• Author / Uploaded
• Riva Choerul Fatihin

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OBJECTIVE Using the information provided in the case, 1.

Develop a high level process map for the Akshaya Patra Foundation using Suppliers Inputs

Process Outputs and Customers (SIPOC). What insights can you draw from the high level process map? 2. Given that VK Hill kitchen needs to serve 87,045 students with 7 tonnes of cooked rice and 16,800 liters of Sambar per day, what is the total time required for cooking Sambar and rice? 3. If the food has to be delivered by 12.00 noon at the farthest school from the Vasanthapura kitchen (which takes 4 hours to reach), what are the lower and upper specification limits for completion of the first batch of Sambar? If the cooking starts at 5.00 am, calculate the corresponding Sigma Score (process capability) assuming that the standard deviation of cooking Sambar is 10 minutes 4. For the upper and lower limits calculated in question 3 for Sambar cooking, calculate the optimal start time for Sambar cooking that will maximize the Sigma score. ANSWER 1. Supplier Raw food suppliers which cooperate with PPP scheme

Input

Raw materials:

Process

Output Mid-day Meal Package for childrens and students

Customer School students and childrens

- Rice Packaging and supporting materials suppliers

- Dhal and pulses - Vegetables and fruits

Cooking utensil dan supporting equipments suppliers Gas company

- Milk and milk products - Groceries - Spices

Electricity company

Cooking utensil and supporting equipments

Transportation services company

Gas, electricity, and water

Water services company

Packaging materials Transport / delivery vehicles

Employees Trained Employee

The Akshaya Praya Foundation stakeholders

2. Cooking Rice process : 30 minutes , Can produces 0.875 ton of rice

Sambar Cooking takes 143 minutes, can produce 8400 litres, if 16800 means need 143x2=286 minutes or 4.8 hours, so total cooking time = Rice cooking + sambar cooking = 0.5 hours + 4.8 hours = 5.3 hours or 318 minutes

3. Lower limit means Delivery come in time (12.00) to most farest school = 4 hours or 240 minutes, Uper Limit means time that need to keep food 60 degree celcius = 6 hours or 360 minutes , If σ = 10 minutes, Cp (process capability)? Cp=

(𝑈𝑆𝐿−𝐿𝑆𝐿) 6𝜎

Cp=

360−240 60

=120/60 = 2

4.

To find optimal time we need happen when Cp min=Cp max

USL-X/3σ = X-LSL/3σ X = (LSL + USL)/2 = (360+240)/2 = 300 minutes for process average or 5 hours

Must keep 60 degree before noon 12.00 is 6 hours, means must finish cook at 6 am. So optimal start time to cook is 6 am – 5 hours = 1 am