Is it possible to have the inside grid surrounded where the outer grid contents flows around the inside grid as what happens in an image float left/right.
---------------------------|
| |
| grid_8 |
| |
|------------| |
| | |
| grid_4 | |
| | |
|------------| |
| |
| |
|--------------------------|
Thank you
try this
http://jsfiddle.net/qCJEF/2/
CSS
.grid_8{
width:960px;
background-color:red;
border:#FFF 1px dashed;
padding:95px 0 110px;
vertical-align:middle;
}
.grid_4{
width:480px;
background-color:blue;
border:#FFF dashed 1px;
padding:100px 0;
margin:100px 0 0 0;
}
Related
I have dataframe like this:
+---+--------------------------------------+-----------+
| | envelopeid | message |
+---+--------------------------------------+-----------+
| 1 | d55edb65-dc77-41d0-bb53-43cf01376a04 | CMN.00002 |
| 2 | d55edb65-dc77-41d0-bb53-43cf01376a04 | CMN.00004 |
| 3 | d55edb65-dc77-41d0-bb53-43cf01376a04 | CMN.11001 |
| 4 | 5cb72b9c-adb8-4e1c-9296-db2080cb3b6d | CMN.00002 |
| 5 | 5cb72b9c-adb8-4e1c-9296-db2080cb3b6d | CMN.00001 |
| 6 | f4260b99-6579-4607-bfae-f601cc13ff0c | CMN.00202 |
| 7 | 8f673ae3-0293-4aca-ad6b-572f138515e6 | CMN.00002 |
| 8 | fee98470-aa8f-4ec5-8bcd-1683f85727c2 | TKP.00001 |
| 9 | 88926399-3697-4e15-8d25-6cb37a1d250e | CMN.00002 |
| 10| 88926399-3697-4e15-8d25-6cb37a1d250e | CMN.00004 |
+---+--------------------------------------+-----------+
I've grouped it with grouped = df.groupby('envelopeid')
And I need to remove all groups from the dataframe and stay only that groups that have messages (CMN.00002) or (CMN.00002 and CMN.00004) only.
Desired dataframe:
+---+--------------------------------------+-----------+
| | envelopeid | message |
+---+--------------------------------------+-----------+
| 7 | 8f673ae3-0293-4aca-ad6b-572f138515e6 | CMN.00002 |
| 9 | 88926399-3697-4e15-8d25-6cb37a1d250e | CMN.00002 |
| 10| 88926399-3697-4e15-8d25-6cb37a1d250e | CMN.00004 |
+---+--------------------------------------+-----------+
tried
(grouped.message.transform(lambda x: x.eq('CMN.00001').any() or (x.eq('CMN.00002').any() and x.ne('CMN.00002' or 'CMN.00004').any()) or x.ne('CMN.00002').all()))
but it is not working properly
Try:
grouped = df.loc[df['message'].isin(['CMN.00002', 'CMN.00002', 'CMN.00004'])].groupby('envelopeid')
Try this: df[df.message== 'CMN.00002']
outdf = df.groupby('envelopeid').filter(lambda x: tuple(x.message)== ('CMN.00002',) or tuple(x.message)== ('CMN.00002','CMN.00004'))
So i figured it up.
resulting dataframe will got only groups that have only CMN.00002 message or CMN.00002 and CMN.00004. This is what I need.
I used filter instead of transform.
I need to analyze Weekly order frequencies over last 1 year period to find out what is the min/max/average frequencies of orders for each product.
whether it is new or old,system should calculate the first occurrence of the order in the year as the starting week of the order. Min order frequency is difference between successive ordering weeks. If the first order is in wk 3 and the second order is in wk6, implies the order frequency is 3 weeks (=>6-3). Orders can be at any week in the past 52 weeks. Average order frequency = (52 - First order week) / no of weeks that have orders.
Attaching the excel for better understanding the issue.
Original image
+---------+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+----------------+-------------------------+-----+-----------------------------------+--+
| Product | wk1 | wk2 | wk3 | wk4 | wk5 | wk6 | wk7 | wk8 | wk9 | wk10 | wk11 | wk12 | wk13 | wk14 | wk15 | wk16 | wk17 | wk18 | wk19 | wk20 | wk21 | wk22 | wk23 | wk24 | wk25 | wk26 | wk27 | wk28 | wk29 | wk30 | wk31 | wk32 | wk33 | wk34 | wk35 | wk36 | wk37 | wk38 | wk39 | wk40 | wk41 | wk42 | wk43 | wk44 | wk45 | wk46 | wk47 | wk48 | wk49 | wk50 | wk51 | wk52 | Order start wk | Order frequency (Weeks) | | | |
+---------+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+----------------+-------------------------+-----+-----------------------------------+--+
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Min | Max | Average | |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | (End wk - Start week)/No of times | |
| SKU 1 | | | | | | | | | y | | y | | y | | y | | y | | y | | y | | y | y | | | y | | y | | y | | y | | | | | | y | | y | | y | | y | | y | | y | | y | | 9 | 1 | 6 | 2.15 | |
| SKU 2 | | | | | | | y | | | | | | y | | | | | | y | | | | | | y | | | | | | y | | | | | | y | | | | | | y | | | | | | y | | | | 1 | 0 | 0 | 7.29 | |
| SKU 3 | | | | | | | | | | | | | | | y | | | | | | | | | | | | | | | | y | | | | | | | | y | | | | | | | | y | | | | | | 15 | 8 | 15 | 9.25 | |
+---------+-----+-----+-----+-----+-----+-----+-----+-----+-----+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+------+----------------+-------------------------+-----+-----------------------------------+--+
So as mentioned #Barry Houdini solves the problem of finding the longest sequence of zeroes separated by ones elegantly here
You only have to change it slightly to check for repeated blank cells separated by 'y'. The only thing is that you don't want to include cells before the first 'y', and (although this isn't clear) may not want to include blank cells after the last 'y'.
The formula for MIN becomes
=MIN(IF((ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))>1)*(ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))<COUNTA(B4:BA4)+1),FREQUENCY(IF(B4:BA4="",COLUMN(B4:BA4)),IF(B4:BA4="y",COLUMN(B4:BA4)))))+1
and the formula for MAX becomes (the same)
=MAX(IF((ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))>1)*(ROW(A$1:INDEX(A:A,COUNTA(B4:BA4)+1))<COUNTA(B4:BA4)+1),FREQUENCY(IF(B4:BA4="",COLUMN(B4:BA4)),IF(B4:BA4="y",COLUMN(B4:BA4)))))+1
where you need to add 1 to make the results agree with the question because #Barry's formula counts numbers of blanks but OP wants interval between two successive y's. An array of ny+1 elements is generated where ny is the number of y's. This is because the FREQUENCY function returns an array with n+1 elements where n is the number of cut points (bins_array in documentation and because the column numbers of cells containing y are used as cut points so there are ny of them.
These are both array formulas and need to be entered with CtrlShiftEnter
The formula for the average is just
=(COLUMNS(B4:BA4)-MATCH("y",B4:BA4,0))/COUNTA(B4:BA4)
Oil Blending
An oil company produces three brands of oil: Regular, Multigrade, and
Supreme. Each brand of oil is composed of one or more of four crude stocks, each having a different lubrication index. The relevant data concerning the crude stocks are as follows.
+-------------+-------------------+------------------+--------------------------+
| Crude Stock | Lubrication Index | Cost (€/barrell) | Supply per day (barrels) |
+-------------+-------------------+------------------+--------------------------+
| 1 | 20 | 7,10 | 1000 |
+-------------+-------------------+------------------+--------------------------+
| 2 | 40 | 8,50 | 1100 |
+-------------+-------------------+------------------+--------------------------+
| 3 | 30 | 7,70 | 1200 |
+-------------+-------------------+------------------+--------------------------+
| 4 | 55 | 9,00 | 1100 |
+-------------+-------------------+------------------+--------------------------+
Each brand of oil must meet a minimum standard for a lubrication index, and each brand
thus sells at a different price. The relevant data concerning the three brands of oil are as
follows.
+------------+---------------------------+---------------+--------------+
| Brand | Minimum Lubrication index | Selling price | Daily demand |
+------------+---------------------------+---------------+--------------+
| Regular | 25 | 8,50 | 2000 |
+------------+---------------------------+---------------+--------------+
| Multigrade | 35 | 9,00 | 1500 |
+------------+---------------------------+---------------+--------------+
| Supreme | 50 | 10,00 | 750 |
+------------+---------------------------+---------------+--------------+
Determine an optimal output plan for a single day, assuming that production can be either
sold or else stored at negligible cost.
The daily demand figures are subject to alternative interpretations. Investigate the
following:
(a) The daily demands represent potential sales. In other words, the model should contain demand ceilings (upper limits). What is the optimal profit?
(b) The daily demands are strict obligations. In other words, the model should contain demand constraints that are met precisely. What is the optimal profit?
(c) The daily demands represent minimum sales commitments, but all output can be sold. In other words, the model should permit production to exceed the daily commitments. What is the optimal profit?
QUESTION
I've been able to construct the following model in Excel and solve it via OpenSolver, but I'm only able to integrate the mix for the Regular Oil.
I'm trying to work my way through the book Optimization Modeling with Spreadsheets by Kenneth R. Baker but I'm stuck with this exercise. While I could transfer the logic from another blending problem I'm not sure how to construct the model for multiple blendings at once.
I modeled the problem as a minimization problem on the cost of the different crude stocks. Using the Lubrication Index data I built the constraint for the R-Lub Index as a linear constraint. So far the answer seems to be right for the Regular Oil. However using this approach I've no idea how to include even the second Multigrade Oil.
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| Decision Variables | | | | | | | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| | C1 | C2 | C3 | C4 | | | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| Inputs | 1000 | 0 | 1000 | 0 | | | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| | | | | | | | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| Objective Function | | | | | | Total | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| Cost | 7,10 € | 8,50 € | 7,70 € | 9,00 € | | 14.800,00 € | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| | | | | | | | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| Constraints | | | | | | LHS | | RHS |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| C1 supply | 1 | | | | | 1000 | <= | 1000 |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| C2 supply | | 1 | | | | 0 | <= | 1100 |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| C3 supply | | | 1 | | | 1000 | <= | 1200 |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| C4 supply | | | | 1 | | 0 | <= | 1100 |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| R- Lub Index | -5 | 15 | 5 | 30 | | 0 | >= | 0 |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| R- Output | 1 | 1 | 1 | 1 | | 2000 | = | 2000 |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| | | | | | | | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| Blending Data | | | | | | | | |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
| R- Lub | 20 | 40 | 30 | 55 | | 25 | >= | 25 |
+--------------------+--------+--------+--------+--------+--+-------------+----+------+
Here is the model with Excel formulars:
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| Decision Variables | | | | | | | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| | C1 | C2 | C3 | C4 | | | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| Inputs | 1000 | 0 | 1000 | 0 | | | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| | | | | | | | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| Objective Function | | | | | | Total | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| Cost | 7,1 | 8,5 | 7,7 | 9 | | =SUMMENPRODUKT(B5:E5;B8:E8) | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| | | | | | | | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| Constraints | | | | | | LHS | | RHS |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| C1 supply | 1 | | | | | =SUMMENPRODUKT($B$5:$E$5;B11:E11) | <= | 1000 |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| C2 supply | | 1 | | | | =SUMMENPRODUKT($B$5:$E$5;B12:E12) | <= | 1100 |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| C3 supply | | | 1 | | | =SUMMENPRODUKT($B$5:$E$5;B13:E13) | <= | 1200 |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| C4 supply | | | | 1 | | =SUMMENPRODUKT($B$5:$E$5;B14:E14) | <= | 1100 |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| R- Lub Index | -5 | 15 | 5 | 30 | | =SUMMENPRODUKT($B$5:$E$5;B15:E15) | >= | 0 |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| R- Output | 1 | 1 | 1 | 1 | | =SUMMENPRODUKT($B$5:$E$5;B16:E16) | = | 2000 |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| | | | | | | | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| Blending Data | | | | | | | | |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
| R- Lub | 20 | 40 | 30 | 55 | | =SUMMENPRODUKT($B$5:$E$5;B19:E19)/SUMME($B$5:$E$5) | >= | 25 |
+--------------------+------+-----+------+----+--+----------------------------------------------------+----+------+
A nudge in the right direction would be a tremendous help.
I think you want your objective to be Profit, which I would define as the sum of sales value - sum of cost.
To include all blends, develop calculations for Volume produced, Lube Index, Cost, and Value for each blend. Apply constraints for volume of stock used, volume produced, and lube index, and optimize for Profit.
I put together the model as follows ...
Columns A through D is the information you provided.
The 10's in G2:J5 are seed values for the stock volumes used in each blend. Solver will manipulate these.
Column K contains the total product volume produced. These will be constrained in different ways, as per your investigation (a), (b), and (c). It is =SUM(G3:J3) filled down.
Column L is the Lube Index for the product. As you noted, it is a linear blend - this is typically not true for blending problems. These values will be constrained in Solver. It is {=SUMPRODUCT(G3:J3,TRANSPOSE($B$2:$B$5))/$K3} filled down. Note that it is a Control-Shift-Enter (CSE) formula, required because of the TRANSPOSE.
Column M is the cost of the stock used to create the product. This is used in the Profit calculation. It is {=SUMPRODUCT(G3:J3,TRANSPOSE($C$2:$C$5))}, filled down. This is also a CSE formula.
Column N is the value of the product produced. This is used in the Profit calculation. It is =K3*C8 filled down.
Row 7 is the total stock volume used to generate all blends. These values will be constrained in Solver. It is =SUM(G3:G5), filled to the right.
The profit calculation is =SUM(N3:N5)-SUM(M3:M5).
Below is a snap of the Solver dialog box ...
It does the following ...
The objective is to maximize profit.
It will do this by manipulating the amount of stock that goes into each blend.
The first four constraints ($G$7 through $J$7) ensure the amount of stock available is not violated.
The next three constraints ($K$3 through $K$5) are for case (a) - make no more than product than there is demand.
The last three constraints ($L$3 through $L$5) make sure the lube index meets the minimum specification.
Not shown - I selected options for GRG Nonlinear and selected "Use Multistart" and deselected "Require Bounds on Variables".
Below is the result for case (a) ...
For case (b), change the constraints on Column K to be "=" instead of "<=". Below is the result ...
For case (c), change the constraints on Column K to be ">=". Below is the result ...
I think I came up with a solution, but I'm unsure if this is correct.
| Decision Variables | | | | | | | | | | | | | | | | |
|--------------------|---------|--------|--------|--------|-------------|--------|--------|--------|--------|--------|--------|--------|---|--------------------------------|----|------|
| | C1R | C1M | C1S | C2R | C2M | C2S | C3R | C3M | C3S | C4R | C4M | C4S | | | | |
| Inputs | 1000 | 0 | 0 | 800 | 0 | 300 | 0 | 1200 | 0 | 200 | 300 | 600 | | | | |
| | | | | | | | | | | | | | | | | |
| Objective Function | | | | | | | | | | | | | | Total Profit (Selling - Cost) | | |
| Cost | 7,10 € | 7,10 € | 7,10 € | 8,50 € | 8,50 € | 8,50 € | 7,70 € | 7,70 € | 7,70 € | 9,00 € | 9,00 € | 9,00 € | | 3.910,00 € | | |
| | | | | | | | | | | | | | | | | |
| Constraints | | | | | | | | | | | | | | LHS | | RHS |
| Regular | -5 | | | 15 | | | 5 | | | 30 | | | | 13000 | >= | 0 |
| Multi | | -15 | | | 5 | | | -5 | | | 20 | | | 0 | >= | 0 |
| Supreme | | | -30 | | | -10 | | | -20 | | | 5 | | 0 | >= | 0 |
| C1 Supply | 1 | 1 | 1 | | | | | | | | | | | 1000 | <= | 1000 |
| C2 Supply | | | | 1 | 1 | 1 | | | | | | | | 1100 | <= | 1100 |
| C3 Supply | | | | | | | 1 | 1 | 1 | | | | | 1200 | <= | 1200 |
| C4 Supply | | | | | | | | | | 1 | 1 | 1 | | 1100 | <= | 1100 |
| Regular Demand | 1 | | | 1 | | | 1 | | | 1 | | | | 2000 | >= | 2000 |
| Multi Demand | | 1 | | | 1 | | | 1 | | | 1 | | | 1500 | >= | 1500 |
| Supreme Demand | | | 1 | | | 1 | | | 1 | | | 1 | | 900 | >= | 750 |
| | | | | | | | | | | | | | | | | |
| | | | | | | | | | | | | | | | | |
| Selling | | | | | | | | | | | | | | | | |
| Regular | 8,50 € | x | 2000 | = | 17.000,00 € | | | | | | | | | | | |
| Multi | 9,00 € | x | 1500 | = | 13.500,00 € | | | | | | | | | | | |
| Supreme | 10,00 € | x | 900 | = | 9.000,00 € | | | | | | | | | | | |
| | | | | | 39.500,00 € | | | | | | | | | | | |
In the below example entity_id and pos are headers and others are values for them.
| Entity_id | pos |
| 16580 | code | count |
| | 056 | 2 |
| | 101 | 4 |
| | 082 | 10 |
How can I create the above format using Spreadsheet::WriteExcel?
my $format = $workbook->add_format(
border => 6,
valign => 'vcenter',
align => 'center',
);
$worksheet->merge_range('A2:A5', '16580', $format);
I am trying to position an inner element inside the root element. The root element is set to be 100% high and 100% wide. However, I want the inner svg to be 10px or so from the edge of the outer svg--all except the right edge, which I would like to be 200px from the outer right edge. I want this to display in a web browser and so the user should be able to resize the browser and hence the outer svg while the inner svg should be able to maintain the correct distance from the outer svg on each side.
Like this:
+OUTER SVG------------------------------------------+
| |
| +INNER SVG----------------+ |
| | | |
| | | 200px |
| | | <---+-----+----> |
| | | |
| | | | 1
| | | | 0
| | | | 0
| | | | %
| | | |
| | | | h
| | | | i
| +-------------------------+ | g
| | h
| |
+---------------------------------------------------+
100% Wide
Is it possible to do this with SVG alone?
If I understand you correctly, this is trivial. SVG is built to have many visual elements in a Z stack, aligned, overlapped, whatever.
Use Inkscape, free and opensource, to create SVGs, and then you can edit the file directly in a text editor if you want to see how SVG is formed. In Inkscape you can use the alignment tools, the grid, the transform-scaling tool, drag and drop, and probably some other tools to accomplish this.