How to calculate the longest process time?

Calculation of the longest processing time

Example: There is a 6/4/F/Fmax problem, and its processing time is shown in the following table. When processing in the order of S = (6, 1, 5, 2, 4, 3), find Fmax.

Solution: List the processing time matrix.

i 6 1 5 2 4 3

Pi 1 3 4 8 6 5 4

Pi2 1 3 7 5 9 3

Pi3 8 7 5 9 6 2

Pi4 3 5 2 4 6 9

According to the formula:

C kSi=max{C(k- 1)Si, C kSi- 1}+ P Sik, calculate the processing time of each line, and finally get the result Fmax=Cmsn.

Fmax=57

2. The optimal algorithm for two machines scheduling problem (Johnson algorithm)

Example: Find the optimal solution of 6/2/F/Fmax shown in the table below.

Arrange Workpiece 2 at position 1.

The workpiece 3 is arranged in the sixth position 2 3.

Arrange the workpiece 5 in the second position 2 5 3.

Arrange the workpiece 6 in the third place 2 5 6 3

Place the workpiece 4 in the fifth position 2 5 6 4 3.

The workpiece 1 is ranked fourth, 2 5 6 1 4 3.

I 2 5 6 1 4 3

ai 1 3 4 5 5 8

bi 2 7 4 7 4 2

Fmax =28 can be calculated from the above table.

3. Optimal algorithm for general n/m/F/Fmax problem.

(1) Palmar algorithm (λI = ∑ [k-(m+1)/2] p ik k =1,2, …, m arranges the workpieces in the order that λ i does not increase) Example: There is a 4/3/F/Fmax problem, and its processing.

Solution: λi= ∑ [k-(3+ 1)/2]P ik, k = 1, 2,3.

λi=-Pi 1+Pi3

Therefore, λ1=-p1+p13 =-1+4 = 3λ 2 =-p21+p23 = = 2+5 = 3λ 3.

Arrange the workpieces in the order that λi does not increase, and get the machining order (1, 2, 3, 4) and (2 1, 3, 4). After calculation, both of them are optimal sequences, Fmax=28 (2) key workpiece method.

Example: there is a 4/3/F/Fmax problem, and its processing time is shown in the following table, which is solved by the key workpiece method.

Solution: As can be seen from the above table, the longest processing time is No.3 workpiece, PI 1

Second, the calculation of production capacity (a), for processing and assembly production, production capacity is a vague concept. Mass production, single variety, can be expressed by the number of specific products; Mass production, few varieties, can be expressed by the number of representative products;

Multi-variety, small and medium-sized batch production can only be expressed by the output of counterfeit products. (2) Representative products

Suitable for: enterprises with similar product structure and technology and multi-variety production. Select the one with the largest total labor as the representative product, and use the output of the representative product to represent the production capacity. Conversion steps: ① Calculate the conversion coefficient of output.

produce

T t k I/ki: I product output conversion coefficient

Ti: I product hourly quota T generation: stands for product hourly quota.

② Convert I product output into representative product output.

Do I know? =→ generate

(3) imaginary products: imaginary products composed of various products according to their total labor ratio, which are suitable for enterprises with different product structures and processes and producing many varieties.

Conversion steps: ① Make various products into an imaginary product T-fake according to their product output ratio: the hourly quota of imaginary products t i :i the hourly quota of I products n i :i the specific planned annual output of I products N: the total annual output of various products.

② Conversion coefficient of I product

wrong

t t k i i /=

③i product output is converted into assumed product output.

Do I know? → = Error

(4) Example: There are four kinds of products: A, B, C and D. See the table for the planned annual output and hourly quota of each product. Now calculate the representative product and the hypothetical product.

Solution: 1 Calculation of representative products: As can be seen from the table, when product C is selected as the representative product, A: 50× 20/40 = 25 (units) B: 100× 30/40 = 75 (units) C: 125 (units) is calculated.

D: 25×80/40 = 50 (pieces) 2. Calculation of hypothetical products

First, calculate the machine man-hour quota of the hypothetical product:

t pj =(50×20+ 100×30+ 125×40+25×80)÷300

=36.67 (working hours at desk)

Then the planned output of each product is converted into the assumed product output A: 50× 20/36.67 = 27b:100× 30/36.67 = 82c:125× 40/36.67 =136.

()

I don't know

=? Suitable horse leave =

D:25×80/36.67 = 55

Third, determine the output and variety of MTS and MTO enterprises.

(1) MTS enterprises that produce goods on the spot (it's too difficult for teachers not to drop them ~ I can't afford to hurt them! )

Multimodal transport operator. Determination of varieties

Example: Orders for products A, B and C have been received. Processing time and available profit are shown in the following table. Ability to work for 40 time units. Which products are the most popular?

Solution: Heuristic algorithm can be used: sort by the value of (profit/processing time) from large to small, that is, give priority to the task with the largest profit per unit processing time, a: 10/ 12 = 0.83 (yuan/hour).

B: 13/8 = 1.63 (yuan/hour)

C: 25/25 = 1 (yuan/hour)

The available priority order is B-C-A C-A. Since the capacity hours are 40, choose B, the remaining capacity hours are 32, and then choose C. The remaining 7 are not enough to handle A, so only B and C can be selected. The result is a profit of 38.

Fourth, the inventory model (multi-cycle inventory basic model)

(A), the cost of inventory

(1) annual holding cost, expressed in c h, is to maintain inventory as the name implies. Including capital cost, depreciation of warehouse and equipment, taxes, insurance, obsolete losses, etc. This part of the cost is related to the value of the goods and the average inventory.

(2) The annual reorder cost, expressed by c r, is related to the number of orders that occurred in the whole year, and generally has nothing to do with the number of orders at one time.

(3) Annual procurement cost (processing cost), expressed in cost price. It is related to the price and the quantity ordered.

(4) Shortage cost. It reflects the loss of sales opportunities, reputation and the loss caused by the impact on production. It is related to the number and frequency of shortages.

If the annual total inventory cost is expressed by C T, the total inventory cost is C T = C H+C R+C P+C S.

(2) Economic order lot size model (EOQ for short) 1. The economic order lot size problem is discussed under the following assumptions:

1) The demand is a known constant, that is, the demand is uniform;

2) Out of stock is not allowed;

3) The order lead time is known and constant;

4) Delivery lead time is zero, that is, instantaneous delivery;

5) The product cost does not change with the batch (no quantity discount).

2. Inventory cost analysis

Total cost = annual storage cost C h++ annual order cost C r++ purchasing cost C P.

Q-the quantity of each order; (to be sought) H-annual storage cost per unit product (yuan/piece/year), where H =p ×h is the capital cost rate or storage cost rate (yuan/piece/year), and p is the unit price of the product, yuan/piece/d-annual demand;

S-the cost of each order (yuan/piece/year)

② Take the derivative of Q in the above formula and make the first derivative equal to zero, and you can get the best order quantity Q*

H-annual storage cost per unit product (yuan/piece/year) D-annual demand; S-the cost of each order (yuan/piece/year)

(3) in this case.

Order point R R=d ×LT

LT-order lead time d-demand rate per unit time

2. Example: Company A buys 8000 pieces of a product every year, and the unit price is 10 yuan. 30 yuan per order, annual interest of funds.

DSH

H DS D

S H DS

H C C R H 2222=

The rate is 12%, and the unit maintenance inventory fee is calculated as 18% of the value of all inventory goods. If the lead time of each order is 2 weeks, try to find out the economic order quantity, the lowest annual total cost, the number of annual orders and the order point.

Solution: p= 10 yuan/piece, D=8000 pieces/year, LT=2 weeks, H =10 *12%+/kloc-0 *18% = 3 yuan/piece.

Therefore, EOQ=

4003

*8000*22==H DS

(piece)

The minimum annual total cost is: t = p * d+(d/q) * s+(q/2) * h.

= 8000 *10+(8000/400) * 30+(400/2) * 3 = 81200 First year order times: n=D/EOQ=8000/400=20.

Order point: R=(D/52)*LT=8000/52*2=307.7 (pieces)

(3), economic production batch method

S-installation cost.

P ... Productivity (pieces/day) D ... Demand rate (pieces/day)

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5.9

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Calculation problems in production and operation management

First, the assembly line sequencing.

1. Calculation of the longest processing time