Supply Chain Analysis Case Study
VerifiedAdded on 2020/03/13
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Case Study
AI Summary
This case study focuses on reducing costs in a cafe by analyzing the supply chain and applying linear programming techniques. It outlines the nutritional requirements for a dish and aims to determine the optimal weight of potatoes and green beans to minimize costs while meeting these requirements. The study includes various reports and sensitivity analyses to evaluate the impact of changing nutritional values on overall costs.

Supply Chain Analysis & Design
Case Study - Reducing Cafe Costs (Linear Programming Problem)
STUDENT _ID
[Pick the date]
Case Study: Reducing Cafe Costs
Given Date
1. Prices
Prices of Potatoes $ 0.30 per pound∨$ 6.61× 10−4 per grams
Prices of Green beans $0.95 per pound∨$ 2.094 × 10−3 per grams
Case Study - Reducing Cafe Costs (Linear Programming Problem)
STUDENT _ID
[Pick the date]
Case Study: Reducing Cafe Costs
Given Date
1. Prices
Prices of Potatoes $ 0.30 per pound∨$ 6.61× 10−4 per grams
Prices of Green beans $0.95 per pound∨$ 2.094 × 10−3 per grams
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2. The respective nutritional requirements in dish
Dish must contain - Protein - 160 grams
Iron - 65 milligrams or 0.065 grams
Vitamin C - 1000 milligrams or 1.00 grams
3. Nutritional content of potato and green beans
Conversion∈grams : 10 oz=28.35 gram
4. Weight ratio (potato to green beans) = 6: 5
5. At least production of dish in café in a week = 12kg (12000 grams )
The main objective is to reduce the cost of café by deciding the optimum weight of two main
ingredient i.e. potato and green beans.
1
Dish must contain - Protein - 160 grams
Iron - 65 milligrams or 0.065 grams
Vitamin C - 1000 milligrams or 1.00 grams
3. Nutritional content of potato and green beans
Conversion∈grams : 10 oz=28.35 gram
4. Weight ratio (potato to green beans) = 6: 5
5. At least production of dish in café in a week = 12kg (12000 grams )
The main objective is to reduce the cost of café by deciding the optimum weight of two main
ingredient i.e. potato and green beans.
1

Linear Programing Model
Assumption: x=weight of potato∧ y=amount of green beans
Objective function
min C= ( 6.61 ×10−4 )∗x+ ( 2.094 ×10−3 )∗y
Subject to constraints
0.014 x +0.0208 y ≥160
( 0.255∗10−5 ) x + ( 1.25∗10−5 ) y ≥ 0.065
( 11∗10−5 ) x+ ( 1.049∗10−4 ) y ≥ 1
x + y ≥ 12000
5 x−6 y ≥0
x , y ≥ 0 (Non- negativity)
(a) Amount of potatoes and green beans that Jane should purchase each week for the dish
casserole in order minimize the ingredient costs.
Solver output
2
Assumption: x=weight of potato∧ y=amount of green beans
Objective function
min C= ( 6.61 ×10−4 )∗x+ ( 2.094 ×10−3 )∗y
Subject to constraints
0.014 x +0.0208 y ≥160
( 0.255∗10−5 ) x + ( 1.25∗10−5 ) y ≥ 0.065
( 11∗10−5 ) x+ ( 1.049∗10−4 ) y ≥ 1
x + y ≥ 12000
5 x−6 y ≥0
x , y ≥ 0 (Non- negativity)
(a) Amount of potatoes and green beans that Jane should purchase each week for the dish
casserole in order minimize the ingredient costs.
Solver output
2
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Answer Report
Sensivity Report
3
Sensivity Report
3
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x ( Amount of potatoes )=8542.7 g
y ( Amount of green beans )=3457.28 g
C ( min )=$ 12.89
(b) Nutrition requirement has been changed: Iron - 100 mg∨0.1 gram and Vitamin C
−500 mg∨0.5 gram
( 11∗10−5 ) x1 + ( 1.049∗10−4 ) x2 ≥ 0.5
( 0.255∗10−5 ) x1+ ( 1.25∗10−5 ) x2 ≥0.1
Solver output
4
y ( Amount of green beans )=3457.28 g
C ( min )=$ 12.89
(b) Nutrition requirement has been changed: Iron - 100 mg∨0.1 gram and Vitamin C
−500 mg∨0.5 gram
( 11∗10−5 ) x1 + ( 1.049∗10−4 ) x2 ≥ 0.5
( 0.255∗10−5 ) x1+ ( 1.25∗10−5 ) x2 ≥0.1
Solver output
4

Answer Report
Sensitivity Report
5
Sensitivity Report
5
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x ( Amount of potatoes ) =7712.08 g∧ y ( Amount of green beans)=6426.73 g
C ( min )=$ 18.56
(c) Green beans are selling at higher price = $ 1.20 per lb (2.64∗10−3 per g)
New cost function
min C= ( 6.61 ×10−4 ) x+ ( 2.64∗10−3 ) y
Solver output
6
C ( min )=$ 18.56
(c) Green beans are selling at higher price = $ 1.20 per lb (2.64∗10−3 per g)
New cost function
min C= ( 6.61 ×10−4 ) x+ ( 2.64∗10−3 ) y
Solver output
6
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Answer Report
Sensivity Report
7
Sensivity Report
7

x ( Amount of potatoes ) =8542.7 g∧ y ( Amount of green beans ) =3457.2 g
C ( min ) =$ 14.77
(d) Amount of potatoes and green beans when
Price of green beans $ 1.20 per lb∨2.64∗10−3 per g
& Iron – 100 mg∨0.1 grams
¿ Vitamin C−500 mg∨0.5 grams
The problem of LPP become
min C= ( 6.61 ×10−4 ) x1 + ( 2.64∗10−3 ) x2
( 0.255∗10−5 ) x1+ ( 1.25∗10−5 ) x2 ≥0.1
( 11∗10−5 ) x1 + ( 1.049∗10−4 ) x2 ≥ 0.5
8
C ( min ) =$ 14.77
(d) Amount of potatoes and green beans when
Price of green beans $ 1.20 per lb∨2.64∗10−3 per g
& Iron – 100 mg∨0.1 grams
¿ Vitamin C−500 mg∨0.5 grams
The problem of LPP become
min C= ( 6.61 ×10−4 ) x1 + ( 2.64∗10−3 ) x2
( 0.255∗10−5 ) x1+ ( 1.25∗10−5 ) x2 ≥0.1
( 11∗10−5 ) x1 + ( 1.049∗10−4 ) x2 ≥ 0.5
8
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Answer Report
Sensitivity Report
9
Sensitivity Report
9
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x ( Amount of potatoes )=7712.08 g∧ y ( Amount of green)=6426.73 g
C ( min )=$ 22.06
(e) According to part (d), it can be concluded that nutrient Iron would create significant
effect on the cost function. By doubling the value of this nutrient (iron), the cost would
enhance by a good number and therefore, it would affect the cost of casserole the most
as compared with the remaining two nutrients.
10
C ( min )=$ 22.06
(e) According to part (d), it can be concluded that nutrient Iron would create significant
effect on the cost function. By doubling the value of this nutrient (iron), the cost would
enhance by a good number and therefore, it would affect the cost of casserole the most
as compared with the remaining two nutrients.
10

11
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