3,000 £750 £3,750
15 Total cost £303,000 Plot cell range F5:I14
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£7,000 EOQ graph
20
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Annual cost
22 £6,000
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24 £5,000
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£4,000
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29 £3,000
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31 £2,000
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£1,000
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36 £0
100 200 300 400 500 600 700 800
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38 Order quantity
39 Holding cost Ordering cost Annual cost
40
41
Figure 8.4 Economic order quantity (EOQ) model.
(Note that this model has been modified)

A B C D E F
1 Example 9.2 - The PROQ Model and Solution to Gizmo's Problem.
2
3 Input Annual Demand 2,100
4 Setup Cost £450.00
5 Unit Cost £30.00
6 Annual production rate 2,500
7 Unit holding cost per year (two options)
8 (i) in £s per year
9 (ii) as % of unit cost 20.0%
10 Annual unit holding cost = £6.00
11
12 Output PROQ 1403.12
13 Production run time,Ro (in weeks) 29.18
14 Optimal cycle time, To (in weeks) 34.74
15 Maximum inventory level 224.5
16 Annual holding cost £673
17 Annual setup cost £673
18 Total cost £64,347
19
20
21 Cell Formula Copied to
22 E10 IF(E9="",E8,E5*E9)
23 E11 IF(E10=0,"Holding cost cannot be zero!","")
24 E12 SQRT(2*E3*E4/E10)*SQRT(E6/(E6 - E3))
25 E13 52*E12/E6
26 E14 52*E12/E3
27 E15 E12*(E6 - E3)/E6
28 E16 0.5*E10*E15
29 E17 E3*E4/E12
30 E18 E16 + E17 + E3*E5
31
Figure 8.5 Production order quantity (PROQ) model.

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A B C D E F G H I
1 Example 9.3 - A Quantity Discount Model for the Wheelie Company
2
3 Input Annual Demand 1,500 User input cells
4 Ordering Cost £80.00 are shaded
5 Unit holding cost per year (two options)
8
9 DISCOUNT TABLE Unit Cost = £10.00 £8.00 £6.00
10 Minimum discount quantity, Mini = 0 1000 2000
11 Annual unit holding cost = £3.00 £2.40 £1.80
12
13 Output Qi = 282.8 316.2 365.1
14 Adjusted order quantities = 282.8 1000.0 2000.0
15 Total costs = £15,849 £13,320 £10,860
16
17 Minimum total cost is £10,860 2
18 Optimal order quantity is 2000.0
19 Cycle time is 69.3 weeks
20
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23 F11 IF($G6="",$G7*F9,$G6) G11:H11
24 F12 IF(F11=0,"Holding cost cannot be zero!","")
25 F13 SQRT(2*$G3*$G4/F11) G13:H13
26 F14 IF(F13>F10,F13,F10) G14:H14
27 F15 $G3*$G4/F14 + 0.5*F14*F11 + $G3*F9 G15:H15
28 F17 MIN(F15:H15)
29 H17 MATCH(F17,F15:H15,0) - 1
30 F18 OFFSET(F18,-4,H17)
31 F19 52*F18/G3
32
Figure 8.6 Quantity discount model for the Wheelie Company.

A B C D E F G H I
1 Example 9.4 - A Delivery Charge Model for the Farmers' Co-operative
2
3 Input Daily Demand (in tonnes) 3.0 User input cells
4 Unit Cost £100.00 are shaded
5 Unit holding cost per day (two options)
6 (i) in £s per day £1.50
7 (ii) as % of unit cost
8
9 DELIVERY TABLE Reorder Cost = £80.00 £130.00 £180.00
10 Maximum delivery quantity, Maxi = 10 20 30
11 Daily unit holding cost = £1.50 £1.50 £1.50
12
13 Output Qi = 17.9 22.8 26.8
14 Adjusted order quantities = 10.0 20.0 26.8
15 Total costs = £332 £335 £340
16
17 Minimum total cost is £332 0
18 Optimal order quantity is 10.0
19 Cycle time is 3.3 days
20
22 F13 SQRT(2*$G3*F9/F11) G13:H13
23 F14 IF(F13<F10,F13,F10) G14:H14
24 F15 $G3*F9/F14 + 0.5*F14*F11 + $G3*$G4 G14:H15
25 F19 F18/G3
26
Figure 8.7 Delivery charge model for the farmers' co-operative.

A B C D E F G H
1 Example 9.5 - An Inventory Model with Shortages Allowed
2
3 Input Annual Demand 12,000
4 Setup/Ordering Cost £50.00 User input cells
5 Unit Cost £25.00 are shaded
6 Holding cost (two options)
9 Shortage cost per unit per year £4.00
10 Unit holding cost per year = £7.50
11
12
13 Output Optimal order size, Qo 678.2
14 Maximum stock level 235.9
15 Back-order size 442.3
16 No. of orders/year 17.7
17 Cycle time 2.9 weeks
18 Annual Costs……..……
19 Setup/ordering cost £884.65
20 Holding cost £307.70
21 Shortage cost £576.95
22 Purchase cost £300,000
23 Total cost £301,769
24
25
27 E10 IF(E8="",E7,E8*E5)
28 E11 IF(E10=0, "Enter a value in either cell E7 or E8!","")
29 E13 SQRT(2*E3*E4*(E9 + E10)/(E9*E10))
30 E14 E9*E13/(E9 + E10)
31 E15 E13 - E14
32 E16 E3/E13
33 E17 52/E16
34 E19 E3*E4/E13
35 E20 0.5*E10*E14*E14/E13
36 E21 E9*(E13 - E14)^2/(2*E13)
37 E22 E3*E5
38 E23 SUM(E19:E22)
39
Figure 8.8 Deterministic model with planned storages.

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Figure 8.8 Deterministic model with planned storages.

A B C D E F G H I J
1 Example 9.6 - An Inventory Model with Storage Space Constraints
2
3 Setup cost £1,500.0 User input cells
4 Holding cost (as % of unit cost) 30.0% are shaded
5
6 Product Demand Unit Space EOQ Average Variable
7 cost (per unit) (Qo) space costs
8 Widget 10,000 £18.00 0.3 2357.0 353.6 £12,728
9 Gadget 8,000 £15.00 0.2 2309.4 230.9 £10,392
10 P 3,000 £10.00 0.15 1732.1 129.9 £5,196
11 Totals = 714.4 £28,316
12
13 Product Demand Unit Space EOQ Average Variable
14 cost (per unit) (Qo) space costs
15 Widget 7054 £18.00 0.3 1979.6 296.9 £12,922
16 Gadget 5643 £15.00 0.2 1939.6 194.0 £10,551
17 P 2116 £10.00 0.15 1454.7 109.1 £5,275
18 Totals = 600.0 £28,749
19
20 Percentage increase in variable costs = 1.53%
21
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23 Scaling Factor = 0.705 (Initially, set Scaling Factor = 1)
24
25
26 Solver Parameters
27 Set Target Cell: E23
28 Equal to: Max
29 By Changing Cells: E23
30 Subject to Constraints: H18 <= 600 = Storage space constraint
31 E23 >=0 = Answer must be positive
32
34 G8 SQRT(2*C8*G$3/(G$4*D8)) G9:G10
35 H8 0.5*E8*G8 H9:H10
36 I8 C8*G$3/G8 + 0.5*G8*D8*G$4 I9:I10
37 H11 SUM(H8:H10) I11
38 Copy range B6:I11 into B13:I18
39 C15 C8*E$23 C16:C17
40 I15 C8*G$3/G15 + 0.5*G15*D15*G$4 I16:I17
41 I20 (I18 - I11)/I11
42
43
Figure 8.9 Multiple-product model with storage space constraint.

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Figure 8.9 Multiple-product model with storage space constraint.

A B C D E F G H I J
1 Example 9.7 - The Newsboy Problem: A Probabilistic Model with Discrete Demand
2
3 Input Unit Cost, C = £3.00
4 Selling Price, S = £5.00 User input cells are shaded
5 Scrap value, V = £0.75
6
7
8 Output <- Probabilities -> <-- Expected -->
9 Indiv. Cumul. profit, EPi
10 Demand, Di Pi CUMi Sales Profit
11 1 10 0.05 1 10 £20
12 2 20 0.1 0.95 19.5 £38
13 3 30 0.15 0.85 28 £52
14 4 40 0.2 0.7 35 £59
15 5 50 0.2 0.5 40 £58
16 6 60 0.15 0.3 43 £48
17 7 70 0.1 0.15 44.5 £32
18 8 80 0.05 0.05 45 £11 4
19
20 Optimal demand, Qo = 40 Maximum profit = £59
21
22
24 E11 SUM(D11:D$18) E12:E18
25 F11 SUMPRODUCT(C$11:C11,D$11:D11) + C11*E12 F12:F17
26 F18 SUMPRODUCT(C$11:C18,D$11:D18)
27 G11 E$4*F11 - E$3*C11 + E$5*(C11 - F11) G12:G18
28 I18 MATCH(H20,G11:G18,0)
29 D20 OFFSET(C10,I18,0)
30 H20 MAX(G11:G18)
31
Figure 8.10 The Newsboy problem - a probabilistic model with discrete demand.
A B C D E F G H I
1 Example 9.8 - A Probabilistic Model with Shortages
2
3 Input Holding cost, H = £40.00 All user input cells
4 Shortage cost, B = £500.00 are shaded
5 B/(B + H) = 0.93
6
7
8 Output <- Probabilities ->
9 Indiv. Sum
10 Demand, Di Pi SUMi
11 1 3 0.4 0.4
12 2 4 0.25 0.65
13 3 5 0.13 0.78
14 4 6 0.11 0.89
15 5 7 0.05 0.94 = Optimal amount
16 6 8 0.04 0.98
17 7 9 0.01 0.99
18 8 10 0.01 1
19
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22 E5 E4/(E4 + E3)
23 E11 SUM(D$11:D11) E12:E18
24 F11 IF(E11>=H$5)," = Optimal amount","")
25 F12 IF(AND(E11<H$5,E12>=H$5)," = Optimal amount","") F13:F18
26
Figure 8.11 Probabilistic model with shortages.

A B C D E F G H I J K
1 Example 9.9 - A Service-Level Model with Variable Demand/ Fixed Lead-Time
2
3 Input - must be in consistent time units
4 Time (day, week, month, year) week Demand is normally-distributed
5 Ordering/Setup Cost £100.00 Mean = 500
6 Unit Cost £10.00 Standard deviation = 60
7 Holding cost (two options) Service Level %, SL = 95%
8 (i) in £s per year Lead Time, Lt = 5 week
10 Unit holding cost per week £0.058 52
11
12 Output
13 Reorder level/point, R 2721.0 Holding cost of safety stock £13
14 Order quantity, Q 1316.6 Holding cost of normal stock £38
15 Safety stock 221.0 Ordering/setup costs £38
16 Total costs per week £89
17
18
20 D8 E4 D10, K8, H16
21 E10 IF(E9="",E8/G10,E9*E6/G10)
22 G10 IF(E4="day",365,IF(E4="week",52,IF(E4="month",12,1)))
23 E11 IF(E10=0,"Enter a value in either cell E8 or E9!","")
24 D13 J5*J8 + D15
25 D14 SQRT(2*J5*E5/E10)
26 D15 ROUNDUP(NORMSINV(J7)*J6*SQRT(J8),0)
27 J13 D15*E10
28 J14 D14*E10/2
29 J15 IF(D14=0,"",E5*J5/D14)
30 J16 SUM(J13:J15)
31
Figure 8.12 Service-level model with variable demand/fixed lead-time.

A B C D E F G H I J
1 Example 9.10 - A Service-Level Model with Fixed Demand/ Variable Lead-Time
2
3 Input - must be in consistent time units
4 Time (day, week, month, year) week Lead-time is normally-distributed
5 Demand 500 Mean = 5
6 Ordering/Setup Cost £100.00 Standard deviation = 1
7 Unit Cost £10.00 Service Level %, SL = 95%
8 Holding cost (two options)
9 (i) in £s per year User input cells are shaded
11 Unit holding cost per week £0.058 52
12
13 Output
14 Lead time 6.6 week Holding cost of normal stock £38
15 Reorder level/point, R 3322.4 Ordering/setup costs £38
16 Order quantity, Q 1316.6 Total costs per week £76
17
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20 D14 J5 + NORMSINV(J7)*J6
21 E14 E4
22 D15 E5*D14
23 D16 SQRT(2*E5*E6/E11)
24 J14 D16*E11/2
25 J15 E6*E5/D16
26 J16 SUM(J14:J15)
27
Figure 8.13 Service-level model with variable demand/variable lead-time.

A B C D E F G H I J K
1 Example 9. 11 - A Periodic Review (i.e. Fixed-Period) Model
2
3 Input - must be in consistent time units Demand is normally-distributed
4 Time (day, week, month, year) day Mean = 40
5 Ordering/Setup Cost £50.00 Standard deviation = 15
6 Unit Cost £10.00 Service Level %, SL = 95%
7 Holding cost (two options) Lead Time, Lt = 8 day
8 (i) in £s per year £20.00 Review Period = 16 day
9 (ii) as % of unit cost Stock On-hand = 60
10 Unit holding cost per day £0.055 365
11
12 Output
13 Reorder level/point, R 1081.0 Holding cost of safety stock £6.63
14 Order quantity, Q 1021.0 Holding cost of normal stock £27.97
15 Safety stock 121.0 Ordering/setup costs £1.96
16 Total costs per day £36.56
17
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20 D13 J4*(J7 + J8) + D15
21 D14 D13 - J9
22 D15 ROUNDUP(NORMSINV(J6)*J5*SQRT(J7+J8),0)
23
Figure 8.14 Periodic review (fixed-period) model.

A B C D E F G H I J
1 Example 9.12 - A Multi-Period Model with Several Constraints
2
3 Input All user input cells are shaded
4 Annual Demand 3,600
5 Ordering Cost £5.00 Output
6 Unit Cost £2.00 EOQ 300.00
7 Unit holding cost per year (two options) Cycle time
8 (i) in £s per year (in months) 1.0
9 (ii) as % of unit cost 20.0% Total cost £7,320
10 Unit holding cost per year £0.40
11
12 Monthly Order Ending Cost per
13 Month Demand Quantity Inventory Period
14 1 240 270 30 £546
15 2 270 330 90 £668
16 3 450 360 0 £725
17 4 210 270 60 £547
18 5 240 270 90 £548
19 6 300 270 60 £547
20 7 330 330 60 £667
21 8 420 360 0 £725
22 9 240 270 30 £546
23 10 330 300 0 £605
24 11 300 300 0 £605
25 12 270 270 0 £545
26 Annual demand = 3,600 £7,274 = Annual cost
27 Objective: Minimize surplus stock = 420
28
29 Note: Switch on the "Assume Linear Model" parameter in the Solver Options dialog box
Figure 8.15 Multi-period model with several constraints.

A B C D E F G H I J
30
31 Solver Parameters
32 Set Target Cell: F27
33 Equal to: Min
34 By Changing Cells: E14:E25
35 Subject to Constraints: E14:E25 >= 270 = Quantity discount constraint
36 E14:E25 <= 360 = Order size - upper limit
37 E14:E25 = int(eger) = Answers must be integer
38 F14:E25 >= 0 = No stockouts allowed!
39
41 I6 SQRT(2*F4*F5/F10)
42 I8 12*I6/F4
43 I9 F4*F5/I6 + 0.5*F10*I6 = F4*F6
44 F10 IF(F9="",F8,F6*F9)
45 F11 IF(F10=0,"Enter a value in eithe cell F8 or F9!","")
46 F14 SUM(E$14:E14) - SUM(D$14:D14) F15:F25
47 G14 F$5 + F14*F$10/12 = E14*F$6 G15:G25
48 D26 SUM(D14:D25) G26
49 F27 SUM(F14:F25)
50
Figure 8.15 Multi-period model with several constraints.

A B C D E F G H I J K L M
1 Example 9.13 - A Simulation Model for Inventory Control
2
3 Demand table Lead-time table
4 <-- Limits --> Dem- <-- Limits --> No. of
5 Lower Upper and Pi Lower Upper days Pi
6 0 0.03 0 0.03 0 0.20 1 0.20 User input
7 0.03 0.08 1 0.05 0.20 0.70 2 0.50 cells are
8 0.08 0.21 2 0.13 0.70 1.00 3 0.30 shaded
9 0.21 0.46 3 0.25 1.00
10 0.46 0.68 4 0.22
11 0.68 0.88 5 0.20 Reorder level = 15
12 0.88 1.00 6 0.12 Order quantity = 30
13 1.00
14
15 Output table
16 Units Begin. RAND Dem- Ending New Lost Lead Recpt.
17 Day Recvd. Invntry. No. and Invntry. Level sales Order? time Day
18 1 30 0.70 5 25 25 0 No
19 2 0 25 0.76 5 20 20 0 No
20 3 0 20 0.84 5 15 15 0 Yes 2 6
21 4 0 15 0.02 0 15 45 0 No
22 5 0 15 0.16 2 13 43 0 No
23 6 30 43 0.85 5 38 38 0 No
24 7 0 38 0.59 4 34 34 0 No
25 8 0 34 0.47 4 30 30 0 No
26 9 0 30 0.85 5 25 25 0 No
27 10 0 25 0.84 5 20 20 0 No
28 11 0 20 0.00 0 20 20 0 No
29 12 0 20 0.47 4 16 16 0 No
30 13 0 16 0.58 4 12 12 0 Yes 2 16
31 14 0 12 0.42 3 9 39 0 No
32 51 0
33
34 Service Level = 100.0%
35
A B C D E F G H I J K L
1 Case Study 9.1 - A Material Requirements Planning (MRP) Model
2
3 The BOM Table
4 Part Number: Description BOM Id. No. of Lead On Planned
5 Level Code Units Time Hand Order User input
6 Table 0 1 1 1 50 Rel. Row cells are
7 Top Assembly 1 1001 1 2 50 25 shaded
8 Table Top 2 2001 1 1 180 35
9 Drawer 2 2002 1 1 200 35
10 Leg Assembly 1 1002 1 1 100 25
11 Legs 2 2003 4 1 250 65
12 Side Rung 2 2004 2 1 50 65
13 Connecting Rung 2 2005 1 1 110 65
14
15
16 The MRP Output Table
17 1
18 Table Lead Time = 1
19 Week Number Overdue 1 2 3 4 5 6 7 8
20 Master Production Schedule 0 0 180 180 100 0 0 0
21 Scheduled Receipts 0 0 0 0 0 0 0 0
22 On Hand 50 50 50 0 0 0 0 0
23 Net Requirements 0 0 130 180 100 0 0 0
24 Planned Order Receipts 0 0 130 180 100 0 0 0
25 Planned Order Releases 0 0 130 180 100 0 0 0 0
26
27 2
28 Top Assembly Lead Time = 2
30 Gross Requirements 0 130 180 100 0 0 0 0
32 On Hand 50 50 20 0 0 0 0 0
36
37 3
38 Table Top Lead Time = 1
42 On Hand 180 20 0 0 0 0 0 0
46
Figure 8.20 MRP model for the kitchen table example.

47 4
48 Drawer Lead Time = 1
52 On Hand 200 40 0 0 0 0 0 0
56
57 5
58 Leg Assembly Lead Time = 1
62 On Hand 100 100 0 0 0 0 0 0
66
67 6
68 Legs Lead Time = 1
72 On Hand 250 130 0 0 0 0 0 0
76
77 7
78 Side Rung Lead Time = 1
81 Scheduled Receipts 10 0 0 0 0 0 0 0 0
82 On Hand 50 0 0 0 0 0 0 0
86
87 8
88 Connecting Rung Lead Time = 1
92 On Hand 110 80 0 0 0 0 0 0
96

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48 Drawer Lead Time = 1
52 On Hand 200 40 0 0 0 0 0 0
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58 Leg Assembly Lead Time = 1
62 On Hand 100 100 0 0 0 0 0 0
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67 6
68 Legs Lead Time = 1
72 On Hand 250 130 0 0 0 0 0 0
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77 7
78 Side Rung Lead Time = 1
81 Scheduled Receipts 10 0 0 0 0 0 0 0 0
82 On Hand 50 0 0 0 0 0 0 0
86
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88 Connecting Rung Lead Time = 1
92 On Hand 110 80 0 0 0 0 0 0
96
Figure 8.20 (cont.)

A B C D E F G H I J K L M
1 Example 9. 14 - A Model for the Part Period Balancing (PPB) Method
2
3 Input Ordering (or Setup) Cost = £200 User input cells
4 Unit holding cost = £1.00 are shaded
5 Economic part period (EPP) = 200
6
7 <----------- Net requirements, REQP, for each period P --------------->
8 REQP 150 100 150 0 50 75 100 25 20
9 Period, P 1 2 3 4 5 6 7 8 9
10
11 Weighted REQi 0 100 300 0 200 375 600 175 160
12 CUMi 0 100 400 400 600 975 1575 1750 1910
13 (CUMi - EPP)/EPP -1.0 -0.5 1.0 1.0 2.0 3.9 6.9 7.8 8.6
14 0.5 1.0 0.5 1.0 1.0 2.0 3.9 6.9 7.8 8.6
15 Order Data = 150 100
16 0 Answer: Place an order for 250 units in period 1
17 New Factor, NFi 0 0 1 2 3 4 5 6 7
18
19 Weighted REQi -150 -100 0 0 100 225 400 125 120
20 CUMi 0 0 0 0 100 325 725 850 970
21 (CUMi - EPP)/EPP -1.0 -1.0 -1.0 -1.0 -0.5 0.6 2.6 3.3 3.9
22 0.5 1.0 1.0 1.0 1.0 0.5 0.6 2.6 3.3 3.9
23 Order Data = 150 0 50
25 New Factor, NFi 0 0 0 0 0 1 2 3 4
26
27 Weighted REQi -150 -100 -150 -1 -50 0 100 50 60
28 CUMi 0 0 0 0 0 0 100 150 210
29 (CUMi - EPP)/EPP -1.0 -1.0 -1.0 -1.0 -1.0 -1.0 -0.5 -0.3 0.1
30 0.1 1.0 1.0 1.0 1.0 1.0 1.0 0.5 0.3 0.1
31 Order Data = 75 100 25 20
33 New Factor, NFi 0 0 0 0 0 0 0 0 0
34
35
36 Copy cell range B11:L17 repeatedly down the spreadsheet, placing the cursor in cells B19,
37 B27…. until the 'New Factor, NFi' row contains nothing but zeros (e.g. see row 33 above).
Figure 8.21 Model for the part-period balancing (PBB) method.
(Note that this model has been modified)

A B C D E F G H
1
2 Product No. in Product Year Cash
3 name stock price flow
4 Gizmo 10 £10.00 1 1
5 Gadget 25 £12.50 2 4
6 Widget 8 £20.00 3 8
7 Sprocket 40 £4.50 4 16
8 5
9 Sample Figure
A B C D E F G H
10
11 Column(B3) = 2 Row(B3) = 3
12 Column(D5:D9) = 4 Row(D5:D9) = 5
13
14 INDEX(B4:D7,2,3) = £12.50 NORMSINV(0.95) =
I
10
11
12
13
14 1.6449

Example 9.1 - An EOQ Model For Bedrock's Problem: Input Cells Are Shaded

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Example 9.1 - An EOQ Model For Bedrock's Problem: Input Cells Are Shaded

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A B C D E F G H I J

1 Example 9.1 - An EOQ Model for Bedrock's Problem

Figure 8.4 Economic order quantity (EOQ) model.

(Note that this model has been modified)

Figure 8.5 Production order quantity (PROQ) model.

Figure 8.5 Production order quantity (PROQ) model.

Figure 8.5 Production order quantity (PROQ) model.

Figure 8.6 Quantity discount model for the Wheelie Company.

Figure 8.7 Delivery charge model for the farmers' co-operative.

Figure 8.8 Deterministic model with planned storages.

Figure 8.8 Deterministic model with planned storages.

Figure 8.9 Multiple-product model with storage space constraint.

Figure 8.9 Multiple-product model with storage space constraint.

Figure 8.11 Probabilistic model with shortages.

Figure 8.12 Service-level model with variable demand/fixed lead-time.

Figure 8.13 Service-level model with variable demand/variable lead-time.

Figure 8.14 Periodic review (fixed-period) model.

Figure 8.15 Multi-period model with several constraints.

Figure 8.15 Multi-period model with several constraints.

Figure 8.20 MRP model for the kitchen table example.

Figure 8.20 MRP model for the kitchen table example.

Figure 8.20 (cont.)

Figure 8.21 Model for the part-period balancing (PBB) method.

(Note that this model has been modified)

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