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8.7E: Exercises

8.7E: Exercises


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Practice Makes Perfect

Solve Proportions

In the following exercises, solve.

Example (PageIndex{31})

(frac{x}{56}=frac{7}{8})

Answer

(49)

Example (PageIndex{32})

(frac{n}{91}=frac{8}{13})

Example (PageIndex{32})

(frac{49}{63}=frac{z}{9})

Answer

(7)

Example (PageIndex{33})

(frac{56}{72}=frac{y}{9})

Example (PageIndex{34})

(frac{5}{a}=frac{65}{11})

Answer

(9)

Example (PageIndex{35})

(frac{4}{b}=frac{64}{144})

Example (PageIndex{36})

(frac{98}{154}=−frac{7}{p})

Answer

(−11)

Example (PageIndex{37})

(frac{72}{156}=−frac{6}{q})

Example (PageIndex{38})

(frac{a}{−8}=frac{−42}{48})

Answer

(7)

Example (PageIndex{39})

(frac{b}{−7}=frac{−30}{42})

Example (PageIndex{40})

(frac{2.7}{j}=frac{0.9}{0.2})

Answer

(0.6)

Example (PageIndex{41})

(frac{2.8}{k}=frac{2.1}{1.5})

Example (PageIndex{42})

(frac{a}{a+12}=frac{4}{7})

Answer

(16)

Example (PageIndex{43})

(frac{b}{b−16}=frac{11}{9})

Example (PageIndex{44})

(frac{c}{c−104}=−frac{5}{8})

Answer

(−frac{5}{8})

Example (PageIndex{45})

(frac{d}{d−48}=−frac{13}{3})

Example (PageIndex{46})

(frac{m+90}{25}=frac{m+30}{15})

Answer

(60)

Example (PageIndex{47})

(frac{n+10}{4}=frac{40−n}{6})

Example (PageIndex{48})

(frac{2p+4}{8}=frac{p+18}{6})

Answer

(30)

Example (PageIndex{49})

(frac{q−2}{2}=frac{2q−7}{18})

Example (PageIndex{50})

Pediatricians prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of a child’s weight. How many milliliters of acetaminophen will the doctor prescribe for Jocelyn, who weighs 45 pounds?

Answer

9 ml

Example (PageIndex{51})

Brianna, who weighs 6 kg, just received her shots and needs a pain killer. The pain killer is prescribed for children at 15 milligrams (mg) for every 1 kilogram (kg) of the child’s weight. How many milligrams will the doctor prescribe?

Example (PageIndex{52})

A veterinarian prescribed Sunny, a 65 pound dog, an antibacterial medicine in case an infection emerges after her teeth were cleaned. If the dosage is 5 mg for every pound, how much medicine was Sunny given?

Answer

325 mg

Example (PageIndex{53})

Belle, a 13 pound cat, is suffering from joint pain. How much medicine should the veterinarian prescribe if the dosage is 1.8 mg per pound?

Example (PageIndex{54})

A new energy drink advertises 106 calories for 8 ounces. How many calories are in 12 ounces of the drink?

Answer

159 calories

Example (PageIndex{55})

One 12 ounce can of soda has 150 calories. If Josiah drinks the big 32 ounce size from the local mini-mart, how many calories does he get?

Example (PageIndex{56})

A new 7 ounce lemon ice drink is advertised for having only 140 calories. How many ounces could Sally drink if she wanted to drink just 100 calories?

Answer

5 oz

Example (PageIndex{57})

Reese loves to drink healthy green smoothies. A 16 ounce serving of smoothie has 170 calories. Reese drinks 24 ounces of these smoothies in one day. How many calories of smoothie is he consuming in one day?

Example (PageIndex{58})

Janice is traveling to Canada and will change $250 US dollars into Canadian dollars. At the current exchange rate, $1 US is equal to $1.01 Canadian. How many Canadian dollars will she get for her trip?

Answer

252.5 Canadian dollars

Example (PageIndex{59})

Todd is traveling to Mexico and needs to exchange $450 into Mexican pesos. If each dollar is worth 12.29 pesos, how many pesos will he get for his trip?

Example (PageIndex{60})

Steve changed $600 into 480 Euros. How many Euros did he receive for each US dollar?

Answer

0.80 Euros

Example (PageIndex{61})

Martha changed $350 US into 385 Australian dollars. How many Australian dollars did she receive for each US dollar?

Example (PageIndex{62})

When traveling to Great Britain, Bethany exchanged her $900 into 570 British pounds. How many pounds did she receive for each American dollar?

Answer

0.63 British pounds

Example (PageIndex{63})

A missionary commissioned to South Africa had to exchange his $500 for the South African Rand which is worth 12.63 for every dollar. How many Rand did he have after the exchange?

Example (PageIndex{64})

Ronald needs a morning breakfast drink that will give him at least 390 calories. Orange juice has 130 calories in one cup. How many cups does he need to drink to reach his calorie goal?

Answer

3 cups

Example (PageIndex{65})

Sarah drinks a 32-ounce energy drink containing 80 calories per 12 ounce. How many calories did she drink?

Example (PageIndex{66})

Elizabeth is returning to the United States from Canada. She changes the remaining 300 Canadian dollars she has to $230.05 in American dollars. What was $1 worth in Canadian dollars?

Answer

1.30 Canadian dollars

Example (PageIndex{67})

Ben needs to convert $1000 to the Japanese Yen. One American dollar is worth 123.3 Yen. How much Yen will he have?

Example (PageIndex{68})

A golden retriever weighing 85 pounds has diarrhea. His medicine is prescribed as 1 teaspoon per 5 pounds. How much medicine should he be given?

Answer

17 tsp

Example (PageIndex{69})

Five-year-old Lacy was stung by a bee. The dosage for the anti-itch liquid is 150 mg for her weight of 40 pounds. What is the dosage per pound?

Example (PageIndex{70})

Karen eats (frac{1}{2}) cup of oatmeal that counts for 2 points on her weight loss program. Her husband, Joe, can have 3 points of oatmeal for breakfast. How much oatmeal can he have?

Answer

(frac{3}{4}) cup

Example (PageIndex{71})

An oatmeal cookie recipe calls for (frac{1}{2}) cup of butter to make 4 dozen cookies. Hilda needs to make 10 dozen cookies for the bake sale. How many cups of butter will she need?​​​​​​​

Solve Similar Figure Applications

In the following exercises, ΔABC is similar to ΔXYZ

Example (PageIndex{72})

side b

Answer

12

Example (PageIndex{73})

side x

​​​​​​​In the following exercises, ΔDEF is similar to ΔNPQ

Example (PageIndex{74})

Find the length of side d.

Answer

(frac{77}{18})

Example (PageIndex{75})

Find the length of side q.

​​​​​​​In the following two exercises, use the map shown. On the map, New York City, Chicago, and Memphis form a triangle whose sides are shown in the figure below. The actual distance from New York to Chicago is 800 miles.

Example (PageIndex{76})

Find the actual distance from New York to Memphis.

Answer

950 miles

Example (PageIndex{77})

Find the actual distance from Chicago to Memphis.

In the following two exercises, use the map shown. On the map, Atlanta, Miami, and New Orleans form a triangle whose sides are shown in the figure below. The actual distance from Atlanta to New Orleans is 420 miles.

Example (PageIndex{78})

Find the actual distance from New Orleans to Miami.

Answer

680 miles

Example (PageIndex{79})

Find the actual distance from Atlanta to Miami.​​​​​​​

Example (PageIndex{80})

A 2 foot tall dog casts a 3 foot shadow at the same time a cat casts a one foot shadow. How tall is the cat?

Answer

(frac{2}{3}) foot (8in)

Example (PageIndex{81})

Larry and Tom were standing next to each other in the backyard when Tom challenged Larry to guess how tall he was. Larry knew his own height is 6.5 feet and when they measured their shadows, Larry’s shadow was 8 feet and Tom’s was 7.75 feet long. What is Tom’s height?

Example (PageIndex{82})

The tower portion of a windmill is 212 feet tall. A six foot tall person standing next to the tower casts a seven foot shadow. How long is the windmill’s shadow?

Answer

247.3 feet

Example (PageIndex{83})

The height of the Statue of Liberty is 305 feet. Nicole, who is standing next to the statue, casts a 6 foot shadow and she is 5 feet tall. How long should the shadow of the statue be?​​​​​​​

Everyday Math

Example (PageIndex{84})

Heart Rate At the gym, Carol takes her pulse for 10 seconds and counts 19 beats.

  1. How many beats per minute is this?
  2. Has Carol met her target heart rate of 140 beats per minute?
Answer
  1. 114 beats per minute
  2. no

Example (PageIndex{85})

Heart Rate Kevin wants to keep his heart rate at 160 beats per minute while training. During his workout he counts 27 beats in 10 seconds.

  1. How many beats per minute is this?
  2. Has Kevin met his target heart rate?

Example (PageIndex{86})

Cost of a Road Trip Jesse’s car gets 30 miles per gallon of gas.

  1. If Las Vegas is 285 miles away, how many gallons of gas are needed to get there and then home?
  2. If gas is $3.09 per gallon, what is the total cost of the gas for the trip?
Answer
  1. 19 gallons
  2. $58.71

Example (PageIndex{87})

Cost of a Road Trip Danny wants to drive to Phoenix to see his grandfather. Phoenix is 370 miles from Danny’s home and his car gets 18.5 miles per gallon.

  1. How many gallons of gas will Danny need to get to and from Phoenix?
  2. If gas is $3.19 per gallon, what is the total cost for the gas to drive to see his grandfather?

Example (PageIndex{88})

Lawn Fertilizer Phil wants to fertilize his lawn. Each bag of fertilizer covers about 4,000 square feet of lawn. Phil’s lawn is approximately 13,500 square feet. How many bags of fertilizer will he have to buy?

Answer

4 bags

Example (PageIndex{89})

House Paint April wants to paint the exterior of her house. One gallon of paint covers about 350 square feet, and the exterior of the house measures approximately 2000 square feet. How many gallons of paint will she have to buy?​​​​​​​

Example (PageIndex{90})

Cooking Natalia’s pasta recipe calls for 2 pounds of pasta for 1 quart of sauce. How many pounds of pasta should Natalia cook if she has 2.5 quarts of sauce?

Answer

5

Example (PageIndex{91})

Heating Oil A 275 gallon oil tank costs $400 to fill. How much would it cost to fill a 180 gallon oil tank?​​​​​​​

Writing Exercises

Example (PageIndex{92})

Marisol solves the proportion (frac{144}{a}=frac{9}{4}) by ‘cross multiplying’, so her first step looks like 4·144=9·a. Explain how this differs from the method of solution shown in Example.

Answer

Answers will vary.

Example (PageIndex{93})

Find a printed map and then write and solve an application problem similar to Example.​​​​​​​

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?


Exercice d’écoute- Lecture- Orthographe- Grammaire

Pour les 8e Texte d’opinion: Les cellulaires interdits dans les écoles!

Qu’en penses-tu? Réponds à cette question ___ lundi 11 mai

3. Orthographe:a et à, ce (c’) et se (s’), c’est et s’est’

4.Grammaire: Futur simple

Devoirs à compléter:

7e année Histoire 7 Projets (Comparaison N-F, affiche publicitaire)

Français: Lecture et écriture

8e année Histoire 8 Projets (Pères de Confédération et Station 1 et Station 2)

Sciences Le rein artificiel

Français: Lecture et écriture

Semaine du 20 au 23 avril

1.Exercice d’écoute 7e & 8e : Afrique Devoir pour mardi 28

2. Écriture 7e: Journaliste de technologie Devoir pour mercredi 29

3. Orthographe: “Leur -leurs”

4. Grammaire: Futur de l’indicatif

Devoirs à compléter:

Histoire 7 Projets (Comparaison N-F et affiche publicitaire)

Histoire 8 Projets (Pères de Confédération et Station 1 )

Lecture indépendante 7&8: Choisis un livre https://stories.audible.com/discovery

Lecture de romans et magazines (voir Google Doc)

Semaine du 14 au 17 avril

5. Devoir de lecture AUDIO- Répondre aux questions de compréhension

2.Orthographe “peu /peut/peux”

4. Histoire 7e : Projet des échanges de fourrures

Histoire 8e : Projet de collimage

5. Devoir de lecture AUDIO- Répondre aux questions de compréhension

Lecture indépendante 7&8: Choisis un livre https://stories.audible.com/discovery

Lecture de roman et magazine (voir Google Doc)

Semaine du 6 au 9 avril

1.Visionnement d’une vidéo Invisible comme Harry sous sa cape !

2.Histoire 7e voir site de classe / Sciences 8e voir site de classe

3.Orthographe ‘La, Là, l’a, l’as

6.Travail à compléter Sciences, Histoire, Français (lecture)

Mercredi 8 avril

4.Lecture indépendante ou écriture (cahier journal)

Projets à compléter :Sciences et Histoire

Lecture indépendante 7&8: Choisis un livre https://stories.audible.com/discovery

Mardi 7 avril

1.Exercice d’écoute Les fausses nouvelles

3.Grammaire Révision du passé composé

4.Sciences 7e Projet des énergies – 8e Réflexion sur la dialyse

Écriture (voir document partagé avec la classe)

Lecture indépendante 7&8: Choisis un livre https://stories.audible.com/discovery

Lundi 6 avril

Histoire 7 & 8

7e: Comparaison de la vie en Nouvelle-France – aujourd’hui // 8e: Les Pères de la Confédération

Sciences 7e année: Les énergies renouvelables // 8e année: Le rein artificiel


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    Network analysis - activity on node examples

    Network analysis example 1997 UG exam

    The table below defines the activities within a small project.

    • Draw the network diagram.
    • Calculate the minimum overall project completion time and identify which activities are critical.
    • What is the slack (float) time associated with each of the non-critical activities.

    In the above network there are a number of options for the completion time for activities C, D and E as shown below:

    For example choosing a completion time for activity E of 6 weeks costs £14,000.

    Formulate the problem of deciding the completion time for C, D and E so as to ensure that the project is completed within 17 weeks as an integer program.

    Solution

    The network diagram is shown below. Note the introduction of a dummy activity I with a duration of zero to represent the end of the project.

    Let E i represent the earliest start time for activity i such that all its preceding activities have been finished. We calculate the values of the E i (i=A,B. I) by going forward, from left to right, in the network diagram.

    To ease the notation let T i be the activity completion time associated with activity i (e.g. T B = 3). Then the E i are given by:

    E A = 0 (assuming we start at time zero)
    E B = 0 (assuming we start at time zero)
    E C = E A + T A = 0 + 2 =2
    E D = max[E A + T A , E B + T B ] = max[0 + 2, 0 + 3] = 3
    E E = max[E C + T C , E D + T D ] = max[2 + 4, 3 + 3] = 6
    E F = E C + T C = 2 + 4 = 6
    E G = E E + T E = 6 + 8 = 14
    E H = max[E F + T F , E G + T G ] =max[6 + 3, 14 + 2] = 16
    E I = E H + T H = 16 + 3 = 19

    Hence the minimum possible completion time for the entire project is 19 weeks i.e. 19 (weeks) is the minimum time needed to complete all the activities.

    We now need to calculate the latest times for each activity.

    Let L i represent the latest start time we can start activity i and still complete the project in the minimum overall completion time. We calculate the values of the L i (i=A,B. I) by going backward, from right to left, in the network diagram. Hence:

    L I = 19
    L H = L I - T H = 19 - 3 = 16
    L G = L H - T G = 16 - 2 = 14
    L F = L H - T F = 16 - 3 = 13
    L E = L G - T E = 14 - 8 = 6
    L D = L E - T D = 6 - 3 = 3
    L C = min[L E - T C , L F - T C ] = min[6 - 4, 13 - 4] = 2
    L B = L D - T B = 3 - 3 = 0
    L A = min[L C - T A , L D - T A ] = min[2 - 2, 3 - 2] = 0

    Note that as a check all latest times are >=0 at least one activity has a latest start time value of zero.

    As we know the earliest start time E i , and latest start time L i , for each activity i it is clear that the amount of slack or float time F i available is given by F i = L i - E i which is the amount by which we can increase the time taken to complete activity i without changing (increasing) the overall project completion time. Hence we can form the table below:

    Any activity with a float of zero is critical. Note here that, as a check, all float values should be >= 0.

    Hence the critical activities are A,B,C,D, E,G and H and the float for the only non-critical activity F is 7 weeks.

    In order to formulate the problem of deciding the completion time for C, D and E so as to ensure that the project is completed within 17 weeks we make use of integer programming.

    Let the x variables (suitably subscripted) represent feasible start times for each activity (>=0 and integer). Note here that we deal with feasible start times in formulating this program rather than (as above) with earliest start times.

    Introduce zero-one variables:
    C i = 1 if a completion time of i (i=4,3,2,1) chosen for activity C, 0 otherwise
    D i = 1 if a completion time of i (i=3,2,1) chosen for activity D, 0 otherwise
    E i = 1 if a completion time of i (i=8,7,6) chosen for activity E, 0 otherwise

    then the constraints relating to the choice of completion time are:

    C 4 + C 3 + C 2 + C 1 =1
    D 3 + D 2 + D 1 = 1
    E 8 + E 7 + E 6 = 1

    i.e. exactly one completion time must be chosen for C, D and E respectively.

    The constraints relating to the network logic are:

    x E >= x C + 4C 4 + 3C 3 + 2C 2 +1C 1

    where 4C 4 + 3C 3 + 2C 2 +1C 1 is the completion time for activity C (remember C 4 , C 3 , C 2 and C 1 are zero-one variables).

    x E >= x D + 3D 3 + 2D 2 +1D 1

    x F >= x C + 4C 4 + 3C 3 + 2C 2 +1C 1

    x G >= x E + 8E 8 + 7E 7 + 6E 6

    x H <= 17 - 3 (to complete in 17 weeks)

    minimise 3C 4 + 7C 3 + 10C 2 +15C 1 + 12D 3 + 16D 2 +25D 1 + 5E 8 +9E 7 +14E 6

    Network analysis example 1996 MBA exam

    A project consists of 8 activities. The activity completion times and the precedence relationships are as follows:

    • Draw the network diagram.
    • Calculate the minimum overall project completion time and identify which activities are critical.
    • If activity E is delayed by 3 days how is the project completion time affected?
    • If activity F is delayed by 3 days how is the project completion time affected?

    Solution

    The network diagram is shown below. Note the introduction of a dummy activity I with a duration of zero to represent the end of the project.

    Let E i represent the earliest start time for activity i such that all its preceding activities have been finished. We calculate the values of the E i (i=A,B. I) by going forward, from left to right, in the network diagram.

    To ease the notation let T i be the activity completion time associated with activity i (e.g. T B = 7). Then the E i are given by:

    E A = 0 (assuming we start at time zero)
    E B = 0 (assuming we start at time zero)
    E C = 0 (assuming we start at time zero)
    E D = E A + T A = 0 + 5 = 5
    E G = max[E A + T A , E D + T D ] = max[0 + 5, 5 + 3] = 8
    E E = max[E B + T B , E C + T C ] = max[0 + 7, 0 + 6] = 7
    E F = E C + T C = 0 + 6 = 6
    E H = max[E E + T E , E F + T F ] = max[7 + 4, 6 + 2] = 11
    E I = max[E G + T G , E H + T H ] = max[8 + 6, 11 + 5] = 16

    Hence the minimum possible completion time for the entire project is 16 days, i.e. 16 days is the minimum time needed to complete all the activities.

    We now need to calculate the latest times for each activity.

    Let L i represent the latest start time we can start activity i and still complete the project in the minimum overall completion time. We calculate the values of the L i (i=A,B. I) by going backward, from right to left, in the network diagram. Hence:

    L I = 16
    L G = L I - T G = 16 - 6 = 10
    L H = L I - T H = 16 - 5 = 11
    L D = L G - T D = 10 - 3 = 7
    L A = min[L D - T A , L G - T A ] = min[7 - 5, 10 - 5] = 2
    L E = L H - T E = 11 - 4 = 7
    L F = L H - T F = 11 - 2 = 9
    L C = min[L E - T C , L F - T C ] = min[7 - 6, 9 - 6] = 1
    L B = L E - T B = 7 - 7 = 0

    Note that as a check all latest times are >=0 at least one activity has a latest start time value of zero.

    As we know the earliest start time E i , and latest start time L i , for each activity i it is clear that the amount of slack or float time F i available is given by F i = L i - E i which is the amount by which we can increase the time taken to complete activity i without changing (increasing) the overall project completion time. Hence we can form the table below:

    Any activity with a float of zero is critical. Note here that, as a check, all float values should be >= 0.

    Hence the critical activities are B, E and H and the floats for the non-critical activities are given in the table above.

    If activity E is delayed by 3 days then as E is critical the project completion time increases by exactly 3 days.

    If activity F is delayed by 3 days then as F has a float of 3 days the project completion time is unaffected.

    Network analysis example 1994 MBA exam

    A project consists of 8 activities named A to H.

    Construct a network so as to satisfy the scheduling requirements shown in the table below.

    Find the least time required to complete the whole project and identify the critical activities.

    How is the project completion time affected if:

    • activity F is delayed by 3 days
    • activity E is delayed by 7 days
    • activity G is finished 7 days early

    Solution

    The network diagram is shown below. Note the introduction of a dummy activity I with a duration of zero to represent the end of the project.

    Let E i represent the earliest start time for activity i such that all its preceding activities have been finished. We calculate the values of the E i (i=A,B. I) by going forward, from left to right, in the network diagram.

    To ease the notation let T i be the activity completion time associated with activity i (e.g. T B = 7). Then the E i are given by:

    E A = 0 (assuming we start at time zero)
    E B = E A + T A = 0 + 3 = 3
    E C = E A + T A = 0 + 3 = 3
    E D = E A + T A = 0 + 3 = 3
    E E = max[E B + T B , E C + T C ] = max[3 + 6, 3 + 7] = 10
    E F = max[E C + T C , E D + T D ] = max[3 + 7, 3 + 5] = 10
    E G = max[E F + T F , E D + T D ] = max[10 + 8, 3 + 5] = 18
    E H = max[E E + T E , E G + T G ] = max[10 + 13, 18 + 11] = 29
    E I = E H + T H = 29 + 6 = 35

    Hence the minimum possible completion time for the entire project is 35 days, i.e. 35 days is the minimum time needed to complete all the activities.

    We now need to calculate the latest times for each activity.

    Let L i represent the latest start time we can start activity i and still complete the project in the minimum overall completion time. We calculate the values of the L i (i=A,B. I) by going backward, from right to left, in the network diagram. Hence:

    L I = 35
    L H = L I - T H = 35 - 6 = 29
    L G = L H - T G = 29 - 11 = 18
    L E = L H - T E = 29 - 13 = 16
    L F = L G - T F = 18 - 8 = 10
    L B = L E - T B = 16 - 6 = 10
    L C = min[L E - T C , L F - T C ] = min[16 - 7, 10 - 7] = 3
    L D = min[L F - T D , L G - T D ] = min[10 - 5, 18 - 5] = 5
    L A = min[L B - T A , L C - T A , L D - T A ] = min[10 - 3, 3 - 3, 5 - 3] = 0

    Note that as a check all latest times are >=0 at least one activity has a latest start time value of zero.

    As we know the earliest start time E i , and latest start time L i , for each activity i it is clear that the amount of slack or float time F i available is given by F i = L i - E i which is the amount by which we can increase the time taken to complete activity i without changing (increasing) the overall project completion time. Hence we can form the table below:

    Any activity with a float of zero is critical. Note here that, as a check, all float values should be >= 0.

    Hence the critical activities are A, C, F, G and H and the floats for the non-critical activities are given in the table above.

    If activity F is delayed by 3 days then as F is critical the project completion time increases by exactly 3 days.

    If activity E is delayed by 7 days then as E has a float of 6 days the project completion time is affected. The project completion time will increase by exactly (7-6) = 1 day to 36 days.

    If activity G is finished 7 days early then as G is a critical activity the overall project completion time will (potentially) be reduced. To see the effect of this change we need to redo the calculation for the earliest times given above but with a completion time for G of 4 days. The only change is in the calculation of E H and E I and these now are:
    E H = max[E E + T E , E G + T G ] = max[10 + 13, 18 + 4] = 23
    E I = E H + T H = 23 + 6 = 29
    Hence the overall project completion time falls to 29 days (a decrease of 6 days).


    8.7E: Exercises

    Welcome to the web companion of the seventh edition of Applied Electromagnetics, developed to serve the student as an interactive self-study supplement to the text.

    The navigation is highly flexible the user may go though the material in the order outlined in the table of contents or may proceed directly to any exercise, module, or technology brief of interest.

    We hope you find this website helpful and we welcome your feedback and suggestions.


    How DMIT report can help

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    Answers

    7.1: States of Matter

    1. solid
    2. gas
    3. gas
    4. solid
    5. liquid and gas
    6. liquid and gas

    Which of the following statements are true? Correct any false statements.

    1. All substances exist as a liquid at room temperature and pressure. at some temperature and pressure .
    2. Water changes from liquid to solid at 32°C °F.
    3. True (although some states are rarely seen for some substances).

    ethanol 78 ° C dimethyl ether (-)24 ° C

    Ethanol has stronger intermolecular forces due to having hydrogen bonding which is not seen in dimethyl ether. The stronger the intermolecular forces, the higher the boiling point.

    As the altitude increases, the boiling point decreases

    1. Lexington, KY (altitude = 978 feet)
    2. New Orleans, LA (altitude = 2 feet) - HIGHEST
    3. Salt Lake City, UT (altitude = 4226 feet) - LOWEST

    7.2: Heat and Changes of State

    1. solid to gas endothermic
    2. liquid to gas endothermic
    3. solid to liquid endothermic
    4. gas to solid exothermic

    Any two of fusion, vaporization, or sublimation.

    Any two of freezing, condensation, deposition.

    Substance (Delta H_) (kJ/mol) (Delta H_) (kJ/mol) (Delta H_) (kJ/mol) (Delta H_) (kJ/mol)
    oxygen, O2 0.44 6.82 (-)0.44 (-)6.82
    ethane, C2H6 2.85 14.72 (-)2.85 (-)14.72
    carbon tetrachloride, CCl4 2.67 30.0 (-)2.67 (-)30.0
    lead, Pb 4.77 178 (-)4.77 (-)178

    (3.0mol ext_2 extleft ( frac<6.01 kJ> ight )=18 kJ)

    1. (655g ext_2 extleft ( frac<1mol><18.02g> ight ) left (frac<-40.7 kJ> ight )=-1.48 imes10^3kJ )
    2. (8.20kg ext_2 extleft ( frac<1000g><1kg> ight ) left ( frac<1mol><18.02g> ight ) left (frac<-6.01 kJ> ight )=-2.73 imes10^3kJ )
    3. (40.0mL ext_3 ext_2 extleft ( frac<0.789g><1mL> ight ) left ( frac<1mol><46.07g> ight ) left (frac<38.56 kJ> ight )=26.4kJ )
    4. (25.0mL ext_3 ext_2 extleft ( frac<0.789g><1mL> ight ) left ( frac<1mol><46.07g> ight ) left (frac<-38.56 kJ> ight )=-16.5kJ )
    1. (9.25kJ left ( frac<6.01kJ> ight )left ( frac<18.02g> ight )=27.7g ext_2 ext)
    2. (9.25kJ left ( frac<40.7kJ> ight )left ( frac<18.02g> ight )=4.10g ext_2 ext)
    3. (9.25kJ left ( frac<38.56kJ> ight )left ( frac<46.07g> ight )=11.1g ext_3 ext_2 ext)
    1. (-15.5kJ left ( frac<-23.35kJ> ight )left ( frac<17.03g> ight )=11.3g ext_3)
    2. (-15.5kJ left ( frac<-6.01kJ> ight )left ( frac<18.02g> ight )=46.5g ext_2 ext)
    3. (-15.5kJ left ( frac<-38.56kJ> ight )left ( frac<46.07g> ight )=18.5g ext_3 ext_2 ext)

    Find the moles of benzene.

    Combine the energy with the moles to calculate the enthalpy of vaporization.

    7.3: Kinetic-Molecular Theory

    Gas particles are much farther from one another than liquid or solid particles.

    Gases have the most ideal behavior at high temperatures (molecules moving more quickly than at low temperatures so less time to interact) and at low pressure (molecules are farther apart from one another than at high pressure).

    1. Molecules are very far apart from one another and are compressible.
    2. Gases are in constant random motion so they collide with the walls of the container.
    3. False. Molecules of the same substance are moving at a range of speeds.
    4. Collisions are elastic. Energy is exchanged but not lost when two particles coll
    5. False. Particles move in a straight line.

    A collision in which no energy is lost.

    1. (1.721atmleft ( frac<760mmHg><1atm> ight )=1308mmHg)
    2. (559 orrleft ( frac<101.3kPa><760 orr> ight )=74.5kPa)
    3. (91.1kPaleft ( frac<1atm><101.3kPa> ight )=0.899atm)
    4. (2320mmHgleft ( frac<1atm><760mmHg> ight )=3.05atm )
    1. (755mmHgleft ( frac<1atm><760mmHg> ight )=0.993atm ) (755mmHgleft ( frac<101.3kpa><760mmHg> ight )=101kPa )
    2. (615mmHgleft ( frac<1atm><760mmHg> ight )=0.809atm ) (615mmHgleft ( frac<101.3kpa><760mmHg> ight )=82.0kPa )

    Closer to the fire, it is warmer and the kinetic energy of the particles (and therefore the average speed) will be greater.

    7.4: The Ideal Gas Equation

    o C o F K
    25 77 298
    37 99 310
    32 90 305
    (-)273 (-)459 0
    27 80 300
    18 65 291

    A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions?

    (PV=nRT)
    (left ( 4.11 atm ight )V=left ( 1.00mol ight )left ( 0.08206frac ight )left ( 300K ight ))
    (V=5.99L)

    (PV=nRT)
    (Pleft ( 2.5L ight )=left ( 2.5mol ight )left ( 0.08206frac ight )left ( 293K ight ))
    (P=24 atm)

    (PV=nRT)
    (left ( 0.979atm ight )left ( 11.2L ight )=n left ( 0.08206frac ight )left ( 328K ight ))
    (n=0.407 mol)

    (PV=nRT)
    (left ( 0.992atm ight )V=left (8.80 mol ight ) left ( 0.08206frac ight )left ( 298K ight ))
    (V=217 L)

    x = volume of one breath of air

    Balloon will have a total of 8 breaths of air (3 original plus 5 additional)

    (PV=nRT)
    (P left (66.8L ight )=left (2.78 mol ight ) left ( 0.08206frac ight )left ( 298K ight ))
    (P=1.02 atm)

    (PV=nRT)
    (left (1.220atm ight ) left (4.3410L ight )=mol left ( 0.08206frac ight )left ( 788.0K ight ))
    (n=0.08190 mol BF_3)

    1. (frac=frac)
    2. (frac=frac)
    3. (frac<1>=frac<1>) or (n_iT_i=n_fT_f)
    4. (P_iV_i=P_fV_f)
    5. (frac=frac)
    6. (frac=frac)
    7. (frac=frac)

    Set the initial pressure = x to calculate the factor of change in terms of x.

    The final pressure is one third of the original pressure.

    Note the final moles is 2.75 because the problem says that 0.75 moles of gas is added to the original amount of 2.00 moles.

    The temperature is doubled so (T_f=2cdot T_i)

    Let (P_i=x) to see the factor the pressure changes.

    The final pressure is twice the initial pressure.

    The volume is tripled so (V_f=3cdot V_i)

    Let (P_i=x) to see the factor the pressure changes.

    The final pressure is one-third of the initial pressure.

    7.5: Aqueous Solutions

    The solute is present in the smaller amount, the solvent is present in the larger amount, and the solution is the combination of the solute and solvent.

    Solutions are a homoogeneous mixture of two or more compounds.

    Endothermic because heat was needed to dissolve the KNO3. Heat present in the solution was consumed by the dissolution process.

    Strong electrolytes completely dissociate into ions in aqueous solution and are conductors of electricity. Weak electrolytes partially dissociate into ions in aqueous solutions are are weak conductors of electricity. Non-electrolytes do not dissociate into ions in aqueous solution and are poor conductors of electricity.

    1. NaCl(s) &rarr Na + (aq) + Cl &ndash (aq)
    2. CoCl3(s) &rarr Co 3 + (aq) + 3Cl &ndash (aq)
    3. Li2S(s) &rarr 2Li + (aq) + S 2&ndash (aq)
    4. MgBr2(s) &rarr Mg 2 + (aq) + 2Br &ndash (aq)
    5. CaF2(s) &rarr Ca 2 + (aq) + 2F &ndash (aq)

    Based on the given information, identify each as a strong, weak, or non-electrolyte.

    1. non-electrolyte
    2. strong electrolyte
    3. strong electrolyte
    4. weak electrolyte
    5. weak electrolyte
    6. non-electrolyte

    7.6: Colloids and Suspensions

    A suspension can be separated from the solvent by filtration while a solution cannot because particles settle out of suspensions but not solutions.

    Particles in a solution are less than 1 nanometer, colloids have particles from 1-1000 nm, and suspensions have particles over 1000 nm.

    The Tyndall effect is the scattering of visible light by particles. The particles in colloids are large enough to scatter light while the particles in solutions are too small to scatter light. Solutions are transparent (we can see through them) because the particles are so small.

    The dispersed phase is present in the smaller amount and the dispersing medium is present in a larger amount.

    1. suspension
    2. colloids and suspensions
    3. solution
    4. colloid
    5. solution
    6. colloids and suspensions
    7. colloid
    8. solutions and colloids
    9. colloids

    7.7: Solubility

    A saturated solution has the maximum amount of solute dissolved. An unsaturated solution does not have the maximum amount dissolved additional solute can be added and will dissolve.

    1. Addition of solute to the solution until no more dissolves.
    2. Removal of solvent such as through evaporation.
    1. The solute and solvent are bot h solids.
    2. The solute is a solid and the solvent is a liquid.
    3. The solute is a gas and the solvent is a liquid.
    4. The solute and solvent are both liquids.
    1. 60 g NH4Cl
    2. 120 g NH4Cl
    3. 31 o C
    4. HCl and KClO3
    5. 90 g
    6. 7 g

    At 90°C, 50 g of KClO3 will dissolve in 100 g of water for a saturated solution. At 20°C, only 10 g of KClO3 is dissolved in 100 g of water for a saturated solution. 40 grams of KClO3 will precipitate out of solution.

    The solubility of a gas in a liquid is the greatest at low temperature and high pressure.


    Semicolons, colons, dashes, quotation marks, Italics (use an underline), and parentheses are added in the following sentences.

    1. The men in question (Harold Keene, Jim Peterson, and Gerald Greene) deserve awards.

    2. Several countries participated in the airlift: Italy, Belgium, France, and Luxembourg.

    3. "There's no room for error," said the engineer, "so we have to double check every calculation."

    4. Judge Carswell--later to be nominated for the Supreme Court--had ruled against civil rights.

    5. In last week's New Yorker , one of my favorite magazines, I enjoyed reading Leland's article "How Not to Go Camping."

    6. "Yes,"Jim said, "I'll be home by ten."

    7. There was only one thing to do--study till dawn.

    8. Montaigne wrote the following: "A wise man never loses anything, if he has himself."

    9. The following are the primary colors: red, blue, and yellow.

    10. Arriving on the 8:10 plane were Liz Brooks, my old roommate her husband and Tim, their son.

    11. When the teacher commented that her spelling was poor, Lynn replied, "All the members of my family are poor spellers. Why not me?"

    12. He used the phrase "you know" so often that I finally said, "No, I don't know."

    13. The automobile dealer handled three makes of cars: Volkswagens, Porsches, and Mercedes Benz.

    14. Though Phil said he would arrive on the 9:19 flight, he came instead on the 10:36 flight.

    15. "Whoever thought," said Helen, "that Jack would be elected class president?"

    16. In baseball, a "show boat" is a man who shows off.

    17. The minister quoted Isaiah 5:21 in last Sunday's sermon.

    18. There was a very interesting article entitled "The New Rage for Folk Singing" in last Sunday's New York Times newspaper.

    19. "Whoever is elected secretary of the club--Ashley, or Chandra, or Aisha--must be prepared to do a great deal of work," said Jumita, the previous secretary.

    20. Darwin's On the Origin of Species (1859) caused a great controversy when it appeared.


    Numbers

    Number types are divided into two groups:

    Integer types stores whole numbers, positive or negative (such as 123 or -456), without decimals. Valid types are int and long . Which type you should use, depends on the numeric value.

    Floating point types represents numbers with a fractional part, containing one or more decimals. Valid types are float and double .

    Even though there are many numeric types in C#, the most used for numbers are int (for whole numbers) and double (for floating point numbers). However, we will describe them all as you continue to read.


    ESMF Releases

    ESMF releases are listed below in reverse chronological order. Links to matching documentation, the release date, release notes, and known bugs are provided for each release.

    IMPORTANT: Starting with release 8.0.0, no distinction is made between "internal" and "public" releases. All releases 8.0.0 and up are considered public. The version number is defined as major.minor.patch , and completely specifies a release. The old "r" and "rpX" suffix, where X is a patch number, is no longer used. Release 7.1.0r was the last release that followed the old versioning scheme and nomenclature.

    Reference Manual for Fortran
    built from Git Develop HEAD
    (html) (pdf)

    Reference Manual for C
    built from Git Develop HEAD
    (html) (pdf)

    User's Guide
    built from Git Develop HEAD
    (html) (pdf)

    NUOPC Layer Reference Manual
    (html) (pdf)

    Building a NUOPC Model Manual
    (html) (pdf)

    Reference Manual for Fortran
    (html) (pdf)

    NUOPC Layer Reference Manual
    (html) (pdf)

    Building a NUOPC Model Manual
    (html) (pdf)

    Reference Manual for Fortran
    (html) (pdf)

    NUOPC Layer Reference Manual
    (html) (pdf)

    Building a NUOPC Model Manual
    (html) (pdf)

    • This release is backward compatible with the last major release update, ESMF 8.0.0, for all the interfaces that are marked as backward compatible in the Reference Manual. There were API changes to a few unmarked methods that may require minor modifications to user code that uses these methods. The entire list of API changes is summarized in a table showing interface changes since ESMF_8_0_0, including the rationale and impact for each change.
    • Some bit-for-bit changes are expected for this release compared to release ESMF 8.0.0 and patch release ESMF 8.0.1. We observe the following impact with Intel compilers using “ -O2 -fp-model precise ”:
      • Fixed a problem that could result in erroneously unmapped destinations when going from a very fine source grid to a coarse destination grid (e.g. 1/20 degree to 10x15 degree). Expected bit-for-bit changes:
        • ESMF_REGRIDMETHOD_CONSERVE_2ND: roundoff level changes in weight values because of a change in the order of calculation.
        • All regrid methods: Missing weights being added for very fine source grid to coarse destination grid regridding cases as this fix comes into play.
        • ESMF_REGRIDMETHOD_BILINEAR: small changes in regridding weights when a destination point exactly matches a source point.
        • ESMF_REGRIDMETHOD_PATCH: small changes in regridding weights when a destination point exactly matches a source point.
        • All regrid methods: Small weight changes when a point in the grid lies at exactly -90.0 latitude.
        • ESMF_EXTRAPMETHOD_CREEP: Small weight changes for the extrapolated destination locations.
        • ESMF_EXTRAPMETHOD_CREEP: Small weight changes for the extrapolated destination locations.
        • The Mesh creation, conservative regridding, and bilinear regridding algorithms when MOAB is active have been optimized to reduce their memory use and expand the size of Meshes they can be used on.
        • Grids can now be explicitly converted to a Mesh when MOAB is active, using the ESMF_MeshCreate() method.
        • Grids can be used to do first-order conservative regridding using MOAB.
        • Grids can be used for bilinear regridding on cell centers or corners using MOAB.
        • The NUOPC API has been simplified through the introduction of semantic specialization labels. The new approach leads to clearer and more concise NUOPC “cap” implementations that do not require specifying an Initialize Phase Definition (IPD) version or using the IPDvXXpY nomenclature when registering methods in the SetServices() method. Existing caps do not have to be re-written or updated, although updating to the new semantic specialization labels is recommended for existing and new NUOPC caps. The older IPD version based approach is supported for backward compatibility.
        • The NUOPC layer now provides features for resource control and handling of threaded components. This mechanism supports mixing of hybrid MPI+OpenMP components with different threading levels and mixing with standard MPI components on the same processing elements (PEs), i.e. cores. It allows each component to fully utilize HPC resources independently. Coupling between threaded components is supported automatically via the standard NUOPC Connectors.
        • The external NUOPC interface that supports interaction between an entire NUOPC application and a layer outside of NUOPC (e.g. a Data Assimilation system) was further refined. The associated prototype code (ExternalDriverAPIProto) has been updated to reflect the current status.
        • The NUOPC Profiling attribute, available in the Driver Metadata, Model Metadata, Mediator Metadata, and Connector Metadata, has been re-implemented. The NUOPC layer profiling features now integrate with the ESMF profiling infrastructure.
        • The NUOPC transfer protocol for geometry objects (Grid, Mesh, LocStream) has been made more efficient. Geometries used for multiple Fields are only transferred once, reducing the initialization overhead associated with the transfer.
        • Several optimizations were implemented in the NUOPC layer to reduce overhead. All applications using NUOPC benefit from these optimizations without requiring code changes.
        • Added the creep_nrst_d value to the extrapMethod NUOPC connection options. This is equivalent to the ESMF_EXTRAPMETHOD_CREEP_NRST_D option in ESMF_FieldRegridStore() discussed below.
        • FieldBundles don't currently enforce that every contained Field is built on the same Grid, Mesh, LocStream, or XGrid object, although the documentation says that this should be so.
        • The packed FieldBundle implementation uses a concatenated string to create a base object. When this string has more than 255 characters, e.g. a large number of Fields with long individual names is packed, the base object is not created correctly resulting in incorrect behavior at the FieldBundle level.
        • When the ESMF regrid weight generation methods and applications are used with nearest destination to source interpolation method, the unmapped destination point detection does not work. Even if the option is set to return an error for unmapped destination points (the default) no error will be returned.
        • The ESMF regrid weight generation methods and applications do not currently work for source Fields created on Grids which contain a DE of width less than 2 elements. For conservative regridding the destination Field also has this restriction.
        • The ESMF regrid weight generation methods and applications do not currently work on Fields created on Grids with arbitrary distribution.
        • The ESMF_GridCreate() interface that allows the user to create a copy of an existing Grid with a new distribution will give incorrect results when used on a Grid with 3 or more dimensions and whose coordinate arrays are less than the full dimension of the Grid (i.e. it contains factorized coordinates).
        • The ESMF_XGrid construction can lead to degenerate cells for cases where the source and destination grids have edges that are almost the same. Often these cells don't produce weights and are benign, but when weights are produced can lead to low accuracy results when transferring data to/from the XGrid.
        • The OpenMP thread count is being reset to 1 within all ESMF components. This affects user code that leverages OpenMP threading inside of components, and uses the OMP_NUM_THREADS environment variable to set the desired number of OpenMP threads. As a consequence the expected speed up from OpenMP threading in user code will not be present.
        • The PETs of all ESMF components, and any potentially created OpenMP threads under such PETs, are pinned to the PE on the respective shared memory node, corresponding to the PET number. As a consequence, even if a user overcomes the OpenMP thread count reset to 1 bug, e.g. by using omp_set_num_threads() API directly, the performance of OpenMP threaded user code is far below that of the expected speed up.
        • The GNU and Intel compilers require GCC>=4.8 for C++11 support (Intel uses the GCC headers). By default ESMF now uses the C++11 standard and cannot be downgraded. If you run into build issues due to the C++11 dependency, you must make sure a GCC>=4.8 is loaded.
        • For GNU compilers GCC>=10.x, the default Fortran argument mismatch checking has become stricter. This results in build failures in some of the code that comes with ESMF. Setting environment variable ESMF_F90COMPILEOPTS="-fallow-argument-mismatch -fallow-invalid-boz", during the ESMF build, can be used as a work around for this issue.
        • On Darwin, with the GNU gfortran+gcc combination, when building MPICH3 from source, it is important to specify the "--enable-two-level-namespace" configure option. By default, i.e. without this option, on Darwin, the produced MPICH compiler wrappers include a linker flag (-flat_namespace) that causes issues with C++ exception handling. Building and linking ESMF applications with MPICH compiler wrappers that specify this linker option leads to “mysterious” application aborts during execution.
        • On Darwin, with the Intel Fortran compiler, command line arguments cannot be accessed from ESMF applications when linked against the shared library version of libesmf. There is no issue when linked against the static libesmf.a version. Setting environment variable ESMF_SHARED_LIB_BUILD=OFF, during the ESMF build, can be used as a work around for this issue.
        • Currently the ESMPy interface to retrieve regridding weights from Python is only supported under the GNU compiler. On all other compilers the method will flag an error.
        • There is an issue with intercepting the MPI calls for profiling on some of the supported platforms. This results in a single FAIL reported for ESMF_TraceMPIUTest.F90. The affected platforms are:
          • Catania: Darwin+GNU+MPICH3
          • Gaea: Unicos+GNU+cray-mpich
          • Discover: Linux+GNU+intelmpi
          • Gaea: Unicos+Intel+cray-mpich
          • Gaea: Unicos+Intel+mpiuni
          • Orion: Linux+GNU+mpiuni

          Reference Manual for Fortran
          (html) (pdf)

          NUOPC Layer Reference Manual
          (html) (pdf)

          Building a NUOPC Model Manual
          (html) (pdf)

          Reference Manual for Fortran
          (html) (pdf)

          NUOPC Layer Reference Manual
          (html) (pdf)

          Building a NUOPC Model Manual
          (html) (pdf)

          • This release is backward compatible with the last release, ESMF 7.1.0r, for all the interfaces that are marked as backward compatible in the Reference Manual. There were API changes to a few unmarked methods that may require minor modifications to user code that uses these methods. A number of new interfaces were added. The entire list of API changes is summarized in a table showing interface changes since ESMF_7_1_0r, including the rationale and impact for each change.
          • Some bit-for-bit changes are expected for this release compared to the last release, ESMF 7.1.0r. We observe the following impact with Intel compilers using "-O2 -fp-model precise":
            • Roundoff level differences in conservative regridding due to an improvement in an area calculation algorithm
            • Roundoff level differences in regridding when used on a Mesh created from a SCRIP format file that contains longitudes =12 PETs are used. This issue leads to intermittent failures in the external_demos tests for the GRIDSPEC_1x1_time_to_C48_mosaic_bilinear case when run on the 16 PET configuration.
            • When regridding from a very fine source grid (e.g. 1/20th degree) to a fairly coarse grid (e.g. 15 degree) using conservative regridding, source and destination fractions can be erroneously less than they should be.
            • When using creep fill extrapolation sometimes the extrapolation will not cross from a piece of a Field on one PET to a piece on another. This can lead to places in the Field not being properly extrapolated to.
            • The exact identity matrix is not returned when using bilinear regridding from a grid that extends all the way to the poles, to an identical grid.
            • The regridding can hang when doing bilinear or patch regridding on the centers of a mesh and the mesh contains a cell whose center is at the exact same position as one of the corners of the cell.
            • Applying the sparse matrix multiplication to cases where the local data allocation is above the 32-bit limit will fail with a memory allocation error. This affects all Regrid(), Redist(), Halo(), and SMM() calls.
            • The ESMF_GridCreate() interface that allows the user to create a copy of an existing Grid with a new distribution will give incorrect results when used on a Grid with 3 or more dimensions and whose coordinate arrays are less than the full dimension of the Grid (i.e. it contains factorized coordinates).
            • Using the ESMF_GridCreate1PeriDim() method to create a grid with a bipole connection on the lower side (typically referring to the southern hemisphere) resulted in no connection there.
            • The ESMF_XGrid construction can lead to degenerate cells for cases where the source and destination grids have edges that are almost the same. Often these cells don't produce weights and are benign, but when weights are produced can lead to low accuracy results when transferring data to/from the XGrid.
            • The ESMF_ArrayCreate() crashes when used with pinflag=ESMF_PIN_DE_TO_SSI or pinflag=ESMF_PIN_DE_TO_SSI_CONTIG from within a component. The crash is from inside MPI with "invalid communicator". The "pinflag" option works correctly from the application level, i.e. in the context of the global VM.
            • Querying the ESMF_DistGridGet() method for "de" or "tile" information for a "localDe" will return incorrect results, and/or crash.
            • ESMF_AttributeWrite() has only been verified to work for ESMF standard Attribute packages. Non-standard Attribute packages may trigger a crash inside the ESMF_AttributeWrite() implementation.
            • For NetCDF installations that have the C and Fortran bindings installed in different locations, a NetCDF enabled build of ESMF does not correctly include the Fortran NetCDF library during linking.
            • When installing ESMF into a location that is shared with other libraries, it can happen that executing the ESMF install target fails with a "permission denied" error.
            • The GNU and Intel compilers require GCC>=4.8 for C++11 support (Intel uses the GCC headers). By default ESMF now uses the C++11 standard. If you run into build issues due to the C++11 dependency, you can either (1) make sure a GCC>=4.8 is loaded, or (2) set ESMF_YAMLCPP=OFF. In the latter case the YAML-dependent features in ESMF will not be available.
            • For GNU compilers GCC>=10.x, the default Fortran argument mismatch checking has become stricter. This will result in build failures. Setting environment variable ESMF_F90COMPILEOPTS="-fallow-argument-mismatch -fallow-invalid-boz", during the ESMF build, can be used as a work around for this issue.
            • On some systems with the PGI compiler, there is an issue with shared memory pointers between PETs on the same SSI. We see failures or crashes for Array tests that exercise this feature (ESMF_ArraySharedDeSSISTest.F90, ESMF_ArrayCreateGetUTest.F90) on the following platforms:
              • Hera/PGI-18.10.1
              • Gaea/PGI-16.5.0
              • Electra/PGI-17.1.0
              • Pleiades/PGI-17.1.0
              • Summitdev/PGI-19.7.0
              • Discover/PGI-14.1.0
              • Discover/PGI-17.7.0

              In the old versioning scheme, used by all the releases below, only releases with an "r" or "rpX" after the version number were considered "public". All other releases were considered "internal". Public releases were shown in colored cells .


              Watch the video: World of Tanks - Patch - E-25 - Limp Penis Tank (May 2022).