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Case Study: RED BRAND CANNERSCase Study: RED BRAND CANNERS

Vice President of OperationsMr. Michell Gorden

ControllerMr. William Copper

Sale ManagerMr. Charles Myers

Production ManagerMr. Dan Tucker

Purpose:

Decide the amount of tomato products to pack at this season.

Tomato Products

Whole Tomato Tomato Juice Tomato Paste

Information:

1. Amount of Tomato: 3,000,000 pounds to be delivered.

Tomato quality:

20% (grade A) × 3,000,000 = 600,000 pounds

80% (grade B) × 3,000,000 = 2,400,000 pounds

(provided by production manager)

2. Demand forecasts & selling prices (provided by sale manager):

Products Whole canned tomato Others

Demand no limitation Refer Exhibit 1

2

lbs.

correction(800,000/18)= 44444.5 Cases

Selling prices has been set in light of the long-term marketing strategy of the company. Potential sales have been forecasted at these prices.

3. Purchasing price & product profitability (provided by controller)

Purchasing price Net profit

6cents/pound Refer Exhibit 2

Product Whole tomato

Tomato juice

Tomato paste

Minimum requirement 8 points 6 points 5 points (without grade A)

5 points9 points

Grade BGrade A

3

0.3

4.0-(1.18+0.24+0.4+0.7) = 1.48

4.5 - (1.32+0.36+0.85+0.65) = 1.32

3.8 -(0.54+0.26+0.38+0.77) = 1.85

5. 80,000 pounds of grade "A" tomatoes are available at 8.5 cents perpound. (provided by the Vice president of operations)

6. Sale manager re-computes the marginal profits (Exhibit 3).

Linear Programming Solutions

(a) How to use the crop of 3,000,000 lbs. of tomatoes?

(b) Whether to purchase an additional 80,000 lbs. of A-grade tomatoes?

Part (a)Formulation:WA = lbs. of A-grade tomatoes in whole.

WB = lbs. of B-grade tomatoes in whole.

JA = lbs. of A-grade tomatoes in juice.

JB = lbs. of B-grade tomatoes in juice.

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解(1) (2) 之聯立方程式

1 CASE = [(0.0932*(3/4)+0.0518*(1/4)]*18

1 CASE = 0.0518* 25= 1.295

PA = lbs. of A-grade tomatoes in paste.PB = lbs. of B-grade tomatoes in paste.

Demand of whole tomatoes ≦ 800,000 lbs. = 44,444.5 ×18 lbs

Demand of tomatoes Juice ≦ 50,000 cases = 50,000 × 20 lbs = 1,000,000 lbs

Demand of tomatoes paste ≦ 80,000 cases = 80,000 × 25 lbs = 2,000,000 lbs

Grade "A" ≦ 600,000 ( 3,000,000lbs × 20% ) = 600,000 lbs.

Grade "B" ≦ 2,400,000 ( 3,000,000lbs × 80% ) = 2,400,000 lbs.

600,000 lbs. - 3WB ≧ 0

WB ≦ 600,000/3 = 200,000

600,000 + 200,000 = 800,000 lbs.

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Quality requirement for whole tomato:

Quality requirement for whole tomato:(0.9×WA + 0.5×WB)/2 ≧ 0.8× (WA + WB)/2 ⇒ WA - 3WB ≧ 0

Quality requirement for tomato juice:

(0.9×JA + 0.5×JB)/2 ≧ 0.6× (JA + JB)/2 ⇒ 3JA - JB ≧ 0

Constraints:

WA WB JA JB PA PBCWA CWB CJA CJB CPA CPB

1 1 ≦ 14,400,0001 1 ≦ 1,000,000

1 1 ≦ 2,000,000

1 1 1 ≦ 600,0001 1 1 ≦ 2,400,000

1 -3 ≧ 03 -1 ≧ 0

800,000 lbs.

Coefficients of Objective Function:

Both Cooper's and Myers' figures (Exhibits 2 and 3) are wrong.

Contribution = selling price - variable cost (excluding tomato cost)

Thus,

CWA = CWB = 1.48/18 = 0.0822

CJA = CJB = 1.32/20 = 0.066

CPA = CPB = 1.85/25 = 0.074

The contribution = $225,340 - $180,000 = $45,340.

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Optimal primal solution

WA WB JA JB PA PB

525,00 175,000 75,000 225,000 0 2×106

Optimal value = 225340

Optimal dual solution

Column 7 8 9 10 11 12 13

Constraint 1 2 3 4 5 6 7

Value 0 0 0.0161 0.0903 0.0579 8.1×10-3 8.1×10-3

Shadow price on constraint 4 = 0.0903

Sensitivity on cost values

variable 1 2 3 4 5 6

Lowerlimit

0.0606 0.0606 -0.0884 1.45333×10-2 -∞ 0.0579

Upperlimit

0.2336 0.5454 0.0876 0.803111×10-2 0.1064 +∞

Currentvalue 0.0822 0.0822 0.066 0.066 0.074 0.074

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constraint 1 2 3 4 5 6 7

Lower limit

700,000 300,000 1.45333×10-6 133,333 2.2×106 -600,000 -200,000

Upper limit +∞ +∞ 2.2×106 1.2×106 2.8×106 46,000 1.4×106

Current value

1.44×107 106 2×106 600,000 2.4×106 0 0

Sensitivity on the right-hand sides

Parametric analysis on constraint 4

Shodowprices 0.0903 0.0876 0.08493 0.0822 0.074 0.066 0

Lower limit

600,000 1,200,000 1,200,000 7,200,000 12,000,000 14,000,000 15,000,000

Upper limit

1,200,000 1,200,000 7,200,000 12,000,000 14,000,000 15,000,000 +∞

Current value 225,340 279,520 279,520 789,120 1,183,680 1,331,680 1,397,680

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Parametric Analysis on the Right-hand side of Constraint 4 (availability of grade A tomato)

const#4 600,000 + ,

600,000133,333 1,200,000 7,200,000 12,000,000

1,183,680

798,120

279,520

225,34008493.0

000,000,6600,509

000,200,1000,200,7)520,279120,789(

=

=

−−

=斜率

0903.0000,600

54180000,600000,200,1

)340,225520,279(

=

=

−−

=斜率

0822.0000,800,4

560,39400,200,7000,000,12)120,789680,183,1(

=

=

−−

=斜率

≤ θ ),0[ ∞∈θ

14,000,00015,000,000

斜率 = 0.074斜率 = 0

斜率 = 0.066

Solve the problem with 680,000 lbs of tomatoes. ( The same conclusion could be reached by inspecting the dual variable of the availability constraint of A-grade tomatoes (constraint 4) in the optimal solution. Since the dual variable $0.0903/lb. > $0.08/lb. And this value is constant for an additional 600,000 lbs. of grade A tomatoes, purchasing 80,000 lbs. will result i a net increase of the contribution.

Part (b)

Linear programming solution with 68,000lbs grade "A" tomatoes

Optimal primal solution

WA WB JA JB PA PB

615,000 205,000 65,000 195,000 0 2×106

Optimal value = 232564

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Net profit of 80,000lbs. A-grade

= (232564-22534)-80,000×0.085 = 7224-6800 = 424

Optimal dual solution

Column 7 8 9 10 11 12 13

Constraint 1 2 3 4 5 6 7

Value 0 0 0.0161 0.0903 0.0579 8×10-3 8×10-3

Part(c) Comparison of Results using Different Objective Coefficients

Myer’s objective function:

CWA = CWB = 1.48/18 = 0.0822, CJA = CJB = 1.32/20 = 0.066,

CPA = CPB = 1.85/25 = 0.074

Correct objective function:

CWA = CWB = 0.01, CJA = CJB = 0.08, CPA = CPB = 0.55

Cooper’s objective function:CWA = CWB = 0.12/18, CJA = CJB = 0.09/20, CPA = CPB = 0.12/25

Net Profit = CWAWA + CWBWB + CJAJA + CJBJB + CPAPA + CPB PB - $180,000

Net Profit = CWAWA + CWBWB + CJAJA + CJBJB + CPAPA + CPB PB

Net Profit = CWAWA + CWBWB + CJAJA + CJBJB + CPAPA + CPB PB

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Myers Cooper Correct

Whole0 800,000 lb 700,000 lb

WA 0 605,000 525,000WB 0 200,000 175,000

Juice1,000,000 0 300,000 lb

JA 250,000 0 75,000JB 750,000 0 225,000

Paste2,000,000 2,000,000 2,000,000

PA 350,000 0 0PB 1,650,000 2,000,000 2,000,000

Total 3,000,000 2,800,000 3,000,000Unused grade-B 0 200,000 lb 0 lb

Objective function ( O ) $48,000 $45,778 $225,340

Fruit cost ( F ) $180,000

Net profit ( O – F ) $45,340

Unallocated or un-covered tomatoes ( U ) $14,000 $12,000 0

O – F - U $34,000 $33,778 $45,340

問題一:如 whole tomato 只供應一大盤商, 售價依購買量而定, 如下之關係:

購買量 x (箱) 0 < x ≤ 100,000 100,000< x ≤ 600,000 600,000 < x ≤ 800,000

單價 $5.00/箱 $4.50/箱 $4.00/箱

100,000 600,000

3,550,000

2,750,000

800,000

500,000

購 買 量/箱

購 買 總 價

註: 例如購買量80,000箱時總價為 $5.00 × 80,000, 購買量700,000箱則總價為$5.00×100,000+$4.50×500,000+$4.00×100,000,

← Solved by using Separable Programming

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問題二:

(a) 以下三個條件至少有一個要成立, 重新建立此問題的數學模式。

(i) whole tomato生產量≥α箱,

(ii) tomato juice生產量≥β箱,

(iii) tomato paste 生產量≥γ箱。

← Solved by using Mixed Integer Programming

(b) 如“whole tomato生產量>α箱”, 則必須”tomato juice生產量≥β箱”與 “tomato paste生產量≥γ箱”, 重新建立此問題的數學模式。← Solved by using Mixed Integer Programming

問題三:

(a)如以市場佔有率考慮,Charles Myers (sale manager) 認

為 tomato paste 至少要生產 P箱(第一優先), tomato

juice 正好是 J箱(第二優先), 總售金額不低於 I ($)(第三

優先), ← Goal Programming

(b) 如何應用preemptive goal programming 建立此問題之數

學模式, 並解釋如何應用Linear Programming 的package

解此問題。← Goal Programming

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問題四:(6-a) 在Red Brand Canner案例中, 如番茄之購買價格依A, B二

種等級而異, A, B,級價格分別為 a, b (cents/lb.), 生產剩餘之

番茄價值為0, 農場番茄最大供應量為 3,000,000 lbs. (20%為

A級), 但可不必全部採購, 在其他條件不變的情況下, 重新

建立此問題的數學模式。← Linear Programming

(6-b) 在Red Brand Canner案例中, 生產剩餘之番茄不論等級可

轉賣價格為 d (cents/lb.), 重新建立此問題的數學模式。

← Linear Programming

問題五:Red Brand Canner案例中, 每一種罐頭須要經過 X, Y 二種機器加工, X 機器(第一階段處理, 如清洗, 燒煮, 攪拌等), Y 機器(第二階段處理, 如包裝, 裝箱等), 依Dan Tucker, production manager 分析人力與機器設備的需求如下:

每箱之人力需求 X 機器之需求 Y 機器之需求

whole tomatoes a1 hours/case a2 hours/case a3 hours/case

tomato juice b1 hours/case b2 hours/case b3 hours/case

tomato paste c1 hours/case c2 hours/case c3 hours/case

由於同一季節有多種罐頭同時生產, 機器設備與人力都有限, 今工廠分配之人力與 X, Y機器可用工時分別為 α, β, γ小時, 其他條件不變的情況下, 重新建立此問題的數學模式。← Linear Programming

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(4-3) 以上模式應用 Linear Programming package 求解得到以下資訊, 今農場可再提供 1,300,000 lbs. 之A級番茄, 單價是 8.5cent/lb., 您的決定是否再購買? 再購買多少lbs? 總共可增加多少利潤? (每一項答案都要說明理由)

WA WB JA JB PA PB

525,000 175,000 75,000 225,000 0 2×106

Optimal value = 225340

Optimal dual solution

Column 7 8 9 10 11 12 13

Constraint 1 2 3 4 5 6 7

Value 0 0 0.0161 0.0903 0.0579 8.1×10-3 8.1×10-3

“Shadow price on constraint 4” = 0.0903

← Linear Programming, sensitivity analysis, parametric analysis

Sensitivity on cost values

variable 1 2 3 4 5 6Lower limit 0.0606 0.0606 -0.0884 1.45333×10-2 -∞ 0.0579Upper limit 0.2336 0.5454 0.0876 0.803111×10-2 0.1064 +∞

Current value 0.0822 0.0822 0.066 0.066 0.074 0.074