SCMD: Assessment 3

Case 1: Waste Recycling Network Optimization

This is a multi-objective linear programming problem that requires balancing transportation cost minimization with recycling maximization.

Decision Variables:

Let xijxij​ = kilotonnes transported from sector ii to recycling site jj
Where: i=1,2,...,10i=1,2,...,10 (sectors), j=1,2,...,5j=1,2,...,5 (recycling sites)

Mathematical Formulation:

Objective Function:
Minimize: Z=2×(−Recycling Amount)+1×(Transportation Cost)Z=2×(−Recycling Amount)+1×(Transportation Cost)

More precisely:

Minimize Z=2×(−∑i=110∑j=15Efficiencyj×xij)+1×(65.90×∑i=110∑j=15Distanceij×xij)Minimize Z=2×(−i=1∑10​j=1∑5​Efficiencyj​×xij​)+1×(65.90×i=1∑10​j=1∑5​Distanceij​×xij​)

Subject to:

1. Supply constraints (all garbage from each sector must be collected):

∑j=15xij=Demandifor i=1,...,10j=1∑5​xij​=Demandi​for i=1,...,10

1. Capacity constraints (sites cannot exceed capacity):

∑i=110xij≤Capacityjfor j=1,...,5i=1∑10​xij​≤Capacityj​for j=1,...,5

1. Non-negativity:

xij≥0for all i,jxij​≥0for all i,j

Recommended Solution Approach:

Given the 50 decision variables and 15 constraints, this requires linear programming software (Excel Solver, Python's PuLP, or specialized OR software).

Key Insights:

• Site 5 has the highest efficiency (60%) but is far from most sectors
• Site 4 has good efficiency (43%) but limited capacity (37 kt)
• Sites near sectors (e.g., Site 1 for Sectors 1-5) have lower transport costs but lower efficiency

The optimal solution will prioritize sending garbage to high-efficiency sites despite higher transport costs, since recycling maximization is weighted twice as important.

Case 2: Brown Onion Inventory Management

Given Data Summary:

• Annual demand D=41,250D=41,250 kg
• Unit cost C=$1.42/kgC=$1.42/kg
• Holding cost rate h=23%h=23%
• Ordering cost S=$48.30S=$48.30
• In-transit holding rate it=11%it​=11%
• Freight rate f=$0.08/kgf=$0.08/kg
• Freight-related charges F=$82/shipmentF=$82/shipment
• Lead time L=1+3=4L=1+3=4 days
• Working days/year = 364

Calculations:

a. Economic Order Quantity (EOQ):

Q∗=2DShC=2×41,250×48.300.23×1.42=3,984,7500.3266=12,202,426≈3,493 kgQ∗=hC2DS​​=0.23×1.422×41,250×48.30​​=0.32663,984,750​​=12,202,426​≈3,493 kg

b. Total Purchasing Cost:

=D×C=41,250×1.42=$58,575=D×C=41,250×1.42=$58,575

c. Total Ordering Cost:

=DQ∗×S=41,2503,493×48.30≈11.81×48.30≈$570.42=Q∗D​×S=3,49341,250​×48.30≈11.81×48.30≈$570.42

d. Total Inventory Holding Cost:

=Q∗2×h×C=3,4932×0.23×1.42≈1,746.5×0.3266≈$570.42=2Q∗​×h×C=23,493​×0.23×1.42≈1,746.5×0.3266≈$570.42

e. Total Transportation Cost (by weight):

=D×f=41,250×0.08=$3,300=D×f=41,250×0.08=$3,300

f. Total Freight-related Cost (by shipment):

=DQ∗×F=11.81×82≈$968.42=Q∗D​×F=11.81×82≈$968.42

g. Total In-transit Holding Cost:

=D×C×it×Transit TimeWorking Days=58,575×0.11×3364≈6,443.25×0.00824≈$53.09=D×C×it​×Working DaysTransit Time​=58,575×0.11×3643​≈6,443.25×0.00824≈$53.09

h. Total Cost for Inventory Plan:

=58,575+570.42+570.42+3,300+968.42+53.09=$64,037.35=58,575+570.42+570.42+3,300+968.42+53.09=$64,037.35

i. Number of Orders:

=DQ∗=41,2503,493≈11.81 orders/year=Q∗D​=3,49341,250​≈11.81 orders/year

j. Ordering Point (Reorder Point):
Daily demand = 41,250/364≈113.3241,250/364≈113.32 kg/day

ROP=Daily Demand×Lead Time=113.32×4≈453.3 kgROP=Daily Demand×Lead Time=113.32×4≈453.3 kg

k. Profit from Inventory Plan:
Revenue = 41,250×3.85=$158,812.5041,250×3.85=$158,812.50

Profit=158,812.50−64,037.35=$94,775.15Profit=158,812.50−64,037.35=$94,775.15

Case 3: Printer Selection Using AHP

Step 1: Normalize Pairwise Matrices

Criteria Comparison Matrix (3×3):

[131/31/311/2321]​11/33​312​1/31/21​​

Sum of columns: [4.333, 6, 1.833]

Normalized matrix:

[0.2310.5000.1820.0770.1670.2730.6920.3330.545]​0.2310.0770.692​0.5000.1670.333​0.1820.2730.545​​

Criteria weights (row averages):

• Price: (0.231+0.500+0.182)/3=0.304(0.231+0.500+0.182)/3=0.304
• Ink Cost: (0.077+0.167+0.273)/3=0.172(0.077+0.167+0.273)/3=0.172
• Printing Quality: (0.692+0.333+0.545)/3=0.524(0.692+0.333+0.545)/3=0.524

Step 2: Normalize Alternative Matrices

Price Comparison (Epson, Canon, HP):
Matrix:

[10.50.333210.76931.31]​123​0.511.3​0.3330.7691​​

Column sums: [6, 2.8, 2.102]
Normalized:

[0.1670.1790.1580.3330.3570.3660.5000.4640.476]​0.1670.3330.500​0.1790.3570.464​0.1580.3660.476​​

Weights: E=0.168, C=0.352, H=0.480

Ink Cost Comparison:
Matrix:

[10.250.33341230.51]​143​0.2510.5​0.33321​​

Column sums: [8, 1.75, 3.333]
Normalized:

[0.1250.1430.1000.5000.5710.6000.3750.2860.300]​0.1250.5000.375​0.1430.5710.286​0.1000.6000.300​​

Weights: E=0.123, C=0.557, H=0.320

Printing Quality Comparison:
Matrix:

[1340.333120.250.51]​10.3330.25​310.5​421​​

Column sums: [1.583, 4.5, 7]
Normalized:

[0.6320.6670.5710.2110.2220.2860.1580.1110.143]​0.6320.2110.158​0.6670.2220.111​0.5710.2860.143​​

Weights: E=0.623, C=0.240, H=0.137

Step 3: Final Weighted Scores

Printer

Price (0.304)

Ink Cost (0.172)

Quality (0.524)

Total Score

Epson

0.168 × 0.304 = 0.0511

0.123 × 0.172 = 0.0212

0.623 × 0.524 = 0.3265

0.3988

Canon

0.352 × 0.304 = 0.1070

0.557 × 0.172 = 0.0958

0.240 × 0.524 = 0.1258

0.3286

HP

0.480 × 0.304 = 0.1459

0.320 × 0.172 = 0.0550

0.137 × 0.524 = 0.0718

0.2727

Recommendation: Epson EcoTank Photo ET-8500 (score: 0.3988)

Case 4: EV Car Selection Using Weighted Sum Model

Normalization and Scoring:

For minimization criteria (Price, Servicing, Charge Time):

Normalized Score=Min ValueActual ValueNormalized Score=Actual ValueMin Value​

For maximization criteria (Power, Drive, Range, Boot):

Normalized Score=Actual ValueMax ValueNormalized Score=Max ValueActual Value​

Calculations:

Criteria

Weight

Polestar 4

Kia EV6

BYD Sealion 7

Tesla Model 3

Price (Min)

7

87,903/87,903=1.000

81,289/87,903=0.925

69,255/87,903=0.788

75,890/87,903=0.863

Servicing (Min)

3

750/750=1.000

1,382/750=0.543

945/750=0.794

1,125/750=0.667

Charge Time (Min)

4

30/18=0.600

18/18=1.000

450/18=0.040

31/18=0.581

Power (Max)

5

200/390=0.513

168/390=0.431

390/390=1.000

366/390=0.938

Drive (Max)

4

2/4=0.500

2/4=0.500

4/4=1.000

4/4=1.000

Range (Max)

6

620/629=0.986

528/629=0.839

456/629=0.725

629/629=1.000

Boot Space (Max)

5

526/594=0.885

490/594=0.825

500/594=0.842

594/594=1.000

Weighted Scores:

• Polestar 4:
  (7×1.000)+(3×1.000)+(4×0.600)+(5×0.513)+(4×0.500)+(6×0.986)+(5×0.885)(7×1.000)+(3×1.000)+(4×0.600)+(5×0.513)+(4×0.500)+(6×0.986)+(5×0.885)
  =7.000+3.000+2.400+2.565+2.000+5.916+4.425=27.306=7.000+3.000+2.400+2.565+2.000+5.916+4.425=27.306
• Kia EV6:
  (7×0.925)+(3×0.543)+(4×1.000)+(5×0.431)+(4×0.500)+(6×0.839)+(5×0.825)(7×0.925)+(3×0.543)+(4×1.000)+(5×0.431)+(4×0.500)+(6×0.839)+(5×0.825)
  =6.475+1.629+4.000+2.155+2.000+5.034+4.125=25.418=6.475+1.629+4.000+2.155+2.000+5.034+4.125=25.418
• BYD Sealion 7:
  (7×0.788)+(3×0.794)+(4×0.040)+(5×1.000)+(4×1.000)+(6×0.725)+(5×0.842)(7×0.788)+(3×0.794)+(4×0.040)+(5×1.000)+(4×1.000)+(6×0.725)+(5×0.842)
  =5.516+2.382+0.160+5.000+4.000+4.350+4.210=25.618=5.516+2.382+0.160+5.000+4.000+4.350+4.210=25.618
• Tesla Model 3:
  (7×0.863)+(3×0.667)+(4×0.581)+(5×0.938)+(4×1.000)+(6×1.000)+(5×1.000)(7×0.863)+(3×0.667)+(4×0.581)+(5×0.938)+(4×1.000)+(6×1.000)+(5×1.000)
  =6.041+2.001+2.324+4.690+4.000+6.000+5.000=30.056=6.041+2.001+2.324+4.690+4.000+6.000+5.000=30.056

Recommendation: Tesla Model 3 (score: 30.056)

Summary of Recommendations:

1. Case 1: Requires linear programming solution prioritizing high-efficiency recycling sites
2. Case 2: EOQ = 3,493 kg, Total Cost = $64,037.35, Profit = $94,775.15
3. Case 3: Select Epson EcoTank Photo ET-8500 (AHP score: 0.3988)
4. Case 4: Select Tesla Model 3 (weighted score: 30.056)

For Case 1, I recommend using optimization software to find the exact distribution xijxij​. For the other cases, all calculations are complete and ready for implementation.