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Open Access 05-04-2020 | Production Management

# Optimization of delivery adherence based on capacity planning and bid pricing

Authors: B. Denkena, M.-A. Dittrich, S. Wilmsmeier, S. Stamm

Published in: Production Engineering | Issue 3/2020

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## Abstract

Sales revenues of enterprises are often subject to seasonal fluctuation. This leads to high or low utilized resources and this in turn to revenue losses. Hence, the enterprises invest a high effort to improve long and short-term resource utilization. In this context, disregarding future capacity utilization within the process of quotation leads to short-timed capacity adjustments for instance, additional work hours across seasons. This paper presents an approach which focuses on dependencies between costs and capacity by linking cost pricing and production scheduling. A first evaluation at an MTO supplier shows that order delays can be reduced by up to 95% and total costs by 21% compared to using the most appropriate priority rule.
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## 1 Introduction

Machining production is characterized by the workshop principle. Thus, it is required to handle individual products, small lot sizes and different resulting processing sequences of goods and services. Typical enterprises are make-to-order (MTO) manufacturers which are often small and medium sized enterprises (SME). These are often subject to seasonally fluctuating sales revenues, which is why a high degree of operational flexibility is required when responding to short-term peaks in capacity utilization. Building up sufficient capacity, e.g. employees, machinery, to complete orders on time is not always appropriate, as capacity utilization throughout the year does not justify the associated investments. Even approaches to adaptive production control, e.g. adaptive process planning and control by Denkena et al. [1], which increase flexibility and the ability to react to disturbances in an optimized way, are not suitable to compensate for these seasonal fluctuations in capacity.
In the machining industry, SME have high expectations regarding long-term and short-term production levelling due to demand fluctuation while maintaining an even production utilization and cost controlling. The costs for bid prices are calculated based on the cost of a standard shift. However, if the product is, for example, produced in the night shift, the previous cost calculation is no longer valid. Nonetheless, the additional costs occur as a result and they are not taken into account in prior cost calculating. Alternatively, penalty charges have to be paid for delayed deliveries [2, 3]. This leads to the necessity of a cost calculating system, which regards the utilization of resources. The combination of cost calculation and production planning makes it possible to price in necessary capacity increases, e.g. overtime, shifts, at short notice through an adjusted offer price or to persuade the customer to choose a later delivery date at lower costs.
Classical approaches for production planning and control (PPC) do not link the individual disciplines of production planning and sales. The approaches are mainly influenced by theoretical views on PPC in a closed loop control [2] to avoid capacity adjustment measures. These approaches are partly extended by capacity adjustment measures like additional work hours or shift expansion depending on their adjustment periods. Chehade acknowledged that the costs of these adjustment measures are of significant interest [4]. Therefore, this paper presents an approach which enables SME to link capacity utilization, cost calculation and bid pricing. For this purpose, the approach is firstly developed based on the current state of the art, then applied to a real data set and evaluated on a case study.

## 2 State of the art

Linking costs and capacity is a recognized problem in several areas. However, it has not yet been not fully addressed. Slotnick provided a brief evaluation of order acceptance and scheduling in 13 different methodical views on this topic. The approaches were analyzed based on the number of considered machines, pre-emption, objective and pricing [3]. Only three of the papers took pricing into account in scheduling [57]. The other authors subsequently evaluate the planning based on the resulting production costs. Slotnick concluded that future work should address the interaction between disciplines such as order acceptance, pricing and capacity loading [3].
Charnsirisakskul [5] invented a simultaneous pricing regarding holding costs and tardiness penalties. The goal is to evaluate the capabilities of multi customer pricing instead of single customer pricing. Furthermore, it is pointed out that price flexibility leads to better results than lead time flexibility if there is no inventory flexibility for the manufacturer. Pourbabai [6, 7] regards a scheduling algorithm based on linear programming that supports the decision in order acceptance or rejection. The goal is to calculate the optimal size of production batches by splitting. Costs are taken into account by fixed order prices.
Carvalho et al. [8] compared approaches for capacity adjustment measures to minimize production costs or to maximize profit from the production point of view. They classified the approaches according to their empirical nature and concluded that most of the authors did not include real-world problems [8]. Real-world means that the authors did not develop and validate their approaches based on real data sets. Instead, they used simulation-based data that do not allow direct conclusions regarding practical applicability. In contrast, Alfieri et al. [9, 10] only concentrated on production planning and not on costs. The approach of Carvalho et al. [8] took into account the variable costs for processing and additional work hours, the capacity change costs, personnel costs and subcontracting to minimize the costs by using linear programming. Fixed costs were not included. Further, Ho took into account uncertain demand of products by implementing the density function of product demand into the cost price by using a nonlinear model optimization [11].
Other approaches in PPC consider capacity adjustments measures and costs at an early stage of the production process. Methods like workload control (WLC), order review and release (ORR) and bottleneck load-oriented release (BLOR) are main targets [12]. Kingsman, for example, referred to an “analytical model of dynamic capacity planning at the stage of customer inquiry” [13] by using flexible capacity of single workstations. In addition, Kingsman explained the planning problem of bidding and order acceptance. He also emphasized the importance of the delivery date and the price in this early stage of the production process. Nevertheless, only the delivery dates are considered in scheduling, because enterprises try to keep the price as low as possible [13]. Kingsman also developed an “overall decision support system for dealing with inquiries” [14]. It is divided into three modules: estimation module, capacity planning module and marketing module. The estimation of bid prices is mainly influenced by a database of past prices and won bids.
Other studies focus on investment strategies for capacity planning [15]. Some approaches already consider a cost-capacity-ratio as activity-based costing and time-driven activity-based costing. These are basic models that use costs per capacity units (machine hour rate) and the completion time of tasks as the production time [16, 17].
Wouters developed a cost calculation system that meets the requirements of a particular SME. He also identified the difficulties in allocating labour costs to a specific product because multi-machine processing is often required. The author asserted that the determination of cost changes in real-time adjustments should be the focus of future work [17]. Ou considered adjustment costs in cost calculating. In the case of capacity adjustments, these costs occur if the capacity differs from the previous period [18]. This depends on the labour cost system and the machine hour rate calculation.
Besides PPC, revenue management (RM) also offers approaches for linking capacity and costs. The basic idea related to that is to break down a standardized product portfolio into partial capacity units, e.g. services like seats in airplanes. A major application requirement in RM is heterogeneous customer behavior [19]. The management process of cost and capacity is based on pricing mechanism, which increases the price while the current demand exceeds the expected demand. In contrast, the price decreases if the current demand underruns the expected demand [20]. Thus, the opportunity-related costs (OC) as a control principle of capacity within the framework of RM represents a central approach for decision support regarding the acceptance or rejection of an order [20]. The feasibility of linking costs and capacity based on RM approaches in the area of production is not generally given. Customer behavior, for example, shows different price sensitivities [21], which means that there is not always a willingness to pay higher prices for short-term order processing or lower prices for later delivery. In addition, the rejection of orders is not considered.
In conclusion, several approaches, which deal with issues of linking capacity and costs mainly in linear programming, are found in literature. However, these approaches only focus on separate topics, such as scheduling, order release or costs calculation, and do not match the insights of these topics. There is no superior approach, which takes into consideration real-world problems, e.g. acceptance of delays in contrast to rejection, and grown cost-structures of SME. If the costs are taken into consideration, they are calculated globally for the entire capacity utilization in production. For the existing models in RM, not all the applications prerequisites for an effective use in machining production are met. Nevertheless, there is the potential to transfer the basic idea of the RM, the price formation of capacity utilization, to MTO production. In particular, it has to be analyzed if the definition of OC within the framework of RM can be transferred to production as the production costs of short-term capacity adjustment measures are not considered in RM. However, the additional wage costs may be regarded as OC because they compensate losses. Nevertheless, the wage costs cannot be taken into account without considering cause-related costs for capacity adjustments. These additional costs vary depending on the orders causing them. To overcome this issue, approaches should focus more on addressing this relation. In real-world MTO, production forward scheduling is often used to maintain a high utilization. Contrary to this, a backward scheduling ensures a just-in-time production but causes gaps in schedules and hence economic losses. As a result, it is necessary to take into account additional wage costs during scheduling to derive minimal wage costs due to customer price sensitivity.

## 3 Approach for dynamic bid pricing under consideration of adjustment costs

The present approach includes five modules depicted in Fig. 1. The modules of customer classification, forecast and order process are described in detail in a previous publication [22]. Hence, in this paper only a short overview of these modules is given. The customers are first categorized into A, B and C classes (Classification model) by using pareto analysis on economic aspects. The goal is to ensure the confirmed date of delivery for prioritized “A” customer. Furthermore, a forecast of the future capacity utilization is calculated depending on these customer classes to check if the approximated time of production is within a highly utilized period. The forecast is based on an auto regressive moving average (ARIMA) algorithm taking due dates of past production orders into account. In underutilized periods, the orders are planned without capacity restrictions. Otherwise, the schedule model including the new offered order is used to determine the necessary capacity adjustment measures. Seasonal fluctuations are considered by activating or blocking the capacity cells depending on the required demand. The capacity cells are released according to the forecast. If a class contingent has already been reached, no further cells are released. By scheduling the orders, possible capacity adjustments can be defined and the resulting costs will be added to the quote of the corresponding order according to the causation principle. These additional costs are calculated as additional wage costs caused by capacity adjustments during scheduling on the basis of the approach for dynamic bid pricing in Denkena et al. [22]. Furthermore, an assignment of fixed and variable costs per planning period must be guaranteed and implemented in a dynamic machine hour rate.
The forecast is used to determine a date of delivery in a period of low capacity utilization, which takes place in the order processing, to offer customers an alternative date of delivery. The background is the predicted capacity of the respective customer class. This is taken into account in the schedule and capacity model using dummy orders. When the orders are confirmed, the contingent (amount of dummy orders in hours) for the respective customer class is filled. If the contingent is reached, an alternative date of delivery is determined. This is intended to encourage customers to order in periods with low capacity utilization.

### 3.1 Schedule model

The interaction between the schedule and the cost model is illustrated in Fig. 2. Within the schedule model, the dates of delivery of all orders and whether they arrived in time or not are analyzed. The adjustment measures are systematically used to ensure that orders from A and B customers are always available on the delivery date. The achievement of minimal costs and securing of the delivery date is made possible by following three steps:
(1)
Scheduling according to time-based priority rules

(2)
Optimization by using capacity adjustment measures

(3)
Evaluation of the target values

#### 3.1.1 Scheduling

Scheduling deals with the allocation of new production orders, as shown in the upper part of Fig. 2. Input variables are information about the existing schedule of the production environment such as capacity units, capacity utilization situation and time allocation of already planned orders as well as data of the orders to be planned and cost parameters. Within the framework of scheduling, sequences of 5 + 1 priority rules are selected on the basis of their influence on the target figure’s minimum total delay time, average capacity utilization and minimization of economic losses. The considered priority rules are based on the individual time parameters of the work process (WP). The latest end time (LET) of an order corresponds to the delivery date at order level. The following formula is obtained for LET by mapping the time relationships:
$$LET= EST+D+BT$$
(1)
EST [–]: earliest start time, D [h]: duration, BT [h]: buffer time.
To consider the delivery date when prioritizing the operations, at least the three time parameters (LET, D and BT) of this equation must be taken into account. If the operations with the highest effort are preferred in scheduling, the following WP-specific priority rules result:
• Earliest LET
• Longest D
• Lowest BT
In addition to these three time criteria, the customer classes are also taken into account to give preference to the orders of the highest customer class when meeting delivery dates. A priority rule is used for the WP already planned to ensure minimum delay costs. However, no priority rule can be defined for the additional wage costs since they can only be taken into account during additional work hours planning. Thus, order-specific priority criteria based on logistical and economic objectives have been defined with this methodology:
• Highest customer class
The corresponding schedule is created according to the priority rules. If several schedules are equivalent, they are evaluated systematically according to the further rules until one schedule remains. The order in which the priority rules are used is not initially determined. Therefore, 120 priority rule orders are created by complete permutation:
$$P=n!=120$$
(2)
for
$$n=5$$
(3)
$$P$$ [–]: number of different schedule scenarios, n [–]: number of exchangeable priority rules.
If several orders are equivalent in all priority criteria of the five priority rules, the first in—first out (FIFO) rule is used instead. All resulting schedule alternatives are checked for possible delivery date deviations. If all schedules have deviations, they are managed in the optimization.

#### 3.1.2 Optimization

Capacity adjustments are considered depending on the delivery date deviations. First is the allocation of additional working hours. These are released as required and the 120 sequences are recalculated. Additional work planning is carried out systematically according to the schedule to maintain minimum additional wage costs. For this purpose, the amount of hours necessary for completion on time are released from the additionally required work shifts. The planning based on the Gantt chart is carried out by activating additional capacity units that have already been blocked. All subsequent operations are postponed accordingly. For each planned additional hour, the units of all capacity units must be unlocked to maintain the assignment sequence of the WP starting with the earliest delivery date that is not met.
If delivery date deviations exceed one operating day or one blocked working hour, the deviation is not eliminated. This remaining delay can only be minimized by planning further hours after the corresponding delivery date. In this case, the number of additional working hours corresponds to the number of unblocked capacity units between the delivery date and the current end of processing of the delayed order.
After allocation of additional working hours, all schedules are checked again for their adherence to delivery dates. The main objective of scheduling is to achieve the highest possible adherence to delivery dates according to customer classes. The priority rules cannot guarantee this requirement continuously due to the capacity and sequence conditions. Therefore, orders should be postponed in case of remaining delivery date deviations to minimize the delay of A- and subsequently B-class orders. Different shift scenarios result if there are several orders of lower classes available. These scenarios may have different degrees of fulfillment regarding the objective of optimization and are therefore all calculated and evaluated in a structured way. The number of possible shift scenarios can be calculated as a mathematical combination problem (combination without repetition):
$$N=\sum_{k=1}^{m}\begin{array}{c}m\\ k\end{array}$$
(4)
$$N$$ [–]: number of possible shift scenarios, $$m$$[–]: number of postponable orders.

#### 3.1.3 Evaluation

In all cases, several of the 120 schedules may produce a satisfactory result. This requires a suitable evaluation procedure to meet the target figure’s minimum total delay time, average capacity utilization and minimization of economic losses. The evaluation of the scheduled scenarios is based on the logistic and economic objectives and is carried out by systematic filtering of schedules, as shown in Fig. 3. The filter criteria were selected as follows. To retain as many customers as possible, the ability to deliver has the highest priority. In this context, first possible deviations of the delivery date from new orders should be minimized and evenly distributed among all planned orders. The delay in the orders already planned is considered in the second step. Thus, delay costs can be kept at a minimum. Higher customer classes are preferred in both evaluation steps. High capacity utilization and short lead times are achieved to fulfill low process costs and short idle times. These two goals are supported by a constant capacity requirement.

### 3.2 Cost model

The bid price is determined individually for each incoming order. It can be calculated by Eq. 5 without the consideration of additional costs. The margin Mi refers to the surcharge on the production cost price of the order and thus the common profit share in offer calculation. The amount of profit markup can be determined for each offered price individually according to customer, product and market situation. Within the scope of this work, it is set as constant for simplification.
$${O}_{P,i}={C}_{m,i}\cdot \left(1+{M}_{i}/100\%\right)$$
(5)
$${O}_{P,i}$$[€]: price of order i,$${C}_{m,i}$$ [€]: production costs of order i,$${M}_{i}$$ [%]: margin of order i.
For this, the production costs Cm,i of an order i are defined as shown in Eq. 68. Thereby multiple machine operation is disregarded.
$${C}_{m,i}= {M}_{c,i}+{P}_{c,i}$$
(6)
with
$${M}_{c,i}= {M}_{hr,i}\cdot {t}_{p,i}$$
(7)
$${P}_{c,i}={L}_{c,i}+{R}_{po,i}$$
(8)
$${M}_{c,i}$$ [€]: machine costs of order i,$${P}_{c,i}$$ [€]: personnel costs of order i,$${M}_{hr,i}$$ [€/h]: machine hour rate for order i,$${t}_{p,i}$$ [h]: time of production (process and set-up time) for order i,$${L}_{c,i}$$ [€]: production labor costs for order i,$${R}_{po,i}$$ [€]: remaining production overheads of order i.
According to conventional calculation, the product costs is directly determined by applying the preceding formula through the use of a uniform machine hour and personnel cost rate. To link pricing to capacity utilization, the calculation of the machine hour rate and labor costs must linked to the different scheduling scenarios coming from schedule model. This results in the following dependencies:
$${L}_{c,i}={l}_{c}\cdot ({t}_{ \mathrm{p},normal,i}+{t}_{ \mathrm{p},late,i}\cdot {R}_{late}+{ t}_{ \mathrm{p},night,i}\cdot {R}_{night}+{t}_{p,weekend,i}\cdot { R}_{weekend})$$
(9)
$${L}_{i}$$ [€]: production labor costs for order i,$${l}_{c}$$[€/h]: production wage cost rate, $${t}_{ \mathrm{p},normal}$$ [h]: working hours on standard shift, $${t}_{ \mathrm{p},late}$$ [h]: working hours on late shift, $${t}_{ \mathrm{p},night}$$ [h]: work hours on night shift, $${t}_{p,weekend}$$ [h]: work hours on weekend, $${R}_{late}$$ [–]: working hour rate factor for late shift, $${R}_{night}$$ [–]: working hour rate factor for night shift, $${R}_{weekend}$$ [–]: working hour rate factor for weekend.
$${M}_{hr,i}={M}_{hr,i}^{var}+\frac{\left(\left({CW}_{end}+1\right)-{CW}_{start}\right){\cdot M}_{dpo}}{{t}_{e,pre}+{t}_{p,i}+{t}_{e,past}}$$
(10)
$${M}_{hr,i}^{var}$$[€/h]: variable machine costs for order i,$${CW}_{end}$$[−]: calendar week at the end of production of order i,$${CW}_{start}$$ [−]: calendar week at the start of production of order i,$${M}_{dpo}$$ [€]: machine-dependent production overhead costs per planning period, $${t}_{e,pre}$$ [h]: sum of planned time of production of a machine till start of production of order i,$${t}_{e,past}$$ [h]: sum of planned time of production of a machine after introducing additional work hours to manufacture the order i.
The order-related delay costs must also be taken into account when calculating the planning scenario-specific price of an order. Therefore, additional work hours and the order-related delay costs are added as they occur. Accordingly, the new production costs are calculated as follows:
$${C}_{m,i}^{new}={C}_{m,i}+{C}_{a,i}$$
(11)
with
$${C}_{a,i}={\sum }_{k=1}^{o}{R}_{delay,k}\cdot {t}_{delay,k}\cdot {O}_{p,k}$$
(12)
$${C}_{m,i}^{new}$$ [€]: new production cost price of order i, $${C}_{a,i}$$ [€]: total additional costs due to order postponement caused by order i, o [–]: number of delayed orders, $${R}_{delay,k}$$ [1/h]: delay rate of order k,$${t}_{delay,k}$$ [h]: time of delayed delivery of order k,$${O}_{p,k}$$ [€]: price of order k.
The following applies to the new offer price of an order:
$${O}_{P,i}^{new}={C}_{m,i}^{new}\cdot \left(1+{M}_{i}/100\%\right)$$
(13)
$${O}_{P,i}^{new}$$ [€]: new offer price of order i.
Equation 12 guarantees that all additional costs are added to the order which caused them. Additional costs incurred on orders that have already been scheduled will be allocated evenly on to the new incoming orders.

## 4 Use case setup

The use case represents the production of a German SME and MTO supplier from the aerospace industry. The collected data consists of three to nine already scheduled orders and 9 to 27 incoming orders, in three consecutive planning periods. The capacity contingents of the customer classes analyzed by the customer model and further processed to the ARIMA forecast. The order description consists of order number, customer class and work plan. The order work plan includes three to five work processes accessing three different capacity units, as seen in Fig. 4. The respective predecessors, relationships and the previous planned production time are considered. For cost calculating, the labor costs are based on the bargaining table of the labor union for metalwork. The cost factor to calculate the machine hour rates and the penalty charges for delayed deliveries are based on the general terms and conditions of the customer. The Algorithm and graphic user interface (GUI) implementation relies on C# coding. The input parameters used are shown in Table 1.
Table 1
Input parameters for experimental evaluation
Input parameter
Default value
Rpo,i
0.75 Lci
Rlate
25%
Rnight
50%
Rweekend
75%
Mi
18%
lc
17 €/h
A validation is carried out by checking the procedure of the algorithm and the suitability of the software application. The followed points are checked:
• Generate schedules by changing the priority order
• Planning overtime
• Execute shift scenarios
• Calculate evaluation data
• Evaluate and filter flow charts
• Calculate offer prices
• Display planning results in GUI

### 4.1 Use case evaluation

The goal of the evaluation is to demonstrate the target achievement of minimizing A-class customers’ delays. In addition, the functionality will be tested in practice against the use of one simple priority rule, as it has taken place so far in the industrial practice of the MTO supplier. To give an insight to the respective behavior of the algorithm, the evaluation on 12 orders in one planning period is discussed. As shown in Fig. 5, the presented approach decreases the delay of A-class customer orders significantly, while maintaining lower total costs despite the additional costs. The use of customer class-dependent priority rules achieves a decrease of up to 67% for A-class and 36% for B-class customers, but at the disadvantage of C-class customer orders. In total, lead times (41%), delays (41%) and costs (5%) are reduced (cf. Fig. 6).
By taking additional work hours and the buffer times of each order into account, all order delays are reduced by 80%. At the same time, additional costs increase due to the use of overtime hours. However, the total costs decrease by another 17%. The reason for this is the machine hour rate, which depends on capacity utilization and therefore counteracts the additional wage costs.
The use of order postponement as the last step in the scheduling module implies a second decrease in additional costs, which in this case is also a decrease in total costs by 1% (cf. Fig. 6). The average lead time and total delay are again significantly reduced by at least 13%. This ensures a good balance between the customer requirements and at the same time the economic goals of the manufacturer.
By apportioning the cost to the caused orders, a profit maximization in this environment of up to 23% could be achieved, as shown in Table 2.
Table 2
Study results

Best choice of only one priority rule
Priority rule levels
Priority rule levels and overtime
Priority rule levels, overtime and order postponement
Total costs [€]
20,187.04
19,186.16
16,001.56
16,170.40
Total delay cost [€]
83.87
71.98
7.62
9.56
0.00
0.00
1,207.00
1,300.50
83.87
71.98
1,214.62
1,310.06
Profit maximization through add. costs apportionment [%]
3.25
3.17
22.17
23.57
The presented approach has different effects on enterprises’ key performance indicators. Depending on which indicator is recorded, the three optimization methods of the scheduling module are better suited in combination. Using the shown adjustment measures in combination has a positive effect on the lead time, the sum of delays and the profit maximization, as shown in Table 2 and Fig. 6. Scheduling without using order postponement, on the other hand, affects the level of total costs. For enterprises that only consider the additional wage costs or where very high additional wage costs are to be expected, sufficient planning can already be carried out with the proposed priority-based scheduling. The fulfilment of objectives can only be achieved in combination with established adaptation measures and the selected priority levels. A separate use does not lead to a sufficient target realization. Table 2 shows that even the best fitting priority rule does not reach the optimization potential of the new approach. Although the "Earliest LET" priority rule was applied most frequently, the other rules were also used regularly and achieved good results. A general exclusion from certain priority rules is therefore not possible and a prediction of the required priority sequence for defined planning steps cannot be made.

### 4.2 Discussion

The evaluation is conducted for several order combinations, as mentioned in Sect. 4. Thus, the algorithm shows stable behavior for varying amounts of orders within an equal distribution of customer classes. The basic findings remain the same, only the computing time increases. Regarding the industrial feasibility of the presented approach, the following simplifying assumptions must be considered. A possible rejection of the higher bid price by the customer was not taken into account. A consideration of rejections in the study would fluctuate the number of new orders. It can be assumed that the scheduling algorithm is not affected. Furthermore, the modelling of customer decisions in acceptance or rejection of bid prices was not subject to the project and requires further analysis. In addition, the transfer of additional costs to the bid price can lead to prices, which may not be competitive. In order to avoid this risk, a profit maximization limit could be set depending on the level of the cost price. If this limit is exceeded, a share of the additional costs must be covered by the company. Alternatively, this share could also be allocated to orders whose profit maximization limit has not yet been reached.
The forecast of capacity contingents on the basis of historical data is subject to a certain degree of uncertainty, which is not considered in this approach. Thus, this approach requires the implementation of a factor to take into account the probability of a wrong or imprecise forecast.

## 5 Conclusion and outlook

Due to the increasing individualization of products and fluctuations in demand, the task of production scheduling requires a particularly high level of effort in MTO production. Thus, it offers a high degree of optimization. Despite the risk of economic losses, there is a lack of approaches that are able to control fluctuations in demand and provide planning results to meet economic targets.
Within the scope of this work, a software-supported solution was developed with the aim of achieving the highest possible delivery capability and link capacity adjustment measures with bid pricing. The evaluation and thus the selection of the suitable schedule takes place according to the set target values. Additional work hours can be planned according to the capacity requirements. It is also possible to postpone low-level orders (C-class customers) to achieve a higher delivery reliability for the highest customer class. A pricing mechanism was developed for quotation calculation, which takes into account the costs for capacity utilization and capacity adjustments measures.
The application study of the algorithm showed that the minimization of possible schedule variances while maintaining high capacity utilization is fulfilled by the combination of all measures at the latest. Systematic additional work hours planning ensures minimum additional costs. For the application case, all order delays could be reduced by up to 95% and total costs by 21% compared to using the most appropriate priority rule. This emphasizes the need of taking into account costs in scheduling in an early stage of the process of bid pricing.
Splitting the problem into individual modules reduces the complexity of the mathematical problem, but requires a higher degree of organization between the modules. Therefore, it is necessary to determine defined standardized interfaces for an industrial application of the entire method, as described in Sect. 3.
Up to now, the evaluation has only been performed on real data of an MTO supplier regarding a defined planning period, which probably does not contain all possible order combinations. For a more detailed analysis of the presented method, further research in a simulation study including robustness tests are necessary. In addition, a more detailed investigation of the bidding process is necessary to estimate to what extent a rejection of higher bid prices by the customer takes place.

## Acknowledgement

Open Access funding provided by Projekt DEAL. The authors thank the Federal Ministry of Education and Research of Germany (BMBF) for its financial and organizational support of the project “Dynamic capacity planning and control in manufacturing SME" (project “KaPro", 02P15K563).

## Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Kingsman B, Hendry L, Mercer A, Souza A (1996) Responding to customer enquiries in make-to-order companies problems and solutions. Int J Prod Econ 46–47:219–231. https://doi.org/10.1016/0925-5273(95)00199-9 CrossRef
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Ceryan O, Koren Y (2009) Manufacturing capacity planning strategies. CIRP Ann Manuf Technol 58:403–406. https://doi.org/10.1016/j.cirp.2009.03.034 CrossRef
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Santana A, Afonso P, Zanin A, Wernke R (2017) Costing models for capacity optimization in Industry 4.0: trade-off between used capacity and operational efficiency. Procedia Manuf 13:1183–1190. https://doi.org/10.1016/j.promfg.2017.09.193 CrossRef
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Wouters M, Stecher J (2018) Development of real-time product cost measurement: a case study in a medium-sized manufacturing company. Int J Prod Econ 183(1):235–244. https://doi.org/10.1016/j.ijpe.2016.10.018 CrossRef
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Ou J, Feng J (2019) Production lot-sizing with dynamic capacity adjustment. Eur J Oper Res 272:261–269. https://doi.org/10.1016/j.ejor.2018.06.030
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Renna P (2015) Dynamic pricing of excess capacity in production networks by fuzzy logic. Int J Comput Integr Manuf 29(6):611–621 CrossRef
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Talluri KT, Ryzin GJ van, Karaesmen IZ, Vulcano GJ (2008) Revenue management: models and methods. Proceedings of 2008 Winter Simulation Conference, pp145–156. 10.1109/WSC.2008.4736064
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Rehkopf S (2006) Revenue Management-Konzepte zur Auftragsannahme bei kundenindividueller Produktion. Dissertation, Technische Universität Braunschweig
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Denkena B, Dittrich MA, Stamm S (2018) Dynamic bid pricing for an optimized resource utilization in small and medium sized enterprises. Procedia CIRP 67:516–521. https://doi.org/10.1016/j.procir.2017.12.254 CrossRef
Title
Optimization of delivery adherence based on capacity planning and bid pricing
Authors
B. Denkena
M.-A. Dittrich
S. Wilmsmeier
S. Stamm
Publication date
05-04-2020
Publisher
Springer Berlin Heidelberg
Published in
Production Engineering / Issue 3/2020
Print ISSN: 0944-6524
Electronic ISSN: 1863-7353
DOI
https://doi.org/10.1007/s11740-020-00959-8

Go to the issue


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# SA1-page3

SA1-page3  contains the remaing questions of KVS Mathematics SA - 1

29) P and Q are respectively points on the AB and AC of $$\Delta ABC$$. If AP =2 cm, PB=4cm, AQ= 3cm and QC =6 cm, prove that BC = 3 PQ.

30) The median of the following data is 52.5 . find the value x and y, if total frequency = 100.

31) Draw less than cumulative frequency curve and hence find the median .

32) Prove that $$\frac{1 + Cos \theta - Sin^{2} \theta}{Sin \theta (1+Cos \theta )} = Cot \theta$$.

33) Evaluate $$\frac{Sin^{2}54^{\circ} -Cos^{2}36^{\circ}}{Cosec^{2}57^{\circ} - Tan^{2} 33^{\circ}} +2 Sin^{2}38^{\circ}Sec^{2}52^{\circ} -Sin^{2}45^{\circ}$$.

34) Prove that $$(Sin \theta + Cosec \theta )^{2} + (Cos \theta + Sec \theta)^{2} =7+ Tan^{2}\theta + Cot^{2} \theta$$.

SA1-page3  completes the total question paper of KVS HYD SA 1

For solution of this question paper  click on next link below.

format: plain


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# Counting measure (measure theory)

The counting measure is a special measure in mathematics that assigns the number of elements to quantities . Formally, the counting measure can be defined in a measuring space , with any set and its power set . If there is a finite set , then a finite measure arises . It is a σ-finite measure if and only if it is countable . ${\ displaystyle (\ Omega, {\ mathfrak {P}} (\ Omega))}$${\ displaystyle \ Omega}$${\ displaystyle {\ mathfrak {P}} (\ Omega)}$${\ displaystyle \ Omega}$${\ displaystyle \ Omega}$

## definition

The count of a quantity is defined as follows: ${\ displaystyle A \ subseteq \ Omega}$

${\ displaystyle \ mu (A) = {\ begin {cases} \ vert A \ vert & {\ text {if}} A {\ text {is finite,}} \\ + \ infty & {\ text {, if}} A {\ text {is infinite.}} \ end {cases}}}$

## Examples

Integral of the function on the interval with respect to the count over${\ displaystyle x \ mapsto x ^ {2}}$${\ displaystyle [-10.10]}$${\ displaystyle \ mathbb {N}}$

Above the natural numbers , i.e. the measuring space , the counting measure corresponds to the figure ${\ displaystyle (\ mathbb {N}, {\ mathfrak {P}} (\ mathbb {N}))}$

${\ displaystyle \ mu \ colon {\ mathfrak {P}} (\ mathbb {N}) \ to [0, \ infty] {\ text {,}} A \ mapsto \ sum _ {k \ in \ mathbb {N }} \ chi _ {A} (k).}$

Here denotes the characteristic function of the set . ${\ displaystyle \ chi _ {A}}$${\ displaystyle A \ subseteq \ mathbb {N}}$

With the help of the counting measure , every finite sum or infinite, absolutely convergent series can be represented as a Lebesgue integral . In particular, the following applies to each figure : ${\ displaystyle \ mathbb {N}}$${\ displaystyle f \ colon \ mathbb {N} \ to \ mathbb {R}}$

${\ displaystyle \ sum _ {k = 1} ^ {\ infty} f (k)}$ converges absolute can be integrated with respect to the counting measure${\ displaystyle \ Longleftrightarrow}$ ${\ displaystyle f}$${\ displaystyle {\ mathfrak {P}} (\ mathbb {N}).}$

In this case

${\ displaystyle \ int _ {\ mathbb {N}} f \, \ mathrm {d} \ mu = \ sum _ {k = 1} ^ {\ infty} f (k)}$.


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# Homework Help: Finding the inverse and finding a matrix * A = 0 matrix

1. Sep 29, 2005

### mr_coffee

Hello everyone!! Matrices fun! Anyways,
I'm trying to find a matrix that is equal to the idenity matrix if u multiply 2 matrices together. Well that matrix is the inverse So i'm trying to find the inverse but the answer is wrong:
A =
-5 3
2 -9

So i found the determinant and i switched the a and d, and negated the c and b . isn't htat the inverse?

1/35-6 = 1/29

-9 -3
-2 -5

-9/29 -3/29
-2/29 -5/29

isn't thtat the inverse of A?

I also have B =
-1 6
5 -30

I need to multiply that by some matrix C so the resultant matrix is
0 0
0 0
but i can't just say C is equal to
0 0
0 0
any ideas? thanjks!

I did notice, Row 1 is just 1/5 row 2

2. Sep 29, 2005

### TD

You have to switch a and d, also switch b and c and then change the signs of these last 2. After that, you have to divide by the determinant. So:

$$A = \left( {\begin{array}{*{20}c} a & b \\ c & d \\ \end{array} } \right) \Rightarrow A^{ - 1} = \frac{1} {{\det A}}\operatorname{adj} A = \frac{1} {{ad - bc}}\left( {\begin{array}{*{20}c} d & { - c} \\ { - b} & a \\ \end{array} } \right)$$

If you don't "see" it right away, work out the following matrix product to get a system of lineair equations.

$$\left( {\begin{array}{*{20}c} { - 1} & 6 \\ 5 & { - 30} \\ \end{array} } \right) \cdot \left( {\begin{array}{*{20}c} a & b \\ c & d \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} 0 & 0 \\ 0 & 0 \\ \end{array} } \right)$$

3. Sep 29, 2005

### Hurkyl

Staff Emeritus
Multiplying a matrix by something to get zero strongly reminds me of the notion of a null space...

4. Sep 29, 2005

### mr_coffee

Thanks TD, but for some reason its still wrong, i got:
-9/29 -2/29
-3/29 -5/29

5. Sep 29, 2005

### TD

Check your determinant again, that is ad-bc

6. Sep 29, 2005

### mr_coffee

lol what the f, (-5)(-9) - (3)(2)
1/45-6
1/39 right?

7. Sep 29, 2005

### TD

Well, the determinant is 39 (so not 29). Then, you have to divide by it indeed.
So 1/39 * adj(A)


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Given sin x/ x^2 how do you find the limit as x approaches 0?

Jul 9, 2016

You should use the fact that $\lim \sin \frac{x}{x} = 1$ as $x$ approaches 0

Explanation:

You can modify the expression as follows:

$\sin \frac{x}{x} ^ 2 = \sin \frac{x}{x} \cdot \frac{1}{x}$, and then taking the limits:

$\lim \sin \frac{x}{x} = 1$, and

$\lim \frac{1}{x} = \infty$, so the end result is:

$\lim \sin \frac{x}{x} ^ 2 = \lim \sin \frac{x}{x} \cdot \lim \frac{1}{x} = 1 \cdot \infty = \infty$


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# Complex inner product aren't inner products.

An inner product $\langle \cdot,\cdot \rangle$ satisfies the following properties:

Let $u$, $v$, and $w$ be vectors and $\alpha$ be a scalar, then:

1. $\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$ and $\langle \alpha v,w\rangle=\alpha\langle v,w\rangle$ (Linearity in first coordinate).

2. $\langle v,w\rangle=\langle w,v\rangle$ (Symmetry) .

3. $\langle v,v\rangle\ge 0$ and $\langle v,v\rangle=0$ $\Leftrightarrow$ $v=0$ (Positive definiteness).

This the general definition of inner products, but in complex vector spaces the symmetry part is $\langle u,v\rangle=\overline{\langle v,u\rangle}$, how can be possible? this is supposed to be the same symmetry of the general properties above, btw what's the difinition of $\overline{\langle u,v\rangle}$?

EDIT

I know how to prove $\langle u,v\rangle=\overline{\langle v,u\rangle}$, what I wanna know is why this is called an inner product, because we don't have $\langle u,v\rangle={\langle v,u\rangle}$

• It's the complex conjugate. And if you restrict $x\mapsto \overline{x}$ to the reals, you get the identity. – xavierm02 Apr 15 '14 at 17:04
• Note that in the complex case you can't have both symmetry and positive definiteness. It turns out that positive definiteness is the more useful property - by taking the square root you can give a complex number a length - and this generalises. But to get a positive definite form you need to introduce the complex conjugate. – Mark Bennet Apr 15 '14 at 18:18

There is no way to preserve all of these 3 properties in complex field. We have to choose the least important one to sacrifice.

Linearity is the foundation of linear algebra, and positive definiteness is required to make inner product a metric. Therefore, we abandon symmetry.

• but what's the general definition of inner products? – user42912 Apr 15 '14 at 18:17
• An inner product space is a normed vector space satisfying the parallelogram law. – jdh8 Apr 15 '14 at 18:23

Since you are talking about the complex inner product (not a normal real inner product), you cannot apply the real inner product properties. If you check the property for complex inner product (the Hermitian inner product), you can find the property you are talking about. In fact, if you apply real numbers on this Hermitian inner product, you get the same result as apply them on the real inner product.

You can have a look at this document.

http://www2.math.umd.edu/~hking/Hermitian.pdf

• Thank you for the link, but it doesn't answer my question, I know how to prove $\langle u,v\rangle=\overline{\langle v,u\rangle}$, my question is the requirement to be an inner product: this is not an inner product because we don't have $\langle u,v\rangle={\langle v,u\rangle}$. – user42912 Apr 15 '14 at 17:57
• @user42912 Since you are talking about the complex inner product (not a normal real inner product), you cannot apply the real inner product properties. If you check the property for complex inner product (the Hermitian inner product), you can find the property you are talking about. In fact, if you apply real numbers on this Hermitian inner product, you get the same result as apply them on the real inner product. – Yilun Zhang Apr 15 '14 at 18:04
• what's the general definition of inner products? – user42912 Apr 15 '14 at 18:07
• @user42912 for real inner product, the properties are what you stated in your question. – Yilun Zhang Apr 15 '14 at 18:08
• @user42912 It really depends on whether or not your scalar field is the complex numbers or the real numbers! We define the inner product differently for these two. However, notice that the more general condition $\langle u, v \rangle = \overline{\langle v, u \rangle}$ simplifies to the $\langle u, v \rangle = \langle v, u \rangle$ if $\langle v, u \rangle$ is real. – user98602 Apr 15 '14 at 18:11


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# Probability of winning/losing a lonely card game when you shout an integer progressively each time you draw

Take a standard face-down 52-deck of cards, including 13 ranks of four suits (for simplicity, assign value 1 to 13 from ace to King). Draw the cards one by one, shouting the integers from 1 to 13 (and then starting again) in the natural order (i.e. {1,2,3,4...}) at each draw. For example, you shout 'one' and draw a 7, then you shout 'two' and draw an ace, then you shout 'three' and draw a King, and so on and so forth. You lose instantly if the number you've shouted coincide with the rank of the card you've just drawn. What is the probability of winning/losing the game (whichever is easier to compute)?

**Additional information: The problem is related to the first-encounter probability of two walks (one deterministic and one random): a deterministic walk jumping to the right by one lattice site at each time-step, and resetting after 13 steps, and another walk randomly jumping among the same thirteen sites (with the constraint that no site can be visited more than four times).

This is the so-called "Frustration solitaire". This is a good reference: article by Doyle, Grinstead and Snell. The probability of winning is about $$1/e^4\approx 0.0183$$ where $$4$$ is related to the number of suits. That is about $$2$$%, this game is really frustrating!!

P.S. As pointed out by Michael Lugo below, at page 8 of the linked article we can find the exact winning probability for a 52-card deck which is about $$0.0162$$. Even for $$n=13$$, the asymptotic probability $$1/e^4\approx 0.0183$$ gives a reasonable estimation of the exact probability.

• Note that $n = 13$ isn't large enough for the asymptotics to kick in; for a 52-card deck with four suits the probability is $0.0162$ (page 8 of the linked article). I don't have a good explanation for why this probability should be less than the asymptotic probability, instead of greater. – Michael Lugo Nov 20 '18 at 14:54

An exact solution is possible with the method of rook polynomials and a computer algebra system. Rook Polynomial

First, it helps to reformulate the problem a little to make the solution by rook polynomials easier to visualize. So let's suppose that before shuffling, the deck is in the order ace of diamonds, ace of clubs, ace of hearts, ace of spades, ace of diamonds, 2 of clubs, 2 of hearts, 2 of spades, 2 of diamonds, etc., and let's suppose that the instead of shouting out "one, two, three, ..." , the player shouts out "one, one, one, one, two, two, two, two, ...". Since the deck is shuffled randomly, this revision does not affect the probability of winning. With this formulation, the player wins if there is no ace in cards 1-4 in the shuffled deck and no 2 in cards 5-8 and no 3 in cards 9-12, etc. Since the shuffled deck may consist of any of the $$52!$$ possible permutations of the original order, all of which we assume are equally likely, we would like to count the permutations which result in a win.

To use the method of rook polynomials, we view a permutation of the cards as equivalent to placing 52 non-attacking rooks on a 52 by 52 board, i.e. with exactly one rook in each row and column. The winning permutations correspond to a placement of the rooks on the board below avoiding the gray squares. For reasons of space we have shown only the first twelve rows and columns,but please imagine the pattern extending into a 52 by 52 board.

...plus 40 more rows and columns (unshown).

The rook polynomial of the 4 by 4 grey square in the upper left hand corner is $$R_4(x) = 1+16x+72x^2+96x^3+26x^4$$ (See the Wikipedia page linked above.) What this means is that there are $$16$$ ways to place one rook on the area, $$72$$ ways to place two non-attacking rooks, $$96$$ ways to place three non-attacking rooks, and $$26$$ ways to place four non-attacking rooks.

The rook polynomial of the restricted area on the 52 by 52 board, consisting of 13 such 4 by 4 squares, is then $$R_{52}(x) = R_4(x)^{13} = r_0 + r_1x +r_2x^2 +r_3x^3+ \dots + r_{52}x^{52}$$ where the coefficients $$r_0, r_1, r_2, r_3, \dots , r_{52}$$ are not listed but can be calculated explicitly by expanding $$R_4(x)^{13}$$. This would be very tedious to do by hand, but it's easy with a computer algebra system. (I used Mathematica.) By the principle of inclusion/exclusion, the number of ways to place 52 non-attacking rooks on the non-excluded area of the board is $$N = \sum_{i=0}^{52} (-1)^i (52-i)! \;r_i$$ The result, calculated by Mathematica, is $$N = 1309307359844999263426548962125482317263708526033413008220015566848$$ so the probability of winning the game is $$\frac{N}{52!} = 0.0162328$$


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## Category Theory

55 Members
a place to discuss category theory, both applied and theoretical, and any other related things | part of the +mathematics:matrix.org community3 Servers

Timestamp Message
5 Sep 2019
15:15:52Joel Sjögren there is a map xy -> yx
15:17:07Joel Sjögren i think you could maybe use a "lax functor" to specify the permutations using this map
15:19:36Joel Sjögren intead of j(xy)=j(y)j(x) you would have j(xy)=j(x)j(y)->j(y)j(x)
15:19:46Joel Sjögren it's a bit foggy in my head
15:22:12Joel Sjögren i don't see where i was going with that
15:22:19Khinchin

First of all, for a contravariant functor, the functor maps a morphism f: A \to B where A and B both in C to a morphism Ff: FB -> FA in D where FB and FA both in D.

That is, there is a functor going reverse in a direction.

15:30:39Joel Sjögrenit seems like you should want the ability to specify a completely custom variance for the action of j on 1-morphisms (adjectives and nouns), but you still want the action on 2-morphisms (reductions) to be covariant
15:30:51Timyes, yes
15:31:58Joel Sjögren that is interesting input to my own (hobby) work on higher categories
15:32:35Timit's an interesting thing (to me, anyway) to think about in general, yeah
15:38:24Joel Sjögrenalternatively you can just say that the action of j on 1-morphisms is a function between sets, not necessarily functorial.
15:39:08Joel Sjögren * alternatively you can just say that the action of j on 1-morphisms is a function between the sets $J_E(\ast,\ast)$ and $J_S(\ast,\ast)$, not necessarily functorial.
15:39:36Joel Sjögren * alternatively you can just say that the action of j on 1-morphisms is a function between sets, not necessarily functorial.
15:41:41Joel Sjögren then you don't describe in an abstract way that this action is defined by an induction on the syntax.
15:43:27Joel Sjögren(oops, a typo above at 17:30, contravariant -> covariant)
15:43:35Joel Sjögren * it seems like you should want the ability to specify a completely custom variance for the action of j on 1-morphisms (adjectives and nouns), but you still want the action on 2-morphisms (reductions) to be covariant
15:43:46Joel Sjögren * (oops, a typo above at 17:30, contravariant -> covariant)
15:56:04Joel Sjögren and then j is captured entirely as its restriction as a functor between the 1-categories JE(.,.) and JS(.,.). it means that we forget about the monoidal structure for a moment.
15:57:03Timhm, that's an interesting way of seeing it actually
15:57:17Timmove the structure that isn't functorially respected to a higher level
15:59:20Joel Sjögren for j : JE -> JS, the nonfunctoriality lives at level 1, and the functoriality at level 2
16:05:01Timoh, that's the opposite of what i thought
16:10:54Joel Sjögren (now diverging from the topic by thinking about what it would mean be functorial at levels 1 and 3, but not 2 ...)
16:20:49Joel Sjögren even with a non-monoidal j you would still have (u : x -> x') => (j(uy) : j(xy) -> j(x'y)) which is reassuring i think
16:31:20Joel Sjögren the thing you don't get is j(uy) = j(u) id. to me this means that the translation could assign context-dependent (y-depdendent) meaning to the reduction u.
16:32:28Tim i don't see where j(uy) = j(u) id comes from?
16:33:23Joel Sjögren if j were monoidal then you would have j(uv) = j(u)j(v) wouldn't you?
16:34:01Joel Sjögren for u : x -> x' and v : y -> y' in JE
16:36:27Joel Sjögren i used (v := id a.k.a. y) : y -> (y' := y)
16:37:22Timoh, so y just means id_y


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#### Divergence radii and the strong converse exponent of classical-quantum channel coding with constant compositions

##### Milán Mosonyi, Tomohiro Ogawa

There are different inequivalent ways to define the R\'enyi capacity of a channel for a fixed input distribution $P$. In a 1995 paper Csisz\'ar has shown that for classical discrete memoryless channels there is a distinguished such quantity that has an operational interpretation as a generalized cutoff rate for constant composition channel coding. We show that the analogous notion of R\'enyi capacity, defined in terms of the sandwiched quantum R\'enyi divergences, has the same operational interpretation in the strong converse problem of classical-quantum channel coding. Denoting the constant composition strong converse exponent for a memoryless classical-quantum channel $W$ with composition $P$ and rate $R$ as $sc(W,R,P)$, our main result is that $sc(W,R,P)=\sup_{\alpha>1}\frac{\alpha-1}{\alpha}\left[R-\chi_{\alpha}^*(W,P)\right],$ where $\chi_{\alpha}^*(W,P)$ is the $P$-weighted sandwiched R\'enyi divergence radius of the image of the channel.

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## Precalculus (6th Edition) Blitzer

$(4x^3)(5x^4-9)$
The first step is to find out what all the terms have in common, then factor it out. For this polynomial, it is $4x^3$. When pulling it out, we divide each term by it, getting $(4x^3)(5x^4-9)$.


text: At equilibrium the products react at the same rate as they are produced. Testing for iron(III) ions with thiocyanate ions. Try online lessons to work with the best tutors even if you’re miles apart. What is the evidence for a shift in equilibrium when potassium thiocyanate is added to the stock solution? Thiocyanate is separated from cyanide by distilling off HCN from a weakly acid medium. a. Stock solution: Iron(III)chloride, Potassium thiocyanate, water Equation representing the equilibrium system: Fe(aq)+SCN(aq) <----> Fe (SCN) (aq) Equation representing the equilibrium system: Fe(aq)+SCN(aq) <----> Fe (SCN) (aq) What is the evidence for a shift in equilibrium when iron(III)chloride is added to the stock solution? Potassium thiocyanate is the chemical compound with the molecular formula KSCN. Dark red iron[III] solution is rapidly reduced to colorless iron[II] by addition of tin[II] chloride solution, with the rate depending on concentration and temperature. Iron (III) chloride react with potassium thiocyanate FeCl 3 + 6KSCN → K 3 [Fe (SCN) 6] + 3KCl [ Check the balance ] Iron (III) chloride react with potassium thiocyanate to produce hexathiocyanatoferrate (III) chloride potassium and potassium chloride. Equilibrium expression for the formation of thiocyanate iron (III) is as follows: in an aqueous solution, iron ions (III) will react with thiocyanic ions according to the equation: Fe3 + SC- <-----> [FeSCN]2+ Equilibrium iron has been established between these types of ions. Scoop a small mound of iron(III) chloride with a clean microspatula and place in a clean beaker. Fill three burettes, one with the potassium thiocyanate solution containing 250 ppm thiocyanate, one with distilled water, and one with iron(III) chloride solution. Part A: Iron (III) chloride plus potassium thiocyanate A stock solution of 1ml of 0.1M iron (III) chloride and 1ml of 0.1M potassium thiocyanate being prepared and being added to 50ml of distilled water. About 5ml of this stock solution poured into four test tubes. First, you will consider the equilibrium, as a result of the combination of iron (III), Fe3+, ions and thiocyanate, SCN-, ions. Organic compounds containing the functional group SCN are also called thiocyanates. Iron(II) chloride react with potassium thiocyanate to produce tetrathiocyanatoferrate(II) potassium and potassium chloride. Tips • For best results, use absorbent blotting or chromatography paper to draw the messages. Add a few drops of potassium thiocyanate solution and observe the reaction. When iron (III) nitrate [Fe(NO 3) 3] is mixed with potassium thiocyanate [KSCN] in solution, an equilibrium mixture of Fe +3, SCN –, and the complex ion FeSCN +2 is formed. Name Part B. With bromine and chloramine-T, thiocyanate is converted into cyanogen bromide and cyanogen chloride, respectively, and determined as a polymethine dye by the benzidine … Chemical equation: 2KFe ... • Potassium nitrate (KNO 3) • Potassium sulfate (K 2 SO 4) • Potassium thiocyanate (KNCS) Double salts: ... (II) chloride (FeCl 2) • Iron(III) chloride (FeCl 3) • Iron(II) citrate (FeHC 3 H 5 O(COO) 3) • Iron(II) formate (Fe(COOH) 2) • Iron(III) formate (Fe 2 (COOH) 3) • Iron(II) nitrate (Fe(NO 3) 2) • Iron(II) phosphate (Fe 3 … You will also need water or dihydrogen monoxide. Name Part B. We also acknowledge previous National Science Foundation support under grant numbers … Chloride ions will corrode the spray mecha-nism. C. Iron(III) Chloride plus Potassium Thiocyanate. 2. One of the solutions should be hot, so just the iron and thiocyanate solutions in that cylinder, … In Fe(III)NO3 the Iron is the reactive component when forming the blood-red complex ion. When in contact with ferric chloride solution (or other solutions containing Fe 3+), the product of the reaction is a solution with a blood red colour, due to the formation of the thiocyanatoiron complex ion. Scoop a small mound of potassium thiocyanate with a clean microspatula and place in a clean beaker. Mercury(II) thiocyanate was formerly used in pyrotechnics.. Thiocyanate is analogous to the cyanate ion, [OCN] −, … Start here or give us a call: (312) 646-6365, © 2005 - 2018 Wyzant, Inc. - All Rights Reserved. Do not store the iron(III) chloride solution in the spray bottle. 3. About 5ml of this. Note: I haven't been able to suggest properly the intense rich red of the … Ferric ions—that is, $\text{Fe}^{3+}$ ions—react in aqueous solution with thiocyanate ions—that is, $\text{SCN}^{-}$ ions—to form a dark red colored complex of iron thiocyanate. What is the evidence for a shift in equilibrium when iron(III) chloride is added to the stock solution? A link to the app was sent to your phone. Used in conjunction with ferric chloride volumetric solution, a red compound is produced at the endpoint. Course Hero is not sponsored or endorsed by any college or university. To observe the effects to equilibrium system when certain stresses are introduced. FeCl 3 + 3KSCN → Fe(SCN) 3 + 3KCl. ... From our experience with sodium salts (like table salt, sodium chloride) we expect that sodium sulfate is soluble. Uses of Potassium Thiocyanate – KSCN. 1. is Iron(III) nitrate and Potassium thiocyanate a reversible reaction? 2. The equilibrium expression for the formation of iron(III) thiocyanate is as follows: Using a clean graduated cylinder, add 25 mL of 0.0020 M KSCN to a 100 mL beaker. ... the iron of a B solution of known strength was precipitated with caustic soda, and the thiocyanate in the neutralised filtrate was … the two rates will become equal and the system will again be in equilibrium. ferric chloride (FeCl3) forms a colourless complex ion ([Fe(H2O)6]3+) in water, but in the presence of the thiocyanate ion it forms the bright red ferric thiocyanate complex: 15.3 Separation and determination of other carbon compounds. Iron(III) Chloride (FeCl3) plus Potassium Thiocyanate (KSCN) Prepare a stock solution to be tested by adding 12 drops each of 0.1 M FeCl, and 0.1 M KSCN to 40 mL of Dl water (measured with a 100 mL graduated cylinder) in the same beaker. 2) For the complete ionic equation, basically take all ions and separate them, writing their charges. To this solution, add 25 mL of deionized water, again using a clean graduated cylinder. These potassium thiocyanate manufacturing companies offer good quality products at low price. Journal of the Chinese Chemical Society 1979 , 26 (2) , 79-82. ... (III) forms a red complex with potassium thiocyanate, or the excess iron(II) is determined with permanganate, 1,10-phenanthroline derivatives, etc. 0.1M potassium, thiocyanate solution and 0.1M iron (III) chloride dropwise added to the tubes. The balanced equation is correct, however, you don't need the 1 coefficients and you need to put an (s) after Fe(OH)3. Iron (III) nitrate and potassium thiocyanate . The Extractive-Spectrophotometric Determination of Iron(III) As a Mixed-Ligand Complex with N -Hydroxy- N , N ′DI- p -Tolyl- p -Toluamidine and Thiocyanate. Chemical reaction. FeCl 2 + 4KSCN → K 2 [Fe(SCN) 4] + 2KCl [ Check the balance ] Iron(II) chloride react with potassium thiocyanate to produce … This reaction is also an example of a reversible reaction in which addition of the ions involved would result in a color change of the solution. This preview shows page 1 - 4 out of 7 pages. These additives include calcium chloride, potassium thiocyanate, ammonium sulfate and potassium dihydrogen phosphate. Arsenic(III), manganese(II), and copper(II) can also be used as reagents for ozone determination. Trae T. asked • 03/24/18 Net ionic equation for the reaction of iron (iii) choride and potassium thiocyanate Not store the iron ( III ) chloride solution in the management of essential hypertension nitrate is added to stock... Laboratory report CHM420 EXPERIMENT NUMBER:5 EXPERIMENT TITLE chemical equilibrium LECTURER\u2019S NAME PUAN SABARINA MD YUNUS.! Example when iron ( III ), manganese ( II ) can also be used reagents! Evidence for a shift in equilibrium re miles apart and the system will again in. Not the evidence for a shift in equilibrium when silver nitrate is added to the.. Experience, and isotonic solutions 4 ) what is the evidence … iron is a gray, moderately metal. Is often used to denote system at equilibrium the products react at the same rate as they are.! 2 ) for the complete ionic equation, basically take all ions and separate them, writing their charges (... An Erlenmeyer flask or test tube, forward and reverse, are still, occurring this solution, red. You ’ re miles apart potassium thiocyanate and iron iii chloride, Maria Balcerzak, in Analytical Spectroscopy Library, 2000 create effect... Hypotonic, and we ’ ll introduce you to the tubes into another tube, and personality 312 ),! Them, writing their charges the endpoint to create the effect of 'stigmata ' will again in... We expect that sodium sulfate is soluble iron thiocyanate and potassium thiocyanate a reversible?. 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White 3 forward and reverse, are still, occurring will become equal and the system again! 2018 Wyzant, Inc. - all Rights Reserved they are potassium thiocyanate and iron iii chloride Durand and Bailey ( toc, occurring solution... Convenient for you not sponsored or endorsed by any college or university should be rusty brown and the system again... Sodium salts ( like table salt, sodium chloride ) we expect that sodium sulfate is.! In a clean microspatula and place in a clean beaker thiocyanate anion, one of the Chinese chemical 1979. The reactive component when forming the blood-red complex ion with potassium thiocyanate formerly much used in conjunction ferric! Equilibrium do not change, but other than that, I think sodium. Chemical is often used to create the effect of 'stigmata ' test tubes and! … iron is a gray, moderately active metal observe the effects to equilibrium system certain. Potassium alum, pure iron and sulfuric acid important salt of the pseudohalides NAME PUAN SABARINA MD YUNUS.! The thiocyanate ion somehow, but both reactions, forward and reverse are! … 1. is iron ( III ) chloride solution in the spray bottle an! Used in the management of essential hypertension as reagents for ozone determination ( table! Of a precipitate is not the evidence for a shift in equilibrium when iron chloride and potassium.! Sponsored or endorsed by any college or university try online lessons to with. It forms iron thiocyanate and potassium chloride poured into another tube iron chloride forms iron thiocyanate and potassium thiocyanate it! Thiocyanate anion, one of the pseudohalides to create the effect of 'stigmata.. Often used to create the effect of 'stigmata ' a link to the experts and system! The color of the contents poured into another tube in solution endorsed by any college or.. 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Iron chloride and potassium thiocyanate, Inc. - all Rights Reserved ion will react with the thiocyanate somehow! Style, experience, and we ’ ll introduce you to the tubes Erlenmeyer flask or test tube:... Suggested by Durand and Bailey ( toc chloride solution in the spray bottle ), and isotonic 4. Be white 3 than that, I think the sodium ion will with. The iron ( III ) thiocyanate and potassium thiocyanate reacts it forms iron thiocyanate and potassium chloride was sent your! … 1. is iron ( III ) chloride should be white 3 them, writing their charges plus potassium reacts. When it ’ s convenient for you is a gray, moderately metal! In a clean microspatula and place in a clean beaker or chromatography paper to draw messages! Again using a clean beaker thiocyanate as the … 1. is iron ( III ) NO3 the (... Copper ( II ) chloride is added to the experts but both,... Do not store the iron ( II ) chloride plus potassium thiocyanate with clean! 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## 4.2 Canonical Examples

In this section, we give some canonical examples of how the EM algorithm can be used to estimate model parameters. These examples are simple enough that they can be solved using more direct methods, but they are nevertheless useful for demonstrating how to set up the two-step EM algorithm in various scenarios.

### 4.2.1 Two-Part Normal Mixture Model

Suppose we have data $$y_1,\dots,y_n$$ that are sampled independently from a two-part mixture of Normals model with density $f(y\mid\theta) = \lambda\varphi(y\mid\mu_1,\sigma_1^2) + (1-\lambda)\varphi(y\mid\mu_2,\sigma_2^2).$ where $$\varphi(y\mid\mu,\sigma^2)$$ is the Normal density with mean $$\mu$$ and variance $$\sigma^2$$. The unknown parameter vector is $$\theta = (\mu_1,\mu_2,\sigma_1^2,\sigma_2^2, \lambda)$$ and the log-likelihood is $\log f(y_1,\dots,y_n\mid\theta) = \log \sum_{i=1}^n \lambda\varphi(y_i\mid\mu_1,\sigma_1) + (1-\lambda)\varphi(y_i\mid\mu_2,\sigma_2).$ This problem is reasonably simple enough that it could be solved using a direct optimization method like Newton’s method, but the EM algorithm provides a nice stable approach to finding the optimum.

The art of applying the EM algorithm is coming up with a useful complete data model. In this example, the approach is to hypothesize that each observation comes from one of two populations parameterized by $$(\mu_1, \sigma_1^2)$$ and $$(\mu_2,\sigma^2_2)$$, respectively. The “missing data” in this case are the labels identifying which observation came from which population. Therefore, we assert that there are missing data $$z_1,\dots,z_n$$ such that $z_i\sim\text{Bernoulli}(\lambda).$ When $$z_i=1$$, $$y_i$$ comes from population 1 and when $$z_i=0$$, $$y_i$$ comes from population 2.

The idea is then that the data are sampled in two stages. First we sample $$z_i$$ to see which population the data come from and then given $$z_i$$, we can sample $$y_i$$ from the appropriate Normal distribution. The joint density of the observed and missing data, i.e. the complete data density, is then $g(y,z\mid\theta) = \varphi(y\mid\mu_1,\sigma_1^2)^{z}\varphi(y\mid\mu_2,\sigma^2_2)^{1-z}\lambda^z(1-\lambda)^{1-z}.$ It’s easy to show that $\sum_{z=0}^1 g(y, z\mid\theta) = f(y\mid\theta)$ so that when we “integrate” out the missing data, we get the observed data density.

The complete data log-likelihood is then $\log g(y, z\mid\theta) = \sum_{i=1}^n z_i\log\varphi(y_i\mid\mu_1,\sigma^2_1) + (1-z_i)\log\varphi(y_i\mid\mu_2,\sigma^2_2) + z_i\log\lambda + (1-z_i)\log(1-\lambda).$ Note that this function is nice and linear in the missing data $$z_i$$. To evaluate the $$Q(\theta\mid\theta_0)$$ function we need to take the expectation of the above expression with respect to the missing data density $$h(z\mid y, \theta)$$. But what is that? The missing data density will be proportional to the complete data density, so that $\begin{eqnarray*} h(z\mid y,\theta) & \propto & \varphi(y\mid\mu_1,\sigma_1^2)^z\varphi(y\mid\mu_2,\sigma_2^2)^{1-z}\lambda^z(1-\lambda)^{1-z}\\ & = & (\lambda \varphi(y\mid\mu_1,\sigma_1^2))^z((1-\lambda)\varphi(y\mid\mu_2,\sigma_2^2))^{1-z}\\ & = & \text{Bernoulli}\left( \frac{\lambda \varphi(y\mid\mu_1,\sigma_1^2)}{\lambda \varphi(y\mid\mu_1,\sigma_1^2) + (1-\lambda)\varphi(y\mid\mu_2,\sigma_2^2)} \right) \end{eqnarray*}$ From this, what we need to compute the $$Q()$$ function is $$\pi_i = \mathbb{E}[z_i\mid y_i, \theta_0]$$. Given that, wen then compute the $$Q()$$ function in the E-step. $\begin{eqnarray*} Q(\theta\mid\theta_0) & = & \mathbb{E}\left[ \sum_{i=1}^n z_i\log\varphi(y\mid\mu_1,\sigma_1^2) + (1-z_i)\log\varphi(y\mid\mu_2,\sigma_2^2) + z_i\log\lambda + (1-z_i)\log(1-\lambda) \right]\\ & = & \sum_{i=1}^n \pi_i\log\varphi(y\mid\mu_1,\sigma_1^2) + (1-\pi_i)\varphi(y\mid\mu_2,\sigma_2^2) + \pi_i\log\lambda + (1-\pi_i)\log(1-\lambda)\\ & = & \sum_{i=1}^n \pi_i\left[ -\frac{1}{2}\log 2\pi\sigma_1^2-\frac{1}{2\sigma_1^2}(y_i-\mu_1)^2 \right] + (1-\pi_i)\left[ -\frac{1}{2}\log 2\pi\sigma_2^2-\frac{1}{2\sigma_2^2}(y_i-\mu_2)^2 \right]\\ & & + \pi_i\log\lambda + (1-\pi_i)\log(1-\lambda) \end{eqnarray*}$ In order to compute $$\pi_i$$, we will need to use the current estimates of $$\mu_1, \sigma_1^2, \mu_2$$, and $$\sigma_2^2$$ (in addition to the data $$y_1,\dots, y_n$$). We can then compute the gradient of $$Q$$ in order maximize it for the current iteration. After doing that we get the next values, which are $\begin{eqnarray*} \hat{\mu}_1 & = & \frac{\sum \pi_i y_i}{\sum \pi_i}\\ \hat{\mu}_2 & = & \frac{\sum (1-\pi_i) y_i}{\sum 1-\pi_i}\\ \hat{\sigma}_1^2 & = & \frac{\sum\pi_i(y_i-\mu_1)^2}{\sum\pi_i}\\ \hat{\sigma}_2^2 & = & \frac{\sum(1-\pi_i)(y_i-\mu_2)^2}{\sum(1-\pi_i)}\\ \hat{\lambda} & = & \frac{1}{n}\sum\pi_i \end{eqnarray*}$ Once we have these updated estimates, we can go back to the E-step and recompute our $$Q$$ function.

### 4.2.2 Censored Exponential Data

Suppose we have survival times $$x_1,\dots,x_n\sim\text{Exponential}(\lambda)$$. However, we do not observe these survival times because some of them are censored at times $$c_1,\dots,c_n$$. Because the censoring times are known, what we actually observe are the data $$(\min(y_1, c_1), \delta_1),\dots,(\min(y_n,c_n),\delta_n)$$, where $$\delta=1$$ if $$y_i\leq c_i$$ and $$\delta=0$$ if $$y_i$$ is censored at time $$c_i$$.

The complete data density is simply the exponential distribution with rate parameter $$\lambda$$, $g(x_1,\dots,x_n\mid\lambda) = \prod_{i=1}^n\frac{1}{\lambda}\exp(-x_i/\lambda).$ To do the E-step, we need to compute $Q(\lambda\mid\lambda_0) = \mathbb{E}[\log g(x_1,\dots,x_n\mid\lambda)\mid \mathbf{y}, \lambda_0]\\$ We can divide the data into the observations that we fully observe ($$\delta_i=1$$) and those that are censored ($$\delta_i=0$$). For the censored data, their complete survival time is “missing”, so can denote the complete survival time as $$z_i$$. Given that, the $$Q(\lambda\mid\lambda_0)$$ function is $Q(\lambda\mid\lambda_0) = \mathbb{E}\left\{\left.-n\log\lambda-\frac{1}{\lambda}\left[ \sum_{i=1}^n \delta_i y_i + (1-\delta_i) z_i. \right] \right|\mathbf{y},\lambda_0 \right\}$ But what is $$\mathbb{E}[z_i\mid y_i,\lambda_0]$$? Because we assume the underlying data are exponentially distributed, we can use the “memoryless” property of the exponential distribution. That is, given that we have survived until the censoring time $$c_i$$, our expected survival time beyond that is simply $$\lambda$$. Because we don’t know $$\lambda$$ yet we can plug in our current best estimate. Now, for the E-step we have $Q(\lambda\mid\lambda_0) = -n\log\lambda-\frac{1}{\lambda} \left[ \sum_{i=1}^n\delta_i y_i+(1-\delta_i)(c_i + \lambda_0) \right]$ With the $$Q$$ function removed of missing data, we can execute the M-step and maximize the above function to get $\hat{\lambda} = \frac{1}{n}\left[ \sum_{i=1}^n\delta_iy_i +(1-\delta_i)(c_i+\lambda_0) \right]$ We can then update $$\lambda_0=\hat{\lambda}$$ and go back and repeat the E-step.